Mr. A. Square’s Quantum Mechanics

Part 1

Bound States in 1 Dimension

The Game’s Afoot

Primarily, the rest of this class is to solve the time independent Schroedinger Equation (TISE) for various potentials

1. What do we solve for? Energy levels

2. Why? Because we can measure energy most easily

by studying emissions and absorptions

2) What type of potentials?

We will start slow– with 1-dimensional potentials which have limited physicality i.e. not very realistic

We then take a detour into the Quantum Theory of Angular Momentum

Using these results, we then study 3-dimensional motion and work our way up to the hydrogen atom….

Some of these potentials seem a little … dumb.

While a particular potential may not seem realistic, they teach how to solve the harder, more realistic problems. We learn a number of tips and tricks.

The Big Rules

1. Potential Energy will depend on position, not time

2. We will deal with relativity later.

3. We will ensure that the wavefunction, , and its first derivative with respect to position are continuous.

4. will have reasonable values at the extremes

Mathematically,

1. ( , , )

2. ( ) ( )

3. ( ) ( )

( ) ( )

4. ( ) ( )

a b

a b

a b

V f x y z or equivalent

x x

d dx x

dx dxx x

and

Re-writing Schroedinger

A more easily form of the Schroedinger equation is as follows:

2 2

2

2 2

2

2

2 2

( )( ) ( ) ( )

2

( )( ( )) ( )

2

( ) 2( ( ) ) ( )

xV x x E x

m x

xE V x x

m x

x mV x E x

x

Infinite Square Well

2

2 2

22

2

0Potential Definition: ( )

,

In order to ensure that is well behaved, we

will let 0 in the region where V=

In the region from (-a, a)

( ) 2(0 ) ( )

2Let

( )

a x aV x

otherwise

x mE x

xmE

k

x

x

22

22

2

( )

( )( ) 0 (obviously sinusoidal function)

( ) cos( ) sin( )

k x

xk x

xx A kx B kx

Apply Boundary conditions

At x=-a, must be zero in order to satisfy our continuity condition; also, at x=a.

cos( ) sin( ) 0

cos( ) sin( ) 0

cos( ) sin( ) 0

cos( ) sin( ) 0

Add these two equations:

2 cos 0

Subtract these two equations:

2 sin 0

A ka B ka

A ka B ka

or

A ka B ka

A ka B ka

A ka

B ka

Even Solutions

Assume A <>0, then B is zero

These are called “even” functions since (-x)=(x)

2 cos 0

3 5, ,

2 2 2

A ka

ka

Odd Solutions

Assume B <>0, then A is zero

These are called “odd” functions since (-x)=-(x)

Evenness or oddness has to do with “parity”

2 sin 0

2 4 6, ,

2 2 2

A ka

ka

Writing a general function for k

2 2 2 2 2

2

, 1, 2,3,2

, 1,2,3,2 8

n

n

nk n

athen

k nE n

m ma

Wavefunctions for Even and Odd

cos , 1,3,5,2

sin , 2,4,6,2

even n

odd n

nA x n

a

nB x n

a

Finding An and Bn

2

2 2

2

2

2 2 2

2

1

cos ?2

1 1sin sin(2 )

2 41 1

cos sin(2 )2 4

cos 12

1

a

a

a

na

a

n na

n

dx

n xA dx

a

x dx x x

x dx x x

n xA dx A a

a

and

B a

1

1

n

n

Aa

and

Ba

Wavefunctions for Even and Odd

1cos , 1,3,5,

2

1sin , 2,4,6,

2

even

odd

nx n

aa

nx n

aa

Wavefunctions

-0.2

-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0.2

-60 -40 -20 0 20 40 60

N=1

N=2

N=3

In the momentum representation

0

1( ) ( ( )) ( )

2

1( ) 2 cos( )cos( )

22

1 1( ) sin sin

2 2 22 2

1( ) sin

2 22

ikxeven even even

a

even

even

odd

k FT x e x dx

n xk kx dx

a

a n nk ka ka

n nka ka

a nk i ka

nka

1sin

22

nka

nka

Matrix Elements of the Infinite Square Well

The matrix elements of position, x, are of interest since they will determine the dipole selection rules.

