Quantum Mechanics in Multidimensions

In this chapter we discuss bound state solutions of the Schrodinger equation

in more than one dimension. In this chapter we will discuss some particularly

straightforward examples such as the particle in two and three dimensional boxes

and the 2-D harmonic oscillator as preparation for discussing the Schrodinger

hydrogen atom. The novel feature which occurs in multidimensional quantum

problems is called “degeneracy” where different wave functions with different

PDF’s can have exactly the same energy. Ultimately the source of degeneracy is

symmetry in the potential. With every symmetry, there is a conserved quantity

which can be used to “label” the states. For rotationally symmetric potentials,

the conserved quantities which serve to label the quantum states are angular

momenta.

Electron in a two dimensional box

By an electron in a two dimensional box, we mean that the potential is zero

within the walls of the box and infinity outside the box. For convenience we will

place the origin at one corner of the box as illustrated below:

V = 0

b

ax

y

V =

V =

V =

V =

To go from the one dimensional to the two dimensional time independent SE

1

we simply take ∂2/∂x2 → ∂2/∂x2 + ∂2/∂y2 and obtain for a particle of mass µ:

− h2

2µ

(∂2

∂x2+

∂2

∂y2

)ψ(x, y) + V (x, y) ψ(x, y) = Eψ(x, y) (1)

For the case under discussion here, the electron is confined within the box, where

we have V (x, y) = 0.

We use a separation of variable technique to solve this partial differential

equation and try product wave functions of the form ψ(x, y) = X(x) × Y (x).

− h2

2µ

(Y∂2X

∂x2+X

∂2Y

∂y2

)= E X Y (2)

We can divide sides of Eq. (2) by X(x)Y (y) to obtain:

(− 1X

h2

2µ∂2X

∂x2

)1

+

(− 1Y

h2

2µ∂2Y

∂y2

)2

= E (3)

Eq. (3) says that the sum of the first term and second term add to a constant

(as opposed to a function). This suggests that each of these terms is equal to a

constant or any functional dependence between them cancels when summed to

give a constant. However the first term can only be a function of x ; while the

second term can only be a function of y. There is therefore no way that the terms

can have a non-constant part but sum to a constant. We therefore conclude that

both terms are individual constants. Lets call these constants ( )1 = Ex, and

( )2 = Ey, and Ex +Ey = E. We then have two ordinary differential equations:

− h2

2µd2X

dx2= Ex X , − h

2

2µd2Y

dy2= Ey Y (4)

d2X

dx2= −k2

x X ,d2Y

dy2= −k2

y Y

2

where E = Ex + Ey =h2

2µ(k2

x + k2y

)(5)

The solutions of Eq. (5) are sinusoidal products of the form:

ψ(x, y) = (α sin (kx x) + β cos (kx x))1 × (γ sin (ky y) + δ cos (ky y))2 (6)

Let us consider the wave function boundary conditions. The wave function must

vanish on the line y = 0, which means ψ(x, y = 0) = 0 or ( )2 = 0 meaning δ = 0.

The wave function must vanish on the line x = 0 which means ψ(x = 0, y) = 0

or ( )1 = 0 meaning β = 0. We thus know we need sine functions for both x and

y:

ψ(x, y) = α sin (kx x) × sin (ky y) (7)

We next consider the boundary conditions on the top and the right of the well.

Since ψ(x = a, y) = 0 , the sin (kx x) piece must have an argument which is an

integral multiple of π which implies kx a = nxπ where nx ∈ 1, 2, 3, .... Because

ψ(x, y = b) = 0 we must have ky b = nyπ where ny ∈ 1, 2, 3, .... We thus have:

ψ(x, y) = α sin (nx π x

a) × sin (

ny π y

b) and E(nx, ny) =

h2π2

2µ

((nx

a

)2+(ny

b

)2)

(8)

Degeneracy

The particle in the two dimensional box has an energy which is controlled

by two integer quantum numbers as opposed to the one dimensional case E =

(h π)2 n2/(2µa2) where a single, integer quantum number (n) controls the energy

level. We get one integer quantum number per dimension. In the case of the

particle in a box, nx +1 is the number of wave function nodes along x and ny +1

is the number of nodes along y. Each set of quantum numbers nx, ny results

in a distinguishable wave function. If once selects a square box with a = b, often

two different sets of quantum numbers nx, ny , with two distinguishable wave

functions, will have the same (or degenerate) energies. It easy to confirm the

below energy level chart for the case of a square, two dimensional box.

3

13E

25E

20E18E17E

(5,1)

10E

a

(1,5)(4,3)(3,4)

(4,2)

26E

a

5E

ha

0

2E

8E2

E =

(3,3)

)( π2m

(2,4)

(1,4)

(2,1)

(4,1)

(3,2)(2,3)

(3,1)(1,3)

(2,2)

(1,2)

(1,1)

2

It looks like the levels are either unique (degeneracy = 1) or pair degenerate

(degeneracy = 2), however if we went up further we would see examples of triply

degenerate and degeneracy = 4 levels. An example of a quadruple degeneracy is

65(h π)2/(2µa2) since it corresponds to 1, 8 , 8, 1 , 4, 7 , and 7, 4. Here

is a plot of the ψ∗(x, y)ψ(x, y) for the 1, 1 ground state.

