1

Chemistry 430b

Curt Wittig: SSC 403, 07368, wittig@usc.edu, officehours: Th 1:30-3:00.

There will be 5 exams. We will cover Chapters 1-12.

Lectures: complement the text; PPT in moderation; moreparticipation that you are used to.

Assignments: This is a big deal. For Wednesday readChapter 1. Problems: 4, 6, 8, 11, 13, 14, 22, 27, 29.

This is a difficult course; you will have to work hard.

Math? Physics? PPT versus board? Volunteers?

TA: Laura Lazarus: llazarus@usc.edu, office (LJS 270)213-740-7024, lab 213-740-1817; office hours: Wed 2-3.

Discussion sessions, SGM 226).

Do not fall behind.

2

Louis de Broglie and quantum theory

Einstein had introduced special relativity and it wasobvious that waves are essential, e.g.,

… waves are needed to reconcile interference.

How can the theory be made quantum mechanical and atthe same time be consistent with special relativity?

ei(kx!" t )

Photon: E = h! (!" ) Get used to !

Relativity: E2 = p2c2 + m2c4; space and time are on equal footing

For m = 0, pc = E (sign!) # p = h! /c = h/$. This is for a photon.

de Broglie: p = h/$ when m % 0

Low energy: E2 = p2c2 + m2c4 ! E = mc2 1+p2c2

m2c4

"#$

%&'

1/2

( mc2 +p2

2m

3

Louis de Broglie #2

ei(kx!" t )

Notice that p =h

!=h

2"

2"

!= !k

kx !"t = (!kx ! !"t) / ! = px !E

cct

#$%

&'(/ !

How does this differ for a particle wave versus anelectromagnetic wave?

Recall space/time equal footing

pxx + pyy + pzz + (iE /c)(ict)

Let's work through astandard wave eqn.

1

v2

!2"

! t2#!2"

!x2= 0

old notation

4

Particle waves, packets, and uncertainty

Let's see what a wave looks like mathematically andhow particle waves differ from electromagnetic waves.

Make a few drawings that depict waves.

Classical waves have ± !.

QM waves have + !.

Why do they differ in this manner?

ei(kx!" t ) Photon: eik(x!ct )

Particle: kx !"t = kx !E

!t = kx !

(!k)2

2m!t = kx ! k2

!t

2m

5

Packets, and uncertainty

What's the deal with packets and how does this relateto uncertainty? For example, if " denotes the spreadof a distribution, how are "k and "x related? Is therean easy way to understand this qualitatively?

….. It is a good idea to spend time thinking about "x"k,"x"p, "E"t, etc. (Fourier, variance, standard deviation).

6

Classical wave equation

Draw a few waves.

!2u

!x2"

1

v2

!2u

! t2= 0 #

1

X

!2XT

!x2=

1

v2

1

T

!2T

! t2= "k2

u(x, t) = A cos kx + Bsin kx( ) C cos!t + D sin!t( )(or exponential equivalent) ! = vk

Boundary conditions:

X = 0 at x = 0 and ax = 0

x = a

Electromagnetic waves

Lasers: many wavelengths (! ' s) between mirrors

Microwave: modest number of ! ' s in 'cavity'

7

Classical wave equation

x = 0

x = a

Relate these expressions

for sinusoidal oscillation.

u(x, t) = C1 cos!t +C2 sin!t( )sin kx

u(x, t) = A cos kx + Bsin kx( ) C cos!t + D sin!t( )

A = 0

D1ei! t + D2 e

"i! t

ka = m!

8

Waves in 2D

!2

!x2+

!2

!x2"#$

%&'u(x, y, t) =

1

v2

!2u

! t2= (k2

XY = A cos kxx + Bsin kxx( ) C cos kyy + D sin kyy( )

!2

1

XYT

!2XYT

!x2+!2XYT

!x2"#$

%&'=

1

XYT

1

v2

!2XYT

! t2= (k2

1

X

!2X

!x2+

1

Y

!2Y

!y2=

1

T

1

v2

!2T

! t2= "k2 #

1

X

d2X

dx2= "kx

2 and 1

Y

d2Y

dy2= "ky

2

Boundary conditions: X(0) = X(! ) = 0;Y (0) = Y (! ) = 0

XY = sin kxx sin kyy

kx = m! /a ky = n! /b

u(x, y, t) = C1mncos!mnt +C2

mnsin!mnt( )(sinm" x/a)(sinn" y/b)

m,n

#

kmn2 = (m! /a)2 + (n! /b)2

9

1D lossy oscillator

m!!x + b !x + kx = 0 !!x + 2! !x +"

02x = 0

x = est ! s2 + 2"s +#02 = 0

s = !" ± i (#02 ! "2 )1/2

x = A1e!" tei# t + A2e

!" te!i# t or C1e!" t cos#t +C2e

!" t sin#t

! 2

10

The Schrödinger eqn is not magic.