The matrix element is defined as

This integral vanishes except for states of opposite parity

It is convenient to specify that n is even parity

m is odd parity

Symbolically using the parity operator,

n n , 1,3,5...

m

m nm x n x dx

n

2 22

m , 2,4,6...

1sin cos

2 2

sin sin2 24

a

a

m

so

m x n xm x n x dx

a a a

m n m n

am x n

m n m n

Several Transitions

2

2 2

2 2

4 402 5

1 1 0 441

4 8 4 162 1 1 2 4 1

9 225

4 24 4 482 3 4 3

25 49

ax

x n x n

a ax x x

a ax x

If I write the wavefunctions as

0 0

1 0

0 02 and 4

0 1

0 0

Then I can write x as a m x n matrix

2

8 160 0 09 2258 240 09 25

240 04 25ˆ16 0 0225

0 0

0

ax

The momentum operator is derived in a similar manner but is imaginary and anti-symmetric

840 0 03 154 120 03 5

120 05ˆ8 0 015

0 0

0

ip

a

Finite Square

The finite square well can be made more realistic by assuming the walls of the container are a finite height, Vo.

Region 1 Region 3

x= -a x=a x

V(x)

V0

Region 2

The Schroedinger Equations

2

02 2

2

2 2

2

02 2

22

2 02

1

2

( ) 2Region 1: ( ) ( ),

( ) 2Region 2: (0 ) ( ),

( ) 2Region 3: ( ) ( ),

2Let

2 ( )Let

Region 1: ( )

Region 2: ( ) cos( ) sin( )

R

x

x mV E x x a

x

x mE x a x a

x

x mV E x x a

xmE

m V E

x Ae

x B x C x

3egion 3: ( ) xx Ae

The Boundary Conditions

1(-a)=2(-a)

2(a)=3(a)

1’(-a)=2’(-a)

2’(a)=3’(a)

The only way to satisfy all four equations is to have either B or C vanish

If C=0, then we have states of even parity

2

2 2 02 2

1

3

2 ( )2Let and

Region 1: ( )

Region 2: ( ) cos( )

Region 3: ( )

cos

sin

Dividing these two:

tan

x

x

a

a

m V EmE

x Ae

x B x

x Ae

at x a

B a Ae

B a Ae

a

If C=0, then we have states of even parity

2 2 02 2

2 2 02

2 ( )2Let and

2Equation 1:

Equation 2: tan

m V EmE

mV

a

If these 2 equations could be solved simultaneously for and , then E could be found.

Two options: Numerically (a computer) Graphically

Even Parity Solutions

0

1

2

3

4

5

6

7

8

9

0 1 2 3 4 5 6 7 8 9

1

2

3

4

5

6

7

8

a

b

c

Each intersection represents a solution to the Schroedinger equation

k=6.8K= 4.2

0

1

2

3

4

5

6

7

8

9

0 1 2 3 4 5 6 7 8 9

1

2

3

4

5

6

7

8

a

b

c

Odd Parity Solutions

Each intersection represents a solution to the Schroedinger equation

Inspecting these graphs

Note that the ground state always has even parity (i.e. intersection at 0,0), no matter what value of Vo is assumed.

The number of excited bound states increases with Vo (the radius of the circle) and the states of opposite parity are interleaved. If Vo becomes very large, the values of approach n/2a This asymptotic approach agrees which the

quantization of energy in the infinite square well

The Harmonic Oscillator

It is useful in describing the vibrations of atoms that are bound in molecules; in nuclear physics, the 3-d version is the starting point of the nuclear shell model

We will solve this problem in two different ways: Analytical (integrating as we have done before) Using Operators

But first we need some definitions

Angular frequency

2 2

2 2 21 1

2 2

kk m

mso

if V kx m x

TISE for SHO

2

2 2

22 2

2 2

22 2

2 2

2 2 22

2 2 2

( ) 2( ( ) ) ( )

( ) 2 1( ) ( )2

( ) 2 1( ) ( ) 0

2

( ) 2( ) ( ) 0

x mV x E x

x

x mm x E x

x

x mE m x x

x

x mE mx x

x

More Definitions

2 22

2 2

2 2 22

2 2 2

22 2

2

2

( ) 2( ) ( ) 0

( )( ) ( ) 0

Let

m mEand

so

x mE mx x

x

xx x

x

Change of Variables

22

2

2

( )( ) ( ) 0

Let

d dq x

dx dq

E

qq q

q

Let q>>

2 2

2

22

2

2 2

2

( )( ) 0

( )