The height of the surface is proportional to the PDF for finding the electron

4

anywhere within the box. The electron tends to populate in the center of the

box and has nodes at the wall boundary as expected.

Here are plots of ψ∗(x, y)ψ(x, y) for the degenerate excited states 2, 1 and

1, 2. We note the presence of an extra node in either the x and y directions. It

is clear that the PDF for these two cases are related by a 90o rotation about the

z axis and hence it is not surprising that they are degenerate with an energy of

5(h π)2/(2µa2).

Electron in a two dimensional harmonic oscillator

Another fairly simple case to consider is the two dimensional (isotropic) har-

monic oscillator with a potential of V (x, y) = 12µω

2(x2 + y2

)where µ is the

electron mass , and ω =√k/µ. The Schrodinger equation reads:

− h2

2µ

(∂2ψ

∂x2+∂2ψ

∂y2

)+

12µ w2

(x2 + y2

)ψ(x, y) = E ψ(x, y) (9)

Following our treatment of the two dimensional box, we insert a product wave

function of the form ψ(x, y) = X(x) Y (y) :

− h2

2µ

(Y∂2X

∂x2+X

∂2Y

∂y2

)+

12µ w2

(x2 + y2

)XY = E X Y (10)

5

We can then divide both sides by X(x) Y (y) and group into x dependent and y

dependent terms:

(− 1X

h2

2µ∂2X

∂x2+

12µ w2 x2

)1

+

(− 1Y

h2

2µ∂2Y

∂y2+

12µ w2 y2

)2

= E (11)

Again Eq. (11) says that the sum of the first term and second term add to a

constant, suggesting that each of these terms is equal to a constant. There is no

way that the terms can have a non-constant part but sum to a constant since the

first term can only depend on x while the second can only depend on y. Calling

these constants ( )1 = Ex, and ( )2 = Ey, we have Ex + Ey = E and the two

differential equations:

− h2

2µ∂2X

∂x2+

12µ w2 x2 X(x) = Ex X(x) , − h

2

2µ∂2Y

∂y2+

12µ w2 y2 Y (Y ) = Ey Y (x)

(12)

The Eq. (12) equations are exactly the same as the Schrodingerrodinger Equa-

tion for the harmonic oscillator which we obtained and solved in the chapter on

Bound States in One Dimension. The energies are:

Ex =(nx +

12

)hω , Ey =

(ny +

12

)hω where nx, ny ∈ 0, 1, 2, .. (13)

The solutions are:

Xnx(x) = Nnx Hnx(ξ) exp (−ξ2/2) where ξ =√µω

hx (14)

Yny(y) = Nny Hny(η) exp (−η2/2) where η =√µω

hy (15)

Putting this all together we have quantized energies of the form:

E = (nx + ny + 1) hω (16)

6

and wave functions of the form:

ψ(ξ, η) = N Hnx(ξ) exp (−ξ2/2) ×Hny(η) exp (−η2/2)

where (ξ, η) =√µω

h(x, y) (17)

We could write the energy spectrum as E = (n+ 1) hω where n ∈ 0, 1, 2....

As shown below the degeneracy of the nth level is just equal to n+ 1.

(0,4)

h

(0,0)

(0,1) (1,0)

(0,2) (1,1) (2,0)

(3,0)

(1,3) (4,0)(2,2) (3,1)5

4

0

(2,1)(1,2)(0,3)

3

ω

h ω

h ω

h ω

h ω

2

As you can see, the spectrum for the isotropic harmonic oscillator is more

degenerate than for the two dimensional square box. We can construct the wave

functions using the explicit forms for the Hermite polynomials:

Table 1: Hermite polynomials

Ho = 1 H3 = 8 ξ3 − 12 ξ

H1 = 2 ξ H4 = 16 ξ4 − 48 ξ2 + 12

H2 = 4 ξ2 − 2 H5 = 32 ξ5 − 160 ξ3 + 120 ξ

The wave functions for these solutions is in Table 1.

This table explicitly shows the energy and wave functions for the ground,

first, and second excited state in both cartesian and polar coordinates. One gets

from cartesian to polar coordinates via x = ρ cosφ and y = ρ sinφ.

