physics 251a advanced quantum mechanics i · physics 251a advanced quantum mechanics i prof.cumrun...

Physics 251a

Advanced Quantum Mechanics I

Prof. Cumrun Vafa

Typesetting: Yichen Shi

Fall 2014, Harvard University

Last updated: December 6, 2014

These are the lecture notes from Professor Cumrun Vafa’s graduate level Quantum Mechanics course given in Fall2014, at Harvard University. If you find errors or have any question, please send an email to [email protected].

1

Contents

1 The History of Quantum Mechanics and Motivation 41.1 1900, Rayleigh . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2 1900, Planck . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.3 1910, Lorentz . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.4 1905, Einstein . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.5 1922, Compton . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.6 1909-1911, Rutherford . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.7 1913, Bohr . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.8 1916, Sommerfeld . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.9 1916, Einstein . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.10 1923, de Broglie . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.11 1923, Schrodinger . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.12 1925, Heisenberg . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.13 1926, Born, Heisenberg, Jordan . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.14 1927, Ehrenfest . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151.15 Probabilistic Interpretation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2 Radially Symmetric Problems 172.1 The Hydrogen Atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182.2 The Laplacian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182.3 L2 and Lz . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222.4 Spherical Harmonics Y ml (θ, φ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242.5 The Radial Part R(r) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262.6 Decay Rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282.7 The Two-Body Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292.8 Three-Dimensional Harmonic Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

3 Principles of Quantum Mechanics 323.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333.2 Principles of Quantum Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

4 Spins 484.1 Addition of Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 534.2 Identical Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 554.3 Electrons and the Nucleus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 564.4 Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 574.5 Crystals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 574.6 Wigner-Eckart Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

4.6.1 Tensorial Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 594.6.2 Isospin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

4.7 Discrete Symmetries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 614.8 Supersymmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

5 Time-independent Perturbation Theory 645.1 Non-Degenerate Perturbation Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 645.2 Degenerate Perturbation Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

5.2.1 Zeeman Effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 665.2.2 Stark Effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

5.3 Variational Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 685.3.1 Virial Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

5.4 Bohn-Oppenheimer Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 705.5 WKB Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

5.5.1 Bohr-Sommerfeld Quantisation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 755.5.2 Barrier Penetration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

2

6 Time-dependent Perturbation Theory 766.1 Sudden Time-Dependence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 776.2 H = H0 +H ′(t) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

6.2.1 Hydrogen Atom Ionisation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 796.3 Fluctuating Perturbation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80

6.3.1 Radiation Absorption and Emission . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 816.4 Adiabatic Perturbation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 816.5 Berry’s Phase . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

3

1 The History of Quantum Mechanics and Motivation

Quantum mechanics started almost exactly in 1900. It started with the explanation of the distributions ofintensities that we obtain in the context of radiations.

1.1 1900, Rayleigh

Lord Rayleigh tried to apply simple ideas from statistical mechanics to radiation interacting with matter. In thecontext of statistical mechanics, we study basic configurations of particles where for each independent configuration,we get some energy related to the Boltzmann constant, i.e., to the temperature. We can thus study the intensityof radiation. In particular, we have some energy density per unit frequency at a given temperature, ρ(ν, T ), i.e.,

ρ(ν, T )dν = energy per unit volume in the frequency range ν and ν + dν at temperature T.

Lord Rayleigh wanted to compute ρ(ν, T ). He started with some cubic box with all sides heated up to temper-ature T . The temperature excites some radiation, i.e., creates some electromagnetic fields. So how much energy,or what kind of frequencies, do we see in this situation? If we take all possible modes, then each mode gives us theenergy of 1

2kT by Boltzmann’s rule.

To find the electromagnetic modes in a box, assume that the box is periodic for simplicity, i.e., we identify thefront and the back, the left and the right, and the top and the bottom sides. Let the box have length L. We cantake the standard form of electromagnetic waves

~E ∼ ~v ei(~k·~x−ωt),

where ~k = 2πL ~n. The allowed modes are such that ~k takes normal mode values because of the periodic boundary

conditions. If we shift x by L in any direction, this guarantees that it comes back to itself.

Now, we want to associate these modes. Take a coordinate system with axes labelled by nx, ny, nz. We haveall the integer points in space representing the modes and they don’t all have the same frequency.

What is the relation among their frequencies? We have the angular frequency

ω = 2πν = |~k|c.

Hence,

ν =|~k|c2π

=|~n|cL.

If we want to look at a particular frequency, take a spherical shell with a particular radius, say |~n|, in the space.Then all particles on the sphere have the same frequency. But the shell is too thin to contain any particles! So weare interested in a particular frequency range, i.e., in the number of modes lying on a shell with some thickness.

4

We’ll first calculate how many modes there are, then use the equipartition theorem from statistical mechanicsto see how much energy each one of them has. The assumption is that L is sufficiently big that this is almost acontinuum. To count one mode per unit cell, write

dnxdnydnz = 4πn2d|~n| = 4π

(νL

c

)2L

cdν,

and the number of modes in the shell

dN = 24πν2

c3L3dν,

where the 2 is due to the two polarisations for each mode. So the number of modes per unit volume

dN

V=

8πν2

c2dν.

This is the kinematics of the number of modes. Next, how much energy does each mode have? Think of themodes as oscillators with kinetic and potential parts. According to the equipartition theorem, we have

E =1

2kT +

1

2kT = kT

for each mode. Hence

ρ(ν, T ) =EdN

V dν=

8πν2

c3kT.

But wait a minute, we have infinitely many modes! If each one of them has energy kT , then the total energy isinfinity!

E

V=

∫ ∞0

ρ(ν)dν =∞.

It’s not that both statistical mechanics and normal modes don’t work, but that there is another part to thestory.

1.2 1900, Planck

Experimentally, if we plot ρ against ν, the curve comes back down smoothly to zero instead of shooting up toinfinity.

Planck looked at the curve and guessed

ρ ∼ ν3#

e#ν − 1,

where the denominator comes from the fact that the curve becomes exponentially suppressed for large ν, andbecause Rayleigh’s ρ ∝ ν2 works for small frequencies, the ν3 is to cancel the factor of ν in the denominator forsmall ν, in order to produce ν2.

5

Now, kT is energy, ν is frequency. How can we get energy out of frequency? We can multiply ν/kT by someconstant to make it dimensionless. Let’s call the constant h, such that

[h] = energy× time.

So he arrived at

ρ(ν) =8πhν3

c3( ehν/kT − 1).

Notice that ehν/kT − 1→ hν/kT as ν → 0, i.e., it matches with Rayleigh’s conclusion for small frequencies.

h is then experimentally measured to have the value 6.6× 10−27 erg sec (using cgs units).

A few months later, Planck wrote another paper to derive the same formula using the assumption that matteronly absorbs or emits radiation in particular chunks proportional to hν.

These exciting discoveries led to the measurement of the Boltzmann constant k, and in turn, the computationof Avogadro’s number NA (through R = kNA, since PV = nRT = NAkT ), the electric charge e (through eNAmeasured by Faraday), me (through e

m from the bending of the electron streams in the presence of the ~B field inthe Cathode Ray).

1.3 1910, Lorentz

Based on Einstein’s work, Lorentz assumed that not only matter comes in chunks proportional to hν, but alsoelectromagnetic radiation. Planck’s formula is correct, but something is wrong with the factor of kT . He startedwith the fact that the probability to be in any particular state of a given energy E is proportional to e−E/kT . Sinceenergy comes in discrete chunks, we count each state easily and the probability to be in a particular state with nunits

P (n) =e−nhν/kT∑∞

m=0 e−mhν/kT,

and the energy of the unit

E =∑

En =

∑∞n=0 nhν e−nhν/kT∑∞m=0 e−mhν/kT

.

Now,∞∑n=0

e−nx =1

1− e−x.

Hence,∞∑n=0

n e−nx = − ∂

∂x

1

1− e−x=

e−x

(1− e−x)2.

Hence the average energy

E =hν e−hν/kT

1− e−hν/kT=

hν

ehν/kT − 1,

which approaches kT as ν → 0.

As we can see, the chunkiness doesn’t matter for small frequencies but does for large ones. It makes sense sincefor hν kT , it’s such a tiny quantity that we seem to have a continuum.

6

1.4 1905, Einstein

Einstein’s photoelectric effect involves shining light to the copper attached to one end of a circuit. As a result,the electrons in the copper would be hit, and we can register a current in the circuit. So what is the voltage weneed to stop the current? Vstop as a function of ν of the light starts at some νmin and increases linearly.

Notice that it has nothing to do with intensity. Classically, when intensity goes up, energy goes up, and theelectrons which absorbed energy should start moving. So it’s strange that up to νmin, no current is registered. TheVstop is only proportional to ν.

Einstein had a basic explanation of it, based on the assumption that radiation also comes in chunks. Theelectrons are bounded by some potential, so to get them out, we need some minimum energy Emin = hνmin, whichis the energy of binding of electrons inside the copper. And if we have some more energy more than the minimum,then that should translate to energy of the electron that’s free. Hence

hν − hνmin = eVstop.

1.5 1922, Compton

Let’s shine light with frequency ν on an electron at rest. Classically, we expect that the radiation coming backto have the same frequency. But that is not the case. Again, the chunkiness in the energy of the electromagneticradiation was included in the explanation.

Think of light as a particle with energy E = hν. In Einstein’s theory, it’s clear that if it were a particle whichgoes at the speed of light, then we’d be in trouble, since we’d need infinite energy. The only resolution is for it tobe a massless particle.

HHHm2γc

4 = E2 − p2c2.

⇒ E = |~p|c.

Take light hiting the electron at rest with ν and coming right back at us with ν′. The electron picks upmomentum pe. With no computation at all, it is clear that ν′ < ν. Momentum conservation gives

pe −hν′

c=hν

c,

i.e.,

pe =h

c(ν − ν′).

Energy conservation giveshν +mec

2 = hν′ +√

(mec2)2 + h2(ν + ν′)2,

i.e.,

ν′ =mec

2

2hν +mec2ν,

7

which approaches ν as h→ 0, i.e., in the classical limit. From c = λν,

λ′ =2h

mec+ λ.

Note that hmec∼ 2.4× 10−10cm is the Compton wavelength of electrons. So for particles, we can associate some

kind of length. Note that now we also have

[h] = length×momentum.

1.6 1909-1911, Rutherford

We used to think that matter was of jelly-type having electrons with negative charges stuck inside. We assumedthat the jelly has some positive charge density to cancel the negative charges.

Rutherford then did a revolutionary experiment, where he shot α++ particles (Helium nucleus) at this “jelly”of gold. He expected that the α++’s would go straight through. To his surprise, some did not, and some even cameback at him! So this directly implied that the positive charge density, that matter, should be localised. So if thekinetic energy of the alpha particles can be cancelled by the potential energy, then

1

2mv2 ∼ 2eZe

r,

where we have 2e because the α++’s are doubly positively charged, Z ∼ 100, and r ∼ 3Z × 10−14cm. So he wasable to measure the size of the nucleus of gold.

However, when charged objects rotate around other charged objects, they radiate energy, and decay. So theorbit becomes smaller and smaller and goes away. And we can easily estimate the time for it to disappear altogether!So this picture is inconsistent.

1.7 1913, Bohr

Now, Bohr thought that if radiation is chunky, then matter might be chunky too. The electrons can only go onspecial orbits. What if angular momentum also comes in chunks? That would explain the specific radii allowed.So suppose

l = rp = n~,

where ~ is some constant with the same units as h. They’re probably related!

An electron from a higher orbit can jump to a lower one, by emitting some energy cancelling the differencebetween the two energies corresponding to the two orbits. Hence,

−Ef + Ei = hνi→f .

So what are the allowed energies? Assume that we have Z units of positive charges here, and electrons aregoing around it in circle.

mv2

r=Ze2

r2⇒ p2r

m= Ze2.

But we also have rp = n~. So we have

r =n2~2

Ze2m.

8

Hence,

En =1

2mv2 − Ze2

r

= −Ze2

2r

= −Z2e4m

2~2

1

n2.

And for Z = 1,

En = −13.61

n2eV.

To find the relation between ~ and h, we go back to the situation where an electron jumped down an orbit,emitting a photon of frequency νi→f .

Ei − Ef = hνi→f = −const.

[1

n2i

− 1

n2f

],

from which we can deduce the spectral lines. We can then use it to compute ~. We do find a difference of a factorof 2π, but that’s not satisfying enough.

Another way to go about it is to use classical results. Of course, quantum mechanics should be reducible toclassical mechanics. Here, for large n, orbits become closer together and we approach the classical continuum. Wewould expect that jumping from one orbit to the one lower corresponds to some ν which matches the frequency ofthe classical orbit.

First note that for n 1,1

(n+ 1)2− 1

n2' − 2

n3.

Now, say we go from n+ 1 to n. Then

En+1 − En = −Z2e4m

2~2

[1

(n+ 1)2− 1

n2

]' Z2e4m

~2

1

n3' hνn,

for n 1. In classical physics,p

2πmr=

v

2πr=

1

T= ν,

which gives

νn =n~

2πmr2=

n~2πm

(n4~4

Z2e4m2

) =mZ2e4

2πn3~3.

Hence

hνn =hmZ2e4

2πn3~3.

Now when we match the results above, we get

h

2π~= 1.

Note that

r =n2

Z0.5× 10−8cm;

v =p

m=n~mr

=1

137

Zc

n,

e2

~c=

1

137.

We know that the speed of electron isn’t the speed of light, so Z 137 is required for this formula to work.

9

1.8 1916, Sommerfeld

Consider electrons going around orbits. We have∮p dx = 2πrp = 2πn~ = nh.

If we look at a classical trajectory in a phase space given by x an p, integral around an orbit is the same ascomputing the area bounded inside the classical trajectory, i.e.,∮

p dx =

∫∫A

dxdp = nh.

So the formula is saying that if we have some classical orbit, the area is quantised in units of h. At some point,they thought that this Bohr-Sommerfeld quantisation rule might be the principle of quantum mechanics. Now weknow that this is just an approximate statement.

By that time, physicists have invented the Hamiltonian formulation of classical physics, in terms of positionsand momenta. Here is a quick reminder of it. With the Hamiltonian

H =p2

2m+ V (x),

we have∂H

∂pi=dxi

dt,

dx

dt=

p

m;

∂H

∂xi= −dpi

dt,

dp

dt= −dV

dx= F.

The first order differential equations can be very useful.

1.9 1916, Einstein

We know that an electron can jump from an orbit of higher energy to a lower one by emitting radiation. Howdo we compute the rate of this decay? If there’s a hydrogen atom in an excited state, how long does it take for itto jump down? Let’s call the decay rate from the higher orbit m to the lower one, n, Am

n.

Einstein, while working on his relativity, had the following remarkable idea. Suppose we put an atom in athermal bath of radiation. Then the reverse process is possible: electrons can jump from the lower energy state nto the upper, m, because of the radiation around. In fact, it’d better be the case since otherwise we won’t havestatistical equilibrium. The number of particles in state m, Nm, compared to that in state n, Nn, is given by

NmNn

=e−Em/kT

e−En/kT= e−

(Em−En)kT .

The decay rate should depend on the density of the radiation around. The frequency νmn of the photon theelectron needs to absorb to jump up is

νmn =Em − En

n.

Then the spontaneous emissionΓsponn→m = Bn

mρ(νmn),

for some constant Bnm. Now, the decay from m to n, which we assigned Am

n to, can happen even withoutradiation. But if there was radiation, would we get something extra? Let’s suppose that the stimulated emission

Γstim→n = Bm

nρ(νmn),

10

for some different constant Bmn.

We now have three unknowns: Amn, Bn

m, and Bmn, but notice that in equilibrium, the number of particles

going from n to m should be the same as that going from m to n. The number of particles that goes from m ton per unit time is Nm multiplied by the sum of the two sources of emission, and he number of particles that goesfrom n to m per unit time is Nn multiplied by the single source. Hence,

Nm(Amn +Bm

n)ρ(νmn) = Nn(Bnmρ(νmn)).

Using our expressions for Nm/Nn, νmn, and recall

ρ(ν, T ) =8πh

c3ν3

ehν/kT − 1.

Then we have

Amn =

(ehνnm/kTBn

m −Bmn) 8πh

c3ν3nm

ehνnm/kT − 1.

Notice that the rate of emission should be temperature independent. It can occur at all temperature! So thetemperature dependence in the denominator should be cancelled by the numerator. Hence

Bnm = Bm

n,

and in turn,

Amn = Bm

n 8πhν3nm

c3.

These are called the Einstein’s A and B coefficients.

1.10 1923, de Broglie

So waves are particles. How about the reverse? de Broglie started with Planck’s relations

E = |~p|c,

which is what we expect since the photon is massless. We can have a wave ∼ ei(~k·~x−ωt). And we also have,

E = ~ω, ~p = ~~k.

Let’s assume these for particles! But ω = c|~k| doesn’t work here since

ω

|~k|=

~ω~|~k|

=E

|~p|=p2/2m

|~p|=|~p|2m6= p

m.

But wave packets have a group velocity! Say we have a wave packet

ψ(~x, t) =

∫ei(kx−ω(k)t)φ(k)dk.

We expand around k0.

ψ(~x, t) = ei(k0x−ω(k0))t

∫ei(x− dω(k)

dk t)∣∣k0

(k−k0)φ(k)dk

= ei(k0x−ω(k0))tf

(x− dω(k)

dk

∣∣∣∣k0

t

)The phase is unimportant. So this is a wave packet which moves with velocity

v =∂ω

∂k=∂(~ω)

∂~k=∂E

∂p=

p

m.

11

Hence it works out if we use the group velocity. So why does the Bohr-Sommerfeld quantisation work? Supposethat we have a particle going around some orbit. Of course, the particle is associated with some wavefunction. Butthe wave is a single-valued function of position. The phase at each point is eikx, so the change in phase after wehave gone all the way around should be

ei∫k dx = e

i~∫p dx = 1.

Hence, ∫p dx = 2πn~ = nh.

The fact that we get integer quantum numbers is analogous to the normal modes of the standing waves in abox, characterised by integers. So the concept of waves explains it all.

1.11 1923, Schrodinger

Schrodinger thought, then expressing ~p as some gradient would bring down a factor of ~k from some plane waveto satisfy ~p = h~k. Similarly, expressing E as some time derivative will bring down a factor of ω to satisfy E = ~ω.Hence,

−i~ ∂

∂xψ = ~pψ; i~

∂

∂tψ = Eψ;

So replace

~p→ i~∇; E → i~∂

∂t

to solve problems! Starting with

H =p2

2m+ V (~x),

we immediately arrive at the Schrodinger equation

i~ =(−i~∇2)

2m+ V (~x)ψ.

For many particles, simply write the Hamiltonian as

H =

N∑i=1

p2i

2mi+ V (~xi),

and takeψ = ψ(~x1, ..., ~xN , t).

Now we can write down any problem in classical mechanics in terms of the wave equation! It was solved for theHydrogen atom, giving the same spectrum as the one Bohr arrived at using heuristic arguments.

12

1.12 1925, Heisenberg

Unlike Schrodinger, Heisenberg didn’t think of the “waves” as the ordinary kind of waves. Suppose a particleis having a transition from energy state m to n. We expect the power of radiation formula

P =2e2x2

3c3

to work. But there are two orbits. By “x”, which one are we referring to? The notion of position should dependon both n and m. Suppose we are going from m to n to k, then we have to have some product of m to n with nto k. So we should have

x2mk =

∑n

xmnxnk,

now called matrix multiplication. Since we are dealing with matrices, order of multiplication matters, and we have

xp− px = i~1 6= 0.

The Hamiltonian, which gives a definite value, should be a diagonal matrix.

[H(x, p)]nm = δnmEn.

He used this definition of energy and the fact that x and p don’t commute to solve the Harmonic oscillator,without any discussion of waves. He found that with

H =p2

2m+

1

2kx2,

the energy eigenstates are

En =

(n+

1

2

)~ω, ω =

√k

m, n = 0, 1, 2, ...

The Heisenberg and the Schrodinger formulations are equivalent in the context of wavefunction being a functionof ~x and t only. We now know that there are other components to quantum mechanics, the quantum numbers suchas the spin, which are not captured by ~x and t. So the Schrodinger formulation is inadequate. Heisenberg’s way ofthinking is more general, and he was correct by considering an abstract space: We can talk about matrices of spin.