However, ( ) ( ) 0 implies that B=0

so

( )

A

q qA

q

qq q

q

then

q Ae Be

q Ae

Constructing a trial solution

2

2 2

2 2 2 2 2

2

2 2

222 2 2 2 2

2

( ) ( )

( )'( ) ( )

( )'' '

q

q q

q q q q q

Let

q e H q

d qe H q qe H q

dq

qe H qe H e H qe H q e H

q

Adding terms

2

2 2 2 2 2 2 2

2

2 2 2

22

2

2 22 2 2 2 2 2 2

2

( ) ( )

( )( ) ( )

'' ' 0

0

'' 2 ' 1 0

q

q q q q q q q

q

Let

q q q e H q

so

qq q q

q

e H qe H e H qe H q e H e H q e H

Obviously e so

H qH H

Sol’ns of H are a power series

1 2

0

2

' '' ( 1)

'' 2 ' 1 0

( 1) 2 1 0

Each of these sums must vanish separately. On the

first term, I will shift everything by +2 so

( 1)

j j jj j j

j

j j jj j j

jj

H a q H ja q H j j a q

so

H qH H

becomes

j j a q ja q a q

j j a q

22

2

( 2)( 1)

( 2)( 1) 2 1 0

jj

j j j

j j a q

Now

j j a ja a

Finding aj’s

2

2

2

( 2)( 1) 2 1 0

For a given ,

( 2)( 1) 2 1 0

1 2

( 2)( 1)

j j j

j j j

j j

j j a ja a

j

j j a ja a

ja a

j j

Note: These equation connects terms of the same parity i.e 1,3,5 or 0,2,4

How to generate aj’s

For odd parity i.e. q1,q3,q5, start with ao=0 and a1<>0

For even parity i.e. q0,q2,q4, start with ao<>0 and a1=0

We will learn an established procedure in a few pages

Too POWERful!

We have a problem, an infinite power series has an infinite value

So we know that the series at large values must approximate exp(-q2/2)

So we must truncate the power series (make it finite)

The easiest way is to make aj+2 vanishThis occurs when -1-2j=0

Making j equal to n

2 1

22 1

1

2

0,1,2,3,...

0

0

n

En

E n

n

If then

no solution because

j

This is a big result, since you know from Modern Physics that we had to get this result.

For a 3-d SHO, E=(n+3/2), 1/2 for each degree of freedom

Summary so far:

2

2 ( )

'' 2 ' 1 0

is a power series:

1

2

q

nn n

n

e H q

where

H qH H

H

H a q

and

E n

The solutions, Hn, to the differential equation are called “Hermite” polynomials.

In the next few slides, we will learn how to generate them

Method 1: Recursion Formula

The current convention is that the coefficient for the n-th degree term of Hn is 2n

E.g. For H5, the coefficient for q5 is 25

Also, recall for n=5, E=(5+1/2) or 11/2

At this eigenvalue, H= a1q+a3q3+a5q5

where a5=32

j+2 j

j+2 j

j j+2

Previously,

( 1 2 )a a

( 1)( 2)

but 2 1

2( )a a

( 1)( 2)

( 1)( 2)a a

2( )

j

j j

n

n j

j j

or

j j

n j

Method 1: Recursion Formula

The current convention is that the coefficient for the n-th degree term of Hn is 2n

E.g. For H5, the coefficient for q5 is 25

Also, recall for n=5, E=(5+1/2) or 11/2

At this eigenvalue, H= a1q+a3q3+a5q5

where a5=32

j+2

j j+2

3

1

5 35

In this case, j+2=5 and n=5, a 32

( 1)( 2)a a

2( )

(3 1)(3 2)a *32=-160

2(5 3)

(1 1)(1 2)a *(-160)=120

2(5 1)

H 32 160 120

j j

n j

q q q

Method 2: Formula of Rodriques

2 2

1n

n q qn n

dH e e

dq

Method 3: Generating Function

The generating function is a function of two variables, q and s, where s is an auxiliary function