Intuitive

Classical mechanics: E = p2 /2m for a free particle.

We have seen that waves can be expressed as e±i! te±ikx in 1D.

Remember deBroglie: p = h/" = (2# /")!.

But 2# /" = k. Is this obvious? Draw a picture.( ) Thus, p = !k.

What needs to be done to waves like $ = e±i! te±ikx to obtain p?

!i!"

"x# = !i!" x# = !i!" x (e±i$ te±ikx ) = !i! (±ik)e±i$ te±ikx

±!k

1D: p = !i!" x

3D: p = !i!"# (gradient)

11

Not magic #2

Consider a single particle.

Classical mechanics: H = pi !qi ! L. ! H is the Hamiltonian.

In simple terms, H corresponds to E.

Using operators: H = p2 /2m + V .

1D: p = !i!" x

3D: p = !i!"# (gradient)

1D: H = (!i!" x )2 /2m + V = ! (!2 /2m)" x

2 +V (x)Carats on the

operators are

often omitted.

! 2

!x2 often written" 2( )

Operate on a wave (1D):

H! = E!

"(!2 /2m)# x2 +V (x)( )! = E!

12

Time dependence

Again, recall classical waves: e±i! te±ikx

We want Op" = E" . Note e±i! t .

(i!# t )e±i! t = ± i2!! e±i! t

!"!

Use e!iEt /!

Thus

i!! t" = H" = #!2

2m$ 2" +V"

1 particle, without spin

We have usedintuition to surmise aresult. There is a lotmore to it than this,but we are OK.

13

Operators and linear algebra

This has precisely-defined momentum (no spatial resolution).

All information about a system is encoded in its wave function. Toobtain the momentum for the above wfn?

ei(kx!" t )

!

!xei(kx"# t ) = ike i(kx"# t )

!i!"

"xei(kx!# t ) = !(i!)(ik)ei(kx!# t ) = !kei(kx!# t )

We have seen that operators are used with functions, e.g., wavefunctions.

momentum eigenvalue

Algebra: A(c1 f1 + c2 f2 ) = c1Af1 + c2Af2

often represents an "observable"

Extensive use will be made of and higherdimensional counterparts.

dx! *(x)A" ! (x)

14

Operators do not in general commute

Let's apply x and p to a wave function ! (x).

p! = "i!! '

xp! = "i! x! '

x! = x!

px! = "i! (! + x! ')

Therefore (xp! px)" = !i! x" '+ i! (" + x" ') = i!"

This is written [ x, p] = i!. It is more common to see [x, p] = i!

This is the cornerstone of non-relativistic QM. Interpretation:it limits the precision of specifying x and p.

! ' "#!#x

$%&

'()

Carats will be droppedunless clarity requirestheir use.

15

Hermitian operators: mysterious ?

Hermitian: dx! *(x) A"# (x) = dx"(x) A! (x)( )*

#

| n! !m | n" !m | A | n"!n |Convenient notation: What are these?

Using this notation, dx! *(x) A"# (x) = $! | A |"%

The Hermitian requirement is !" | A |#$ = !# | A |" $*

!" i | A |" j # = !" j | A |" i #* or ! i | A | j # = ! j | A | i #* or Aij = Aji*( )

Eigenvalues a : Aij = aj ! i | j " and A ji = ai ! j | i " # A ji* = ai

*! i | j "

Subtracting: 0 = Aij $ A ji* = (aj $ ai )! i | j " ! ai = ai

*

Eigenvalues of a Hermitian operator are real. Non-degenerateeigenvalues have orthogonal eigenfunctions.

16

Dirac (bra and ket) notation

bra's and ket's

This will be abstract for a while. Do not be discouraged. Onceyou "get it," you will not go back to the less compact notation.

If operators commute, the corresponding observables can beknown simultaneously with arbitrarily high precision. This is veryimportant. For a set of commuting operators, we can, in principle,find wave functions that are simultaneous eigenfunctions ofthese operators.

The concept of Hermitian operators is usually vague when oneencounters it for the first time. The most important fact is thatthe eigenvalues of Hermitain operators are real.

| n!!n |

17

Hermitian matrices

H! = E! ! = c1"1 + c2"2 # H c1"1 + c2"2( ) = E c1"1 + c2"2( )

Multiply from left by "1* and integrate: d$ "i*H" j

V% = Hij and d$ "i*" jV% = i j

To see how this works, let's go through an example.