Let’s quickly review Schrodinger’s approach. He started with

Hψn(~x, t) = Enψn(~x, t)

And he defined inner products

〈ψ|φ〉 =

∫ψ∗(~x, t)φ(~x, t) d3x,

and

〈ψ|ψ〉 =

∫|ψ|2d3x

is a positive definite quantity. Then the Heisenberg matrices, written in terms of Schrodinger’s approach, are

xmn =

∫ψ∗m(~x, t)xφn(~x, t) d3x.

The statement of matrix multiplication

(x2)mk =∑n

xmnxnk

makes sense turns out to be a statement about the completeness of space of eigenfunctions. Notice that∫ψ∗m(~x, t)(xp− px)ψn(~x, t) d3x = i~

∫ψ∗mψnd

3x.

So for anything we’re computing, we can replace xp− px with i~.

13

1.13 1926, Born, Heisenberg, Jordan

We can write the Hamiltonian in terms of mode expansions as

H =∑~k

(a2~k

+ ω2~ka2~k

).

This is basically an infinite set of Harmonic oscillators already solved by Heisenberg. The energy

E =

(N~k +

1

2

)~ω,

where the N is the level of excitation. But what is the meaning of the wavefunction ψ(~x, t)? Schrodinger knew thatthe energy is proportional to |ψ(~x, t)|2. So ψ(~x, t) somehow represents the amount of wave. He made an obviousstatement that

|ψ(~x, t)|2 = the amount of particle at ~x, t.

But Bohr pointed out the caveat that as time goes on, waves spread out to infinity. So are you telling me thatthe particle at a later time is everywhere? We actually measure the particle at one place only! So the interpretationcame after the formulation of quantum mechanics. How about the probability instead of the “amount”? i.e.,

|ψ(~x, t)|2 = the probability density for a particle to be found at ~x, t,

or,|ψ(~x, t)|2d3x = the probability for a particle to be in the volume element d3x.

This is the probabilistic interpretation of quantum mechanics. Suddenly, probability came into physics! Eventoday, the interpretation of quantum mechanics is still controversial.

Now, regarding probabilities, we know that the average value of some quantity O,

〈O〉 =∑i

PiOi.

Hence,

〈f(x)〉 =

∫|ψ|2f(~x) d3x.

Of course, a particle must be somewhere. So we have the normalisation condition∫|ψ(~x, 0)|2f(~x) d3x = 1.

It’d better be the case that the probability doesn’t change in time, i.e.,∫|ψ(~x, t)|2f(~x) d3x = 1.

In other words, we require that

d

dt

∫|ψ|2d3x =

∫ [(∂

∂tψ∗)ψ + ψ∗

(∂

∂tψ

)]d3x = 0.

Notice that (− ~2

2m∇2 + V

)ψ = i~

∂

∂tψ;(

− ~2

2m∇2 + V

)ψ∗ = −i~ ∂

∂tψ∗.

The potential cannot be complex since the energy is real. So we have

d

dt

∫|ψ|2d3x =

∫ [~

2im(∇2ψ∗)ψ − ψ∗ ~

2im∇2ψ

]d3x.

14

But since the wavefunction is normalised, the ψ at infinity vanishes. So we can use integration by parts toobtain

d

dt

∫|ψ|2d3x =

∫ [− ~

2im(∇ψ∗)∇ψ + (∇ψ∗) ~

2im∇ψ]d3x = 0.

1.14 1927, Ehrenfest

Ehrenfest started with an easy computation. For O with no explicit time dependence,

d

dt〈O〉 =

d

dt

∫ψ∗Oψ d3x =

∫ [Hψ∗Oψ−i~

+ψ∗OHψ

i~

]d3x =

1

i~

∫[−ψ∗H(Oψ) + ψ∗OHψ] .

We introduce here the commutator[A,B] = AB −BA.

Hence we can writed 〈O〉dt

=1

i~

∫ψ∗[O, H]ψ,

or,d 〈O〉dt

=1

i~〈[O, H]〉 .

So the rate at which the average of a quantity changes depend on the average value of its commutator with theHamiltonian. If O was time-dependent, then

d 〈O〉dt

=1

i~〈[O, H]〉+

⟨∂O∂t

⟩.

This is known as the Ehrenfest theorem. It’s basically the statement that if something commutes with theHamiltonian, it’s a conserved quantity. So the Hamiltonian tells us what doesn’t change with time: it is a generatorof time translations. We can think of the time derivative as being replaced as

d

dt· 7→ 1

i~[ · , H].

Here are the properties of the commutator.

[A,B] = −[B,A];

[A, λB] = λ[A,B];

[A,B + C] = A(B + C)− (B + C)A = [A,B] + [A,C];

[A,BC] = ABC −BAC +BAC −BCA = [A,B]C +B[A,C].

Now,

ψ∗[x, p]ψ = ψ∗x(−i~)∂

∂xψ + i~

∂

∂x(xψ) = i~.

We work in one-dimension for simplicity. Using the Ehrenfest theorem, we have

d 〈x〉dt

=1

i~〈[x,H]〉

=1

i~

⟨[x,

p2

2m] +

[x, V (x)]

⟩=

1

i~1

2m〈[x, p]p+ p[x, p]〉

=〈p〉m.

15

Now,

〈[p, V (x)]〉 =

∫ψ∗[(−i~ ∂

∂x(V ψ)

)− V

(−i~ ∂

∂xψ

)]dx = −i~

⟨dV

dx

⟩.

Similarly,

d 〈p〉dt

=1

i~〈[p,H]〉

=1

i~〈[p, V (x)]〉

= −⟨dV

dx

⟩.

So we have recovered Newton’s laws in an average sense. Satisfying Schrodinger equation implies that thingswork, in an average sense.

1.15 Probabilistic Interpretation

First, here’s a note on our notation.

〈ψ, φ〉 = 〈ψ|φ〉 =

∫ψ∗φd3x;

We’ll stick with physicists’ Dirac notation 〈ψ|φ〉 instead of the mathematicians’ 〈ψ, φ〉 notation in this class.Notice that

〈ψ|φ〉 = 〈φ|ψ〉.So for a physical operator O, 〈O〉 being real implies∫

ψ∗Oψ d3x = 〈ψ|O|ψ〉 != 〈ψ|O|ψ〉 =

∫ψ∗O∗ψ d3x.

We thus have the definition of a Hermitian operator. An operator A is Hermitian if, in mathematicians’ notation,

〈φ,Aψ〉∗ = 〈Aψ, φ〉 != 〈ψ,Aφ〉 ,

or in physicists’ notation (which could be confusing as to which side A acts on),

〈φ|A|ψ〉 != 〈ψ|A|φ〉 .

So for real expectation values, we must have Hermitian operators correspondingly.

Now, suppose A = a for a particular wavefunction, where a is some number, an outcome of some measurement.Then computations of the deviation of A in that state is zero. But that just means that the average is zero, i.e.,

〈A− a〉 = 0.

The expectation value of any powers of (A − a) must also be zero in that particular state. Let’s see theimplications of ⟨

(A− a)2⟩

= 0.

We have, noting again that A is Hermitian and a is real,

0 =⟨ψ|(A− a)2|ψ

⟩=

∫ψ∗(A− a)(A− a)ψ d3x

=

∫[(A− a)ψ]∗(A− a)ψ d3x

≡∫φ∗φd3x.

16

So the integrand must be zero, i.e.,|φ|2 = 0.

Hence, (A− a)ψ = 0, or,Aψ = aψ,

i.e., ψ is an eigenvector of A with eigenvalue a. The only values that we can possibly measure are these eigenvalues.For every Hermitian A, we can expand ψ as a linear combination of eigenstates of A.

ψ =∑m

cmψm, Aψm = λmψm,

where the cm’s are complex numbers. And we claim that

Prob(A = λm) = |cm|2.

Proof:

〈A〉 = 〈ψ|A|ψ〉 =∑n,m

cnc∗m〈ψm|A|ψn〉.

But since A is hermitian, A∗ = A acting on the bra gives

〈ψm|A|ψn〉 = λm〈ψm|ψn〉,

and A acting on the ket gives〈ψm|A|ψn〉 = λn〈ψm|ψn〉.

So if λm 6= λn, then 〈ψm|ψn〉 = 0. We assume that all eigenvalues are distinct. Later we’ll discuss that wecan always choose the bases for the vector spaces such that the distinct eigenvectors are orthogonal. And we cannormalise them by the condition that each one of them has norm one, i.e., 〈ψn|ψn〉 = 1.

So we assume that m = n. Hence,

〈A〉 =∑n

|cn|2λn.

But we know from basic probability that 〈A〉 =∑n P (A = λn)λn. Hence,

P (A = λn) = |cn|2.

This is Bohr’s probabilistic interpretation. Now, take ψ as the state we are in, and represent a particular statewith ψm. Then

|cm|2 = | 〈ψm|ψ〉 |2.

Proof: Let ψ =∑n cnψn. Then

〈ψm|ψ〉 =∑n

〈ψm|cnψn〉 = cm 〈ψm|ψm〉 = cm.

This is called the overlap of ψ with ψm. The important point here is that if we measure the state ψ and find itto be ψm, then it’s no longer ψ. So the wavefunction abruptly jumps, or, collapses because of a measurement!

2 Radially Symmetric Problems

The universe is spherically symmetric; many interesting problems in physics have spherically symmetry.

17

2.1 The Hydrogen Atom

Take the hydrogen atom. Its Hamiltonian is given by

H =~p2

2m+ V (r),

where the potential part depends on the radius only, not the angles. When we talk about “solving problems”, wemean finding solutions to

Hψ(~x, t) = i~∂ψ(~x, t)

∂t.

This can be done in two steps. Firstly, find eigenvectors and eigenvalues of H, i.e.,

Hψi(~x) = Eiψi(~x).

We only need to consider a time-independent ψ(~x) since for a time-dependent ψ(~x, t), we can always write

ψ(~x, t) =∑i

ci(t)ψi(~x),

so thatHψ(~x, t) =

∑i

ci(t)Hψi(~x) =∑i

ci(t)Eiψi(~x)

and

i~∂ψ

∂t= i~

∑i

∂ci(t)

∂tψi(~x).

Hence,

i~∂ci(t)

∂t= Eici(t),

and we solve for ci(t) asci(t) = ci(0) e−iEit/~.

Therefore,

ψ(~x, t) =∑i

ci(0) e−iEit/~ψi(~x).

All we need to know is what the state is like at t = 0, i.e., the initial condition ψ(~x, 0) =∑i ci(0)ψi(~x). But we

can simply use the overlap to findci(0) = 〈ψi|ψ(t = 0)〉 .

Hence all time-dependent problems reduce to solving

Hψi(~x) = Eiψi(~x).

2.2 The Laplacian

Using the Laplacian ∇2, we write

H = − ~2

2m∇2 + V (r).

Our conventions for spherical coordinates are

x = r sin θ cosφ; y = r sin θ sinφ; z = r cos θ,

with 0 < θ < π and 0 < φ < 2π.

18

The statement of spherical symmetry corresponds to the system’s independence of θ and φ. Our goal now isto understand what the ∇2 is in spherical coordinates. The angular momentum will be a main player. Recall theinvariant line element

ds2 = dx2 + dy2 + dz2

= dr2 + r2dθ2 + r2 sin2 θdφ2

≡ gijdxidxj .

Note that gij is the inverse of gij such that gijgjk = δik. (Repeated indices are summed over.) In sphericalcoordinates,

gij =

1 0 00 r2 00 0 r2 sin2 θ

.

Now, the Laplacian is what allows integration by parts such as∫|∇ψ|2d3x = −

∫ψ∇2ψd3x.

Setting√g ≡√

det g, we start by looking at the volume element in∫|∇ψ|2 dV =

∫|∇ψ|2√g d3x

=

∫∂iψ∂jψg

ij√g d3x

= −∫ψ

1√g∂i(g

ij√g∂jψ)√g d3x.

In general coordinates,

∇2 =1√g∂i(g

ij√g)∂j .

Substituting in√g = r2 sin θ for the spherical coordinates case, we have

∇2 =1

r2 sin θ

[∂r(r

2 sin θ)∂r + ∂θ sin θ∂θ + ∂φ1

sin θ∂φ

]=

1

r2∂rr

2∂r +1

r2 sin θ∂θ sin θ∂θ +

1

r2 sin2 θ∂2φ

≡ ∇2r +

1

r2∇2θ,φ.

Claim:1

r2∂rr

2∂r =

(1

r∂rr

)2

.

19

Proof: It’s important to remember that we have to act ∂r on everything on the right as well as pass it through tothe very right (to act on some state). We have(

1

r∂rr

)(1

r∂rr

)=

(1

r∂rr

)(∂r +

1

r

)=

1

r(∂r(r∂r)) +

1

r∂r

(r

1

r

)= ∂2

r +2

r∂r =

1

r2∂rr

2∂r.

Further, clearly (1

r∂rr

)2

=1

r∂2rr.

We’ll be using these interchangeably. Let’s now consider the Laplacian in physics terms. How do we think of

p2 = −~2∇2

in problems with radial symmetry? Classically,

~p = pr r + ~p⊥ = pr r + r × ~p,

which gives

p2 = |r · ~p|2 +

∣∣∣∣~r × ~pr∣∣∣∣2 = p2

r +L2

r2.

We see some similarities between this expression and our expression for ∇2. It too consists of a radial piece andan angular piece. Notice first that there is no ordering ambiguity in

~L = ~r × ~p

because of the nice cross product property (the position vector and the momentum along this particular positionvector never get multiplied together, e.g., Lx = ypz − zpy). Hence we can use it in quantum mechanics. For theradial piece, classically,

pr = r · ~p.

But the dot product is unlike the cross product: there is ordering ambiguity in terms like xpx. Hence we dothe most sensible thing of defining

pr =1

2(r · ~p+ ~p · r).

The first term1

2r · ~p =

1

2

~r

r(−i~∇) = − i~

2

∂

∂r,

and the second term1

2~p · r =

1

2(−i~)∇ · ~r

r,

where

∇ · ~rr

= ∂xx

r+ ∂y

y

r+ ∂z

z

r=

(1

r− x2

r3

)+

(1

r− y2

r3

)+

(1

r− z2

r3

)=

3

r− r2

r3=

2

r.

Hence1

2~p · r = − i~

2

(2

r+~r

r· ∇)

= −i~(

1

r+

1

2∂r

).

Finally, we have the form of pr as a result of the non-commuting nature of ~r · ~p,

pr =1

2(r · ~p+ ~p · r) = −i~

(1

r+ ∂r

)= −i~1

r∂rr.

Now we deal with the angular piece, which is crucial to our understanding of angular momentum. Since r2

commutes with L2 (a fact we’ll prove later), we can write

~L · ~Lr2

.

We have the commutation relations[Li, Lj ] = i~εijkLk.

20

In fact, we claim that for any vector Vj ,

[Li, Vj ] = i~εijkVk,

i.e., L is a generator of rotation of V . For example,

[Li, xj ] = i~εijkxk.

Proof: Writing Li compactly asLi = εijkxjpk = −i~εijkxj∂k,

we have[Li, xl] = −i~εijk[xj∂k, xl] = −i~εijkxjδlk = −i~εijlxj = i~εiljxj .

Another example is[Li, pj ] = i~εijkpk.

Proof:[Li, pl] = (−i~)2εijk[xj∂k, ∂l] = (−i~)2εijk(−δjl∂k) = i~εilk(−i~∂k) = i~εilkpk.

Theorem: If two commuting vector operators Vi and Wj transform as vectors according to

[Li, Vj ] = i~εijkVk, [Li,Wj ] = i~εijkWk.

then their cross product also transforms as a vector.

Proof: We have(~V × ~W )i = εijkVjWk.

Hence,

[Li, (~V × ~W )l] = εljk[Li, VjWk]

= εljk([Li, Vj ]Wk + Vj [Li,Wk])

= i~εljk(εijmVmWk + VjεikmWm)

= i~[(δliδkm − δlmδki)VmWk + (δlmδji −δliδjm)VjWm]

= i~(−VlWi + ViWl)

= i~εilk(~V × ~W )k.

Indeed, this theorem implies that ~L = ~r × ~p satisfies [Li, Lj ] = i~εijkLk. So does ~L commute with scalars?

[Li, VjVj ] = [Li, Vj ]Vj + Vj [Li, Vj ] = i~εijkVk, Vj = 0,

since εijk is antisymmetric under the exchange of j and k but the anti-commutator Vk, Vj is obviously symmetricunder the exchange of these indices. Hence we have relations such as

[Li, r2] = 0, [Li, L

2] = 0.

Let’s now return to the Hamiltonian

H =p2

2m+ V (r).

Since H consists of scalars only, we have[Li, H] = 0.

This is equivalent to the statement of rotational symmetry. So we also have

[H,L2] = 0.

Notice that Li cannot contain a ∂r term so [Li, r] = 0. We now want to establish

−~2∇2 =L2

r2− ~2

r2∂rr

2∂r.

21

Proof:

L2

r2= −~2

r2εijkεilmxj∂kxl∂m

= −~2

r2(δjlδkm − δjmδkl)xj∂kxl∂m

= −~2

r2(xj∂kxj∂k − xj∂kxk∂j)

= −~2

r2(r2∇2 + xj∂j − 3xj∂j − xjxk∂j∂k)

= −~2

r2(r2∇2 − 2r∂r − r2∂2

r )

= −~2

(∇2 − 1

r2∂rr

2∂r

).

2.3 L2 and Lz

Now, suppose we have operators Ai and Aj . If

[Ai, Aj ] = 0,

then there can be a function ψ which is the eigenfunction of both of the operators simultaneously. Suppose

[Ai, Aj ] 6= 0,

then,AiAjψ = Ai(ajψ) = ajaiψ = aiajψ,

butAjAiψ = Aj(aiψ) = aiajψ

as well. Hence we require [Ai, Aj ] = 0. Given this, our [H,L2] = 0 gives

L2ψ = αψ.

Then analogously, does [H,Li] = 0 give Liψ = βψ? No, since the Li’s don’t commute among themselves. Wecan however pick a direction and take

Lzψ = βψ.

Let’s now substitute α for L2. Our Hamiltonian becomes[− ~2

2m

(1

r

d

drr

)2

+α

2mr2+ V (r)

]ψ = Eψ.

Notice that this eigenvalue problem only fixes the radial part. Let’s now explore the reason why the eigenstatesof the Li’s fix the angular part. We have Lz = xpy − ypx = −i~(x∂y − y∂x).

Claim:Lz = −i~∂φ.

Proof:

∂φ = (∂φx)∂x + (∂φy)∂y + (∂φz)∂z

= −r sin θ sinφ∂x + r sin θ cosφ∂y

= −y∂x + x∂y.

Hence [H,Lz] = 0 implies thatH(−i~∂φ)ψ = −i~∂φ(Hψ),

22

i.e.,∂φH = 0.

We also haveLxψ = i~(sinφ∂θ + cot θ cosφ∂φ)ψ;

Lyψ = i~(− cosφ∂θ + cot θ sinφ∂φ)ψ.

Notice that there is no r−dependence at all. We can check that these are correct by choosing, for example,φ = π/2 to get Lxψ = i~∂θψ. As we can see, in the coordinates we choose, Lx and Ly are much messier than Lzto work with.

We’ll now proceed to find values of α and β in a quick way. We can take V to be anything that’s not singular.Here, wlog, take it to be 0. Then [

− ~2

2m

(1

r

d2

dr2r

)+

α

2mr2

]ψ = Eψ.

E is not singular, therefore there is no reason for the wavefunction to be singular at the origin either. We cantake ψ = rl so the RHS goes like rl. But then the LHS goes like rl−2, so they have to cancel. We have

− ~2

2m

1

r

d2

dr2rl+1 +

α

2mrl−2 = Erl,

or, (− ~2

2m(l + 1)l +

α

2m

)rl−2 = Erl.

This immediately implies the eigenvalues

α = ~2l(l + 1), l ∈ Z≥0.

l has to be a non-negative integer since if we do a power series expansion of a function to some power, havingnegative numbers or fractions as the power will give terms in the series which blow up at the origin, i.e., we’llencounter singularities which render inconsistency.