To use this function, take the derivative n times and set s equal to 0

22

For Hermite polynomials, the generating function is

( , ) qs sS q s e

Method 4: Recurrance Relations

Hn+1=2*q*Hn-2*n*Hn-1

We know that H0=1 and H1=2q

Ideally suited for computersAlso, the derivative with respect to q isH’n =2*n*Hn-1

Method 5: Table Look Up

0

1

22

33

4 24

5 35

6 4 26

1

2

4 2

8 12

16 48 12

32 160 120

64 480 720 120

H q

H q q

H q q

H q q q

H q q q

H q q q q

H q q q q

Other Methods

Continued FractionsSchmidt Orthogonalization

Integrals with Hermite Polynomials

2

2

2 2

2 2 2

2

*

-

2 2

-

2

-

Recall

( )

We know that 1

1

From Rodriques formula, ( 1)

we use 1 factor of H to subsitute this

1= ( 1)

q

n n n

n n

qn n

nn q q

n n

nq n q q

n nn

mq x C H q e

dq

C e H dq

dH e e

dq

dC e e e H dq

dq

We know that the biggest term of Hn is 2nqn

Hn~2nqn

H’n~2n*n*qn-1

H’’n~2n*n*n-1*qn-2

Hn-’=2n*n!*qo=2n*n!

Going back and using the n-th derivative

2 2 2

2

2

2

-

2

-

2 2

-

4

First integrate this

1= ( 1)

By parts for eight different times,

1=

and 2 !

1= 2 ! 2 !

1

2 !

nq n q q

n nn

nq

n nn

nn

n

n q nn n

n n

dC e e e H dq

dq

dC e H dq

dq

d Hn

q

C n e dq C n

Cn

Now, we have everything

2

4

2

4

1

2 !

1( ) ( )

2 !1

( )2

n n

q

nn

n

Cn

q H q en

E n

Transition Matrix Elements

2

2

4

n+1 1

11

, 1 , 1

1

2 !

Using the recurrance formula

H 2 2

2

11

2

qm n m n

j j

n n

qnm n m n

m n m n

m q n C C H q H e dq

Cj

qH nH

Hm q n C C H nH e dq

m q n n n

SHO via Operators (or via Algebra)

While the analytical solution is one way to solve this problem, using operators

1. Teaches us the operator method which will be required when we study the quantum theory of angular momentum as well as quantum field theory

2. Gives us another tool in our toolbox for problem solving.

Many Slides ago,

22

2

2

( )( ) ( ) 0

Let

d dq x

dx dq

E

qq q

q

Which is good Schroedinger notation but what about Dirac?

In Dirac notation,

22

2( ) 0q n

q

Where |n> is the nth state of the oscillator

Even More Definitions,

22

2

Now, using the commutator,

[q,p]=i

Proof: where ( )

[ , ]

[ , ]

[ , ]

and furthermore,

p

dp i

dq

f q

d d d dq p qp pq i q q iq i q i

dq dq dq dq

q p i

q p i

d

dq

A new way to TISE

22

2

2 2

2 2

( ) 0

0

dq n

dq

p q n

p q n n

This is TISE in operator form

A necessary aside

2 2

We can write;

if [ , ] 0

i i

Gosh, more definitions?

†

Let

2

2

q ipa

q ipa

It turns out that

† 2 2

† 2 2

† 2 2

Let

1[ , ]

2but [ , ]

so

11

21

12

aa q p i q p

q p i

aa q p

a a q p

Let’s solve for q2+p2

† 2 2

† 2 2

† † 2 2

† †

11

21

12

Adding these together, we get

Subtracting these , we get

1

aa q p

a a q p

a a aa q p

aa a a

A crazy form of TISE

† †a a aa n n

Theorem 8

a+ is called the “raising operator”

Proof of Thm 8

2 2 † †

2 2 † † † † †

† † †

† †

2 2 † † † † †

2 2 † † † †

2 2 † † † †

( ) ( )

( ) ( )

Recall [ , ] 1 i.e. 1

1

( ) ( 1 )

( ) ( ( 1))

( ) ( ( 1 1))

p q n a a aa n

p q a n a aa aa a n

a a aa a a

aa a a

p q a n a aa a a a n

p q a n a aa aa n

p q a n a aa a a n

2 2

2 2 † † † †

2 2 † † † †

2 2 † † 2 2 † 2 2 †

2 2 † † †

2 2

†

( ) ( ( 1 1))

( ) ( ( 2))

