5. The Harmonic Oscillator

Consider a general problem in 1D

•Particles tend to be near their minimum•Taylor expand V(x) near its minimum

•Recall V’(x0) = 0•Constant term is irrelevant•We can arbitrarily choose the minimum to be x0 = 0

•We define the classical angular frequency so that

All Problems are the Harmonic Oscillator 21

2H p V x

m

210 0 0 02 0V x V x V x x x V x x

2 20

1 1

2 2H p V x x

m

2 2 21 1

2 2H p m x

m

Raising and Lowering Operators 5A. The 1D Harmonic Oscillator

• First note that V() = , so only bound states• Classically, easy to show that the combination mx + ip has simple behavior• With a bit of anticipation, we define

• We can write X and P in terms of these:

2 2 21 1

2 2H P m X

m

2 2

m Pa X i

m

†

2 2

m Pa X i

m

†

2X a a

m

†

2

mP i a a

Commutators and the Hamiltonian

• We will need the commutator

• Now let’s work on the Hamiltonian

2 2 21 1

2 2H P m X

m

†, ,2 22 2

m P m Pa a X i X i

m m

†

2X a a

m

†

2

mP i a a

1, ,

2 2

mX iP iP X

m

1

2i i i i

1

†, 1a a

2 22 † 2 †1 1

2 2 2 2

mH i a a m a a

m m

2 2† †14 a a a a

† †12 aa a a † 1

2H a a

† †1aa a a

† †12 1a a a a

Raising and Lowering the Eigenstates

• Let’s label orthonormal eigenstates by their a†a eigenvalue

If we act on an eigenstate with a or a†, it is still an eigenstate of a†a :• Lowering Operator:

• Raising Operator:

• We can work out theproportionality constants:

†a a a n

† 12H a a

12nE n

n n nH E nmn m

† †1aa a a

† 1aa a n

†a a n n n

†aa a a n

† †a a a n † †a aa n † † 1a a a n

1n a n

†1n a n

1a n n

† 1a n n

2a n †n a a n n n n n 1a n n n

† 1a n † 1a a n

n

nn

n † 1 1a n n n n n

• It is easy to see that since ||a|n||2 = n, we must have n 0.• This seems surprising, since we can lower the eigenvalue indefinitely

• This must fail eventually, since we can’t go below n = 0– Flaw in our reasoning: we assumed implicitly that a|n 0

• If we lower enough times, we must have a|n = 0 ||a|n||2 = 0• Conclusion: if we lower n repeatedly, we must end at n = 0

– n is a non-negative integer • If we have the state |0, we can get other states by acting with a†

– Note: |0 0

12nE n

1 1 2n a n n a n n

2a n n1a n n n † 1 1a n n n

0,1,2,n

†11n a n

n †1

0!

nn a

n

2†1

21

a nn n

What are the possible eigenvalues

• Sometimes – rarely – we want the wave functions• Let’s see if we can find the ground state |0: 0 0a

The Wave Functions (1)

02 2

m PX i

m

002 2

m dx x

m dx

0 0

d mx x x

dx

0

0

d mxdx

2

0ln2

m xC

2

0 exp2

m xx N

• Normalize it:

2

01 x dx

22 m xN e dx

2

Nm

2

40 exp

2

m m xx

†10

!

nn a

n

The Wave Functions (2)• Now that we have the

ground state, we can get the rest

2

40 exp

2

m m xx

2

41

exp2 2 2!

n

n

m m d m xx x

m dxn

• Almost never use this!– If you’re doing it this

way, you’re doing it wrong

†

2 2

m iPa X

m

†

2 2

m da x x x

m dx

n = 3n = 2n = 1n = 0

Working with the Harmonic Oscillator 5B. Working with the H.O. & Coherent States

• It is common that we need to work out things like n|X|m or n|P|m• The wrong way to do this:

• The right way to do this:

*7 67 6X x x x dx

Abandon all hope all ye who enter here

1a n n n † 1 1a n n n

†

2X a a

m

†

2

mP i a a

7 6X †7 62

a am

7 6 5 7 7

2m

7

2m

Sample ProblemAt t = 0, a 1D harmonic oscillator system is in the state

(a) Find the quantum state at arbitrary time(b) Find P at arbitrary time

1 1 12 22

0 0 1 2i

niE tn n

n

t c e

12nE n

12i n t

nn

t c e n

1 1 10 1 22 22

, ,c c i c

3 512 2 21 1 1

2 220 1 2i t i t i tt e ie e

Sample Problem (2)