Recall that in the probabilistic interpretation, the probability of finding a particle goes like the square of thewavefunction, i.e., r2l. This means that near the origin, there is a very high vanishing power if l is large, i.e., thereis a smaller and smaller probability of finding the particle at the origin at large angular momenta.

Next, we tackle−i~∂φψ = βψ.

We can easily write down the general solution

ψ = eimφf(r, θ), β = m~, m ∈ Z.

m has to be an integer since the wavefunction is single-valued. Since L2 = L2x + L2

y + L2z, we also have β2 ≤ α.

Hence,β2 = m2~2 ≤ ~2l(l + 1),

or,m2 ≤ l(l + 1), |m| ≤ l.

So because of the periodicity of the wavefunction, the angular quantum numbers are quantised.

23

2.4 Spherical Harmonics Y ml (θ, φ)

Let’s now consider a unit sphere, representing ψ(θ, φ). Is there any way to think about arbitrary functions onthe sphere other than using θ and φ? Yes, we can always have

rlf(θ, φ) = xiyjzk,

i.e., some monomial subject to the condition x2 + y2 + z2 = 1. But each x, y, z gives an r if we consider their formin spherical coordinates. Hence i+ j + k = l, i.e., the degree of the monomial is l.

Degree (l) Polynomial m0 1 01 x, y, z ±1, 02 x2, y2, z2, xy, xz, yz (but with constraint x2 + y2 + z2 = 1) ±2, ±1, 0

We can setx± ≡ x± iy = r e±iφ sin θ.

Now let’s do the counting. The monomials are basically the integer points inside this triangle.

This gives 12 (l+ 2)(l+ 1) terms. But x2 + y2 + z2 = 1 tells us that if we are able to go two degrees lower, we’re

including something that we should not have included. Hence we have, for a given l, the degeneracy

1

2(l + 2)(l + 1)− 1

2l(l − 1) = 2l + 1.

Notice that if we writexn+

+ xn−− ,

then m = n+ − n−.

Let’s write down some of the famous spherical harmonics, the Y ml (θ, φ)’s. We normalise them such that whenintegrated over the angular piece, ∫

|Y ml (θ, φ)|2 sin θ dθdφ = 1.

• l = 0:1√4π.

• l = 1:

- m = 1:

−√

3

8πsin θ eiφ;

24

- m = 0: √3

4πcos θ;

- m = −1: √3

8πsin θ eiφ.

Now, notice that∇2 : deg(l)→ deg(l − 2).

But recall the cancellation of the rl−2 terms in the radial part of the Schrodinger equation above. Hence,

∇2(rlY ml (θ, φ)

)= 0.

And we have (∂2x + ∂2

y + ∂2z

)∑αijkx

iyjzk = 0.

Take ax2 + by2 + cz2. Then a+ b+ c = 0 since

∇2(ax2 + by2 + cz2) = 0.

• l = 2:

- m = 2: √15

32πx2

+ =

√15

32πsin2 θ e2iφ;

- m = 1:

−√

15

8πx+z = −

√15

8πsin θ cos θ eiφ.

- m = 0: √5

16π(x+x− − 2z2) =

√5

16π(3 cos2 θ − 1).

- m = −1: √15

8πx−z =

√15

8πsin θ cos θ e−iφ.

- m = −2: √15

32πx2− =

√15

32πsin2 θ e−2iφ.

Let’s writeY ml (θ, φ) = eimφPml (θ),

where the Pml (θ)’s are the Legendre polynomials. Now using the angular part of the Laplacian, we get

− 1

sin θ

d

dθsin θ

dPmldθ

+m2

sin2 θPml = l(l + 1)Pml .

Now, under parity transformation,θ → π − θ, φ→ φ+ π.

Hence,Y ml (π − θ, φ+ π) = (−)lY ml (θ, φ).

Also note that ∫(Y ml )∗Y m

′

l′ dΩ = δl,l′δm,m′ .

In summary, we see that there is a one-to-one correspondence between the Y ml (θ, φ)’s and some monomial f(xi)in cartesian coordinates subject to the condition

∑i x

2i = 1 (i.e., that it’s of degree l), satisfying ∇2f(xi) = 0.

25

2.5 The Radial Part R(r)

Let us now tackle the R(r) part inψ ≡ R(r)Y ml (θ, φ),

for all spherically symmetric problems. Cancelling the Y ml (θ, φ)’s from both sides, we have[− ~2

2m

(1

r

d

drr

)2

+~2l(l + 1)

2mr2+ V (r)

]R(r) = ER(r).

Define

R(r) ≡ u(r)

r.

Then we have [− ~2

2m

d2

dr2+ Veff(r)

]u(r) = Eu(r),

where

Veff(r) =~2l(l + 1)

2mr2+ V (r).

This is like a one-dimensional problem. There are a few things to note here. Firstly, r ≥ 0. Secondly, u(r) ∼ rl+1

as r → 0. Also recall our normalisation condition∫|ψ|2r2dΩdr =

∫|RY ml |2r2dΩdr = 1.

We had ∫|Y ml |2dΩ = 1.

Hence, ∫ ∞0

|u|2dr =

∫ ∞0

r2|R|2dr = 1.

We’ll specialise to solve the hydrogen atom problem from this point onwards.

This problem looks radically different for the l = 0 and the l 6= 0 cases, since l = 0 gives α = 0 and we simplyhave the V (r) < 0 curve as our effective potential. Can we have infinitely negative energy? Classically, yes, butthe quantum theory is smarter. Let’s solve our problem, looking at the E < 0 bound state for now.[

− d2

dr2+l(l + 1)

r2− 2mZe2

~2r

]u(r) = −κ2u(r),

where

κ ≡√−2mE

~2,

26

i.e.,

E = −~2k2

2m.

Let ρ ≡ κr. In terms of this dimensionless quantity, we have[− d2

dρ2+l(l + 1)

ρ2− ξ 1

ρ

]u(ρ) = −u(ρ),

where

ξ ≡ 2mZe2

κ~2.

We have the conditions that as ρ→ 0,u ∼ ρl+1,

and as ρ→∞,d2u

dρ2∼ u,

which givesu ∼ A e−ρ,

where the eρ term vanishes since u must be normalisable. So in general,

u(ρ) ∼ ρl+1 e−ρF (ρ).

We have the conditions that F (0) ∝ 1, and that F (∞) cannot grow too fast. In fact,

u(ρ) ∼ ρl+1 e−ρ

is a solution! But let’s find something more general by substituting u(ρ) ∼ ρl+1 e−ρF (ρ) back into the Schrodingerequation above. This gives

F ′′ − 2

(1− l + 1

ρ

)F ′ +

ξ − 2(l + 1)

ρF = 0.

At ρ = 0, since F (0) ∝ 1 we have ξ = 2(l + 1). Let’s now use the usual power series method to find recursionrelations, i.e., we write

F (ρ) =

∞∑k=0

akρk, a0 6= 0.

This givesak+1

ak=−ξ + 2(l + 1 + k)

(k + 1)(k + 2(l + 1)).

So for large k,

ak+1 ∼2

kak.

Hence

ak ∼2k

k!,

and we have

F (ρ) ∼∞∑k=0

(2ρ)k

k!= e2ρ,

but this is a non-renormalisable solution! We need some fine-tuning here. A fallacious assumption we made wasthat ak 6= 0 for any k. The only way for it to be 0 such that we don’t get an exponentially growing solution is forthe numerator to vanish at some stage. Let n ≡ l + 1 + k > l, then we require

ξ = 2n.

F (ρ) is a polynomial of degree k = n− l − 1. These are the Laguerre polynomials,

F (ρ) = L2l+1n−l−1(2ρ).

27

Recall our expressions relating ξ, κ and E. Substituting ξ = 2n, we have

κ =1

na,

where the Bohr radius

a =~2

mZe2=

0.5× 10−8

Zcm.

And we have

E = −m2Z2e4

2~2

1

n2= −13.6Z2

n2eV.

Putting everything together, we have found that

ψ(r, θ, φ) = Y ml (θ, φ)L2l+1n−l−1(2κnr)× (κnr)

l e−κnr,

where for a given l, m ranges from −l to l, and n = l + 1, l + 2, .... The wavefunction is normalised such that∫|ψ|2r2dΩ = 1.

So for a given energy level, how many states have that energy? For any n, we have n2 being the degeneracydue to the possible values of (l,ml) that we can choose. In fact, the degenerative states will all split. We’ll comeback to this when we discuss spins, and later, the Runge-Lenz symmetry.

2.6 Decay Rates

Now, let’s compute the decay rates. The technology we have is still insufficient, but we can have the formulafirst and get back to it when we talk about perturbative theory. The classical formula for power,

P =2

3

e2|~x|c3

=dE

dt.

When going from an initial to a final state,

Ei − Ef = ~ω.

And the decay rate

Γ =1

~ωP =

2ω3e2

3c3~|~xif |2,

where|~xif |2 = | 〈f |~x|i〉 |2 + | 〈i|~x|f〉 |2,

and a “real” derivation involving Einstein’s A and B coefficients will give us

Γ =4ω3e2

3c3~(| 〈f |x|i〉 |2 + | 〈f |y|i〉 |2 + | 〈f |z|i〉 |2

).

This already gives us some selection rules. Firstly, ni > nf , otherwise decay cannot happen. What are thepossible initial and final l’s? The formula tells us that x acting on |i〉 gives |f〉, so we claim that ∆l = ±1, and∆m = 0,±1. Recall the states’ representation as monomials of degree l. If we multiply them by x, we are onedegree higher. But remember our constraint relating them with the two-degree lower states. Hence the ∆l = ±1.Now, recall that m comes from ∼ e±imφ. z has no φ dependence, and x± iy ∝ e±imφ. Hence ∆m = 0,±1. We’lltackle the selection rules in detail later.

28

2.7 The Two-Body Problem

Up till now, we’ve been using the mass of the electron in the hydrogen atom problems. Let the nucleus havemass m1, and the electron m2, then the problem has two degrees of freedom. Then, the centre of mass positionand momentum

~xcm =m1~x1 +m2~x2

m1 +m2, ~pcm = ~p1 + ~p2,

and the relative position~xrel = ~x1 − ~x2.

Is the relative momentum~prel = ~p1 − ~p2?

Not really. In quantum mechanics, ~p is defined through its dependence on ~x, i.e., ~x in ~p = −i~∇ should be ~xrel.We want

[~xcm, ~pcm] = i~, [~xrel, ~prel] = i~;

[~xrel, ~pcm] = 0 = [~xcm, ~prel].

So we can take

~prel = µ

(~p1

m1− ~p2

m2

), µ =

m1m2

m1 +m2.

The Hamiltonian~p2

1

2m1+

~p22

2m2+ V (|~r1 − ~r2|) ≡ Hcm +Hrel,

where

Hcm =~p2

cm

2(m1 +m2), Hrel =

~p2rel

2µ+ V (~rrel).

And the wavefunctionψ(~x1, ~x2) ≡ ψ1(~xcm)ψ2(~xrel),

whereψ1(~xcm) = ei

~kcm·~xcm ,

and ψ2(~xrel) is the ψ(~r) that we had been computing. The total energy

E =~p2

cm

2(m1 +m2)+ Erel, ~pcm = ~~kcm,

where Erel is our good old −13.6Z2

n2 eV.

2.8 Three-Dimensional Harmonic Oscillator

The harmonic oscillator sets the stage for us to tackle quantum mechanics. Assuming spherical symmetry sothat the x acts as our r in earlier problems,

H =~p2

2m+

1

2kx2 ≡ ~p2

2m+

1

2mω2x2.

It is still solvable using the Y ml ’s and the radial equation, but it turns out that there’s something simpler thatwe can do. First separate the three-dimensional problem into three one-dimensional problems, writing

ψ(x1, x2, x3) ≡ ψ1(x1)ψ2(x2)ψ3(x2).

Then Hψ = Eψ gives

H1ψ1(x1)ψ2(x2)ψ3(x2) + ψ1(x1)H2ψ2(x2)ψ3(x2) + ψ1(x1)ψ2(x2)H3ψ3(x2) = (E1 + E2 + E3)ψ1ψ2ψ3.

29

Now, ~p2

2m + 12mω

2x2 motivates us to think of

|p+ ix|2 = (p+ ix)(p− ix).

Define

a ≡ 1√2m~ω

(p− imωx), a† ≡ 1√2m~ω

(p+ imωx).

Classically, i.e., without taking non-vanishing commutators into consideration,

a†a =1

~ω

(p2

2m+

1

2mω2x2

).

Note that a†a is dimensionless.

Claim:

H = ~ω(a†a+

1

2

).

Proof:

RHS =p2

2m+

1

2mω2x2 +

1

2ω[x, p] +

1

2~ω = H.

It’s easy to find that[a, a†] = 1;

[H, a†] = ~ωa†, [H, a] = −~ωa.

Hence,[H, a]† = −[H, a†].

Given an eigenstate of H, called |α〉, then

H(a†|α〉) = [H, a†]|α〉+ Ea†|α〉 = (E + ~ω)a†|α〉,

andH(a|α〉) = [H,α]|a〉+ Ea|α〉 = (E − ~ω)a|α〉.

Hence a†|α〉 and a|α〉 are eigenstates of H too, with eigenvalues E + ~ω and E − ~ω, shifting E up and downby ~ω, respectively. That’s why a† and a are referred to as creation and annihilation operators.

Can we have energy being raised and lowered indefinitely? Infinite negative energy is impossible since∣∣a|α〉∣∣2 = 〈α|a†a|α〉 ≥ 0

gives

〈α|H|α〉 = E = ~ω⟨α|a†a|α

⟩+

1

2~ω 〈α|α〉 ≥ 1

2~ω.

The problem can be solved if we reach a zero state at some point. We can set

a|α〉 ≡ 0,

such that

H|α〉 = ~ω(a†a+

1

2

)|α〉 =

1

2~ω.

To find the ground state ψ, take

0!= aψ =

1√2mω~

(−i~∂x − imωx)ψ,

which gives us

ψ = C exp

[−mωx

2

2~

].

30

Notice that ψ is a Gaussian with width√

~/mω. To find C, simply normalise the state according to∫ ∞−∞|ψ|2dx = 1.

And we have

C =(mω~π

) 14

.

The ground state is unique, and we’ll denote it with |0〉 from now on. So we have

H|0〉 =1

2~ω|0〉.

Now let’s go up. We have

Ha†|0〉 =3

2~ωa†|0〉.

But can a†|0〉 = 0 as well? No, since

a(a†|0〉) = ([a, a†] + a†a)|0〉 = |0〉.

Since a acting on a†|0〉 is non-trivial, a†|0〉 cannot be zero. And in turn, all the higher states obtained througha† are non-trivial. Moreover, we can always come back to the same lower state using a. Let’s define

|n〉 ≡ c(a†)n|0〉.

Then

H|n〉 =

(n+

1

2

)~ω|n〉.

Claim:a|n〉 ∝ |n− 1〉.

Proof: We can commute the a to the very right, passing through the a†’s using [a, a†] = 1 continuously. This gives

a(a†)n|0〉 = a†aa† · · · a†|0〉+ a† · · · a†|0〉 = · · · = n(a†)n−1|0〉.

The ground state is normalised such that〈0|0〉 = 1.

We want to demand〈n|n〉 = |c|2

⟨0|an(a†)n|0

⟩ != 1.

We have ⟨0|an(a†)n|0

⟩= n

⟨0|an−1(a†)n−1|0

⟩= · · · = n! 〈0|0〉 .

Hence,

|n〉 =1√n!

(a†)n|0〉.

Now, we know that |0〉 is of the form e−αx2

. Then using our definition of a† in terms of operators x and p, wehave

a†|0〉 = bx e−αx2

,

(a†)2|0〉 = (b′x2 + b′′) e−αx2

,

etc. In fact, we have

|n〉 = Pn(x) e−αx2

,

where Pn(x) are the Hermite polynomials. They have the property that the parity of n is the same as the parityof Pn(x), i.e.,

Pn(−x) = (−)nPn(x).

Claim:a†|n〉 =

√n+ 1|n+ 1〉.

31

Proof:

a†(

1√n!

(a†)n|0〉)

=(a†)n+1

√n!|0〉 =

√n+ 1

(a†)n+1√(n+ 1)!

|0〉 =√n+ 1|n+ 1〉.

Similarly,a|n〉 =

√n|n− 1〉.

Now we briefly discuss a powerful theorem, called Wick’s theorem. Firstly, notice that since a + a† ∝ p anda− a† ∝ x, any number of x’s and p’s sandwiched between states |n〉 and |m〉 can be put into the form⟨

0| · · · (a†)m|0⟩.

If the number of a’s and the number of a†’s sandwiched between the |0〉’s don’t match, then the whole expressionvanishes. Otherwise, we can use Wick’s theorem to do fast computations. For example, given a state⟨

0|aa†aaa†a†|0⟩,

we can contract pairs of left a’s and right a†’s according to

〈0|aa†aaa†a†|0〉,

or,

〈0|aa†aaa†a†|0〉.

Hence, ⟨0|aa†aaa†a†|0

⟩= 2.

Let’s come back to our oscillator problem.

H = ~ω(a†1a1 + a†2a2 + a†3a3 +

3

2

)= ~ω

(n1 + n2 + n3 +

3

2

),

where we call n ≡ a†a the number operator.

Now, we should be able to rewrite everything in terms of spherical harmonics, since we are dealing with aspherically symmetric problem.

[ht]

State (l) Number of Ways l|000〉 1 0|100〉 3 1

|200〉 and |110〉 3 + 3 = 6 2 and 0

3 Principles of Quantum Mechanics

32

3.1 Preliminaries

Quantum mechanics postulates that physical states can be represented as vectors in the Hilbert space, sooperations on Hilbert spaces will be a focus of our discussion. But there are a few things that we have to getaquatinted with before we get there. Let’s begin with the concept of vector space. Take

~v, ~w ∈ V.

Then for a ∈ C, we have operations~v + ~w ∈ V, a~v ∈ V.

Of course, we can take~v = ax~ex + ay~ey + az~ez = (ax, ay, az),

and likewise for ~w, such that

~v + ~w =∑i

ai~ei +∑j

bj~ej =∑i

(ai + bi)~ei; c~v = c∑i

(ai~ei) =∑i

(cai)~ei.

But we aren’t satisfied with this kind of description. The reason is that choosing some specific vectors ~ei isn’t aconvenient thing to do in quantum mechanics. We want to think about vectors more abstractly, without referenceto specific bases. Different experiments use different ~ei’s so we want to be more general.

We’ll omit the arrows on vectors from now on. The collection of vectors

v1, ..., vr

are linearly independent ifr∑i=1

aivi = 0 ⇔ ai = 0 ∀i.

V is said to be n−dimensional if the maximum dimension of linearly independent vectors is n. If there isn’tsuch an n, we call the vector space infinite-dimensional. For now, we will assume a finite n. An n−dimensionalvector space has a basis

B = v1, ..., vn,

where each vi are linearly independent. Existence is obvious by the definition of a basis; regarding uniqueness...

Claim: Once we have chosen such a basis, then for every v ∈ V , we can always write it uniquely as

v =

n∑i=1

aivi, ai ∈ C.

Proof: Suppose bi 6= ai for some i, then take

v =

n∑i=1

aivi

along with

v =

n∑i=1

bivi.

If we subtract the former from the later, then we end up with

ai = bi.

There are infinitely many bases that we can choose from by simply taking linear combinations of the vi’s. If wehave v1, ..., vn and w1, ..., wn, then

vi =

n∑j=1

aijwj ,

33

which gives

v =

n∑i=1

aivi =

n∑i,j=1

aiajiwj =

n∑i=1

bjwj ,

where aji = Aji, A being an n× n matrix, and

bj ≡n∑j=1

ajiai.

So the same vector can be written in different bases. If we take

v ≡

a1

...an

, w ≡

b1...bn

,

then notice that even though we are talking about the same vector v, the column vectors are different. The relationbetween them is simply

w = Av.

Note that we are talking about any linearly independent vectors. Orthogonality, etc do not need to hold forthese statements to be true.

We next move to inner products. Take v, w ∈ V . The interesting properties of the inner product are

v · w = w · v, v · v ≥ 0 (= 0 iff v = 0).