( ) ( ( 2)) ( ( ) ) (2)

( ) ( 2) ( 2)

( ) 1 ( 2) 1

1

p q

n

p q a n a aa a a n

p q a n a aa a a n

p q a n a p q n a p q n a n

p q a n a n a n

but

p q n n

Thus

a n n

Theorem 9

a is called the “lowering operator”

Proof of Thm 9

See “Proof of Thm 8” and put minus signs in the appropriate places

2 22 1a n n q p n

Theorem 10

1

2 2

†

11

2

2 2

Proof

First: 0

0

11 0

2

11 0

21 QED

q p

n n

n a a n

n q p n

so

Consequences

We could start with state |n> and lower it until we reached a ground state i.e. a(a|n>)=(-4)|n> etc.

Theorem 11

In the ground state, =1

2 2

†

†

11

2

Proof

First: 0 0

0 0

0 0

11 0 0

21

q p

o

o

a

a a

a a

Now, let’s find the wavefunction of the ground state (hint: it must agree with what we found earlier Hoexp(-q2/2)

2

0 0

0 20 0

0

0 0

0 0 or 0 02

dUse our definition of p=-i

dq

0 0

0 0 (Let me go to S. notation)

q

a

q ipq ip

dq

dq

dq

dq

dq

dq

dq dq A e

A0 is found by

normalizing the wavefunction and this is exactly in agreement with the analytical method.

Now other states, use a+ on |0> and raise, raise, raise

2

2

21

2

1q

n q

n

dA q e

dq

or

dn A q e

dq

An is found by normalization

Calculating Transition Matrix Elements

†

R L

††

† 2

† 2

† 2

2 2

2 2

2 2

Let

1

1

where A and A are just normalization constants

Since

1

Adding them, we get our old friend ( )

2 1

1

Adding these relation

R

L

R

R

R

L

R L

R L

a n A n

a n A n

a a

n a A n

n a a n A

n aa n A

n a a n A

q p

A A n

A A

2 2

ships

1R LA n A n

So the matrices look like

, 1

†, 1

†

1

0 1 0 0 0 0

0 0 2 1 0 0

0 0 0 0 2 0

m n

m n

m a n n

m a n n

so

a a

The matrices of q and p are

† †

2 2

0 1 0 0 1 0

1 0 2 1 0 21

2 20 2 0 0 2 0

a a a aq p i

so

iq p

What about matrices of q2 and p2 ?

2 2

2 2

1 0 2 1 0 2

0 3 0 0 3 01 1

2 22 0 5 2 0 5

(2 1)

q p

p q n I I

The Dirac Delta

The Dirac delta, (x), is used to represent a point particle (x)=0 if x<>0 (x)= if x=0

But what keeps it finite, is the normalization ( ) 1x dx

The Dirac Delta cont’d

If it is at x=a (x-a)=0 if x<>a (x-a)= if x=a

( ) 1

( ) ( ) ( )

x a dx

f x x a dx f a

TISE for the Dirac Delta

2

2 2

2

2 2

( )

( ) 2( ( ) ) ( )

( ) 2( ( ) ) ( ) 0

Let V A x

x mA x E x

x

x mA x E x

x

Some Boundary Conditions

At x=||, =0The big problem lies at x=0

It turns out that is continuousBut ’ is discontinuous

X=0

Technically, this is infinitesimally thick. So the derivatives change from + to – in an infinitesimal distance

We could integrate over a small space

X=0

X=+X=-

So the idea is to integrate from – to and then let go to zero

Mathematically

2

2 20

Let ' represent the discontinuity

2lim ( ) ( ) ( ) 0

The differential integrates to the first derivative,

the energy term goes to zero under limit condition

( )

mdx V x x dx E x dx

x

V x

2

2

2

2

2( ) (0)

'