7

Table 2: unnormalized 2-d harmonic oscillator wave functions

(nx, ny) E ψ ψ

(0,0) hω exp (−γρ2/2) exp (−γρ2/2)

(1,0) 2hω x exp (−γρ2/2) ρ cosφ exp (−γρ2/2)

(0,1) 2hω y exp (−γρ2/2) ρ sinφ exp (−γρ2/2)

(1,1) 3hω x y exp (−γρ2/2) ρ2 cosφ sinφ exp (−γρ2/2)

(2,0) 3hω(2 γ x2 − 1

)exp (−γρ2/2)

(2γρ2 cos2 φ− 1

)exp (−γρ2/2)

(0,2) 3hω(2 γ y2 − 1

)exp (−γρ2/2)

(2γρ2 sin2 φ− 1

)exp (−γρ2/2)

where γ =µω

h

The below figure illustrates the two degenerate states ψ(1, 0) and ψ(0, 1).

Cylindrical Symmetry

We note that the two dimensional, isotropic harmonic oscillator has has cylin-

drical symmetry both in the potential and boundary condition. By a cylin-

drically symmetric potential, we mean that in polar coordinates (ρ, φ) where

8

ρ =√x2 + y2 and φ = tan−1 (y/x), the potential has no φ dependence but

rather is a function of ρ only:

V =12µω2 (x2 + y2) =

12µω2 ρ2 (20)

Given that the potential is cylindrically symmetric, it is surprising that the ψ(1, 0)

and ψ(0, 1) PDFs depicted above break this symmetry by orientating their two

probability antinodes along the x or y axis. Because the ψ(1, 0) and ψ(0, 1)

are degenerate, any linear combination of them will also be a stationary state

with energy 2hω. The ψ± = ψ(1, 0) ± iψ(0, 1) combinations have cylindrically

symmetric PDFs.

ψ+ ∝ N (x+ iy) e−γ(x2+y2)/2 , ψ− ∝ (x− iy) e−γ(x2+y2)/2 (21)

Some very simple computation shows that we can write these states in polar

coordinates as:

x± iy = ρ(cosφ± i sinφ) = e±iφ

Thus

ψ±(ρ, φ) = Nρ e−γρ2/2 e±iφ (22)

where it is clear that ψ∗ψ is a function of ρ only since the azimuthal (φ) depen-

dence only affects the phase of of the wavefunction. The PDF for either the ψ±combination is illustrated below:

9

We note that the original ψ(1, 0) and ψ(0, 1) were real, since it is always

possible with a real potential to have a real stationary state. These real wave

functions carry no current density. The combinations ψ± , on the other hand,

are complex stationary states, with position dependent phases. As such the wave

functions carry current densities which (as you can show in the exercises) circulate

along the φ direction. This situation seems to be a bit contradictory. Since ψ± are

stationary states they have a static PDF, and yet there is a probability density

current flow.

The way around this apparent contradiction is that the electron described

by ψ± is moving in a direction which leaves its PDF constant. Since ψ± wave

function has an azimuthally symmetric PDF, the simplest motion would be that

the electron is rotating with an angular velocity of Ω about the z axis. Classically

a charged disk spinning about its symmetry axis, has a constant charge density,

but still carries an electrical current density. This current density would be in

the φ direction which is the same direction which applies to our wave function.

Angular momentum

Given that ψ± wave function involves a circulating electron current which

10

classically would correspond to a rotating charge density, it is not surprising that

the ψ± wave function describes an electron with angular momentum about the

symmetry (z) axis. Classically the angular momentum L about a point O is

related to its momentum ( p) and displacement ( r from the point O according to L = r × p. This means (after converting to momentum operators):

Lz = xpy − y px =h

i

(x∂

∂y− y

∂

∂x

)(23)

One can convert this operator form to polar coordinates by making use of of the

chain rule to transform the Cartesian derivatives to polar derivatives.

∂

∂x=∂ρ

∂x

∂

∂ρ+∂φ

∂x

∂

∂φ,

∂

∂y=∂ρ

∂y

∂

∂ρ+∂φ

∂y

∂

∂φ

ρ =√x2 + y2 , φ = tan−1 (y/x)

∂ρ

∂x=x

ρ,∂φ

∂x=

− sin2 φ

y

∂ρ

∂y=y

ρ,∂φ

∂y=

cos2 φx

(24)

In the exercises you will find it straightforward to assemble the pieces in Eq. (24)

to obtain the fairly elegant result:

Lz =h

i

(x∂

∂y− y

∂

∂x

)=h

i

∂

∂φ(25)

As the below calculation shows, the ψ± wave functions are eigenstates of Lz

with eigenvalues of Lz = ±h: To recapitulate, the original ψ(0, 1) and ψ(1, 0) were

purely real wave functions with a zero current density, which broke the cylindrical

symmetry of the potential. Since these two wave functions were degenerate we

could combine them in the form ψ± = ψ(1, 0)± iψ(0, 1) to form two alternative

11

wave functions whose PDF exhibits cylindrical symmetry. Because these wave

functions have position dependent complex phases they have either clockwise and

counter-clockwise current flow. We saw that the ψ± wavefunctions are eigenstates

of Lz and thus have unique values of Lz = ±h. Each of the original ψ(0, 1) and

ψ(1, 0) wave functions can in turn be written as linear combinations of ψ± with

equal probability of being in either ψ± state. This means that half the time they

will be measured to have Lz = h and half the time they will be measured to have

Lz = −h.