(b) Find P at arbitrary time

3 512 2 21 1 1

2 220 1 2i t i t i tt e ie e

P t P t †12i m a a

3 5 3 51 12 2 2 2 2 2†1 1 1 1 1 1 1

2 2 2 2 22 20 1 2 0 1 2i t i t i t i t i t i ti m e ie e a a e ie e

3 51

2 2 23 51

2 2 2

3 52 2

31 122 21 1 1 1

2 2 22 1 12 2

1 2 30 1 2

0 1

i t i t i t

i t i t i t

i t i t

e ie ei m e ie e

ie e

1 1 1 1 12 2 2 2 2 2 2 2 2

i t i t i t i ti m ie ie ie ie 1 122

i t i tm e e

cosP m t

1a n n n

† 1 1a n n n †

2

mP i a a

Coherent StatesCan we find eigenstates of a and a†?•Yes for a and no for a†

•Because a is not Hermitian, they can have complex eigenvalues z– Note that the state |z = 1 is different from |n = 1

•Let’s find these states:

•Act on both sides with m|:

•Normalize it

a z z z

0n

n

z c n

0 0

n nn n

z c n c a n

0

1nn

c n n

0 0

1n nn n

z c m n c n m n

1 1m mzc c m 11

mm

zcc

m

2 31 1 1 11 0 2 1 0 3 2 02 2 3 6

, , ,c zc c zc z c c zc z c 0

!

n

n

zc c

n

00 !

n

n

zz c n

n

2

2

0 !

nz

n

zz e n

n

Comments on Coherent StatesThey have a simple time evolution•Suppose at t = 0, the state is•Then at t it will be

When working with this state, avoid using the explicit form•Instead use:•And its Hermitian conjugate equation:

•Recall: these states are eigenstates of a non-Hermitian operator– Their eigenvalues are complex and they are not orthogonal

•These states roughly resemble classical behavior for large z– They can have large values of X and P – While having small uncertainties X and P

a z z z

00t z 2

2

0 !

nz

n

zz e n

n

2 10 22 0

0 !

ni n tz

n

zt e e n

n

2102 2

00

1

!

ni t z i t

n

e e z e nn

12

0i t i te z e

† *z a z z

0z z

Sample ProblemFind X for the coherent state |z

† *a z z z z a z z

22 2X X X

†

2X z a a z

m

*

2z z z z

m

*

2X z z

m

22 †

2X z a a z

m †2 † † 2

2z a a a aa a z

m † † 1aa a a

2 †2 † 22 12

X z a a a a zm

†

2X a a

m

2 *2 * 22 12

X z z z zm

2*2 * 2 *2 12 2

z z z z z zm m

2m

2X

m

All Problems are the Harmonic Oscillator 5C. Multiple Particles and Harmonic Oscillator

• Consider N particles with identical mass m in one dimension

• This could actually be one particle in N dimensions instead• These momenta & position operators have commutation relations:• Taylor expand about the minimum X0. Recall derivative vanishes at minimum

• A constant term in the Hamiltonian never matters • We can always change origin to X0 = 0. • Now define: • We now have:

21 2

1

1, , ,

2

N

i Ni

H P V X X Xm

,i j ijX P i

0

0 0 01 1

1

2

N N

i i j ji j i j

V V V X X X XX X

XX X X

0ij

i j

k VX X

X

2

1 1 1

1 1

2 2

N N N

i ij i ji i j

H P k X Xm

ij jik k

Solving if it’s Diagonal

• To simplify, assume kij has only diagonal elements:

• We define i2 = ki/m:

• Next define• Find the commutators:

• Write the Hamiltonian in terms of these:• Eigenstates and Eigenenergies:

2

1 1 1

1 1

2 2

N N N

i ij i ji i j

H P k X Xm

2 2

1 1

1 1

2 2

N N

i i ii i

H P k Xm

2 2 2

1 1

1 1

2 2

N N

i i ii i

H P m Xm

2 2

ii i

Pma X i

m

† † †, , 0 , ,i j i j i j ija a a a a a

† 12

1

N

i i ii

H a a

1 2, , , Nn n n 1 2

1, , 2

1N

N

n n n i ii

E n

1 , , 1

1

, ,N i

N

n n n n ii

x x x

• Note that the matrix made of kij’s is a real symmetricmatrix (Hermitian)

Classically, we would solve this problem by finding the normal modes• First find eigenvectors of K:

– Since K is real, these are real eigenvectors• Put them together into a real orthogonal matrix

– Same thing as unitary, but for real matrices• Then you can change coordinates:• Written in terms of the new

coordinates, the behavior is much simpler.• The matrix V diagaonalizes K

• Will this approach work quantum mechanically?