H11c1 + H12c2 = Ec1 1 1 + Ec2 1 2 = Ec1 Likewise for !2* from the left.

H11c1 + H12c2 = Ec1

H 21c1 + H 22c2 = Ec1

H11 ! E H12

H 21 H 22 ! E

"

#$

%

&'c1

c2

"

#$

%

&' = 0

detH11 ! E H12

H 21 H 22 ! E

"

#$$

%

&''= 0 ( E = 1

2H11 + H 22( ) ± 1

4(H11 ! H 22 )2 + H12H 21

The !i ' s are

basis functions.

Unless H12H 21 is positive real, there is a problem. Specifically, the eigenvalues

need to be real. Therefore H12 = H 21*

volume element (any dimension) ! ij in this case

18

Matrix multiplication does not in general commute

1 3

4 2

!

"#$

%&0 1

2 3

!

"#$

%&=6 10

4 10

!

"#$

%&

0 1

2 3

!

"#$

%&1 3

4 2

!

"#$

%&=

4 2

14 12

!

"#$

%&

Even with symmetric matrices:

1 1

1 2

!

"#$

%&2 3

3 2

!

"#$

%&=5 5

8 7

!

"#$

%&

2 3

3 2

!

"#$

%&1 1

1 2

!

"#$

%&=5 8

5 7

!

"#$

%&

M1M2 !M2M1 =0 -3

3 0

"

#$%

&'

How does this relate to [x, p]? It is straightforward to writematrices for x, p, etc. (discussed later).

Note: Trace is unaffected: TrAB = TrBA

19

Particle in a box

free waves

1D: eikx

3D: ei!k !!r

Boundary conditions determine the constants (A, B, …) and the allowedvalues of the "wave vector" k. In other words, the boundaries determinethe quantization conditions (quantum numbers). For example, ka = n# yieldsthe energies of states with different n values:

particle in a box

1D: Asin kx + B cos kx

3D: Asin kxx + B cos kxx( ) C sin kyy + D cos kyy( ) E sin kzz + F cos kzz( )

E =p2

2m=(!k)2

2m=!2 (n! /a)2

2m=

h2

8ma2n2

0 a

Correspondence principle: use large quantum numbers and add wavestogether in order to achieve localization.

Note the quadratic dependence.

What are the "expected" (i.e., average) values of quantities like x and p ?

!1 | x | 2"!1 | x | 1"

!2 | x | 2" !2 | p | 2"

!1 | p | 1" !1 | p | 2"

!1 | x | 3" !1 | p | 3"

Big boxes: entropy

Small boxes: QM systems

Test-type Q's

Matrixelements

20

Boxes in 2D and 3D

2/a sin(n! x/a)2D

" #"" 2/a 2/b sin(n! x/a)sin(m! y/b)3D

" #"" ...

2D: Each state has 2 quantum numbers2D

! "!! | m,n#

What if a = b = c ? Spherical shell: r = a ? (separate radialand angular parts using spherical coordinates)

1D: ! n = 2/a sin(n" x/a) # | n$

dx ! n '* ! n" = (2/a) dx sin(n '# x/a)sin(n# x/a)" = $n 'n

!n ' | n"

Try to visualize this.

An extremely useful tool is the "density of states." Calculatedn /dE in 1D, then in 2D and 3D. This shows how the numberof states per unit energy interval varies.

n is wave function

!n ' | n" is integration

(inner product)

21

Uncertainty relations

The extent to which observables cannot be known independentlyof one another can be calculated. Consider the particle in a box.

0 a

!a/2 0 a/2

To get ! x = x2 " x2

it is easiest to relabel the x-axis:

After some minor math: ! x =a

2"n" 2n2

3# 2

$%&

'()

1/2

Therefore ! x! p =a

2"n" 2n2

3# 2

$%&

'()

1/2

n"!a

=1

2!

" 2n2

3# 2

$%&

'()

1/2

n = 1: ! x! p = 1

2!(1.14)

Only for Gaussian distributions is equality obtained: ! x! p = 1

2!

x = 0

! p = p2 " p

2= p2 Using p = ! (n# /a) gives ! p = n#! / a

positive= 0 here

22

Postulates and things of this nature

The so-called wave function has a privileged status. We need to be carefulwith words. The term "wave function" usually refers to functions like theones we have encountered so far. However, these functions do not includespin, which is essential for a complete description. Therefore, we shallagree to use the term "wave function" with the understanding that spin isgoing along for the ride.

Note that spin does not have a spatial wave function … a point we willdiscuss later. Note also that Postulate 1 on page 116 is not entirely correctin that it ignores spin.