We’ve been using the dot product notation. Dot products are a special case of inner products. They areproducts of vectors in Rn, but we need to deal with complex numbers here, and we want our inner product ofsomething complex to spit out real numbers. We’ll first use mathematician’s notation 〈 ·, · 〉 for inner products.Physicists like to use the Dirac notation 〈 · |· 〉. In order to have

〈v, v〉 ∈ Rn, 〈v, v〉 ≥ 0,

we simply need the requirement〈w, v〉∗ = 〈v, w〉 .

The inner product is bilinear, but we’ll encounter a contradiction if we impose the condition naively. Forexample, take

z∗ ≡ 〈av, w〉∗ = (a 〈v, w〉)∗ = a∗ 〈w, v〉

but also,z∗ ≡ 〈av, w〉∗ = 〈w, av〉 = a 〈w, v〉 .

To overcome the contradiction, define

〈av, w〉 ≡ a∗ 〈v, w〉 , 〈v, aw〉 ≡ a 〈v, w〉 .

Then we have bilinearity.

〈a1v1 + · · ·+ anvn, b1w1 + · · ·+ bnwn〉 =

n∑i,j=1

〈vi, wj〉 .

From this point onwards we use the Dirac notation. Take

〈vi|vj〉 ≡ Bij .

Then〈vj |vi〉 ≡ B∗ij = Bji.

We say that B is Hermitian ifB = (Bt)∗,

34

where we call |vi〉〈vi| the identity transformation. Now, |vi〉 represents a column vector with all entries 0 exceptthe ith row, and 〈vi| represents a row vector with all entries 0 except the ith column. Then, clearly

10...0

(1 0 · · · 0)

+

010...0

(0 1 0 · · · 0

)+ · · · = 1 =

∑j

|vj〉〈vj |.

We now move to the discussion of the Schwartz inequality. For real vectors, starting with

v · w = |v||w| cos θ,

since | cos θ| ≤ 1, we have|v · w|2 ≤ |v|2|w|2.

We want a complex vector version of this, i.e., we want to show that

| 〈v|w〉 |2 ≤ 〈v|v〉〈w|w〉 .

Proof: Take

|w′〉 ≡ |w〉 − 〈v|w〉 |v〉〈v|v〉

.

Then

0 ≤ 〈w′|w′〉 =

(〈w| − 〈w|v〉〈v|

〈v|v〉

)(|w〉 − 〈v|w〉 |v〉

〈v|v〉

)= 〈w|w〉 − 2

〈w|v〉〈v|v〉

+〈w|v〉〈v|v〉〈v|v〉〈v|v〉

.

Hence,〈v|w〉〈w|v〉 ≤ 〈v|v〉〈w|w〉 .

Finally, we can discuss the Hilbert space H, which is a vector space over C with this inner product. Our mainoperation on H is linear transformation T such that

T : H → H,

T (av + bw) = aT (v) + bT (w).

In fact, given an orthonormal basis v1, ...vn, then

T (vi) =

n∑j=1

aijvj ,

or in Dirac notation,

T |vi〉 =

n∑k=1

aik|vk〉.

So we have〈vj |T |vi〉 = aji,

andT |v〉 =

∑i

aiT |vi〉 =∑i,j

aiaji|vj〉 ≡∑j

bj |vj〉 ≡ |w〉.

Therefore, if V and W have some bases, then every linear map from V to W can be represented by a matrix.In this context, we see that the identity transformation

1 =∑i

|vi〉〈vi|

is a linear transformation, corresponding to the identity matrix.

36

Next, we define the adjoint operator, which is a linear transformation. Define⟨w|T †|v

⟩≡ 〈v|T |w〉.

Indeed, the inner product alone defines the adjoint operation. So we have⟨vi|T †|vj

⟩= (T †)ij = 〈vj |T |vi〉 = T ∗ji.

Hence,T † = (T t)∗.

Since the operations of transposition and complex conjugation commute, we have

(T †)† = (((T t)∗)t)∗ = 1.

Products in linear transformations are straightforward.

(T1T2)ij = 〈vi|T1T2|vj〉 =

n∑k=1

〈vi|T1|vk〉〈vk|T2|vj〉 =

n∑k=1

(T1)ik(T2)kj =

n∑k=1

(T t2)jk(T t1)ki = (T t2Tt1)ji.

So we also have(T1T2)† = T †2T

†1 .

Next, we move to the self-adjoint (Hermitian) operators, defined by

T † = T,

or,T ∗ij = Tji.

So diagonal entries are real, off-diagonal opposites are complex conjugates of each other.

Now, a transformation T is called diagonalisable iff there exists a basis such that T is diagonal, with itseigenvalues being the diagonal entries. Also, if T is diagonalisable, then there exists a basis |ei〉 of eigenvectors|ei〉 such that

T |ei〉 = λi|ei〉, i = 1, ..., n.

The Hermitian operator is always diagonalisable, and in physics, we almost always deal with transformationswhich are diagonalisable, although not all matrices are diagonalisable.

Claim: For any matrices, we can find at least one eigenvector and one eigenvalue such that

T |v〉 = λ|v〉.

Proof: Go to a basis T and takedet(T − λI) ≡ Pn(λ).

Now, if the determinant of a matrix is zero, then the columns forming the matrix are linearly dependent. Thenthere is a zero eigenvector. Suppose Pn(λ∗) = 0, i.e., det(T − λ∗I) ≡ detM = 0. Then we have (T − λ∗I)|v〉 = 0,i.e.,

T |v〉 = λ∗|v〉.

We now apply this fact to a Hermitian operator T = T †. In an orthonormal basis, we have

T |e1〉 = λ|e1〉,

i.e., λ0 ?...0

10...0

=

λ0...0

37

Since T = (T ∗)t, we see that λ must be real and that the first row, apart from the λ entry, must be zero. Whatabout the leftover (n− 1)× (n− 1) matrix? Notice that a transformation acting on e2 to en brings back a linearcombination of e2 to en, but nothing from the e1 direction. So we can repeat our first step in the second step,restricting to the part of the matrix without the first row or column and claiming the existence of one eigenvalueand eigenvector. So by induction, we can make this matrix completely diagonal. Hence every Hermitian matrix isdiagonalisable, and we have orthonormal basis elements.

Claim: For T Hermitian, givenT |ei〉 = λi|ei〉;

T |ej〉 = λj |ej〉,

if λi 6= λj , then 〈ei|ej〉 = 0.

Proof:〈ei|T |ej〉 = λj〈ei|ej〉.

Since T † = T and λ is real, we have

〈ei|T |ej〉 = 〈ei|T |ej〉 = λ∗i 〈ej |ei〉 = λi〈ei|ej〉.

Hence, if λi 6= λj ,〈ei|ej〉 = 0.

The opposite is not necessarily true. If all eigenvalues are different, then the eigenvectors are unique up tophases.

Claim: Suppose that for Hermitian matrices A,B, such that

[A,B] = 0.

This is equivalent to saying that there exists a basis in which both are simultaneously diagonalisable.

Proof: Assume that A,B have distinct eigenvalues λ1, ...λn, α1, ...αn respectively. If A|ei〉 = λi|ei〉, then

BA|ei〉 = λi(B|ei〉) = A(B|ei〉).

But we have showed that the eigenvector corresponding to the eigenvalue λi is unique. Hence B|ei〉 must beproportional to |ei〉. So |ei〉 is an eigenvector for B as well.

Suppose now that not all eigenvalues are distinct, i.e., A can be diagonalised such that

A = diag(λ1, λ1, λ2, λ2, λ2, λ3, · · · ).

This implies that B is block diagonal. Now, restrict B to subspaces since the subspaces do not mix amongthemselves. In every subspace, choose a basis such that B is diagonal.

In fact, we can prove by induction that if we have a bunch of Hermitian operators which commute with eachother, i.e., for any pair i, j, we have

[Ai, Aj ] = 0,

then there exists an orthonormal basis such that all the Ai’s are diagonal.

We can label the basis as|λA1

1 λA22 · · ·λ

Akk,α〉,

representing a simultaneous eigenstate of these operators Ai’s labelled by their corresponding eigenvalues. Thenif we act, say, A2 on it, we immediately see that we can pick out λ2. To deal with repeated eigenvalues, i.e., theleftover degeneracies, add an extra label, say, α. Notice that we can write

1 =∑λ

|λA11 · · ·λ

Akk 〉〈λ

A11 · · ·λ

Akk |.

38

Now we come to the concept of unitary operators. These linear transformations are the most beautiful part ofquantum mechanics since they realise symmetries. Symmetries must be unitary.

Claim: Unitary operators preserve the inner product 〈φ|ψ〉.

Proof: For all |φ〉, |ψ〉,〈φ|U†U |ψ〉 = (〈φ|U)Uψ〉 = 〈φ|ψ〉.

Now, choose a basis |ei〉. Then

〈ei|ej〉 = 〈ei|U†U |ej〉 = δij = (U†U)ij .

HenceU†U = 1.

Claim: A unitary transformation can always be diagonalised; all diagonal elements are phases, i.e., we can put Uinto the form

U = diag(

eiθ1 , · · · , eiθk).

Proof: Recall that every matrix has at least one eigenvector. In other words, there is a basis in which we can write

U =

λ0 ?...0

.

Hence, U†U = 1 implies that|λ| = 1.

Similarly,

UU† = 1 =

λ α1 · · · αn−1

0... ?0

λ 0 · · · 0α∗1... ?

α∗n−1

.

Hence,

|λ|2 +

n−1∑i=1

|αi|2 = 1.

But we have already shown that |λ|2 = 1. Hence each αi must be zero. We can then restrict to the one-dimensionlower subspace and finish the proof by induction.

We briefly mention the anti-linear transformation. The most important example is the time reversal operatorT . It takes t 7→ −t in

ψE(x, t) ∼ eiEt~ ψn(x).

We can write every unitary matrixU = exp(iA),

where A is Hermitian. If we takeA = diag(θ1, · · · , θn).

Then,eiA = diag( eiθ1 , · · · , eiθn).

Can it be written uniquely in this way? No. We can have a family of unitary operators labelled by real numbera, such that

U(a) = eiA(a), U(0) = 1.

Suppose A(a) = aA, i.e.,U(a) = eiaA.

39

Then we say that A generates U . And we have

U(a)U(b) = eiaA eibA = ei(a+b)A = U(a+ b).

since A is Hermitian. The punchline is: Hermitian operators generate symmetries.

Now, takeU : H → H.

Through U , for some operators Ti’s we obtain an equivalence between

〈ψ|T1 · · ·Tn|φ〉 = 〈ψ′|T ′1 · · ·T ′n|φ′〉,

if we define|φ′〉 = U |φ〉; 〈ψ′| = 〈ψ|U†; T ′i = UTiU

†,

since〈ψ′|T ′1 · · ·T ′n|φ′〉 = 〈ψ|U†(UT1U

†) · · · (UTnU†)U |φ〉.

We next discuss the notion of tensor products. In physics, if we want to have more degrees of freedom, we cantensor product two Hilbert spaces together, combining their degrees of freedom. Take H1 with dimension n1 andH2 with dimension n2. Then

H ≡ H1 ⊗H2

has dimension n1n2. Take |vi〉 ∈ H1, |wj〉 ∈ H2, then

|vi〉 ⊗ |wj〉 ≡ |viwj〉 ∈ H.

And we have(a|vi〉, |wj〉) ∼ (|vi〉, a|wj〉).

Take T : H1 → H1 such thatT |vi〉 = aki|vk〉.

ThenT |viwj〉 = aki|vkwj〉.

Note also that(〈u′| ⊗ 〈w′|)(|u〉 ⊗ |w〉) = 〈u′|u〉〈w′|w〉.

Using the tensor product, we can construct huge matrices using smaller ones, which is very useful. The niceproperties of matrices are usually inherited as well.

In physics, we often use the notationT1 ≡ T1 ⊗ 1 : H1 → H1;

T2 ≡ 1⊗ T2 : H2 → H2;

T1 ⊗ T2 ≡ T1T2.

SoT1 + T2 ≡ T1 ⊗ 1+ 1⊗ T2.

Now, we can think of discrete situations as a finite-dimensional Hilbert space and a continuous one as an infinitedimensional Hilbert space. Take

ψ : R→ C.

We are interested in the physical states |ψ〉 such that

〈ψ|ψ〉 <∞.

How do we approach these in the context of the Hilbert space? We discretise. Say we have a function ψ(x).Let |x〉ε denote the function which is zero outside of its width ε centred at x. We also take its height to be 1

ε . Wecan now write an arbitrary

|ψ〉 =∑i

ψ(xi)ε|xi〉ε =

∫dxψ(x)|x〉,

40

One more thing to note is the overlap

〈φ|φ〉 =

∫〈φ|x〉dx〈x|ψ〉 =

∫φ∗(x)ψ(x)dx.

Below we list some properties of the Dirac-delta function.

• δ(x) = 0, x 6= 0,∫∞−∞ δ(x)dx = 1.

•∫∞−∞ δ(x)f(x) dx = f(0) for a non-zero function f(x).

•∫∞−∞ δ′(x)f(x) dx = −

∫δ(x)f ′(x) dx+ δ(x)f(x)|∞−∞ = −f ′(0), since δ(x 6= 0) = 0.

•∫δ(ax)d(ax) = 1 ⇒

∫δ(ax)dx = 1

a .

• δε(x) = 12π

∫∞−∞ dk eikx.

Proof: Using the regulator e−ε2k2/4, we have∫ ∞

−∞

dk

2πeikx−

ε2k2

4 =

∫ ∞−∞

dk

2πe−

ε2

4 (k− 2ixε2

)2− x2

ε2 =e−

x2

ε2

2π

√4π

ε2=

1√πε

e−x2

ε2 = δε(x).

Now, in three-dimensions, we simply have

〈x′y′z′|xyz〉 = 〈x′|x〉〈y′|y〉〈z′|z〉.

So〈~r|~r〉 = δ(x− x′)δ(y − y′)δ(z − z′) = δ3(~r − ~r′).

We can also write

|ψ〉 =

∫ψ(x, y, z)|xyz〉 dxdydz,

and in general,

|ψ〉 =

∫ψ(~r1, ~r2, ...)|x1y1z1 x2y2z2 · · · xnynzn〉 d3r1 · · · d3rn.

Note that we are only taking tensor products of the degrees of freedom of spaces.

3.2 Principles of Quantum Mechanics

• For any physical system, we introduce a Hilbert space H. H depends on the system we are interested in.

• The state of a physical system is represented by |ψ〉 ∈ H. It’s normalisable and we take 〈ψ|ψ〉 = 1.

• Physically measurable quantities correspond to Hermitian operators acting on H. The result of measurementof Ai yields only the eigenvalues of Ai, namely, λi. Moreover,

P (Ai = λi) = |〈λi|ψ〉|2.

Now, given |ψ(0)〉, what is |ψ(t)〉? The time evolution operator is unitary and we have

|ψ(t)〉 = U(t)|ψ(0)〉,

and there is only one unitary operator for all states. A time evolution operator is generated by the Hermitianoperator Hamiltonian. We focus on the time-independent Hamiltonian for now. So we have

U(t) = eiHt/~.

42

Hence,∂

∂t|ψ(t)〉 =

∂U(t)

∂t|ψ(0)〉 = − i

~HU(t)|ψ(0)〉 = − i

~H|ψ(t)〉.

And we arrive at

i~∂

∂t|ψ(t)〉 = H|ψ(t)〉.

We have been and will be discussing the Schrodinger picture, where states depend on time and operators aretime-independent, as in

〈ψ(t)|A1 · · ·An|ψ(t)〉.

There is also the Heisenberg picture, where operators depend on time and states are time-independent, as in

〈ψ(0)|A1(t) · · · An(t)|ψ(0)〉.

The two pictures are equivalent by the isomorphism between the two Hilbert spaces, since we can go from oneto the other through a unitary operator. As |ψ(t)〉 = U(t)|ψ(0)〉, we also have

|ψ(0)〉 = U†(t)|ψ(t)〉.

Along withAn(t) = U†AnU,

we have〈ψ(0)|A1(t) · · · An(t)|ψ(0)〉 = 〈ψ(t)|UU†A1U · · ·U†AUU†|ψ(t)〉 = 〈ψ(t)|A1 · · ·An|ψ(t)〉.

In the Hamiltonian formulation of classical physics, we have the Poisson brackets

f(x, p), g(x, p)PB =∑i

(∂f

∂xi

∂g

∂pi− ∂g

∂xi

∂f

∂pi

),

giving usx, pPB = 1.

Quantum mechanically, we have the corresponding commutator relation of

[x, p] = i~.

Now, we can think of translation as a unitary operator. Given x|x′〉 = x′|x′〉, with the unitary translationoperator U(a), we have

x(U(a)|x′〉) = (x′ + a)U(a)|x′〉,

i.e.,U−1(a)xU(a)|x′〉 = (x′ + a)|x′〉.

Hence, U(0) is the identity operator, and since we are dealing with translation, we require U(a)U(b) ∼ U(a+b),which gives us

U(a) = e−iaA/~.

We say that A generators translation, and demand

eiaA/~x e−iaA/~!= x+ a

for small a. But if we simply compute, we end up with

eiaA/~x e−iaA/~ = x+ia

~[A, x] +O(a2).

We therefore demand that[A, x]

!= −i~.

It is only natural at this point to set A ≡ p since

[x, p] = i~.

43

Hence, the momentum operator p = −i~ ddx generates translation, and we take

U(a) = eipa/~ = e−addx .

Note that the reason why we have such pairs as x, p is because one generates the other. We have

〈x|U(a)|ψ〉 =(〈x|U†(a)

)|ψ〉 = 〈x− a|ψ〉 = ψ(x− a) = U(a)ψ(x) =

∑n

1

n!

(−a d

dx

)nψ(x).

And altogether, we have the commutation relations

[xi, pj ] = i~δij ; [pi, pj ] = 0,

since schematically,

U1(ε1)U2(ε2)U−11 (ε1)U−1

2 (ε2) ∼ eε1A1 eε2A2 e−ε1A1 e−ε2A2 = 1+ ε1ε2[A1, A2] +O(ε3)!= 1.

Next, we can write〈x|p′〉 = ψp′(x).

Along withp|p′〉 = p′|p′〉,

we have

〈x|p|p′〉 = −i~∂ψp′

∂x= p′ψp′(x).

Then,ψp′(x) = C eip

′x/~.

We want 〈p|p′〉 = δ(p− p′), so to find the normalisation constant C, compute∫ ∞−∞

dx 〈p|x〉〈x|p′〉 =

∫ ∞−∞

dxC∗C ei(p′−p)x/~

= |C|22πδ

(p′ − p~

)= 2π|C|2~ δ(p′ − p).

Hence,

〈x|p〉 =eipx/~√

2π~, 〈p|x〉 =

e−ipx/~√2π~

,

and

|p〉 =

∫ ∞−∞

dx′eipx

′/~√

2π~|x′〉; |x〉 =

∫ ∞−∞

dp′e−ip

′x/~√

2π~|p′〉.

Next, given〈x|S〉 = ψ(x), 〈p|S〉 = φ(p),

how can we related the two?

〈p|S〉 =

∫ ∞−∞〈p|x〉〈x|S〉 dx =

∫ ∞−∞

e−ip′x/~

√2π~

ψ(x) dx = φ(p);

〈x|S〉 =

∫ ∞−∞〈x|p〉〈p|S〉 dp =

∫ ∞−∞

eip′x/~

√2π~

φ(p) dp = ψ(x).

Note also that we can write

〈x1x2x3|p1p2p3〉 = 〈x1|p1〉〈x2|p2〉〈x3|p3〉 = (2π~)−32 ei~p·~r/~ = 〈~r|~p〉.

44

Now we consider the spread ∆ψA of an observable corresponding to the operator A in a state ψ. WithA ≡ 〈ψ|A|ψ〉, we have

(∆ψA)2 ≡ 〈ψ|(A− A)2|ψ〉 = 〈ψ|A2 − 2AA+ A2|ψ〉 = A2 − 2AA+ (A)2.

Hence,(∆ψA)2 = 〈ψ|A2|ψ〉 − 〈ψ|A|ψ〉2.

We now prove the general statement of the uncertainty principle

∆A∆B ≥ 1

2|〈ψ|[A,B]|ψ〉|,

which implies the famous

∆x∆p ≥ ~2.