2' (0)

mAx dx

dxx

so

mA

Using an ansatz

Let

( ) for x<0

( ) for x>0

x

x

x

x

x Ce

x Ce

dCe

dxd

Cedx

2 2

2 22

2 4

in the limit as goes to zero

' 2

at the origin,

2 22 (0)

d dC C

dx dxC C C

mA mAC C

mA m A

Plugging into the TISE

22

2

22

2

2

2 20 for x 0

2 22 2

22

Let

( ) for x<0

( ) for x>0

No matter which I choose

( ) 2( ( ) ) ( ) 0

2 20

In any case,

20

x

x

x

x

x x x x

x Ce

x Ce

dCe

dx

dCe

dx

x mA x E x

x

mE mECe Ce Ce Ce

mE

Solving for E

22

2 22

4

2 2

4 2

2

2

20

Previously,

20

2

mE

m A

m A mE

mAE

Note that energy is a single value; based on the potential amplitude, A

Find

2

2

2

2

2

For compactness, we write

( )

( ) 1

1

( )

2

x

mAx

x Ce

x dx

C mAC

so

mAx e

mAE

The Uniform Force Field

V(x)=G*x, x>0 V(x)=0, x<0 G is a positive constant equal to the gradient

of the potential Function has several physical examples:

An electric charge in an uniform field near an impermeable plate

A tennis ball dropped down an elevator shaft (hence the name of the quantum bouncer)

TISE

2

2 2

2

2 2

( ) 2( ) ( )

( ) 2( ) ( ) 0

Let V Gx

x mGx E x

x

x mGx E x

x

A change of variables3

2

32

32

22 2

32 2 2

2

2 2

22

32 2 2 2

32

2( )

2

2

2

( ) 2( ) ( ) 0

2 2 20

2

E mGLet z x

GE z

xG mG

d d dz mG d

dx dz dx dz

d mG d

dx dz

so

x mGx E x

xbecomes

mG d mG E z mE

dz G mG

2 2 32

32 2 2

2

2

2 20

0 which is called Airy's function

mG d mGz

dz

dz

dz

Airy’s Functions

More about Airy

32

Ai( )

(0) 0

For large z, Ai( )

Now at x=0 (where the floor is), =0

E 2at x=0, z=-

G

z

o

C z

z e

mGz

The root of the matter

The “roots” of a function are the values where a function is equal to zero

We solve the previous equation for E

We set zo equal to the roots of the Airy function, n

n Root

1 2.3381

2 4.08794

3 5.52055

4 6.7867

5 7.94413

6 9.02265

7 10.04017

8 11.00852

9 11.93601

10 12.82877

32

2n n

mGE G

Graphically,

General Features of 1-D bound states

1. vanishes at |x|=2. 1-d Bound states are non-degenerate

By degenerate, 2 states have equal energy

3. Wave functions for a 1-d bound state can be constructed so that it is real

4. For a 1-d bound state, <p>=05. If H is symmetric, the wave functions are

eigenfunctions of the parity operator6. The Schroedinger Equation can be

converted into an integral equation.

Proof of #2: Nondegeneracy of 1-d bound states

1 2

21

1 12 2

22

2 22 2

1 22

1 2

Let 2 states ( and ) have same energy E

2( ) ''

2( ) ''

'' ''2( )

d mV E

dx

d mV E

dxmV E

1 2 1 2

1 2 1 2

1 2 1 2 1 2 1 2

1 2 1 2

1 2 1 2

1 2 1 2

'' '' 0

Add a zero, ' ' ' '

'' ' ' ' ' '' 0

( ' ) ' ( ') ' 0

' ' a constant, C

As x goes to zero, so does the

first derivative of both functions so C=0

' ' 0

1 21 2

1 2

' '* these states merely differ

by a constant

D

Proof of #3: Wavefunctions can be constructed to be real

2

2 2

2 ** *

2 2

*

*

- -

2( ) ''

2( ) ''

by proof of #2,

Since

1 1

d mV E

dx

d mV E

dx

D

dx D dx D

Proof of #4: <p>=0

*

- -

2

- -

-

But ( ) 0

0

d dp dx dx

i dx i dxd d

p dx dxi dx i i dx

dp dx p

i dxso

p

Proof of #5: Wavefunctions are eigenfunctions of parity operator

1

1 2

1

( ) ( ) where is the parity operator

since the Hamiltonian is symmetric and

1

But the nondegeneracy rule means that

since E is the same for both and

where

x x

H H

H E

H E

k

2 2

1

k is some constant

1 which are the only eigenvalues for parity

k

k

Expansion on #6:

2

If there is some value of x (say x=a) where

both (a) and '(a) are known, then

2( ) ( ) '( ) ( )

where y and z are dummy variables.

yx

a a

mx a x x V z E dz dy