The Hamiltonian in Polar Coordinates

One can solve for the wave functions in cylinderically symmetric systems using

the direct separation of variable approach in polar coordinates in essentially the

same way as we used separation of variables in cartesion coordinates (Eq. (1)

- Eq. (8)). Here we will use a slightly different approach taking advantage of

some insights based on classical orbit theory. We begin by writing Schrodinger’s

equation in polar coordinates. Using the same sort of tedious procedure described

by Eq. (24) one can show one can write ∇2 in polar coordinates which allows us

to write the kinetic energy operator T = − h2

2µ∇2 as:

T = − h2

2µ∇2 = − h

2

2µ

(∂2

∂x2+

∂2

∂y2

)= − h

2

2µ

(1ρ

∂

∂ρ

(ρ∂

∂ρ

)+

1ρ2

∂2

∂φ2

)(26)

The underlined portion can be written in terms of the angular momentum oper-

ator:

T = − h2

2µ

1ρ

∂

∂ρ

(ρ∂

∂ρ

)+

L2

2µρ2

The entire Hamiltonian is then:

H =

[− h

2

2µ

1ρ

∂

∂ρ

(ρ∂

∂ρ

)]radial

+

L2

2µρ2

centifugal

+ V (ρ)

eff pot

(27)

We have written the Hamiltonian in a form inspired by classical mechanics: For

the a classical orbit of a satellite about the earth the classical energy expression

12

is given by:

E =12µ(r2 + (rφ)2

)+ V (r)

Here r is the radial velocity and rφ is the tangential velocity. The angular

momentum, given by L = µ vφ r = µ r2 φ, is a constant of motion. This allows

us to express φ in terms of L, insert it in Eq. (28) to obtain an expression in r

only:

φ =L

µr2, E =

12µr2 +

[(L2

2 µr2

)centrifugal

+ V (r)

]eff pot

(28)

The quantum effective potential expression Eq. (27) is identical to the classi-

cal effective potential expression Eq. (28). The centrifugal barrier piece of the

effective potential plays the pivotal role in preventing a satellite from crashing

into the earth as first realized by Isaac Newton. We illustrate this role for a

V = −K/r coulomb potential well (as a classical model of the atom):

µeffL 2

2 r2

r

=r

dominated by centrifugal

PerigeeApogee

E

potential

+ -K

Veff

dominated byCoulomb

V

potential

A general principle of both quantum mechanics and ordinary mechanics is

sometimes called Noether’s Theorem which says that every symmetry can be

associated with a conserved quantity. By symmetry, we generally mean a sym-

metry of the classical or QM Hamiltonian. In particular the spacial symmetries

of the Hamiltonian such as translation or rotation symmetry are associated with

13

with conservation of linear or angular momentum. If , for example, a potential

has translational invariance, the potential is independent of position and has no

gradients and therefore no force acts. A particle with a given momentum P will

continue traveling with this momentum in a straight line according to Newton.

Similarly a central force law about an origin, will lead to a cylindrically sym-

metric or spherically symmetric potential which will be the same independent

of rotation. A particle under the interaction of a central force, experiences no

torques and therefore has a constant angular momentum. If the Hamiltonian

has rotation symmetry, the angular momentum operator will commute with the

Hamiltonian ([H, L] = 0), and it is possible to find wave functions which are si-

multaneous eigenfunctions of commuting operators. If [H, L] = 0 the stationary

states can always be arranged as eigenstates of the angular momentum operator

which in turn implies that stationary states can be arranged to have a unique

(non-varying) value of angular momentum. Hence the classical statement that

angular momentum is conserved for the case of particle acted upon by a cen-

tral force has an exact quantum mechanical analogy : the stationary states can

be chosen to have unique angular momentum. Since this angular momentum

is “constant” we can assign each stationary state a conserved angular momen-

tum quantum number which helps specify or “label” the state. Inspired by the

separation of variable solutions we used for the particle in the two-dimensional

box or two-dimensional harmonic oscillator, we assume that for a cylindrically

symmetric potential, we are looking for cylindrically symmetric wave functions

are of the form ψ = R(ρ) Φ(φ).