Normalization is a "no-brainer." It would be a pretty strange theory if theintegral over all space of the probability density for a particle did notreturn unity.

Let's see how this works by taking combinations of functions that areeach normalized separately … on the board.

A surprisingly tricky case is the waves we began with: eikx.

23

Operators, eigenvalues, eigenfunctions

p = !i!

"" x = x H = kinetic plus potential energies

Angular momentum:

!L =!r !!p

A! n = an! n

What do we measure? Eigenvalues? Average values? How are theserelated? What is the average value of an operator? Let's go througha few examples to see what is going on.

What are eigenvalues and eigenfunctions?

Eigenvalue problems have been around for hundreds of years.

A particle confined by V (x) resultsin discrete states. Labels such as Enand $n are used. Values of n labelthe eigenstates.

x = 0

V (x)

H! = E! gives many ! n " | n #

carats vs. arrows?

24

Schrödinger equation

i!! "

! t= H" Use " =# (x)$(t) and separate variables.

Notice that the Schrödinger equation is first order in timederivative and second order in space derivative. It is sometimesreferred to as a diffusion equation rather than a wave equation.

i!!"#

! t= H"# $ i!"

!#

! t= #H" $

i!

#

!#

! t=

1

"H" = E

H! = E! and i!"#

" t= E#

! " e#iEt /!

The Schrödinger eqn is for non-relativistic systems, which is OKfor most chemistry.

Eigenfunctions ! n

H! n = En! n "n # e$iEnt /!

25

Expansions and bases

x = 0

A whole family of | n ! is obtained, i.e., different values of n.

|1!

| 2 !

| 3 !

| 4 !With commuting

observables, there

are several labels

(quantum numbers):

| n1,n2,n3,...!.

The set of functions obtained from H! = E! comprise what is called a

complete set, or a basis. These can be used to expand the most general

wave function that can exist in the physical space defined by the potential

(boundaries).

Can you think of systems in which this arises? Rotations? Vibrations?

26

Commutation

Let's examine this using the board.

[A,B] = 0

AB! = BA! [H , B] = 0

[x, p] = i!

A! nA= anA! nA

two independent oscillators

27

Recall the classical harmonic oscillator

1D harmonic oscillator: F = !kx"V (x) = 1

2kx2

x = 0

m!!x + kx = 0 yields e±i! t or cos!t

and sin!t... satisfies initial conditions! 2 = k /m

The oscillator can be "driven," e.g.,consider near resonant excitation. Classical turning points: ± xm

When T = 0, E =1

2kxm

2

Expand V (x) around x = 0: k = (! 2V / !x2 )x=0

... harmonic for small displacements.

phase !

parabolaV (x)

With x = A cos!t + Bsin!t,

x(0) = x0 , !x(0) = 0 " B = 0, A = x0 .

28

Classical oscillator: details

Equation of motion: m!!x + b !x + kx = Fei! t " !!x + 2# !x +!02x = Aei! t

This includes loss and a sinusoidal driving force.

x = x0ei! t " (#! 2 + i2$! +!0

2 )x0ei! t = Aei! t

x0 =!A

" 2 ! i2#" !"02=

!A

(" ! i# + !" )(" ! i# ! !" ) where !" 2 ="0

2 ! #2

But ! can be near + !! or – !! . Taking the former, x0 "#A / 2 !!

(! # i$ # !! )

How can we interpret this strange looking expression?

| x0 |2

29

Quantum harmonic oscillator

x = 0

! 2 = k /m !i

"

2[x, p]

!!

2

H = !! a†a + 1

2( )“zero-point”energy

H =kx2

2+

p2

2m=!

m!2

x2 +p2

2m!"

#$%

&'= !!

m!2!

x"

#$

%

&'

2

+p

2m!!

"

#$%

&'

2()*

+*

,-*

.*Factor this.

a† a

Verify this.

!!m!2!

x " ip1

2m!!

#

$%

&

'(

m!2!

x + ip1

2m!!

#

$%

&

'(

30

Interesting properties

H = !! a†a + 1

2( )

It is easy to show that [a,a†] = 1 and that [H ,a] = !!" a. Brute force (next slide):

[H ,a]! = "!# a! = Ha! " aH! = Ha! " Ea!

! H (a" ) = (E # !$ )(a" )

Therefore a" is proportional to an eigenfunction with eigenvalue E # !$ .

Likewise for a† and E + !$ . Eigenvalues are separated by !$ ! E = !$ v + 1

2( ).

The ! eigenvalue is E

The a! eigenvalue is E " !#

Raising and lowering operators

m!2!

x " ip1

2m!!

#

$%

&

'(

m!2!

x + ip1

2m!!

