Lecture 22: Coherent States Phy851 Fall 2009

Lecture 22:
Coherent States
Phy851 Fall 2009
Summary
memorize
• Properties of the QM SHO:
P2 1
H=
+ mω 2 X 2
2m 2
A=
h
mω
λ=
1 X
λ 
1 X
λ 
†
A
=
−
i
P
+
i
P



h 
h 
2λ
2λ
λ
(A + A† ) P = −i h A − A†
X=
2
2λ
€
(
)
1

H = hω  A† A + 
2

H n = hω (n + 1 / 2) n
n =
A n = n n − 1 A† n = n + 1 n + 1
ψ n (x) =
[
π 2 n n! λ
]
−1/ 2
H n ( x / λ) e
−
( A† )
n!
x2
2 λ2
€ n −1
2 x
ψ n (x) =
ψ n−1 (x) −
ψ n−2 (x)
n λ
n
€
ψ 0 (x) =
[
πλ
]
−1/ 2 −
e
x2
2 λ2
€
ΔX = λ n + 1 / 2 ΔP =
€
€
[
ψ1 (x) = 2 π λ
]
−1/ 2
h
n +1/ 2
λ
x
2 e
λ
n
−
x2
2 λ2
0
What are the `most classical’ states
of the SHO?
• In HW6.4, we saw that for a minimum
uncertainty wavepacket with:
λosc
Δx =
2
λosc =
h
Mωosc
The uncertainties in position and
momentum would remain constant.
• The interesting thing was that this was
true independent of x0 and p0, the initial
expectation values of X and P.
• We know that other than the case x0=0
and p0=0, the mean position and
momentum oscillate like a classical
particle
• This means that for just the right initial
width, the wave-packet moves around
like a classical particle, but DOESN’T
SPREAD at all.
‘Coherent States’
• Coherent states, or as they are sometimes
called ‘Glauber Coherent States’ are the
eigenstates of the annihilation operator
Aα =α α
α α =1
– Here α can be any complex number
– i.e. there is a different coherent state for every
possible choice of α
– (Roy Glauber, Nobel Prize for Quantum Optics
Theory 2005)
• These states are not really any more
‘coherent’ then other pure states,
– they do maintain their coherence in the
presence of dissipation somewhat more
efficiently
• In QM the term ‘coherence’ is over-used and
often abused, so do not think that it always
has a precise meaning
• Glauber Coherent States are very important:
– They are the ‘most classical’ states of the
harmonic oscillator
– They describe the quantum state of a laser
• Replace the number of ‘quanta’ with the number
of ‘photons’ in the laser mode
– They describe superfluids and super-conductors
Series Solution
• Let us expand the coherent state onto energy
eigenstates (i.e. number states)
∞
α = ∑ cn n
n =0
• Plug into eigenvalue equation:
Aα =α α
∞
∞
n =0
n =0
A∑ cn n = α ∑ cn n
∞
∑c
n =0
∞
n
n n − 1 = α ∑ cn n
n =0
• Hit from left with 〈m|:
∞
m
→
∑c
n =0
∞
∑c
n =0
∞
n
n n − 1 = α ∑ cn n
n =0
∞
n
n m n − 1 = α ∑ cn m n
n =0
cm +1 m + 1 = α cm
Continued
α
cm +1 =
cm
m +1
• Start from:
α
cm =
cm −1
m
c 0 = ! (α )
– The constant N(α) will be used at the end for
normalization
€few iterations:
• Try a
α
α
c1 =
c0 =
! (α )
1
1
α
α2
c2 =
c1 =
! (α )
2
2 ⋅1
3
α
α
€
c3 =
c2 =
! (α )
3
3⋅ 2 ⋅1
α
α4
€
c4 =
c2 =
! (α )
4
4 ⋅ 3⋅ 2 ⋅1
• So clearly by induction we have:
€
€
αn
cn =
! (α )
n!
Normalization Constant
αn
cn =
! (α )
n!
• So we have:
∞
αn
α = ! (α )∑
n
n= 0 n!
€
• For normalization we require:
1= α α
€
∗m
∞
α αn
= ! (α ) ∑
mn
m!n!
n= 0
2
m= 0
∞
2n
α
2
= ! (α ) ∑
n!
n= 0
€
2
= ! (α ) e
α
2
• Which gives us:
€
! (€α ) = e
−
α
2
2
α =e
α
−
2
2
αn
n
∑
n!
n =0
∞
Orthogonality
• Let us compute the inner-product of two
coherent states:
2
α β =e
α +β
−
2
2
∗m
α βn
mn
∑
m!n!
n =0
∞
m =0
2
=e
−
α +β
2
∑
2
(α β )
∗
∞
n!
n= 0
2
=e
€
• Note that:
€
e
−
α +β
− α−β
2
2
n
2
+α ∗ β
=e
(
)
− α ∗ − β ∗ (α − β )
(
=e
2
− α +β
= αβ
2
+a ∗ β + β ∗α
)
2
• So coherent states are NOT orthogonal
– Does this contradict our earlier results
regarding the orthogonality of eigenstates?
€
Expectation Values of Position
Operator
• Lets look at the shape of the coherent
state wavepacket
– Let
ψ α ( x) = x α
X = ∫ dxψ α∗ ( x) xψ α ( x)
– Better to avoid these integrals, instead
lets try using A and A† :
X = α
λ
A + A† ) α
(
2
– Recall the definition of |α〉:
€
Aα =α α
α A† = α ∗ α
λ
X =
α A α + α A† α )
(
2
λ
∗
=
α
α
α
+
α
αα )
(
2
λ
=
α + α∗)
(
2
€
€
€
X = 2 λ Re{α }
Expectation Value of Momentum Operator
• We can follow the same procedure for the
momentum:
h
α ( A − A† ) α
2λ
2h
†
=
α
A
α
−
α
A
α
(
2iλ
P = −i
€
€
€
€
2h
∗
=
α
−
α
(
)
2iλ
2h
P =
Im{α}
λ
X = 2 λ Re{α }
• Not surprisingly, this gives:
€
α=
1 1
λ