An eigenfunction of Lz obeys the eigenvalue equation:

Lz ψ = Lz ψ orh

i

∂

∂φψ = Lz ψ

Since Lz only operates on φ ,h

i

∂

∂φΦ = Lz Φ (29)

14

where Lz is just a constant eigenvalue. Re-arranging this we have:

∂ΦΦ

= Lzi

h∂Φ (30)

Integrating both sides we have:

lnΦ =iφLz

h→ Φ = ei Lzφ/h (31)

Eq. (31) has an important boundary condition, if one goes from φ to φ+ 2π one

is back to the same space point hence: Φ(φ+ 2π) = Φ(φ) Hence

exp(iLzφ

h

)= exp

(iLz(φ+ 2π)

h

)= exp

(iLzφ

h

)× exp

(2iπ

[Lz

h

])(32)

The only way that Eq. (32) can be satisfied is if Lz/h is an integer which we will

call m. Hence we in general we write

Φ = ei m φ where Lz = m h , m = 0,±1,±2,±3, . . . (33)

Hence in general , the eigenstates of a cylindrically symmetric Hamiltonian can

be written as:

ψ = R(ρ) exp(imφ) (34)

Lets reconsider previous example :

ψ± = N (x± iy) e−γ(x2+y2)/2 = Nρ e−γρ2/2 e±iφ

This is of the expected form given by Eq. (34) with m = ±1 and R(ρ) =

Nρ e−γρ2/2. As you will shown in the exercise these wave functions satisfy the

Hamiltonian of Eq. (27).

H

[ρ exp

(−γρ

2

2

)]=

− h

2

2µ

(1ρ

∂

∂ρ

(ρ∂

∂ρ

))+(m2

2µρ2+

12µω2ρ2

)[ρ exp

(−γρ

2

2

)]

= 2hω[ρ exp

(−γρ

2

2

)](35)

as long as |m| = 1 and γ = µω/h.

15

Three Dimensions

Particle in a 3-d box

The reasoning for a particle in a two dimensional box can be easily extended

to three dimensions.

ψ(x, y) = N sin (nx π x

a) × sin (

ny π y

b) × sin (

nz π z

c) (36)

E(nx, ny, nz) =h

2µ(k2

x + k2y + k2

z

)=h2π2

2µ

((nx

a

)2+(ny

b

)2+(nz

c

)2)

(37)

where nx,y,z ∈ 1, 2, 3, .... The above wave functions are valid if the lower left

corner of the box is at the origin. Three quantum numbers are required in 3-d

and the system has a great deal of degeneracy for the case of a cube (a = b = c).

There will be even more degeneracy for the case of a spherically symmetric

potential. The 3 D harmonic oscillator will have a potential of the form: V =

(1/2) µ ω2(x2 + y2 + z2). The solution described by Eq. (9) - Eq. (17) can be

totally recycled with the addition of the Z(z) terms. The 3 D energy is described

by:

E =(nx + ny + nz +

32

)hω (38)

where nx, ny, nz ∈ 0, 1, 2.... In scaled variables the solution in Cartesian coordi-

nates is:

ψ(ξ, η) = N Hnx(ξ) exp (−ξ2/2) ×Hny(η) exp (−η2/2) ×Hnz(ζ) exp (−ζ2/2)

where (ξ, η, ζ) =√µω

h(x, y, z) (39)

This table explicitly shows the energy and wave functions for the ground

and first excited state in both Cartesian and spherical coordinates in unscaled

coordinates.

16

Table 3: 3-d harmonic oscillator solutions

(nx, ny, nz) E ψ ψ

(0,0,0) 3hω/2 exp (−γr2/2) exp (−γr2/2)

(1,0,0) 5hω/2 x exp (−γr2/2) r sin θ cosφ exp (−γr2/2)

(0,1,0) 5hω/2 y exp (−γr2/2) r sin θ sinφ exp (−γr2/2)

(0,0,1) 5hω/2 z exp (−γr2/2) r cos θ exp (−γr2/2)

(1,1,0) 7hω/2 x y exp (−γr2/2) r2 cosφ sinφ sin2 θ exp (−γr2/2)

(1,0,1) 7hω/2 x z exp (−γr2/2) r2 cosφ sin θ cos θ exp (−γr2/2)

(0,1,1) 7hω/2 y z exp (−γr2/2) r2 sinφ sin θ cos θ exp (−γr2/2)

(2,0,0) 7hω/2

(0,2,0) 7hω/2

(0,2,2) 7hω/2

where γ =µω

h

This table is constructed using the following relationships between spherical

and Cartesian coordinates.

y = θ

z

θsin φ

Spherical coordinates

z =

r = x + y + z2 2 2

x

y

r

φ

r sin

r cos θ

r sin θ cos φx =

r θ

r cos θ

r sin θ

φ

17

As we shall shortly, these cartesian wave functions can be rearranged to satisfy

the Hamiltonian for a spherically symmetric potential.

Spherically Symmetric Hamiltonian

Except for the form of the radial kinetic energy, the Eq. (40) Hamiltonian is

very similar to the Eq. (27) Hamiltonian:

H =−h2

2µ

1r2

∂

∂r

(r2∂

∂r

)radial

+

L2

2µr2+ V (r)

eff pot

(40)

Another important difference is L2 involves all three component of L in Eq. (40)

but involves L2z alone in Eq. (27). In particular L2 operates on both θ and φ ,

where as Lz operates on φ alone.