#

$%

&

'(

a†

a

algebra

31

Interesting properties #2

It is easy to show that [a,a†] = 1 and that [H ,a] = !!" a. Let's do this.

[a,a†] =m!2!

x + ip1

2m!!

"

#$

%

&' ,

m!2!

x ( ip1

2m!!

"

#$

%

&'

)

*++

,

-..

=

m!

2!x2 " x2( ) +

1

2m!!( p2 " p2 )+

i

!( px " xp)

= 0 + 0 !

i

![x, p]

=m!2!

x + ip1

2m!!

"

#$

%

&'

m!2!

x ( ip1

2m!!

"

#$

%

&' (

m!2!

x ( ip1

2m!!

"

#$

%

&'

m!2!

x + ip1

2m!!

"

#$

%

&'

= 1

[H ,a] = !! (a†a + 1

2),a"

#$% = !! a†a,a"# $% = !! a†aa & aa†a( ) = !! (a†a & aa† )a = &!! a.

32

Interesting properties #3

Start with the expressions: a† v = c+ v +1 and v a = c+

* v +1

Taking the inner product gives:

v aa† v = v a†a +1 v = (v +1) v v = c+*c+ v +1 v +1 =| c+ |2

1

Choose c+ = v +1

a†a = number of quanta

Carrying out the analogous calculation for c! and c!* gives c! = v.

(phase convention)

Do you see how this works? ! v

! v+1

[a,a†] = 1 1

a† v = v +1 v +1

a† v = v v !1

33

Quantum harmonic oscillator #2

m!

2!x" 0 + !

d" 0

dx

1

2m!!= 0 #

m!

!x" 0 +

d" 0

dx= 0 #

d" 0

" 0

= $m!

!x dx

a! 0 =m"2!

x + ip1

2m"!

#

$%

&

'(! 0 = 0

To obtain the vibrational wave functions, start with an eqn forthe lowest energy (ground) vibrational state $0:

This is solved right away, yielding:

! 0 = N exp "m#2!

x2$%&

'()

In other words, lowering the ground state must give zero.

The ground state wfn has been obtainedwithout solving a 2nd order differential eqn.

"Gaussian"

($0 can’t get lower)

For the experts: What does this sayabout the momentum distribution?

34

Quantum harmonic oscillator #3

! 0 = N exp "m#2!

x2$%&

'()

Obtain N from normalization:

! 0 ! 0 = 1= dx N02 exp "

m#!

x2$%&

'()"*

+*

+

N0 = m! /"!( )

1/4

! 0 = N0 exp "m#2!

x2$%&

'()

The rest of the wfns are obtained by

applying the raising operator, e.g.,

! 1 * a†! 0

Let's see how this works.

Also:

a†! v = v +1! v+1

a! v = v ! v-1

a†! 0 =m"2!

x #!

2m"!

$

$x

%

&'

(

)*! 0 + N0 x exp #

m"2!

x2,-.

/01

35

Using raising and lowering operators

a =m!

2!x + ip

1

2m!!

a† =m!

2!x " ip

1

2m!!

add andsubtract

a + a† =2m!

!x

a " a† =2

m!!ip

x =!

2m!a† + a( )

p = im!!

2a† " a( )

v x2 v =

!

2m!v (a† + a)(a† + a) v =

!

2m!v a†a + aa† v =

!

2m!(2v +1)

Therefore v

1

2kx2 v =

1

2!! (v +

1

2)

Likewise v

p2

2mv =

1

2!! (v +

1

2)

T = U

for a harmonic oscillator

We can form matrices for x and p. Therefore, by matrix multiplication,we can form the matrix for any polynomial comprised of terms of theform x

n p m. In most cases p

m only appears by itself as p 2. However,

polynomials in x are common.

k =! 2m

Work this out.

36

Practice

x =!

2m!a† + a( )

p = im!!

2a† " a( )

Let's build matrices based on v x v ' and v p v '

v x v ' v p v '

0 x 1 0 p 1

[x] =!

2m!

0 1 0 0

1 0 2 0

0 2 0 3

0 0 3 0

"

#

$$$$$

etc.

etc. etc.

[ p] =m!!

2

0 "i 0 0

i 0 "i 2 0

0 i 2 0 "i 3

0 0 i 3 0

#

$

%%%%%

etc.

etc. etc.

Work out [x, p].

Work out [H ] = 1

2m[ p][ p]+ k

2[x][x]

2D: V =

1

2kxx2 +

1

2ky y2 + ! xy

37

Practice: 2D

Independent: ! xy = 0

" =" vx(x)" vy

( y) = vx vy = vx ,vy

Set up a basis.