Proof: Set|ψA〉 ≡ (A− A)|ψ〉;

|ψB〉 ≡ (B − B)|ψ〉.

The Schwartz inequality tells us that

|〈ψA|ψB〉|2 ≤ 〈ψA|ψA〉〈ψB |ψB〉,

whereLHS = |〈ψ|AB − AB|ψ〉|2 ≡ |C|2 ≥ |ImC|2.

But

ImC =〈ψ|AB − AB|ψ〉 − 〈ψ|BA− AB|ψ〉

2i=

1

2i〈ψ|[A,B]|ψ〉.

Hence,

|〈ψA|ψB〉|2 ≥1

4|〈ψ|[A,B]|ψ〉|2.

But〈ψA|ψA〉 = 〈ψ|(A− A)2|ψ〉 = (∆A)2.

Hence we arrive at

∆A∆B ≥ 1

2|〈ψ|[A,B]|ψ〉|.

Now, just like how the momentum operator generates spatial translation, the Hamiltonian operator generatestime translations. In this sense, quantum mechanics is ready to made special relativistically invariant. Spatialtranslations and time translations are both realised through unitary operators, and since

[H,U ] = 0,

we say that U is a symmetry of our problem. The fact that it’s generated by A implies that

[H,A] = 0.

Claim: [H,A] = 0 is equivalent to saying that A is conserved.

Proof: TakeU(t) = e−iHt/~.

Then,∂U

∂t= − i

~HU ;

∂U†

∂t=i

~U†H.

So we have Ehrenfest’s theorem

d

dt〈ψ(t)|A|ψ(t)〉 =

d

dt〈ψ(0)|U−1(t)AU(t)|ψ(0)〉 =

1

i~〈ψ(t)|[A,H]|ψ(t)〉.

45

Hence[H,A] = 0 ⇔ A is conserved.

In this case, U(t) only acts on the energy and will not change the eigenvalue of A. So conservation law followsfrom commutation with H.

We can also have discrete symmetries. For example, in the Harmonic oscillator case, parity operator

P : x→ −x

is a symmetry. The discrete symmetries are also given by unitary operators which preserve norms. Here

U2 = 1,

and the only possible eigenvalues are α = ±1. So without solving the harmonic oscillator problem, we alreadyknow that the eigenstates are even or odd functions of x.

Now, consider some physical system with Hamiltonian H. What are all the symmetries of the problem? Everysymmetry can be represented by some unitary operator Uα such that

UαHU−1α = H.

This is the symmetry statement. The symmetries of a system form a group. Now, the (unitary) realisation orrepresentation of a group with operation ∗ satisfies

U(g1)U(g2) = U(g1 ∗ g2);

U(1G) = I.

-

Aside: Just to be mathematically clear, a representation D(G) of a group G is a linear group action v 7→ D(g)von a vector space V , by invertible transformations. D(g) ∈ GL(dim(V )) ∀ g ∈ G, and by linearity,

D(g)(av1 + bv2) = aD(g)v1 + bD(g)v2.

To be a group action D must satisfyD(g1g2) = D(g1)D(g2).

Also note thatD(1) = 1, D(g−1) = D(g)−1,

(These are not properties of unitary representations only!)

Another point is that we must choose a basis for V to have explicit matrix representations.

-

For every group element, we have a representation of it acting on the symmetries of the Hilbert space, i.e., thereis a homomorphism between the two. If U2 = 1, then there is a basis

U = diag(+1, · · · ,+1,−1, · · · ,−1).

Now, if g1 ∗ g2 6= g2 ∗ g1, thenU(g1)U(g2) 6= U(g2)U(g1).

So the corresponding representations of the symmetries in the quantum theory will not be simultaneouslydiagonalisable. Then we can choose a basis which is as close to diagonalisation as possible, i.e., we can blockdiagonalise.

Next, we have used without proof that〈ψ|A|ψ〉 = A.

46

Proof: Take

|ψ〉 =∑i

ci|ai〉.

Then〈ψ|A|ψ〉 =

∑i,j

(〈aj |cj)A(ci|ai〉) =∑i,j

cic∗j 〈ai|A|aj〉 =

∑j

|cj |2aj = A.

In the context of statistical mechanics, we also have probabilities. And in quantum physics, we can also takea thermal bath. How do we think about the intrinsic probability of a quantum system together with thermalprobability? For every state |ψ〉, we introduce an operator

ρψ = |ψ〉〈ψ|.

Then,〈ψ|An · · ·A1|ψ〉 = Tr(|ψ〉〈ψ|A1 · · ·An) = Tr(ρψA1 · · ·An).

Note that ρψ is a projection operator since 〈ψ|ψ〉 = 1, giving

ρ2ψ = ρψ.

For any projection operator P , its eigenvalue too satisfies λ2 = λ. Hence we can have the basis

P = diag(1, · · · , 1, 0, · · · , 0).

And from∑i |ei〉〈ei| = 1, we have ∑

i

Pei = 1.

If we don’t know which state we are in, but only the statistical probability pα to be in a state, we write

ρ =∑α

|ψα〉pα〈ψα|.

Then we have ∑α

pα〈ψα|A1 · · ·An|ψα〉 = Tr(ρA1 · · ·An).

Note that since∑α pα = 1 and pα ≥ 0, we have

Trρ = 1,

and we can choose the basisρ = ρ† = diag(p1, · · · , pn).

This is what we call a mixed state of a quantum system. A pure state is one where, for example,

ρ = diag(1, 0, · · · , 0).

In statistical mechanics, using the eigenstates of energy, the density matrix

ρ = |Eα〉 e−βEα〈Eα|.

Then,

Tr(ρA) =

∑α〈Eα|A|Eα〉 e−βEα∑

α e−βEα.

How do we translate objects like entropy S in statistical mechanics to quantum mechanics? Suppose we have anN−dimensional Hilbert space, i.e., we take N of |ei〉’s. If we don’t know which state we are in, then what shouldρ be?

ρ = diag

(1

N, · · · , 1

N

)

47

satisfies Trρ = 1. This is when we have a maximal amount of entropy, since all possible situations are equally likelyhere. This system has N configurations, so by definition

S = lnN.

So we can takeS = −Tr(ρ ln ρ)

to be the quantum entropy associated with a state, which is always positive. Note also that for the pure state case,we have S = 0.

4 Spins

Around 1925, physicists were trying to figure out the spectrum of atoms. The splitting lines showed that therewere more states than expected. So it was suggested that there’s something between l = 0 (which has one state) andl = 1 (which has three states). They called this extra degree of freedom spin. Spin is basically the representationof rotation in a quantum system.

The rotation group is given by the transformations

Xi → RijXj .

In two dimensions for example, we can take(xy

)=

(cos θ − sin θsin θ cos θ

)(xy

).

And the group is abelian in the sense that

R(θ1)R(θ2) = R(θ1 + θ2).

It’s not the case in higher dimensions. In three dimensions, we can take

Rxy(θ) =

cos θ − sin θ 0sin θ cos θ 0

0 0 1

.

Then,[Rxy, Ryz] 6= 0,

i.e., they are not simultaneously diagonalisable. Rotations R have the property

RtR = 1.

Notice thatx→ Rx, y → Ry.

(Ry)tRx = ytx.

Let’s expandR = 1 + iεijTij , Rt = 1+ iεijTij .

ThenTij = −Tji

since RtR = 1. So Tij is an antisymmetric matrix. For example, in three dimensions,

T12 =

0 −i 0i 0 00 0 0

.

48

Hence antisymmetric matrices generate rotations. And it has the algebra

[Tij , Tkl] = −iTilδjk + iTkjδil + iδjlTik − iδikTlj .

The number of generators in n dimensions is

1

2n(n− 1)

since they are anti-symmetric.

-

Aside: For those who like math, here we give a relevant discussion on Lie groups and Lie algebras.

A Lie group is a group with the structure of a smooth manifold such that group operation G×G→ G, (g1, g2) 7→g1g2 and inverse G→ G, g 7→ g−1 are smooth maps between manifolds.

The Lie algebra g of a Lie group G is the tangent space T1G to G at the identity element 1 ∈ G. Withoutconsidering the Lie group, we can give the definition that Lie algebra is a vector space g equipped with an anti-symmetric bilinear operation [ , ] : g× g→ g which satisfies the Jacobi identity.

Now, in class we talked about the rotation group in three-dimensions, which is nothing but the Lie group SO(3).Here, from a more or less mathematical perspective, we discuss SO(3) and its Lie algebra so(3), as well as SU(2)and su(2), which will soon be useful.

SO(3) = 2×2 real matrices O with OTO = 1, det(O) = 1.

Let’s consider a curve O(t) = 1 + tX +O(t2) ∈ SO(3) with O(0) = 1. OTO = 1 gives

d

dt[OTO]t=0 = [OTO +OT O]t=0 = OT + O = 0.

detO = 1 gives no extra constraint since all elements in SO(3) near the identity have det = 1. Hence

so(3) = vector space of 3×3 real antisymmetric matrices.

Now, let so(3) have bases

T1 =

0 0 00 0 −10 1 0

, T2 =

0 0 10 0 0−1 0 0

, T3 =

0 −1 01 0 00 0 0

.

Then,(Ta)bc = −εabc.

Hence,

[Ta, Tb]cd = (Ta)ec(Tb)de − (Tb)ec(Ta)de

= εaceεbed − εbceεaed= δadδcb −δabδcd − δbdδca +δbaδcd

= εeabεedc

= −εabeεecd= εabe(Te)cd.

Therefore we have the Lie bracket structure

[Ta, Tb] = εabcTc.

Next,SU(2) = 2×2 complex matrices U with U†U = I, det(U) = 1.

49

Consider a curve U(t) := 1+ tZ +O(t2) ∈ SU(2) with U(0) = 1. U†U = 1 gives

d

dt[U†U ]t=0 = [U†U + U†U ]t=0 = U† + U = Z† + Z = 0.

Also, in matrix form,

U(t) =

(1 + tZ11 tZ12

tZ21 1 + tZ22

).

detU = 1 gives1 + t(Z11 + Z22) +O(t2) = 1 + tTr(Z) +O(t2) = 1

⇒ Tr(Z) = 0.

Hencesu(2) = vector space of 3 × 3 complex antihermitian traceless matrices.

Let su(2) have bases

Ta = −1

2iσa

where σa are the Pauli matrices, which have the property σaσb = δabI + iεabcσc.

⇒ [Ta, Tb] = −1

4(σaσb − σbσa) = −1

2iεabcσc = εabcTc.

Since the Lie bracket structure for su(2) and so(3) are the same, the Lie algebras are isomorphic. In fact, SU(2)is the universal covering group of SO(3) so their Lie algebras must be isomorphic. The isomorphism is induced bythe universal covering map.

-

We continue our discussion of the quantum theory. Recall that we had e−iaJ/~. Since the exponential of J/~generates unitary transformations, we can take for the representation of rotation in Hilbert space,

Tij 7→ Jij/~,

i.e., Jij/~ should satisfy the same algebra as Tij . In three dimensions,

Ji =1

2εijkJjk.

Hence, by the symmetry of rotations,[Ji, Jj ] = i~εJk.

Explicitly, we can take

J1 =

0 0 00 0 −i0 i 0

, J2 =

0 0 i0 0 0−i 0 0

, J3 =

0 −i 0i 0 00 0 0

.

Now, for any vector V j ,[Ji, Vj ] = i~εijkVk.

We want to represent the rotation of a vector, i.e.,

U−1(R)ViU(R) = RijVj ,

where U−1(R) and U(R) are unitary operators and Rij is a number. So we have both

[Ji, Lj ] = i~εijkLk

and[Li, Lj ] = i~εijkLk.

50

Hence,[Ji − Li, Li] = 0.

We defineSi = Ji − Li

to be the extra piece that rotations could include, and call it spin. We have

[Si, Sj ] = [Ji − Li, Jj − Lj ] = i~εijkSk.

So how do we represent such objects as Li, Si, or Ji, which satisfy such commutation relations, in the Hilbertspace?

Let’s pause for a moment to again consider the hydrogen atom. We are saying that it has the possibility of theextra piece of spin. Since [S,L] = 0, S has no angular degree of freedom and is independent of the physical space.So we need to tensor our theory with a new Hilbert space, i.e., H → ⊕α,jHαj . In other words, since S has newdegrees of freedom, we need to represent it.

Now, recall our maximally commuting[J2, J3] = 0.

It’s useful to think in terms of them. Define

J± ≡ J1 ± iJ2.

Then,[J3, J±] = ±~J±.

Recall that J3|m〉 = m~|m〉. Hence,

J3(J+|m〉) = (J3J+ − J+J3 + J+J3)|m〉 = ~(m+ 1)J+|m〉.

Similarly,J3(J−|m〉) = ~(m− 1)J−|m〉.

Note that if we assumeJ2|m,α〉 = ~2α|m,α〉,

where α doesn’t change, then there exists

J+|mf,α〉 = 0, J−|mi,α〉 = 0,

since〈m,α|J2|m,α〉 = ~2α ≥ m2~2.

In other words, if α is fixed, then m cannot be too big. α is determined by J2 and J± cannot shift it, since

[J2, J±] = 0.

Next, we haveJ2

1 + J22 = (J1 ± iJ2)(J1 ∓ iJ2)± i[J1, J2].

Hence, we can write in two waysJ2 = J±J∓ ∓ ~J3 + J2

3 .

Hence,J2|mf , α〉 = ~2α|mf , α〉 = (J−J+ + ~J3 + J2

3 )|mf , α〉,

giving~2α = ~2(mf +m2

f ).

Similarly,J2|mi, α〉 = ~2α|mf , α〉 = (J+J− − ~J3 + J2

3 )|mi, α〉,

51

giving~2α = ~2(m2

i −mi).

So we havemi(mi − 1) = mf (mf + 1).

We require that mi < mf , so we take the solution

mf = −mi ≡ j, j ∈ Z or Z +1

2,

since mf = mi + k = −mf + k, k ∈ Z gives mf = k/2. Now, there are 2j + 1 states in the ladder. We havecomputed that

J2 = ~2α = ~2j(j + 1).

We’ll from now on denote states as|m, j〉, − j ≤ m ≤ j.

So we haveJ3|m, j〉 = m~|m, j〉;

J2|m, j〉 = ~2j(j + 1)|m, j〉.

Notice that if j is an integer (half-integer), then m is also an integer (half-integer). Now, j3 is a generation ofrotation around the z−axis, so if we want to act a rotation of 2π on the state, then write

e2πiJ3/~|m, j〉 = e2πim|m, j〉.

So if m is a half-integer, the state only comes back up to a minus sign. This didn’t happen for, say, the hydrogenatom.

J±|m, j〉 = C|m± 1, j〉.

Then,〈m, j|J∓J±|m, j〉 = |C|2〈m± 1, j|m± 1, j〉 = |C|2

ButJ∓J± = J2 − J2

3 ∓ ~J3.

Hence,|C|2 = ~2[j(j + 1)−m(m± 1)].

The phase of C is fixed such that

J±|m, j〉 = ~√j(j + 1)−m(m± 1)|m± 1, j〉.

In summary, we have shown that every time we have an algebra of angular momentum, we can find a ladder-likestructure. Of course, a Hilbert space can have ladders of many different sizes. Each of the ladders is called anirreducible representation (irrep). So we are decomposing our Hilbert space into irreps of the rotation group.

-

Aside: Let D be a representation of G acting on a vector space V . D is reducible if there exists a proper, invariantsubspace W ⊂ V , i.e., there exists a subspace W such that D(G)W = W . If no such subspace exists, then D is anirreducible representation.

-

We briefly discuss the Stern-Gerlach experiment, which originally determined the existence of spins. A beamof material is passed through a region of strong magnetic fields, and split into two. We have

H =p2

2m+ α~S · ~B,

52

where we assume interaction between the ~S and ~B. Ehrenfest’s theorem says

dOdt≡ d

dt〈ψ|O|ψ〉 =

i

~〈ψ|[H,O]|ψ〉

We have the unsurprisingdx

dt= 〈 ~p

m〉.

Next,dpidt

=i

~α〈[~S · ~B(x),−i~ ∂

∂xi]〉,

i.e.,

md2xidt2

= −α〈∂i ~B · ~S〉.

Suppose the beam is in the z−direction, and the gradient is also in the z−direction. Then we get a beam splitdepending on whether ~S is upward or downward. So basically, the spin causes a force which acts on the beam tosplit it.

4.1 Addition of Angular Momentum

We refer to expressions like~J = ~S + ~L

as the addition of angular momentum. Let’s now take general

J = J ′ + J ′′.

A general state can be written as|j′m′, j′′,m′′〉 ⊂ Hj

′,j′′

and the Hilbert space has (2j′ + 1)(2j′′ + 1) dimensions. Now, it’s easy to count the ladders. If we were givendifferent ladders with different sizes, we can decide immediately just by looking at the spectrum of j3’s what thepossible values of j are. We have

J3 = J ′3 + J ′′3

such thatJ3|j′m′, j′′,m′′〉 = ~(m′ +m′′)|j′m′, j′′,m′′〉.

Let’s check the following cases.

• j′ = 0, j′′ arbitrary gives j = j′′;

• j′ = 12 , j′′ arbitrary gives j = (j′′ + 1

2 )⊕ (j′′ − 12 );

• general j′, j′′, where j′ > j′′ gives

j′ + j′′ j′ + j′′ − 1 j′ + j′′ − 2 · · · j′ − j′′ · · · −j′ + j′′ · · · −j′ − j′′# of states 1 2 3 2j′′ + 1 2j′′ + 1 1

i.e.,j = j′ ⊗ j′′ = (j′ + j′′)⊕ (j′ + j′′ − 1)⊕ · · · ⊕ (j′ − j′′) = ⊕|j′−j′′|≤j≤j′+j′′j.

Note thatdim(LHS) = (2j′ + 1)(2j′′ + 1) =

∑|j′−j′′|≤j≤j′+j′′

(2j + 1) = dim(RHS)

53

Expressed in terms of the Clebsch Gordon coefficients Cj′j′′ ,

|j′m′, j′′m′′〉 → |j,m〉 =∑

Cj′j′′(j,m;m′,m′′)|j′, j′′〉.

So for example,|j = j′ + j′′,m = j′ + j′′〉 = |j′, j′′〉,

i.e., Cj′j′′ = 1. We now find the next state.

|j′ + j′′, j′ + j′′ − 1〉 = J−|j′ + j′′, j′ + j′′〉= (J ′− + J ′′−)|j′, j′′〉

= ~√j′(j′ + 1)− j′(j′ − 1)|j′ − 1, j′′〉+ ~

√j′′(j′′ + 1)− j′′(j′′ − 1)|j′, j′′ − 1〉.

However, in order to normalise, we need to divide the expression by

~√

(j′ + j′′)(j′ + j′′ + 1)− (j′ + j′′ − 1)(j′ + j′′).

Next, the Hamiltonian

H = α~S · ~L =α

2[(~S + ~L)2 − S2 − L2] =

α

2[J2 − J ′2 − J ′′2],

where J2, J ′2, J ′′2 have eigenvalues ~2j(j + 1), ~2s(s+ 1), ~2l(l + 1) respectively. Notice that

[J2, J ′2] = [(J ′ + J ′′)2, J ′2] = [J′2 +J ′′2 + 2J ′ · J ′′, J ′2] = 2[J ′i , J

′2]J ′′ = 0

since J ′i is the generator of rotation around the ith axis, and is a scalar. Hence, we can go from a basis where

J ′2, J ′′2, J ′3, J′′3

are diagonal to one whereJ ′2, J ′′2, J2, J3

are diagonal.

We now discuss the Pauli matrices. They are special since the spin of the electron s = 12 . 2S + 1 = 2 so there

are two states. Hence,S3 = (~/2,−~/2).

Note the notations|+ 1/2〉 = | ↑〉 = (1, 0);

| − 1/2〉 = | ↓〉 = (0, 1).

Then,

S3 =~2σ3, σ3 =

(1 00 −1

).

We have

S+| ↓〉 = ~√

1

2

3

2− 1

2

−1

2| ↑〉 = ~| ↑〉, S+| ↑〉 = 0.

Similarly,S−| ↑〉 = ~| ↓〉, S−| ↓〉 = 0.

Recall that

S1 =1

2(S+ + S−), S2 =

1

2i(S+ − S−).