The L2 is based on all three components:

L2 ≡ L2x + L2

y + L2z where :

Lx =h

i

(y∂

∂z− z

∂

∂y

), Ly =

h

i

(z∂

∂x− x

∂

∂z

)

Lz =h

i

(x∂

∂y− y

∂

∂x

)(41)

These three Cartesian components are based on the classical expression L = r× p.By using the chain rule technique one can show:

L2 = −h2

(1

sin2 θ

∂2

∂φ2+

1sin θ

∂

∂θ

(sin θ

∂

∂θ

))(42)

Hence L2 operates on both φ and θ. The eigenfunctions that satisfy Eq. (40)

can be written as:

ψ(r, θφ) = N R(r) × Y m (θ, φ) (43)

The Y m (θ, φ) functions are called spherical harmonics and they are simulta-

neous eigenfunctions of L2 and Lz. If two operators have simultaneous eigenfunc-

tions the two operators must commute. It is easy to verify that [Lz, L2] = 0 by

18

direct computation. In fact [Lx, L2] = [Ly, L

2] = 0 as well. On the other hand,

none of the three components of L mutually commute–that is :

[Lx, Ly] = ihLz , [Ly, Lz] = ihLx , [Lz, Lx] = ihLy (44)

By the reasoning of the chapter on Quantum Measurements this means that

it is possible to find simultaneous eigenfunctions of either L2 and Lz , or L2 and

Ly , or L2 and Lx. But it is not possible to find simultaneous eigenfunctions of

the three operators L2 and Lz and Ly.

Before using the spherical harmonics we should summarize their properties.

Because of lack of time, we will not be able to prove these important proper-

ties but unfortunately will just have time to inventory them. There is a rather

beautiful and elegant way of proving these properties using angular momentum

ladders. These proofs appear in my Physics 386 Notes.

Properties of Spherical Harmonics

The eigenvalues for the spherical harmonics are defined by the and m quan-

tum numbers:

L2 Y m (θ, φ) = h2 (+ 1) Y m

(θ, φ) , Lz Ym (θ, φ) = h m Y m

(θ, φ) (45)

Both and m are integer quantum numbers with ∈ 0, 1, 2 . . . and m ∈0,±1,±2, . . . but − ≤ m ≤ . The condition that − ≤ m ≤ enforces the

condition the square of the z component of L should be smaller than the squared

of the total length of L or 〈L2z〉 < 〈L2

x+L2y +L2

z〉. Our quantum number condition

insures this since m2 < (+ 1).

From our experience with Lz we already know that the azimuthal part of

the wavefunction is of the form exp (imφ) which means that the spherical wave

function must be of the form:

Y m (θ, φ) = P m

(θ) eimφ (46)

The P m (θ) functions are called associated Legendre polynomials.

19

Here is a table of some of the low lying spherical harmonics:

Table 4 : Spherical Harmonics

Y 00 Y 0

1 Y ±11√

14π

√34π cos θ ∓

√38π e±iφ sin θ

Y 02 Y ±1

2 Y ±22√

516π

(3 cos2 θ − 1

) ∓√

158π e±iφ sin θ cos θ

√1532π e±2iφ sin2 θ

L2 Y m = (+ 1)h2Y m

Lz Ym

= mhY m

Quantum Numbers

We are now in a position to flesh out Eq. (43). Our wave functions are of

the form ψ(r, θφ) = N R(r)× Y m (θ, φ). It we insert this form into Eq. (40), we

can essentially factor out the Y m (θ, φ) piece leaving a differential equation in R

alone. The Y m (θ, φ) factor only affects the centrifugal barrier term:

H R(r) Y m =

(−h2

2µ

1r2

∂

∂r

(r2∂

∂r

)+

L2

2µr2+ V (r)

)R(r) Y m

=

(−h2

2µ

1r2

∂

∂r

(r2∂

∂r

)+

ˇh2 (+ 1)2µr2

+ V (r)

)R(r) Y m

= E R(r) Y m

Hence

(−h2

2µ

1r2

∂

∂r

(r2∂

∂r

)+

ˇh2 (+ 1)2µr2

+ V (r)

)R(r) = E R(r) (47)

We thus see that both the E of the state and the form of R will depend in

general on the quantum number. In three dimensional problems there are three

quantum numbers. The radial wave function and energy will thus depend on

a third quantum number n as well. As illustrated in Eq. (47), because of the

20

symmetry of the potential, the radial wave function cannot depend on m which

describes the orientation of L.

The entire wave function will therefore involve a radial wave function R(r)

multiplying the spherical harmonic.

Ψ = Rn(r) Y m (θ, φ) (48)

In general the energy of a spherically symmetric state will depend on just n

and quantum numbers as does the radial wave function. Wave functions with

different m quantum numbers will necessarily be degenerate.

Normalization

The constant factors such as 1/√

4π for Y 00 are connected with the normal-

ization condition. For the case of spherical harmonics, the usual one dimensional

normalization condition:

1 =

+∞∫−∞

dx ψ∗(x)ψ(x)

is replaced by an integral over the solid angle of a sphere. Recall that the solid

angle is the area “swept out” by moving through dθ and dφ on a unit sphere.