V =

1

2kxx2 +

1

2ky y2 + ! xy

This is more involved, and significantly more important, than the1D case, even when the 1D case has considerable anharmonicity.It is our first foray into the couplings that always exist betweenvibrational degrees of freedom in a polyatomic molecule.

vx ,vy = 0,0 , 0,1 , 0,2 ... 1,0 , 1,1 , 1,2 ,...

This is usually done in order of increasing energy.

CH 4 problems: 1,2,4,7,8, 14-19,26,27,29,33

38

2 oscillators

1

2kxx2 +

1

2ky y2 + ! xy

0,0 0,1 1,0 1,1 0,2 2,0 1,2 2,1 ...

0

0

0

0

0

0

0

0

!

"

#########

$

%

&&&&&&&&&

x y =!

2mx! x

ax† + ax( )

!

2my! y

ay† + ay( ) = K ax

† + ax( ) ay† + ay( )

First let's fill in the places where the matrix elements arenonzero. Then we can put in the numbers.

39

2 oscillators

1

2kxx2 +

1

2ky y2 + ! xy

0,0 0,1 1,0 1,1 0,2 2,0 1,2 2,1 ...

0 0 0 • 0 0 0 0

0 0 • 0 0 0 • 0

0 • 0 0 0 0 0 •

• 0 0 0 • • 0 0

0 0 0 • 0 0 0 0

0 0 0 • 0 0 0 0

0 • 0 0 0 0 0 •

0 0 • 0 0 0 • 0

!

"

#########

$

%

&&&&&&&&&

x y = K ax

† + ax( ) ay† + ay( )

Simplification is sometimes possible by reordering thecolumns and rows.

From this example, could you work out the case of x 2y ?

40

Special problems

2. Given H =p2

2m+

1

2kx2 + ax :

(i) sketch the potential for a few values of a/k;

(ii) express the H matrix for the 5 lowest harmonic oscillator basis functions.

1. Use a harmonic oscillator basis to obtain 4 x 4 matrices for x and p.

x11 x12 x13 x14

x22 x23 x24

x33 x34

x44

!

"

###

$

%

&&&

p11 p12 p13 p14p22 p23 p24

p33 p34p44

!

"

###

$

%

&&&

Is xp Hermitian?

Watch out for“edge effects.”

etc. etc.

41

Rotation

2D: particle-on-a-ring

Angular momentum:

p = !i!"

"(a#)= !

i!

a

"

"#

T = !!2

2mea2

" 2

"#2

Schrödinger eqn:

!2

2mea2

! 2"

!#2+ E" = 0

d 2!

d"2+ m2! = 0 # eim"

m2 = 2mea2E /!2

E =!2

2mea2m2

Why is there no zero point energy? Compare this to the particle-in-a-box.

!L =!r !!p

a

normalization: 1

2!

m ! mass

me = mass

! 2 =1

r

"

"rr"

"r

#

$%&

'(+

1

r 2

" 2

")2

42

Angular momentum

Lz = (!r !!p)z = "i"

#

#$

When the continuity boundary condition is applied to eim! , we find that m

must be an integer: ...,"2,"1,0,1,2,... Applying Lz to eim! gives eigenvalues

m!. Both angular momentum and energy are conserved # E,m .

!L =!r !!p

a

Lz = xpy ! ypz Lx = ypz ! zpy Ly = zpx ! xpz

!L =!r !!p

Lx , Ly!" #$ = ( ypz % zpy ),(zpx % xpz )!" #$ =

ypz zpx ! zpx ypz ! ypz xpz + xpz ypz

!zpyzpx + zpxzpy + zpyxpz ! xpz zpy

"

#$%

&'

Terms 3-6 cancel, leaving:

ypz zpx ! zpx ypz + zpyxpz ! xpz zpy = (zpz ! pz z)(xpy ! ypx ) = i!(xpy ! ypx ) = i!Lz

43

Angular momentum #2

[Lx , Ly ] = i!Lz

+ cyclic permutations Amazing !!

!L !!L = i"

!L

Also works for spin. Howcan this be justified?

Terms 3-6 cancel, leaving:

ypz zpx ! zpx ypz + zpyxpz ! xpz zpy = (zpz ! pz z)(xpy ! ypx ) = i!(xpy ! ypx ) = i!Lz

What happpens if we take Lx ± iLy . Wavefunctions?

We need to express the angular momentumcomponents in terms of r, !, and ". Thisinvolves some geometry and algebra.

44

Miserable algebra

Rectangular and spherical coordinates: x = r sin! cos" y = r sin! sin" z = r cos!

dx = sin! cos" dr + r cos! cos" d! # r sin! sin" d"

dy = sin! sin" dr + r cos! sin" d! + r sin! cos" d"

dz = cos! dr # r sin! d!