Hence,

S1 =~2σ1, σ1 =

(0 11 0

);

S2 =~2σ2, σ2 =

(0 −ii 0

).

54

Note the properties of the sigma matrices that

σ2 = 1, σiσj = iεijkσk.

Let’s consider rotation around the θ axis.

Claim:

ei~θ·~σ/~ = cos

θ

21− i sin

θ

2(σ · θ).

Proof: Suppose ~θ = θz. Then

eiθσ3/2 =

∞∑n=0

1

n!

(−iθ σ3

2

)n=∑n even

1

n!

(θ

2i

)n+∑n odd

1

n!

(θ

2i

)nσ3 = cos

θ

21− i sin

θ

2σ3.

Now, every particle that has mass has internal spin S = 0, 1/2, 1, ... Fermions have half integer spin; bosons haveinteger spin. An even number of fermions gives bosons, any number of bosons give bosons, and an odd number offermions with any number of bosons give fermions. You can have arbitrary spins for the massive particles but...

For massless particles, you cannot take all possible values for S3. For example, for a spin-1 particle, you canonly have the states of ±1, not 0. This corresponds to the photon, which only has two helicities, ±1, correspondingto left- or right-handed polarisations: there is no “no-handed” polarisation. Also, the graviton has spin-2 but if wehave a massless particle of spin greater than 2, we won’t have a unitary quantum theory. So the massless case isvery limited.

4.2 Identical Particles

Now, all electrons in the universe are the same. Suppose we have a state of n identical particles, then

|ψ〉n ∈ H1 ⊗H1 ⊗ · · · ⊗ H1,

where H1 denotes the Hilbert space of one particle and we have n copies of it. Let |vi1〉 be a basis for H1. Take theunitary operator Pij such that

Pij |vα11 · · · v

αii · · · v

αjj · · · v

αnn 〉 = |vα1

1 · · · vαjj · · · v

αii · · · v

αnn 〉.

The eigenvalues are ±1. For bosons corresponding to (H⊗n1 )symm, Pij = 1 and for fermions corresponding to(H⊗n1 )anti-symm, Pij = −1 for all i, j. We list a few k−particle cases below, along with the basis for the Hilbertspace corresponding to the k particles.

• 1 fermion: |vi11 〉, dim = N ;

• 2 fermions: |vi11 vi22 〉 − |v

i21 v

i12 〉, dim =

(N2

);

• 2 bosons: |vi11 vi22 〉+ |vi21 v

i12 〉, dim = 1

2N(N + 1). So we can take the wavefunction

ψ(~x1, ~x2) =∑α,β

(ψα(~x1)ψβ(~x2)± ψα(~x2)ψβ(~x1)) .

• 3 fermions: Take i, j, k = 1, ..., N , dim =

(N3

).

det

|αi1〉 |αi2〉 |αi3〉|αj1〉 |αj2〉 |α

j3〉

|αk1〉 |αk2〉 |αk3〉

= |ijk〉+ |kij〉+ |jki〉 − |jik〉 − |ikj〉 − |kji〉 (∗).

So for example, P12(∗) = −(∗).The antisymmetric property of fermions corresponds to the Pauli exclusion principle.

55

4.3 Electrons and the Nucleus

We have the Hamiltonian which includes the Coulomb attraction between the electron and the nucleus, as wellas the repulsion between pairs of electrons.

H =

Z∑i=1

(p2i

2m− Ze2

|~ri|

)+∑i<j

e2

|~ri − ~rj |.

The repulsion term is hard to deal with, so we don’t consider it for the moment. Then, we have a diagonalHamiltonian

H =

Z∑i=1

(p2i

2m− Ze2

|~r1|

)=

Z∑i=1

Hi.

Taking|ψ〉 = |E1,α1 , E2,α2 , · · · , En,αn〉,

we see that we get the eigenstates by simply writing H as the individual Hi’s.

H|ψ〉 =n∑i=1

Ei|ψ〉.

For the hydrogen, we can write|E1, α1〉 = |n1, l1,m1, s = ±1/2〉.

Notice that we are dealing with fermions, so the labels must be distinct. Let’s look at some of the lowest energystates (n = 1). Hydrogen has

|n = 1, l = 0,m = 0, s = ±1/2〉

where the first three quantum numbers tell us that we are in the 1s1 states. Helium has

|1, 0, 0, 1/2; 1, 0, 0,−1/2〉 − |1, 0, 0,−1/2; 1, 0, 0, 1/2〉,

which, with normalisation, can be written as

1√2

(| ↑↓〉 − | ↓↑〉).

In a two-dimensional Hilbert space, we cannot make a three-part state antisymmetric. So we go to 2s1, etc.The Hilbert space is decomposed in terms of the tensor product of the spatial degrees of freedom and the spinones. Now, the magic numbers 2, 8, 18, ... come from adding the number of states for each l as follows, where 2corresponds to the two spin states.

• l = 0: 1× 2 states;

• l = 1: 3× 2 states;

• l = 2: 5× 2 states; etc.

Are there magic numbers associated with the nuclei? Let’s try to look at it in terms of the harmonic oscillator.

• |000〉s: 1× 2 states;

• |100〉p: 3× 2 states;

• |200〉 and |110〉: 6× 2 states; etc.

So the magic numbers here are 2 (He), 8 (16O), 20 (40Ca), 40, ... No! In fact, we should get 28 instead of 40. A fewcomplications arise due to the spin-orbit coupling.

56

4.4 Thermodynamics

Pi ∝ exp

[−(EikT− µN

kT

)].

For bosons,

ZB =∑n

e−nx =1

1− e−x,

and

NBi =

∑iNi e−Ni(Ei−µ)/kT∑j e−Nj(Ej−µ)/kT

=1

e(Ei−µ)/kT − 1.

For fermions,ZF = 1 + e−(Ei−µ)/kT ,

and

NFi =

e−(Ei−µ)/kT

1 + e−(Ej−µ)/kT=

1

e(Ei−µ)/kT + 1.

Note that for the Bose-Einstein condensation, we take T = 0, then all bosons are in the same quantum state.So together they behave like one gigantic particle. Notice that for the fermionic case,

NFi =

1, E < µF

0, E > µF,

where µF is the Fermi energy.

4.5 Crystals

A crystal is an amalgamation of atoms with some periodicity structure. The Hamiltonian commutes with thetranslation operator, i.e.,

U(n1~L1 + n2

~L2 + n3~L3)H = HU(n1

~L1 + n2~L2 + n3

~L3).

So we can choose a basis of eigenstates of the Hamiltonian, for which both the Hamiltonian and this operatorare diagonal simultaneously. The H can be very complicated, but if the crystal has exact symmetry under thistranslation, the we can use it to analyse the system without solving it. The translation operators should themselvescommute due to the symmetry, so

U(~L1)U(~L2) = U(~L2)U(~L1).

We can diagonalise them to some phase, i.e.,

U(~L1)| 〉 = e−iα1 | 〉,

and similarly for U(~L2) and U(~L3). Hence, there exists some state |~α, β〉 such that

U(n1~L1 + n2

~L2 + n3~L3)|~α, β〉 = ei~n·~α|~α, β〉.

So without solving anything, we know that

H|~α, β〉 = E(~α, β)|~α, β〉.

Recall that the momentum operator generates translations. So the ~αi’s seem like momentum, except that theyare defined from −π to π. Let’s take

αi ≡ Liqi/~.

57

The states are called Bloch waves. Now let’s forget the ~ and take

ψ(x) = ei~q·~x = ei~q·Ljnjψ → ein·αψ

as ~x → ~x + n · ~L. In other words, acting on the state with translation gives a phase times itself. Also note thatshifting

~q → ~q +mj~Qj ,

which givesαl = ~q · ~Ll → (~q +mj

~Qj) · ~Ll,is equivalent to taking

αl → αl + 2πδjlmj = αl + 2πml.

So we have a dual space made of ~Q1, ~Q2, ~Q3. You can specify any ~Q inside this region, and the region is calledthe Brillouin zone. In this case we have the energy eigenstates such that

H|~q, β〉 = Eβ(~q)|~q, β〉.

Hence, for each value of ~q, inside the Brillouin zone, we have an infinitely many energy eigenstates, and eachof them is called an energy band. And the band structure is crucial in characterising conductors, semi-conductors,and insulators.

4.6 Wigner-Eckart Theorem

Suppose we have some J which satisfies the commutation relations of angular momentum, and suppose we havesome state |j,m, α〉. Let’s take also |j′,m′, β〉. Now, if j′ 6= j, m′ 6= m, then the inner product between the twovanishes. If they were equal, then...

Claim:〈j,m, β|j,m, α〉 = 〈j,m− 1, β|j,m− 1, α〉.

Proof: We have

|j,m− 1, α〉 =J−

~√j(j + 1)−m(m− 1)

|j,m, α〉;

|j,m− 1, β〉 =J−

~√j(j + 1)−m(m− 1)

|j,m, β〉.

Let’s write the latter as

|j,m, β〉 =J+

~√j(j + 1)−m(m− 1)

|j,m− 1, β〉.

So we also have

〈j,m, β| = 〈j,m− 1, β| J−

~√j(j + 1)−m(m− 1)

.

Hence,

〈j,m− 1, β|j,m− 1, α〉 = 〈j,m− 1, β| J−

~√j(j + 1)−m(m− 1)

|j,m, α〉 = 〈j,m, β|j,m, α〉.

In fact, we can generalise it to

〈j,m, β|O|j,m, α〉 = 〈j,m− 1, β|O|j,m− 1, α〉,

as long as [O, ~J ] = 0. It is clear from the manipulation we did above. Next, take O = eiHt/~. Then [H, ~J ] = 0

gives [O, ~J ] = 0. Recall how probability is represented. The Wigner-Eckart theorem basically tells us that theprobabilities

|〈j,m, β|O|j,m, α〉|2 = |〈j,m− 1, β|O|j,m− 1, α〉|2.We’ll soon see some remarkable consequence.

58

4.6.1 Tensorial Operators

Take the dipole moment operator ~d = q~x, which corresponds to l = 1 in the hydrogen wavefunction. We canalso have the quadruple tensor modes xixj− 1

3δijx2, which corresponds to l = 2, etc. The decay rate is proportional

to the matrix element of the dipole moment,〈β|~d|α〉.

To find out more about it, let’s first take

Omj , − j ≤ m ≤ j

to be an operator that’s like a state. We have

~x =

(x3,

1√2

(x1 + x2),1√2

(x1 − x2)

).

Then here

O01 = x3, O±1

1 =1√2

(x1 ± x2)

are the tensorial operators, i.e., the collection of the three operators. And we have

[J3,Omj ] = m~Omj .

We also have[J±,Omj ] = ~

√j(j + 1)−m(m± 1)Om±1

j .

Recall thatJ3 = −i~(x1∂2 + x2∂1) = ~(x+∂+ − x−∂−).

Then,J+ = ~(−x+∂3 + 2x3∂−);

J− = ~(x−∂3 − 2x3∂+).

Let’s write〈j′′,m′′, β|Omj |j′,m′, α〉 = f(j, j′, j′′,m,m′,m′′).

Can we compute f? We can say that it is a function of β, α,O. Let’s denote the reduced matrix element as

〈β||O||α〉.

Now, for any state |∗〉 considerOmj |∗〉

and|j,m〉 ⊗ |∗〉.

We’ll establish a correspondence between the two. Take

J3(Omj |∗〉) = J3Omj −Omj J3 +Omj J3|∗〉 = m~Omj |∗〉+Omj J3|∗〉.

On the other hand,J3|j,m〉 ⊗ |∗〉 = m~|j,m〉 ⊗ |∗〉+ |j,m〉 ⊗ J3|∗〉.

Thus we see the correspondence. Put it in another way, we can define

Omj |0〉 ≡ |j,m〉, ~J |0〉 = 0.

Then, commutation is equivalent to action in the sense that

[ ~J,Omj ]|0〉 = ~JOmj |0〉 = ~J |j,m〉,

and we haveOmj |j′,m′, α〉 ↔ |j,m〉 ⊗ |j′,m′, α〉.

59

Hence, we can think of

〈j′′,m′′, β|Omj |j′,m′, α〉 ≡ 〈j′′,m′′, β|j,m, j′,m′, α〉 = Cjj′(j′′,m′′;m,m′)〈β||O||α〉.

Now, let’s take

|j′′,m′′, α) ≡∑m,m′

Cjj′(j′′,m′′;m,m′)Omj |j′,m′, α〉.

Then, because Cjj′ is a real unitary matrix, i.e., C−1jj′ = Ctjj′ , we have

Omj |j′,m′, α〉 =∑j′′,m′′

Cjj′(j′′,m′′;m,m′)|j′′,m′′, α).

Hence,

〈j′′,m′′, β|Omj |j′,m′, α〉 =∑j′′,m′′

Cjj′(j′′,m′′;m,m′)〈j′′,m′′β|j′′, m′′, α).

And if j′′ 6= j′′ then the whole thing vanishes, so we don’t need to sum. Then,

〈j′′,m′′, β|Omj |j′,m′, α〉 = Cjj′(j′′,m′′;m,m′)〈j′′,m′′β|j′′,m′′, α).

Note that consequently,|〈β|~d|α〉|2 = 〈j2,m2, ∗|O0,±1

j=1 |j1,m1, ∗〉,

which doesn’t vanish only if|j2 − j1| ≤ 1.

Hence we have the selection rule|∆l| ≤ 1.

4.6.2 Isospin

Now, an argument based on quantum gravity says that there cannot be any exact symmetries other thanrotation symmetries (like spin) and gauge symmetries (conservation of electric charge).

Since u and d have similar masses, p = uud and n = udd have similar masses and interactions. This gives riseto an approximate SU(2) symmetry called isospin. And there is also an SU(3) flavour symmetry involving the u,d and s quarks. Such approximate symmetries simplify particle classification and properties.

-

Aside: Since our main aim here is to provide a concrete example of the Wigner-Eckart theorem, for more detailson isospin, particle contents, and relations like 3⊗ 3 = 8⊕ 1, where the in class discussion was not self-contained,the typesetter decided to just recommend Chapter 9 of the following set of notes, which goes into details of therepresentation of su(3). The root diagrams are especially useful to learn, if you have the time. Further, you canlook up Young tableau for the box diagrams (two horizontally placed next to each other: symmetric; two verticallystacked: anti-symmetric, etc) Prof. Vafa used in class.

http://www.damtp.cam.ac.uk/user/examples/3P2.pdf

-

We consider the decay processes∆++ → π+ + p,

which has spin(3/2, 3/2)→ (1, 1) + (1/2, 1/2);

60

and∆+ → π+ + n,

which has spin(3/2, 1/2)→ (1, 1) + (1/2,−1/2).

Using the fact that the unitary operator commutes with the isospin, the probability of decaying in time t isgiven by,

|〈j′′,m′′, α|U(t)|j,m, j′,m′, β〉|2

|〈j′′, m′′, α|U(t)|j, m, j′, m′, β〉|2=|Cjj′(j′′,m′′;m,m′)|2

|Cjj′(j′′, m′′; m, m′)|2=

1

1/3= 3,

for top decay to bottom decay. This is the power of the Wiger-Eckart theorem: even if there’s a piece of unknowninformation, they are the same in the numerator and denominator and cancels as we only need to know the Cjj′ ’s.

4.7 Discrete Symmetries

• Parity

In the context of the harmonic oscillator, we discussed parity. The parity operator and its eigenvalue satisfy

P 2 = 1, P = ±1.

So we can decompose our states in terms of even and off parities. Take the dipole vector ~d. Then

P ~dP−1 = −~d,

or,P ~d = −~dP.

What’s the probability for a state to go to another state through dipole radiation? We have

〈+α|~d|+ β〉 = 〈+α|P−1~dP |+ β〉 = −〈+α|~d|+ β〉 = 0.

Similarly,〈−α|~d| − β〉 = 0.

Next, take〈n′, l′,m′|~d|n, l,m〉.

We know that |l− l′| ≤ 1. In fact, we also have the condition l 6= l′, since by considering spherical harmonics,we have

P |n, l,m〉 = (−1)l|n, l,m〉.

So if the two states have the same parity, we get zero. Therefore

|l − l′| = 1.

Unfortunately, parity is not a symmetry of nature. The Hamiltonian of the universe is not parity invariant,since only the left-handed electrons interacts. Parity is a good symmetry of the electromagnetic and thestrong interactions, but not a good one of the weak interactions.

• Time Reversal

Entropy aside, time reversal is still not a symmetry of nature. This has to do with anti-unitary operator

T : ψ(t)→ ψ∗(t).

61

• Charge Conjugation

Again not a symmetry of nature,C : e− → e+.

However, as predicted by relativistic quantum field theory, CTP together is a symmetry of nature.

4.8 Supersymmetry

We discuss supersymmetry in order to solve hydrogen atom! Weinberg uses what’s called the Runge-Lenz vectorsymmetry to do this alternate calculation but we’ll proceed with supersymmetry.

Suppose we have a Hilbert space and an operator of order 2, such as the parity operator P ≡ (−)F , which haseigenvalues ±1. We decompose the Hilbert space as

H = HB ⊕HF ,

where B corresponds to +1 and F , −1. The operator commutes with the Hamiltonian H, i.e.,

H(−)F = (−)FH.

Now, lets take the operatorsQ : HB → HF ;

Q† : HF → HB .

We can diagonalise(−)F = diag(−1, ...,−1, 1, ..., 1);

H = diag(E1, ..., En, E′1, ..., E

′n).

Then for some matrix A which may or may not be square, we can write

Q =

(0 A0 0

); Q† =

(0 0A† 0

),

such that(−)FQ = −Q(−)F ; (−)FQ† = −Q†(−)F .

We call Q the supercharge. Now we arrive at the SUSY Hamiltonian, that can be decomposed as follows.

H = QQ† +Q†Q.

It’s easy to check that[Q,H] = 0.

Notice that〈ψ|H|ψ〉 = |Q|ψ〉|2 + |Q†|ψ〉|2 ≥ 0.

So if E = 0, thenQ|ψ〉 = 0 = Q†|ψ〉.

If E > 0, thenH|ψ〉 6= 0,

and either or both of Q|ψ〉 or Q†|ψ〉 6= 0. So we can diagonalise the Hamiltonian as

H = diag(0, ..., 0, E1, ..., Ek, 0, ..., 0, E′1, ..., E

′k).

Let’s take a bosonic state that’s an eigenstate of H, i.e.,

H|ψB〉 = E|ψB〉.

62

Then since E 6= 0 and Q†|ψB〉 = 0, we must have

Q|ψB〉 6= 0.

Since [H,Q] = 0, we also haveHQ|ψB〉 = QH|ψB〉 = EQ|ψB〉.

As long as the energy is greater than zero,

Q†Q|ψB〉 ∝ |ψB〉.

So the state comes back to itself up to normalisation. Notice that QQ† is almost the Hamiltonian. In fact, wedo get an eigenstate of the Hamiltonian since the Q†Q piece kills |ψB〉. Hence, for every positive eigenvalue, all theeigenstates are paired up. For each bosonic one, we get a fermionic one, i.e., the number n of bosonic and femionicexcited states equal, but there could be a mismatch for the ground states. In fact, by considering the diagonalisedmatrix (−)F , it is easy to see that

nB |E=0 − nF |E=0 = Tr(−)F .

This is called the Witten index, and provides a link between quantum mechanics and topology. Note that evenif we allow A to depend on some parameter so that our Hamiltonian is changed, Witten index doesn’t change, sincefor example, if a bosonic state goes up one level, the corresponding fermionic state must go up one level too.

Now let’s discuss an example. Consider the Hilbert space

H : ψ(x)⊗(s =

1

2

).

So suppose we are talking about a spin-1/2 particle in one-dimension. Take

(−)F = −σ3, σ3 = (1,−1).

Take Q and Q† as defined before, and

A =~√2m

d

dx+W (x);

A† = − ~√2m

d

dx+W (x),

where W (x) is real and the momentum-looking term doesn’t have the factor of i so it is anti-hermitian. Then,substituting the values into H, we have

H = QQ† +Q†Q = diag(AA†, A†A).