2

1 1 1

1 1

2 2

N N N

i ij i ji i j

H P k X Xm

11 12

21 22

k k

K k k

i i iK v k v

1 TV V

i j jij

X X V

1 2V v v

What if it’s Not Diagonal?

1

2

0

0

T

k

K V KV k

• Define new position and momentum operators as

• Because V is orthogonal, these relationsare easy to reverse

• The commutation relations for these are:

• We now convert this Hamiltonian to the new basis:

2

1 1 1

1 1

2 2

N N N

i ij i ji i j

H P k X Xm

1 TV V

,i j ji i j jij j

X X V P PV

Does this Work Quantum Mechanically?1

2

0

0

T

k

K V KV k

,i ij j i ij jj j

X V X P V P

, ,i k j ji l lkj l

X P X V PV ji lk jlj l

V V i Tij jk

j

i V V T

iki V V iki

2i ij j ik k

i i j k

P V P V P Tji ik j k

j k i

V V P P jk j kj k

P P 2j

j

P

ij i j ij ik k jl li j i j k l

k X X k V X V X Tki ij jl k l

k l i j

V k V X X kl k lk l

K X X

The procedure:•Find the eigenvectors |v and eigenvalueski of the K matrix•Use these to construct V matrix•Define new operators Xi’ and Pi’•The eigenstates and energies are then:

Comments:•To name states and find energies, all you need is eigenvalues ki

•Don’t forget to write K in a symmetric way!

2

1 1 1

1 1

2 2

N N N

i ij i ji i j

H P k X Xm

The Hamiltonian Rewritten:1

2

0

0

T

k

K V KV k

2 2i j

i j

P P

ij i j kl k li j k l

k X X K X X 2k k

k

k X 2 2

1 1

1 1

2 2

N N

i i ii i

H P k Xm

1 2, , , Nn n n

1 2

1, , 2

1N

N

n n n i ii

E n

i ik m

Sample ProblemName the eigenstates and find the

corresponding energies of the Hamiltonian

2 2 2 2 211 2 0 1 1 2 22

15 8 5

2H P P m X X X X

m

2

1

1 1

1

2

1

2

N

ii

N N

ij i ji j

H Pm

k X X

1 2, , , Nn n n

1 2

1, , 2

1N

N

n n n i ii

E n

i ik m

• Find the coefficients kij that make up the K matrix

• NO! Remember, kij must be symmetric! So k12 = k21

• Now find the eigenvalues:

2 2 211 0 22 0 12 05 5 8 ?k m k m k m 2

12 21 04k k m

20

5 4

4 5K m

5 4

0 det4 5

2 25 4

5 4 2 21 0 2 0, 9k m k m

1 0

2 0

,

3

The states and energies are: ,n m

1 10 02 23nmE n m

0 3 2nmE n m

It Isn’t Really That Complex 5C. The Complex Harmonic Oscillator

• A classical complex harmonic oscillator is a system with energy given by

Where z is a complex position• Just think of z as a combination of two real variables:• Substituting this in, we have:

• We already know everythingabout quantizing this:

• More usefully, write them interms of raising and loweringoperators:

• The Hamiltonian is now:

* 2 *E mz z m z z

12

z x iy

2 2 2 2 2 21 1 1 12 2 2 2E mx my m x m y

, , ,x yx P m y P m x X y Y

† †, ,2 2x x x xx i a a x a a

m m

† † 1x x y yH a a a a

Working with complex operators• Writing z in terms of a and a†

• Let’s define for this purpose

• Commutation relations:

• All other commutators vanish• In terms of these,

• And the Hamiltonian:

† †12

† †12

1,

22

1,

22

x y x y

x y x y

z x iy a ia a iam

z x iy i a ia a iam

† † † † † †1 12 2x y x y x y x ya a a a a ia a ia a ia a ia

12 x ya a ia

† † †12, ,x y x ya a a ia a ia † 2 †1 1

2 2, ,x x y ya a i a a 12 1 1 1

† †

* † * †

, ,2 2

, ,2 2

z a a z i a am m

z a a z i a am m

† 2 †x x y ya a i a a † †

x x y ya a a a

† † 1x x y yH a a a a † † 1H a a a a

The Bottom Line• If we have a classical equation for the energy:• Introduce raising/lowering operators with commutation relations

• The Hamiltonian in terms of these is:

• Eigenstates look like:

• For z and z* and theirderivatives, we substitute:

• This is exactly what we will need when we quantize EM fields later

† †, , 1 , all other vanisha a a a

† †

* † * †

, ,2 2

, ,2 2

z a a z i a am m

z a a z i a am m

† † 1H a a a a

* 2 *E mz z m z z

,n n , 1n nE n n