Solve these for dr,d! , and d" :

dr = sin! cos" dx + sin! sin" dy + cos! dz

d! = r#1(cos! cos" dx + cos! sin" dy # sin! dz)

d" = (r sin! )#1(# sin" dx + cos" dy)

We can now write :

!

!x=!r

!x

!

!r+!"

!x

!

!"+!#

!x

!

!#= sin" cos#

!

!r+

1

rcos" cos#

!

!"$

sin#

r sin"

!

!#

!

! y= sin" sin#

!

!r+

1

rcos" sin#

!

!"+

cos#

r sin"

!

!#

!

!z= cos"

!

!r$

sin"

r

!

!"

Tedious

A picture helps.

help

45

Slightly less miserable algebra

Lz = xpy ! ypz = !i! x"

" y! y

"

"x

#

$%&

'(

Using eqns in the box: Lz = !i!" /"#

which comes as no surprise.

Now form L± = Lx ± iLy .

Further algebra yields: L± = ! e± i! ±"

"#+ icot#

"

"!$

%&'

()

Raising and lowering operators have been introduced. Let’s work this out.

! x = sin" cos#!r +1

rcos" cos#!" $

sin#

r sin"!#

! y = sin" sin#!r +1

rcos" sin#!" +

cos#

r sin"!#

! z = cos"!r $sin"

r!"

Lz = xpy ! ypz Lx = ypz ! zpy Ly = zpx ! xpz

from previous slide

46

Raising and lowering

L+L! = (Lx + iLy )(Lx ! iLy ) = Lx2 + Ly

2 ! i[Lx , Ly ] = Lx2 + Ly

2 + !Lz

Therefore: L2 = L+L! + Lz2 ! !Lz

[L+ , L! ] = 2!Lz

[L± , Lz ] = " !L±

[L2 , Lz ] = 0

[L2 , L± ] = 0 (L+Lz ! Lz L+ ) l,m = !!L+ l,m

L+Lz l,m ! Lz L+ l,m = !!L+ l,m

m!L+ l,m ! Lz L+ l,m = !!L+ l,m

Lz L+ l,m = (m+1)!L+ l,m

Lz L+ l,m{ } = (m+1)! L+ l,m{ }

Therefore L+ changes l,m into l,m+1 , i.e. L+ l,m = C+ l,m+1

The raising operator is established. An analogous procedure showsthat L– is the lowering operator. The parameter m goes from itsmost negative value to its most positive value in steps of 1; therange must be symmetric: mmin = –mmax.

All of this stuffis worked outusing algebra.

47

More results

l,m L+L! l,m = l,m L2 ! Lz

2 + !Lz( ) l,m

[L+ , L! ] = 2!Lz

[L± , Lz ] = " !L±

[L2 , Lz ] = 0

[L2 , L± ] = 0

When m = mmax, L+ cannot raise the state to one with a higher m, so itgives zero. Likewise, when m = mmin, L– cannot lower the state to one witha lower m, so it gives zero. Thus we have expressions for the eigenvaluesof L2 in terms of mmin and mmax.

l,m L!L+ l,m = l,m L2 ! Lz

2 ! !Lz( ) l,m

l,mmax L!L+ l,mmax = 0 = l,mmax L2 l,mmax ! mmax (mmax +1)!2

l,mmin L+L! l,mmin = 0 = l,mmin L2 l,mmin ! mmin (mmin !1)!2

With l = mmax the eigenvalue of L2 is l(l +1)!2. We also get C+ and C!

l,m L!L+ l,m = l(l +1)! m(m+1)( )!2 =| C+ |2

C±= l(l +1)! m(m ±1) !

48

Summary of the algebra

[L+ , L! ] = 2!Lz

[L± , Lz ] = " !L±

[L2 , Lz ] = 0

[L2 , L± ] = 0

[Lx , Ly ] = i!Lz

L± = Lx ± iLy

L± l,m = C± l,m ±1

C±= l(l +1)! m(m ±1) !

L2 l,m = l(l +1)!2 l,m

Lz l,m = m! l,m

The cone pictureis invaluable.

This works for integer and half-integer angular momentum.

x

y

z

49

Wave functions

L+ (!eil" ) = 0 = !ei" #(!eil" )

#$+ icot$

#(!eil" )

#"%

&'(

)*

x

y

z

eil!"#

"$= l#cot$ eil! %

d#

#= l

cos$

sin$d$ = l

d(sin$ )

sin$

The solution of

d!