X
+
i
P


h
2 λ

)
Variance in Position
• Now let us compute the spread in x:
X2
λ2
† 2
= α
A+ A ) α
(
2
λ2 2
†
†
† †
= α
A
+
AA
+
A
A
+
A
A )α
(
2
• €Put all of the A ’s on the right and the A† ‘s on
the left:
– This is called ‘Normal Ordering’
€
λ2 2
†
† †
= α
A
+
2A
A
+
1+
A
A )α
(
2
λ2 2
∗
∗2
=
α + 2α α + 1+ α
2
(
λ2
∗ 2
=
α + α ) +1
(
2
€
(
λ€
(α + α ∗ )
X =
2
€
ΔX =
X
2
− X
)
2
λ
=
2
)
X2 = X
2
λ2
+
2
Exactly the same variance as
the ground state |n=0〉
Momentum Variance
• Similarly, we have:
P
2
h2
† 2
= − 2 α (A − A ) α
2λ
h2
= − 2 α ( AA − AA† − A† A + A† A† ) α
2λ
€
– Normal ordering gives:
P
€
2
h2
= − 2 α ( AA − 2A† A −1+ A† A† ) α
2λ
h2
= − 2 α ( AA − 2A† A −1+ A† A† ) α
2λ
€
2
h2
= − 2 α 2 − 2α ∗α + α ∗ −1
2λ
(
€
)
h2
∗ 2
= − 2 (α − α ) −1
2λ
2h
∗
P =
α
−
α
(
)
2iλ
(
€
)
€
P
2
h2
= P + 2
€ 2λ
2
ΔP =
2
P − P
2
h
=
2λ
Minimum Uncertainty States
• Let us check what Heisenberg Uncertainty
Relation says about coherent states:
ΔX =
ΔP =
X
2
− X
P2 − P
2
2
=
λ
=
2
h
2λ
λ h
ΔXΔP =
2 2λ
ΔXΔP =
h
2
• So we see that all coherent states (meaning
no matter what complex value α takes on)
are Minimum Uncertainty States
– This is one of the reasons we say they are
‘most classical’
Time Evolution
• We can easily determine the time evolution of
the coherent states, since we have already
expanded onto the Energy Eigenstates:
– Let
ψ (t = 0) = α 0
– Thus we have:
ψ (0) = e
ψ (t ) = e
=e
– Let
=e
α
−
2
2
α
−
2
2
α 0n
n
∑
n!
n =0
∞
α 0n −iω ( n +1/ 2 )t
e
n
∑
n!
n =0
∞
α
−
− i ωt / 2
2
e
−iωt / 2
e
−
α
2
2
2
α 0n −iω n t
e
n
∑
n!
n =0
∞
∞
∑
−iω t n
(α e )
n= 0
0
n!
n
α (t ) = α 0 e − iω t
€
ψ (t ) = α (t )
By this we mean it remains in a
coherent state, but the value of
the parameter α changes in
time
Why ‘most classical’?
• What we have learned:
– Coherent states remain coherent states as time
evolves, but the parameter α changes in time
as
α (t ) = α 0 e − iω t
– This means they remain a minimum
uncertainty state at all time
– The momentum and position variances are the
same as the n=0 Energy eigenstate
– Recall that:
X = 2 λ Re{α }
2h
P =
Im{α}
λ
€
– So we can see that:
1 x
λ 
α 0 = €  0 + i p0 
h 
2λ
x0 = α (t ) X α (t )
p0 = α (t ) P α (t )
– We already know that <X> and <P> behave as
classical particle in the Harmonic Oscillator, for
any initial state.
x(t ) = x0 cos(ωt ) +
p0
sin(ωt )
ω
p (t ) = p0 cos(ωt ) − ωx0 sin(ωt )
Conclusions
• The Coherent State wavefunction looks
exactly like ground state, but shifted in
momentum and position. It then moves as a
classical particle, while keeping its shape
fixed.
– Note: the coherent state is also called a
‘Displaced Ground State’