5. the harmonic oscillator
DESCRIPTION
5. The Harmonic Oscillator. All Problems are 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’ ( x 0 ) = 0 Constant term is irrelevant We can arbitrarily choose the minimum to be x 0 = 0 - PowerPoint PPT PresentationTRANSCRIPT
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