!= l

d(sin" )

sin" is: ! = sinl"

l, l ! eil" sin l # $ obtain l, l %1 etc. by lowering, for example,

1,1 is the spherical harmonic Ylm(# ,") = Y1

1(# ,") = % 38&

ei" sin#

Now apply L% to % 38&

ei" sin#. This will give Y10 and Y1

%1.

50

2.4.09: Day of examples

! (x) f (t)"! n(x)e# i$nt

Time-dependence, Schrödinger eqn, Fourier transformation, etc.

! (x) = 1

2"dk eikx#(k)

$%

%

&

plane waves

momentum amplitude

!(k) = 1

2"dx e# ikx$ (x)

#%

%

&

! (x) = 1

2"dk eikx 1

2"dx 'e# ikx '! (x ')

#$

$

%#$

$

%

=1

2!dx ' dk eik (x" x ')

"#

#

$%

&''

(

)**

"#

#

$ + (x ')

2! " (x # x ')Also E and t. This is

Fourier analysis.Dirac

51

2.4.09: Day of examples #2

4.33

!1 = Aeik1x + Be" ik1x

Work out the amplitudes for reflected and transmitted waves.

Use continuity of amplitude and momentum at the boundary.

! 2 = Ceik2x +De" ik2x( )

V0

0

x = 0

| B |2

| A |2=

k1 ! k2

k1 + k2

"

#$%

&'

2

| C |2

| A |2=

4k12

(k1 + k2 )2

52

2.4.09: Day of examples #3

4.33

!1 = Aeik1x + Be" ik1x

! 2 = Ceik2x +De" ik2x( )

V0

0

x = 0

R =| B |2

| A |2=

k1 ! k2

k1 + k2

"

#$%

&'

2

T !| C |2

| A |2=

4k12

(k1 + k2 )2

T =p2 | C |2

p1 | A |2=

4k1k2

(k1 + k2 )2

What is going on? How can we understand this? Make plots

53

2.4.09: Day of examples #4

T =p2 | C |2

p1 | A |2=

4k1k2

(k1 + k2 )2What is going on? How can weunderstand this? Make plots.Apply to Schrödinger eqn.

probability current density: !

i!

2m" *#" !" #" *( )

!!2

2m"#"$ +V$ = i! "$ % !

!2

2m$ *"#"$ +V$ *$ = i!$ * "$

!" # *!#( )$ !#( ) !# *( )

Subtract the complex conjugate of eqn 1 from eqn 1 to get:

eqn 1

!!2

2m"# $ *"$ !$"$ *{ } = i! $ * "$ + "$ *$( )

54

2.6.09: pre-exam #1

Problem 4.4 deals with discontinuity. Let's examine a similar case.

!!2

2m

d" '

dx+V (x)" = E"

Integrate:

d(! ')

"#

#

$ =2m

!2dx V (x)" E( )! (x)

"#

#

$ =2m

!2dxV (x)! (x)

"#

#

$

! '(")#! '(#") = #2m

!2dxV0a$ (x)! (x)

#"

"

%

This is the discontinuity. Now get the wave functions.

!V0 a

0

Area is fixed.

55

2.6.09: pre-exam #2

!V0 a

0

! '(")#! '(#") = #2m

!2dxV0a$ (x)! (x)

#"

"

%

! LHS = Aeikx + Be" ikx

! RHS = Ceikx

A+ B = C

ik( A! B)! ikC = (2mV0a / !2 )C = "C

ik( A! B) = C(ik + ")

ik A! (C ! A)( ) = C(ik + ")

2ikA = C(2ik + ")

! '(")#! '(#") = #

2mV0a

!2! (0)

! RHS =k

k " i# /2eikx

56

2.6.09: pre-exam #3

What is the natural frequency? k1

k2

m

Take a 1D oscillator with loss (proportional to velocity) Assume amodest amount of dissipation. At t = 0 the particle is located at theequilibrium position. Its velocity is v0. What is x(t) for the particle?

Operators: particle in a box, particle on a ring, harmonic oscillator

1 p 2 and things like this. Be careful about the "basis" functions.

Non-degenerate eigenfunctions of a given Hamiltonian are orthogonal.

57

2.6.09: pre-exam #4

Momentum distributions for states of: particlein a box, particle on a ring, harmonic oscillator.

[L2 , Lx ] [Lx

2 , Ly ] [Lx Ly , Lz ] and so on

[x, p] = i! Show how this works in "momentum space."

Tunneling: The probability is 10-4. The mass is changedby a factor of two. What is the probability?

Tunneling: A particle-wave is incident at E = V0. Theprobability of transmission is 0.1. The length of thebarrier is doubled. What is the transmission probability?