It is most easily visualized by considering the area of an infinitesimal rectangle

which is swept out first by dθ followed by sweeping out dφ as illustrated below:

21

θdθ

r=1

r=1

z

r=1 sin

y

x

z

x

θ= dθd Ω

d

sin

dφdφ

φdΩd

φ

θ

dθ

d

θd

φ

sin

sin θ

dθ

θsin

θsin

2π∫0

dφ

π∫0

(sin θ dθ)|Y m (θ, φ)|2 = 1 (49)

where we use the fact that an element of spherical area (dΩ) is given by dΩ =

sin θ dφ dθ as illustrated above. Lets check this for Y 00 which is just a constant.

The area of a sphere is Ω = 4π and hence to ensure the normalization given Eq.

(49) we need |Y 00 (θ, φ)|2 = 4π or Y 0

0 (θ, φ) = 1/√

4π.

As discussed in the chapter on Quantum Measurements, two eigenfunc-

tions with two different eigenvalues are orthogonal. This very useful result means

that the spherical harmonics obey:

2π∫0

dφ

π∫0

(sin θ dθ) (Y m′′ )∗Y m

= δ′ δm′m (50)

The symbol δij , called the Kronecker Delta, is defined below:

δij =

(1 if i = j

0 if i = j

)(51)

Hence the orthogonality condition described by Eq. (50) means that area integral

22

over (Y m′′ )∗Y m

vanishes unless both ′ = and m′ = m.

Building Spherically Symmetric Harmonic Oscillator Wave Functions

Exploiting our experience with the two dimensional harmonic oscillator we

can obtain φ symmetric combinations of the three degenerate first excited state

wave functions as follows:

ψ± = (1, 0, 0)± i(0, 1, 0) = N(sin θ e±iφ

)r e−γr2/2

ψ3 = (0, 0, 1) = N (cos θ) r e−γr2/2 (52)

Comparing the wave functions in Eq. (52) to our spherical harmonic table

(Table 4) we see:

ψ± = N r exp(−γr2

2) Y ±1

1 (θ, φ) , ψ0 = N r exp(−γr2

2) Y 0

1 (θ, φ) (53)

Hence these three wave functions all have = 1 and have m values of +1,−1, 0

and are degenerate with an energy of 5h/2 Although there isn’t any need to do

so, one could check that these wave functions are eigenfunctions Lz and L2 by

direct computation. For example

Lzρ e−γρ2/2 Y 0

1 = Lzρ e−γρ2/2 e±iφ = ρ e−γρ2/2 h

i

∂

∂φe±iφ = ±h

(ρ e−γρ2/2 e±iφ

)(54)

and

L2 Y ±1 = L2 sin θ e±iφ = −h2

(1

sin2 θ

∂2

∂φ2+

1sin θ

∂

∂θ

(sin θ

∂

∂θ

))sin θ e±iθ

= −h2

(− e±iθ

sin θ+e±iφ

sin θ∂

∂θ(sin θ cos θ)

)

= −h2

(− e±iθ

sin θ+e±iφ

sin θ(cos2 θ − sin2 θ

))= 2h2

(sin θ e±iφ

)(55)

The below vector picture gives a graphical summary of the angular momentum

state of ψ± and ψ0.

23

0

2 h

2 h

z

1 h

-1 h

h2L

The angular momentum lies on a cone with Lz = ±h or a disk with Lz = 0.

The magnitude of√L2 =

√2h2 =

√2h. The Lx and Ly values are uncertain

since ψ± are not eigenstates of these operators.

Extracting the angular momentum contents of wave functions

As described in Quantum Measurements, the orthonormality condition

(Eq. (50) ) allows one to expand any arbitrary angular wave function f(θ, φ) as

a sum of spherical harmonics.

f(θ, φ) =∑

m=∑m=−

am Y m (θ, φ)

where am =

2π∫0

dφ

π∫0

(sin θ dθ) (Y m )∗f(θ, φ) (56)

The number am is the amplitude for finding the electron with an angular wave

function f(θ, φ) in an angular momentum state (,m) that is with L2 = (+1) h2

and Lz = mh.

24

When evaluating solid angle integrals, I find the following trick to be most

useful.

d cos θdθ

= − sin θ → sin θ dθ = −d cos θ (57)

This means that one can replace the sin θ dθ part of the solid angle integral by

an integral over −d cos θ. We also have to be careful to convert the limits of

integration.

π∫0

sin θ dθ → −cos π∫

cos 0

d cos θ =

1∫−1

d cos θ (58)

We thus obtain the very useful formula

π∫0

dθ sin θ

2π∫0

dφ ≡1∫

−1

d cos θ

2π∫0

dφ (59)

Example

Assume that an angular wave function for an electron in a spherically sym-

metric potential is of the form

ψ ∝ x2

r2= N sin2 θ cos2 φ

Find the probability the electron has zero angular momentum = 0.