So HF and HB don’t mix. We call the spin-up states fermion and the spin-down states boson. We have

AA† = − ~2

2m

d2

dx2+

~√2m

W ′(x) +W 2;

A†A = − ~2

2m

d2

dx2− ~√

2mW ′(x) +W 2.

Notice that they are of the formp2

2m+ V (x).

So we have two particles with different potentials, and yet share the same energy spectrum, except for theground state. So let’s consider the ground states. Q|ψB〉 = 0 corresponds to(

~√2m

d

dx+W (x)

)ψB = 0.

63

Then,

ψB = C exp

[−√

2m

~

∫W (x)dx

];

ψF = C exp

[√2m

~

∫W (x)dx

].

The solutions only make sense if they are normalisable. Suppose

W (x) = Pn=even

∫→ xodd.

Then Tr(−)F = 0 and so it is not normalisable. On the other hand, suppose

W (x) = Pn=odd

∫→ xeven.

Then Tr(−)F = ±1 and so it is normalisable. Refer to problem set 5 for more details!

5 Time-independent Perturbation Theory

The most general Hamiltonian in a quantum system is not solvable, so we’ll discuss a few approximationtechniques as follows.

• Degenerate and non-degenerate perturbation theories, where we take

H = H0 + εH1, ε 1;

• Variational method, which involves guessing wavefunctions and making parameters better;

• Bohn-Oppenheimer approximation, or “integrating out fast variables”, where “fast” refers to, for example,the fast-moving electrons;

• WKB approximation, where the potential of the system doesn’t vary too much, i.e.,

V (x) ' V0.

5.1 Non-Degenerate Perturbation Theory

We start with some eigenstates

H0|i〉 = E(0)i |i〉.

We want to solve(H0 + εH1)(|i〉+ δ|i〉) = (E

(0)i + δEi)(|i〉+ δ|i〉).

We expand the variations in power series of ε as

δ|i〉 = ε|i〉(1) + ε2|i〉(2) + · · · ;

δEi = εE(1)i + ε2E

(2)2 + · · · .

Then,

(H0 + εH1)(|i〉+ ε|i〉(1) + ε2|i〉(2) + · · · ) = (E(0)i + εE

(1)i + ε2E

(2)2 + · · · )(|i〉+ ε|i〉(1) + ε2|i〉(2) + · · · ).

64

Suppose that for all ε,∞∑i=0

εiV (i) =

∞∑i=1

W (i)εi,

then for all i,V (i) = W (i).

• ε(0):H0|i〉 = E

(0)i |i〉.

• ε(1):H0|i〉(1) +H1|i〉 = E

(0)i |i〉

(1) + E(1)i |i〉.

• ε(2):H0|i〉(2) +H1|i〉(1) = E

(0)i |i〉

(2) + E(1)i |i〉

(1) + E(2)i |i〉.

Now, if we don’t worry about the overall normalisation, then by a choice of normalisation (i.e., rescaling), wecan always assume that the only term proportional to the original state we started with, |i〉, is the first term, andall other terms are orthogonal to |i〉. So we can write

|i〉(l)〉 =∑j 6=i

a(l)j |j〉,

where all |i〉(l)’s are orthogonal.

Let’s now act on the ε(1) case with the bra 〈i|. Then

〈i|H0|i〉(1) + 〈i|H1|i〉 = E

(1)i .

Hence, the first order change in energy of the ith state is the expectation value of perturbation in that state.We next act on it with 〈j|, where j 6= i. Then

〈j|H0|i〉(1) + 〈j|Hi|i〉 = E(0)i 〈j|i〉

(1) + E(1)i

〈j|i〉.

Since 〈j|H0 = E(0)j 〈j|, we have

〈j|i〉(1) =〈j|H1|i〉

E(0)i − E

(0)j

.

Similarly, we act on the ε(1) case with the bra 〈i|. Then,

〈i|H1|i〉(1) = E(2)i .

From our expression of 〈j|i〉(1), we can write

|i〉(1) =∑j 6=i

|j〉〈j|H1|i〉E

(0)i − E

(0)j

.

Hence,

E(2)i =

∑j 6=i

〈i|H1|j〉〈j|H1|i〉E

(0)i − E

(0)j

=∑j 6=i

|〈i|H1|j〉|2

E(0)i − E

(0)j

.

Now, notice that for the ground state, E(0)i < E

(0)j , so the denominator is negative, and so,

E(2)gs ≤ 0.

Suppose that the states all have close energies. Then we can write

E(2)gs '

1

−∆

∑j 6=i

|〈i|H1|j〉|2 =1

−∆

(|〈i|H2

1 |i〉| − |〈i|H1|i〉|2)

=1

−∆

(〈H2

1 〉 − |〈H1〉|2).

65

Next, the second order equation with 〈j| acting on it gives

〈j|H0|i〉(2) + 〈j|H1|i〉(1) = 〈j|E(0)i |i〉

(2) + 〈j|E(1)i |i〉

(1) +〈j|E(2)

i |i〉,

i.e.,

E(0)i 〈j|i〉

(2) + 〈j|Hi|i〉(1) = E(0)i 〈j|i〉

(2) + E(1)i 〈j|i〉.

Hence we have, substituting our expression for |i〉(1),

〈j|i〉(2) =1

E(0)i − E

(0)j

∑k 6=i

〈i|Hi|k〉〈k|H1|i〉E

(0)i − E

(0)k

− 〈i|H1|i〉〈j|H1|i〉E

(0)i − E

(0)j

.Note however, that we don’t always have good convergence properties, so going to higher order terms might

not work out. Also, if we take things small, then the first few terms should be enough.

5.2 Degenerate Perturbation Theory

The previous formulae tell us that if E(0)i = E

(0)j , then we have infinitely big answers. The assumption of small

ε expansion is wrong.

Take the famous Mexican hat potential. The ground state (say, at x2 + y2 = 1, z = 0, z denoting the energy)is degenerate. Now, if we place a particle at the ground state y = 1, x = z = 0 and tilt the hat down a bitalong the y−axis towards the +x direction, then the particle will roll all the way to x = 1, y = 0. So even if theperturbation is tiny, the effect is dramatic if we have degeneracy. To fix this problem, we can start at the positionx = 1, y = z = 0, so that the change of position is 0, although energy goes down.

Let’s go to the second last step of

〈j|i〉(1)(E(0)i − E

(0)j ) = 〈j|H1|i〉.

Compute H1 in the subspace of the Hilbert space where the eigenvalues are equal. If we diagonalise it, then allcomponents of the state, when being acted upon by H1, are going to give results orthogonal to each other. So wedon’t need to worry about the degenerate subspace anymore. Diagonalising H1 corresponds to putting the particleat the right starting place!

Let’s now discuss some examples.

5.2.1 Zeeman Effect

The Zeeman effect was observed in the 1890s, when atoms are put in a magnetic field. We have

δH = −~m · ~B,

where, classically,

~m = − e

2mc~L.

And we have something similar for the spin part. What’s surprising is that the actual spin part is about doubleof what we thought it was, due to a factor of gs = 2.002 · · · , which was first obtained through the Dirac equation.(You’ll see the whole story when you take QFT!) It turns out that

δH =e ~B · L2mc

+ gee ~B · ~S2mc

.

66

Suppose we have an electron in an atom like the hydrogen atom, so we have j = l ± 12 . Let’s compute the first

order perturbation of energy.

δE(1) = 〈ψ|δH|ψ〉 =e

2mc〈j,m, l, n|a~L+ b~S|j,m, l, n〉 · ~B.

We can put the coefficients a and b into a ratio between the two that can be pulled outside. It doesn’t dependon m thanks to the Wigner-Eckart theorem. Then,

〈j,m, l, n|a~L+ b~S|j,m, l, n〉 = gn,j,l〈j,m, l, n| ~J |j,m, l, n〉,

where 〈j,m, l, n| ~J |j,m, l, n〉 is nothing but the Clebsch-Gordon coefficients multiplied by the reduced matrix ele-

ment, and similarly for ~L. We next dot it into ~J to get

〈j,m, l, n|a~L · ~J + b~S · ~J |j,m, l, n〉 = gn,j,l〈j,m, l, n| ~J · ~J |j,m, l, n〉,

where~L · ~J = ~L · ~S + ~L2 =

1

2[(~S + ~L)2 − ~S2 − ~L2] + ~L2 =

1

2[ ~J2 − ~S2 + ~L2],

which gives the eigenvalue~2

2[j(j + 1)− s(s+ 1) + l(l + 1)] .

Similarly, ~S · ~J gives~2

2[j(j + 1) + s(s+ 1)− l(l + 1)] .

Taking s = 12 and putting things together, we have

gn,j,l =(a− b)l(l + 1) + 3

4 (b− a) + (a+ b)j(j + 1)

2j(j + 1).

Notice that it is in fact independent of n, so with a = 1, b = ge, we have

gj,l = 1 + (ge − 1)j(j + 1)− l(l + 1) + 3

4

2j(j + 1).

We call gj,l the Lande g-factor. Indeed, we can always make such a replacement, and for small ~B field,

δE(1) = gj,le

2mecm~B3.

If ~B is much bigger than the hyperfine structure, i.e., the effect of ~B is much greater than that of ~S · ~L, thenstarting from

H = HHydrogen +~S · ~Lf(r) +

e

2mec(~L+ ge~S) · ~B,

we can simply go back to the basis of the hydrogen atom where we can take the eigenstates of the originalHamiltonian along with replacing (~L+ ge~S) · ~B with its eigenvalue

(me + gems)~B.

5.2.2 Stark Effect

Take the perturbed HamiltonianH = H0 − e ~E · ~x,

since φ = − ~E ·~x. Naively, the first order perturbation vanishes because of parity. Specifically, ~x→ −~x under parity,but the states together is invariant under parity. However, if we have a degenerate case, then we have different

67

l’s. When we diagonalise the Hamiltonian, we get mixed terms which have the same energy. Here’s an example tomake this concrete.

Take n = 1, then we only have l = 0, and j = 1/2. Now, take n = 2. Then we can have l = 0, which givesj = 1/2 or l = 1, which gives j = 1/2, j = 3/2. So j = 1/2 corresponds to both l = 0 and l = 1, which could inprinciple mix. When computing matrix elements, we see that we have off-diagonal pieces involving l = 0 and l = 1that need to be diagonalised. We ignore the spin for the moment and take E3 for simplicity. Then,

〈j = 1/2, l = 0,m = 0 or l = 1,m = 0,±1|x3|j = 1/2, l = 0,m = 0 or l = 1,m = 0,±1〉.

We want to compute the expectation value of H1 in the matrix element. Let the first three columns or rows ofa matrix represent the l = 1,m = 1,−1, 0 cases, and the last, l = 0,m = 0.

00 0

α0 0 α∗ 0

The big block of zero follows from our parity argument. The zero entries at (1,4), (2,4), (4,1), (4,2) are due to

the Wigner-Eckart theorem, which gives

〈l = 1,m = ±1|x3|m = 0, l = 0〉 = 0.

If we write down the wavefunction, it’s not difficult to compute α explicitly, but we’ll just write down theeigenvalues here as E = ±α.

5.3 Variational Method

Take any normalised state |ψ〉, not necessarily an eigenstate of the Hamiltonian. We’ll have

〈ψ|H|ψ〉 ≥ Egs,

since the average energy cannot be lower than the ground state energy. So this method is about getting an upperbound for the lowest energy. Take

|ψ〉 =∑n

an|En〉.

such that ∑n

|an|2 =∑n

Pn = 1.

〈ψ|H|ψ〉 =∑n

PnEn ≥ Egs.

Normalising,〈ψ|H|ψ〉〈ψ|ψ〉

≥ Egs,

and if ψ = ψ(~α), we have the minimum

E(~α) =〈ψ(~α)|H|ψ(~α)〉〈ψ(~α)|ψ(~α)〉

,∂E

∂~α= 0.

68

5.3.1 Virial Theorem

The Virial theorem is an application of the variational method. There is a classical analogue, but we deal withthe quantum version here. Take

V (α~ri) = αkV (~ri).

For the hydrogen, k = −1; for the harmonic oscillator, k = 2. Suppose we have a bound state which is a groundstate, |ψ〉.

Claim:

〈ψ|T |ψ〉 =k

2〈ψ|V |ψ〉.

Proof: Given some ground state wavefunction, if we vary it, the energy should go up, and the original point mustbe critical. Take the trial wavefunction

ψ(~ri, α) ≡ ψ0(α~ri).

Then,dE(α)

dα

∣∣∣α=1

= 0.

Now,

〈V 〉 =

∫d3~riψ

∗(~ri)V (~ri)ψ(~ri)∫d3~riψ∗(~ri)ψ(~ri)

=

∫d3~riψ

∗0(α~ri)V (~ri)ψ0(α~ri)∫

d3~riψ∗0(α~ri)ψ0(α~ri)=

1

αk〈V 〉0,

since as ~ri → α~ri, we can write

V (~ri) =1

αkV (α~ri).

Similarly,

〈T 〉 =

∫d3~riψ

∗0(α~ri)

(− ~2

2mi∇2i

)ψ0(α~ri)∫

d3~riψ∗0(α~ri)ψ0(α~ri)= α2〈T 〉0,

since ∇i is with respect to ~ri and if we want it to be with respect to α~ri, we have to replace it with α∇i. Hence,

E(α) = α2〈T 〉0 +1

αk〈V 〉0

So,dE(α)

dα

∣∣∣α=1

= 2α〈T 〉0 −k

αk+1〈V 〉0

!= 0.

Hence,2〈T 〉0 − k〈V 〉0 = 0.

Indeed, this result agrees with the hydrogen atom and the harmonic oscillator. In fact, it is true for the excitedeigenstates of the Hamiltonian too (homework).

Above we derived the Virial theorem through the variational method. There is an easier derivation. We have,with |ψ〉 an eigenstate of H,

〈ψ|[Q,H]|ψ〉 = E〈Q〉 − E〈Q〉 = 0.

Now take an operatorQ = ~x · ~p,

where in position space,~p = −i~~x∂x,

and in momentum space,~p = i~∂pp = const. + i~p∂p.

Then to find [Q,H] with the Hamiltonian

H =p2

2m+ V (x),

69

we compute the kinetic part as[p∂p, p

2] = 2p2.

Notice that we are basically concluding that the degree of p2 is 2. And for the potential part, by assumption,the degree is k, or more explicitly,

[~x∂x, V ] = kV.

Hence, ⟨2i~

p2

2m− i~kV

⟩= 0,

i.e.,2〈T 〉 − k〈V 〉 = 0.

5.4 Bohn-Oppenheimer Approximation

The Bohn-Oppenheimer approximation is based on that heavy things should be left untouched at first approx-imation. So if we have a molecular problem, we first solve the problem involving electrons only, then involve thenucleus. In summary, we split a problem into two steps:

1. Solve the problem for fast (light) variable.2. Include more massive particles.

Let’s take ~Ri, M to be the position and mass of the nucleons, and ~ri, m to be those of the electrons. We assumethat M is infinite and take

H =∑i

~P 2i

2M+∑i

~p2i

2m+ V (~r, ~R) ≡

∑i

~P 2i

2M+ H.

So we now solveHψα(~ri, ~Rj) = Eα(~R)ψα(~ri, ~Rj).

The energy spectrum is of course, location-, i.e., ~R−dependent. We can view Eα(~R) as some effective potential

Vα,eff(~R). Now, we can always make the expansion

ψ(~r, ~R) =∑α

fα(~R)ψα(~r, ~R).

Hence,

Hψ =

(∑i

~P 2i

2M i+ H

)∑α

fα(~R)ψα(~ri, ~Ri) = Eψ.

Since there is no derivative with respect to ~R within H, assuming small ∂Rψα(r,R), and in particular,

1

f∂Rf

1

ψα∂Rψα,

then we simply have

Hψ =

(∑i

~P 2i

2M i+ Eα(~R)

)fα(~R)ψα(~ri, ~Ri) = Eψ.

Hence, (∑i

~P 2i

2M i+ Eα(~R)

)fα(~R) = Eiαf

iα(~R),

or,Hf iα(~R)ψα(~r, ~R) = Eiαf

iαψα.

70

For the second step, we approximate the minimum of the potential as some quadratic potential like the harmonicoscillator. The approximation is only good for energy close to the minimum. Take the atomic energy

E ∼ e2

a, a =

~2

me2,

or,

E ∼ me4

~2= mc2

(e2

~c

)2

= mc2α2.

We are interested in computing the vibrational energy of the molecules. So let’s consider the spring case

E ∼ kx2, x2 ∼(

~2

me2

)2

.

Notice the the only length scale is the one involving the Bohr radius. Then the spring constant

k ∼ E

(~2/me2)2=me4

~2

m2e4

~4.

Hence, the vibrational energy

Evib ∼√

k

M~ =

√m

Mmc2α2.

So comparing to the atomic energy, we have a suppression factor of√m/M for the vibrational case. Now, even

if we freeze the molecules, there is still angular degrees of freedom. We have the rotational energy

Erot ∼L2

2I=

~2

Ma2=

~2

M

m2e4

~4=m

Mmc2α2.

So the suppression factor is even larger in the rotational case. Let’s now return to the assumption of small∂Rψα(r,R). How much can R vary before a wavefunction gets exponentially suppressed? We want

ψ(r +R)− ψ(r) ∼ R

a

∂ψ

∂r,

R

a 1.

We have

R =

√Evib

k∼ ~2

m4/3M1/4e2.

We want to compare R with the atomic scale a = ~2/me2. We have

R

a∼(mM

)1/4

∼ 0.15 < 1,

so the approximation is just ok, not amazing.

5.5 WKB Approximation

WKB approximation is about solving problems assuming that the potential varies very slowly, and the percentagechange in potential should be tiny. It is most elegantly seen in the context of the Feynman path integral.

-

Aside: Here, just for fun, we present the derivation of the path integral method, as we are already familiar withall the necessary tools. Setting ~ ≡ 1 for convenience, let’s look at

〈xf , tf |xi, ti〉 = 〈xf | e−iH(tf−ti)|xi〉.

71

If we chop the path into N intervals of length ε, then using the completeness relation

1 =

∫d3x |x〉〈x|,

we can write

〈xf , tf |xi, ti〉 =

∫d3x

N−1∏k=1

〈xf | e−iHε|xN−1〉〈xN−1| e−iHε|xN−2〉 · · · 〈x1| e−iHε|xi〉.

Now, using the completeness relation

1 =

∫d3p

(2π)3|p〉〈p|,

we can write, for a typical term,

〈x′| e−iHε|x〉 =

∫d3p

(2π)3〈x′|p〉〈p| e−iHε|x〉,

where〈x′|p〉 = e−ip·x

′,

and for H = p2

2m + V (x),

〈p| e−iHε|x〉 = e−iε[p2

2m+V (x)]〈p|x〉 = e

−iε[p2

2m+V (x)]

eip·x.

⇒ 〈x′| e−iHε|x〉 =

∫d3p

(2π)3e−ip·(x

′−x) e−iε[p2

2m+V (x)]

=

∫d3p

(2π)3e−iε[p· x′−xε + p2

2m+V (x)]

≡∫

d3p

(2π)3e−iε[p2

2m+V (x)+p·x].

If we put things together now, we’ll have

〈xf , tf |xi, ti〉 =

∫ N−1∏k=1

d3xk

N∏l=1

d3pl(2π)3

e−iε[p2f

2m+V (xf )+pf ·xN−1

]

× e−iε[p2N−12m +V (xN−1)+pN−1·xN−2

]× · · · × e

−iε[p21

2m+V (x1)+p1·xi].

But we can first let p := p−mx, which gives

p2

2m+ p · x =

p2

2m−

1

2m2p ·mx+

1

2mx2 +p · x−mx2 =

p2

2m− 1

2mx2.

Hence

〈x′| e−iHε|x〉 =

∫d3p

(2π)3e−iε

p2

2m eiε(12mx

2−V (x))

=1

(2π)3

[2πm

iε

]3/2

eiε(12mx

2−V (x))

=[ m

2πiε

]3/2eiεL(x,x).

And now putting things together, we have

〈xf , tf |xi, ti〉 = limε→0

[ m

2πiε

]3N/2 ∫ N−1∏k=1

dxk eiS(xf ,xi).

72

-

The idea of the WKB approximation corresponds to the de Broglie wavelength

λ =~p,

where we see that in the limit ~→ 0, we have λ→ 0, which means that everything’s almost constant in the limit.So we are dealing with some kind of classical approximation. Start with some free particle in one-dimension.

ψ(x) = A eipx/~ ≡ A eiS(x)/~,

since p and x are taken to be classical and p = p(x). S(x) corresponds to the classical action, and we can take apower series expansion

S(x) = S0(x) + ~S1(x) + ~2S2(x) + · · ·

Hence we have(− ~2

2m

d2

dx2+ V (x)

)ei[S0(x)+~S1(x)+~2S2(x)+··· ]/~ = E ei[S0(x)+~S1(x)+~2S2(x)+··· ]/~.

Working out the derivatives, we have[− ~2

2m

((i

~

)2(∂S

∂x

)2

+i

~∂2S

∂x2

)+ V (x)

]= E.

Hence, (∂S

∂x

)2

− i~∂2S

∂x2= 2m(E − V (x)) = p2(x).

But if ~ is small, we can ignore the second term. So we get

∂S

∂x' ±p(x),

i.e.,

S(x) = ±∫ x

dx′ p(x′).

And hence we haveψ(x) = A e±i

∫ x dx′ p(x′)/~.This is the first order WKB approximation. Let’s now go beyond linear approximation and write a systematic

one. Using the expansion of S(x), we have(∂xS0 + ~∂xS1 + ~2∂xS2 + · · ·

)2 − i~ (∂2xS0 + ~∂2

xS1 + · · ·)

= p2(x).

We immediately see that, to the zeroth power in ~,

(∂xS0)2 = p2(x).

Hence,

S0(x) = ±∫ x

dx′ p(x′).

The higher order terms must vanish since the RHS doesn’t contain ~. So we write down order by order theequations. First order:

2∂xS0∂xS1 − i∂2xS0 = 0,

where the only unknown is S1, and second order:

2∂xS0∂xS2 + (∂xS1)2 − i∂2xS1 = 0,

73

where the only unknown is S2, etc. Let’s first deal with the first order piece. We have

∂xS1 =i∂2xS0

2∂xS0=i

2∂x log(∂xS0).

Hence,

S1 =i

2log(∂xS0) + c.

So to first order, we have

ψ = eiS0/~+i~S1/~ = A e±i∫dx p/~− 1

2 log p(x) = A1√p(x)

e±i∫dx p/~.

Suppose we have particles of momenta p moving in one-dimension from a lower potential to a higher one. Wehave

ψ = A eipx/~,

where

|ψ|2 ∝ |A|2 =number

length≡ n,

the number density. And the flux

J ' n · v = |A|2v =|A|2p(x)

m,

where p(x) is not constant as potential is changing. But the number of particles cannot change. So we must have|A|2p(x) together being independent of x, which makes sense since we had A/

√p(x) in our expression for ψ. Let’s

now deal with the second order piece. As above, we easily arrive at

S2(x) =1

2

m∂xV

p3(x)− 1

4

∫dxm2(∂xV )2

p5.

We have to make sure that ~S2 is small, i.e.,

m∂xV

~p3 1.

When p(x) = 0, i.e., at turning points, it is not small. Recall that de Broglie wavelength

λ(x) =h

p(x).

So the wavelength of the object we are computing varies according to p(x), and we need it to not to vary rapidly.The amount the wavelength changes over one wavelength is given by

δλ

λ= ∆x∂xλ(x)

1

λ= λ∂xλ(x)

1

λ

? 1.

We have

λ2 =h2

p2=

h2

2m(E − V (x)).

Sodλ2

dx= −h

2

p4

dp2

dp= −h

2

p4

(−2m

dV

dx

),

i.e.,2λdλ

dx= 2m

h

p

h

p3

dV

dx.

Hence indeed we haveδλ

λ 1.

74

So in WKB, there is no percentage wavelength change as we move one wavelength. On the other hand, if E < V ,then

2m(E − V (x)) = p2(x),

so p(x) is imaginary. And the exponential exp[±i∫dx p/~

]is real. Let’s define

k ≡ 1

~√

2m(V (x)− E).

Then, we have an exponentially rising and falling expression

ψ ∼ 1√κ

e±∫k dx.

Now, we have solved the problem in individual pieces but cannot connect them together due to the turningpoints. But we just need some connecting formulas, which are solutions of

− ~2

2m

d2ψ

dx2+ αxψ = Eψ.

The solutions to these are called Airy functions. For an exponentially rising V (x) where it goes above E at thepoint x2, use

2√k

cos

(∫ x2

x

k dx− π

4

)↔ 1√

ke−∫ xx2k dx

;

1√k

sin

(∫ x2

x

k dx− π

4

)↔ − 1√

ke∫ xx2k dx

.

And for an exponentially falling V (x) where it goes below E at the point x1, use

1√k

e∫ xx1k dx ↔ 2√

kcos

(∫ x

x1

k dx− π

4

);

− 1√k

e∫ xx1k dx ↔ 1√

ksin

(∫ x

x1

k dx− π

4

).

WKB approximation is especially applicable to the Bohr-Sommerfeld quantisation and barrier penetration.

5.5.1 Bohr-Sommerfeld Quantisation

Take a potential which rises, then oscillates, then either rises or falls. If it rises, then it won’t be normalisable.So we adjust the energy so that we’ll only have falling solutions, i.e., bound states. That’s the quantisation ofenergy. The condition will end up being the Bohr-Sommerfeld quantisation. We have

2√k

cos

(∫ x2

x

k dx− π

4

)=

2√k

cos

(−∫ x2

x

k dx+π

4+

∫ x2

x

k dx

),

by∫ CA

=∫ BA

+∫ CB

and cosx = cos(−x). So we require

−∫ x2

x1

k dx+π

4= −π

4+ nπ,

i.e., ∫ x2

x1

k dx =

(1

2+ n

)π.

Now, p = ~k. Hence,

2

∮p dx =

∫ x2

x1

p dx = ~∫ x2

x1

k dx = 2~(

1

2+ n

)π = h

(n+

1

2

).

And it relates to the energy sincep =

√2m(En − V (x)).

So the Bohr-Sommerfeld quantisation is justified.

75

5.5.2 Barrier Penetration

Take region I to have V < E, V rising, region II to have V > E, V rising then falling, and region III to haveV < E, V falling. The boundary between I and II occurs at x1 and that between II and III occurs at x2. Now,particles come from the left, and we want to know what comes out on the right. We start from the very right, thenwhen we get to region II,

ψII =1√k

e∫ xx2k dx − 2i√

ke−∫ xx2k dx

.

Now taker ≡ e

∫ x2x1

k dx.

Then,

ψI =r−1

√k

sin

(∫ x

x1

k dx+π

4

)− 4ir√

kcos

(∫ x

x1

k dx+π

4

).

So the reflection coefficient

R =

∣∣∣∣4r − r−1

4r + r−1

∣∣∣∣2 =

∣∣∣∣1− r−2/4

1 + r−2/4

∣∣∣∣2 ∼ (1− 2r−2

4

)2

∼ 1− r−2.

And the transmission coefficientT = r−2 = e

−2∫ xx1k dx

.

6 Time-dependent Perturbation Theory

The time evolution of a state corresponds to the statement

i~∂t|ψ(t)〉 = H(t)|ψ(t)〉.

Recall that for time-independent H,

|ψ(t)〉 = U(t)|ψ(0)〉 = eiHt/~|ψ(0)〉.

When H = H(t), we require thati~∂tU(t) = H(t)U(t).

Let’s first define the time-ordering operator T , such that for any function f(t, ~x),

Tf(t1, ~x)f(t2, ~x) =

f(t1, ~x)f(t2, ~x), t1 > t2

f(t2, ~x)f(t1, ~x), t1 < t2.

Now, it’s important which side H(t) gets pulled down to by differentiation, so we must use the time-orderingoperator T to write

U(t) = T e−i∫ t0H(t′)dt′/~,

where, reminiscent of the path integral derivation we presented in the last section, we chop the time interval from0 to t into tiny pieces of length ε to write

e−i∫ t0H(t′)dt′/~ = · · · eiH(2ε)ε/~ eiH(ε)ε/~ eiH(0)ε/~.

The approximation techniques for the time-dependent cases that we’ll be discussing are the following.

• Sudden time-dependence, where we have the Hamiltonian H0 for t < 0 and a different one, H1, for t > 0,where neither is time-depednent;

• H = H0 +H ′(t), where H ′(t) is tiny. Here the frequency doesn’t have to be small but the amplitude is small;

76

• Fluctuating perturbation, where we deal with classical thermal fluctuations. Following from above, we canwrite H ′(t) = H ′f(t), where we can use Fourier transform to write f(t) = eiωt + e−iωt;

• Adiabatic perturbation, where H(t) is almost constant, having weak t−dependence. Here the amplitudedoesn’t have to be small but the frequency is small.

6.1 Sudden Time-Dependence

We already know that for t, t0 < 0, we evolve with

e−iH0(t−t0)/~;

and for t, t0 > 0, we evolve withe−iH1(t−t0)/~.

For t = 0, there is no evolution since it’s one point in time. The only matching boundary condition is that thestates must be the same. Say we have

H0|n〉 = En|n〉;

H1|n′〉 = En′ |n′〉.

How do we write the former in the |n′〉 basis? Note that we can always write for t < 0,

|ψ〉 =∑n

an eiEnt/~|n〉.

Next, using

|n〉 =∑n′

|n′〉〈n′|n〉,

we have|ψ(0)〉 =

∑n

an|n〉 =∑n,n′

an〈n′|n〉|n′〉.

Hence for t > 0,

|ψ(t)〉 =∑n

an|n〉 =∑n,n′

an〈n′|n〉 e−iEn′ t/~|n′〉.

6.2 H = H0 +H ′(t)

TakeH = H0 +H ′(t),

where H ′(t) is tiny. To first order we can ignore the second term and write

H0|n〉 = En|n〉.

Take ωn = En/~. Then we write

|ψ(t)〉 =∑n

e−iωntan(t)|n〉

such that an 6= an(t) when H ′(t) = 0. So we want to solve

i~∂t|ψ(t)〉 = (H0 +H ′(t))|ψ(t)〉,

whereLHS =

∑n

[i~(−iωn) e−iωntan(t) + i~ e−iωntan(t)

]|n〉,

77

andRHS =

∑n

[En e−iωntan(t) + e−iωntan(t)H ′(t)

]|n〉.

The first terms from both sides cancel, but we cannot naively drop the sum over n since H ′(t), which acts on|n〉, is not diagonal. Let’s act on the left with 〈m|. Then,

〈m|∑n

i~ e−iωntan(t)|n〉 = 〈m|∑n

e−iωntan(t)H ′(t)|n〉.

Hence,

i~ e−iωmtam =∑n

e−iωntan(t)〈m|H ′(t)|n〉.

This is an exact equation with no approximation. Now, we start with perturbation theory where H ′(t) is small.We assume that an(t) is small, which is consistent with small H ′(t). So we can approximate with an(t) ' an(0).Then,

i~am(t) =∑n

an(0)〈m|H ′(t)|n〉 e−i(ωn−ωm)t,

which is integrated to

am(t) = am(0)− i

~∑n

an(0)

∫ t

0

〈m|H ′(t′)|n〉 e−i(ωn−ωm)t′dt′.

For an(0) = δin and m 6= i,

am(t) = − i~

∫ t

0

〈m|H ′(t′)|i〉 e−i(ωi−ωm)t′dt′.

Now, going back to the more general case and let’s take H ′(t) to be of the form

H ′(t) = U e−iωt + U† eiωt.

Hence,

am(t) = am(0)− i

~∑n

an(0)

[〈m|U |n〉

∫ t

0

e−iωt′−i(ωn−ωm)t′dt′ + 〈m|U†|n〉

∫ t

0

eiωt′−i(ωn−ωm)t′dt′

]= am(0)− i

~∑n

an(0)

[〈m|U |n〉 e−i(ωn−ωm+ω)t − 1

−i(ωn − ωm + ω)+ 〈m|U†|n〉 e−i(ωn−ωm−ω)t − 1

−i(ωn − ωm − ω)

].

Now, take an(0) = 0 for all n 6= i, and take ai(0) = 1. Also set m = f . Then,

af (t) = − i~

[〈f |U |i〉

2 sin(ωi − ωf + ω) t2ωi − ωf + ω

e−i(ωi−ωf+ω) t2 + 〈f |U†|i〉2 sin(ωi − ωf − ω) t2

ωi − ωf − ωe−i(ωi−ωf−ω) t2

].

For ω ∼ ωf − ωi, first notice that

limx→0

sinx

x= 1,

and away from 0, it’s oscillatory near the x−axis. Using this, we see that the U term above, instead of the U†

term, is the dominating one. Hence we have

Pi→f (t) = |af (t)|2 =4

~2|〈f |U |i〉|2

sin2(ωf − ωi − ω) t2(ωf − ωi − ω)2

.

Claim:

limt→∞

2 sin2(αt/2)

πα2t= δ(α).

Proof: First notice that for α away from 0, the expression approaches infinity as t→∞. Now, using the substitutionu = αt/2, we easily see that ∫ ∞

−∞

2 sin2(αt/2)

πα2tdα = 1.

78

Hence,

Pi→f (t) =4

~2|〈f |U |i〉|2πt

2δ(ωf − ωi − ω).

Since E = ~ω, we arrive at

Pi→f (t) =2π

~|〈f |U |i〉|2tδ(Ef − Ei − E).

Hence, the decay rate

Γi→f (t) =dPi→fdt

=2π

~|〈f |U |i〉|2δ(Ef − Ei − E).

Questa formula importantissima si chiama regola d’oro di Fermi. In general applications, the delta function isintegrated out for a final state as a continuum of states. Notice that the dimension works out here since

[δ] = E−1, [〈〉] = E2, [~−1] = TE−1.

Now, a particle is a continuum of states. Take normalised eigenstates

ψ~n =1√V

ei~p·~x/~,

where V = L3, L being the sides of a periodic cube. So if ~p = ~~k = ~ 2π~nL , then summing over final states

corresponds to ∫d3n =

∫L3d3p

(2π~)3= V

∫d3p

(2π~)3.

Hence,

dΓi→f =2π

~V d3p

(2π~)3

∣∣〈 1√V

ei~p·~x/~|U |i〉∣∣2δ(Ef − Ei − E)

=2π

~d3p∣∣〈 1

(2π~)3/2ei~p·~x/~|U |i〉

∣∣2δ(Ef − Ei − E).

6.2.1 Hydrogen Atom Ionisation

Now let’s consider the example of ionising a hydrogen atom. We send an electrical wave (whose energy is highenough) through an electron so that the electrical field pushes the electron free. Take

H ′(t) = e ~E · ~x,

where~E = ~E0 e−iωt + ~E∗0e

iωt, ~E0 = E x3.

Notice that this is in the formH ′(t) = U e−iωt + U† eiωt,

with U = eEx3. Take the initial state |i〉 = |n = 1, l = 0,m = 0〉 to be the ground state

ψgs =1√πa3

e−r/a.

Then,

〈 1

(2π~)3/2ei~p·~x/~|U |i〉 =

1√πa3

∫e−r/a

e−i~p·~x/~

(2π~)3/2eEx3 d

3x.

Its computation without Mathematica is quite possible! It mainly involves is the computation of∫ei~p·~x/~f(r)d3x =

∫eipr cos θ/~f(r)r2dr sin θdθdφ

=

∫eikr cos θf(r)r2dr(−d cos θ)dφ

=

∫2π

ikreikr cos θ

∣∣∣1−1f(r)r2dr

=

∫4π

ksin(kr)f(r)r dr.

79

Then by differentiating the above with respect to k3, using dkdk3

= k3

k , and setting k3 = k cos θ, we have aftersome algebra,

〈 1

(2π~)3/2ei~p·~x/~|U |i〉 =

−8√

2ieE cos θ

π~3/2k5a5/2.

Now, since the delta function imposes energy-momentum conservation, p is fixed up to some angle. So usingdΩ = sin θdθdφ, we have

d3p δ(Ef − Ei − E) = ~3d3k δ(Ef − Ei − E) = ~3k2dΩdk δ

(~2k2

2m− ~ω

).

Hence,dΓ

dΩ=

256e2E2m cos2 θ

π~3k9a5.

6.3 Fluctuating Perturbation

Here, H ′(t) includes thermal fluctuation, i.e., we are taking classical reasons into account. Using results fromthe previous case, we have for f 6= i,

af (t) = − i~

∫ t

0

〈f |H ′(t′)|i〉 ei(ωf−ωi)t′dt′.

Hence,

|af |2 =1

~

∫ t

0

∫ t

0

〈f |H ′(t1)|i〉〈i|H ′(t2)|f〉 ei(ωf−ωi)(t1−t2) dt1dt2.

Let t ≡ t1 − t2. We assume that

gif (t) = 〈f |H ′(t1)|i〉〈i|H ′(t2)|f〉 =

∫ ∞−∞

g(ω) e−iωtdω.

Then,

|af |2 =1

~

∫dω

∫ t

0

∫ t

0

dt1dt2 g(ω) e−iω(t1−t2) ei(ωf−ωi)(t1−t2)

=1

~

∫dω g(ω)

[∫ t

0

dt′ ei(ωf−ωi−ω)t′]2

=1

~

∫dω g(ω)

4 sin2(ωf − ωi − ω) t2(ωf − ωi − ω)2

=1

~

∫dω g(ω)

πt

2δ(ωf − ωi − ω).

Hence,

Γ =d|af |2

dt=

2π

~2

∫ ∞−∞

gif (ω)δ(ωf − ωi − ω)dω.

And finally,

Γ =2π

~2gif (ωf − ωi).

80

6.3.1 Radiation Absorption and Emission

Now we consider the example of absorption and stimulated emission of radiation. Take

H ′(t) = e ~E(t) · ~x.

Assume that the fluctuations of the electric field have an average of the form

Ei(t1)Ej(t2) = δij

∫ ∞−∞P(ω) e−iω(t1−t2)dω.

Since LHS is real and symmetric under the exchange of (t1, i)↔ (t2, j), we have

P(ω) = P(−ω).

The rate of transition is thus

Γi→f =2π

~2e2|〈f |~x|i〉|2P(ωf − ωi).

Next, the average energy density of radiation

ρ =~E2 + ~B2

8π=

~E2

4π.

Since in three-dimensional space δii = 3,

ρ =3

4π

∫ ∞−∞P(ω)dω =

3

2π

∫ ∞0

P(ω)dω.

And using ω = 2πν,ρ(ν) = 3P(2πν).

Hence,

Γi→f =2πe2

3~2|〈f |~x|i〉|2ρ(νfi).

6.4 Adiabatic Perturbation

Take a Hamiltonian H(s) where the parameter s = s(t) is a slowly changing function of t, i.e.,

ds(t)

dt 1

For example, we can takeH = ασ · ~B,

such thats(t) = (Bx(t), By(t), Bz(t)).

We have the energy eigenstatesH(s(t))|n(s)〉 = En(s)|n(s)〉.

Now, take U(s) such that

U(s)|n(0)〉 =∑n

|n(s)〉〈n(0)|n(0)〉 = |n(s)〉.

UsingH(s) = U−1(s)HU(s),

we haveδijEi(s) = 〈j(0)|H(s)|i(0)〉 = 〈j(0)|U−1H(s)U |i(0)〉 = 〈j(s)|H(s)|i(s)〉 = δijEi

81

Note that sµ and Aµ play the role of space-time coordinates and vector potential, respectively, and the trans-formation above is the analog of a gauge transformation. Of course,

∮A is gauge-independent. So for every state

in a quantum mechanical problem, if we take it around a loop of parameters, the state comes back to itself up toa phase, called Berry’s phase. On the other hand, a multidimensional case with many states gives a non-abelianversion of the gauge field.

Now, as in the adiabatic case, take an eigenstate |n(s(t))〉. If the t variation is small, then the state will stay asan eigenstate of the Hamiltonian. Take the evolution of this state, an(t)|n(s(t))〉. Then,

an(t) = exp

[i

∫ t

0

En(s(t′))

~dt′]

exp

[−i∫ t

0

〈n| ddt|n〉],

where the second term is associated with Berry’s phase. It is a useful concept in condensed matter!

83

physics 251a advanced quantum mechanics i · physics 251a advanced quantum mechanics i prof.cumrun...

Documents