25

Lets begin by normalizing the wave function.

N−2 =

2π∫0

dφ

+1∫−1

d cos θ sin4 θ cos4 φ

=

2π∫0

dφ cos4 φ×+1∫

−1

d cos θ(1 − cos2 θ

)2=

3π4

× 1615

→ N =

√54π

The φ integral can be done analytically through two applications of the well

known substitution:

cos2 φ =(

1 + cos 2φ2

)

which leave vanishing integrals over full cycles of trig functions. The integral over

cos θ is just an integral over a polynomial in cos θ

We next evaluate a 00 using Eq. (59):

a 00 =

2π∫0

dφ

+1∫−1

d cos θ(

1√4π

)∗ √ 54π

sin2 θ cos2 φ

= (1√4π

)∗√

54π

2π∫0

dφ cos2 φ

+1∫−1

d cos θ(1 − cos2 θ

)

=√

54π

2π∫0

dφ cos2 φ×+1∫

−1

d cos θ(1 − cos2 θ

)=

√5

4π× π × 4

3=

√59

where we used the facts that

2π∫0

dφ cos2 φ =2π2

= π

26

+1∫−1

d cos θ(1 − cos2 θ

)=[cos θ − 1

3cos3 θ

]+1

−1

= 2 − 23

=43

Hence the probability of finding this electron with zero angular momentum is

|a 00 |2 = 5/9.

Important Points

1. We applied the separation of variables technique to the time independent

Schrodinger Equation in Cartesian coordinates for both the particle in an

infinite box and the isotropic harmonic oscillator for the case of two and

three dimensions. The three dimensional expressions are:

Particle in a box

ψ(x, y) = N sin (nx π x

a) × sin (

ny π y

b) × sin (

nz π z

c)

E(nx, ny, nz) =h

2µ(k2

x + k2y + k2

z

)=h2π2

2µ

((nx

a

)2+(ny

b

)2+(nz

c

)2)

where nx,y,z ∈ 1, 2, 3, .... The above wave functions are valid if the lower left

corner of the box is at the origin.

Isotropic Harmonic Oscillator

ψ(ξ, η) = N Hnx(ξ) exp (−ξ2/2) ×Hny(η) exp (−η2/2) ×Hnz(ζ) exp (−ζ2/2)

where (ξ, η, ζ) =√µω

h(x, y, z) and nx,y,z ∈ 0, 1, 2, ..

To go from three to two dimensions one simply drops the nz terms.

27

2. It is often possible to get degeneracies in the above expressions where several

sets of nx, ny, nz quantum numbers give exactly the same energy. This

degeneracy is connected with the symmetry of the system. The maximum

degeneracy in the infinite box occurs for a cube.

3. When the two dimensional harmonic oscillator is specified by Cartesian

quantum numbers nx, ny, the PDF is generally not azimuthally sym-

metric although such wave functions are real. One can build azimuthally

symmetric, complex wave functions of the form Rn,m(rho) exp (imφ) which

involve position dependent phases. These position dependent phases imply

circulating probability currents which give the electron a non-zero angular

momentum (Lz).

4. It is possible to choose stationary states which are simultaneous eigenstates

of angular momentum since the angular momentum operator commutes

with the Hamiltonian in the case of a cylindrically symmetric or spheri-

cally symmetric potential. In two dimensions such eigenstates of angular

momentum have a φ dependence of the form ψ ∝ exp (imφ) where m is an

integer.

5. In three dimensions, the stationary states spherically symmetric potentials

can simultaneous eigenfunctions of both L2 and Lz. Because [Lx, Lz] = 0

and [Ly, Lz] = 0 they cannot also be eigenfunctions of Lx or Ly. The oper-

ators L2 and Lz only operate on angular part of the wave function. Their

eigenfunctions are called spherical harmonics, Y m (θ, φ) with L2 eigenval-

ues (+ 1)h2 and Lz eigenvalues of mh. The and m quantum numbers

are integers, ∈ 0, 1, 2, .. , m ∈ 0,±1,±2... but − ≤ m ≤ +.

6. The spherical harmonics involve powers of sin θ or cos θ which increase with

increasing times a φ dependence of the form exp (imφ). In order to insure

a single valued phase at the poles, the spherical harmonics include at least

one factor of sin θ whenever m = 0 which causes them to vanish at the

poles.

28

7. The spherical harmonics are orthonormal in the sense:

2π∫0

dφ

+1∫−1

d cos θ (Y m′′ )∗Y m

= δ′ δm′m

which allows one to expand any wavefunction in terms of spherical harmon-

ics via:

f(θ, φ) =∑

m=∑m=−

am Y m (θ, φ)

where am =

2π∫0

dφ

+1∫−1

d cos θ (Y m )∗f(θ, φ)

The amplitudes am determine the probability of finding an electron in an

(,m) angular momentum state.

29