Transcript Document
Harmonic oscillator and coherent states
1.
2.
3.
4.
Energy eigen states by algebra method
Wavefunction
Coherent state
The most classical quantum system
Reading materials:
1. Chapter 7 of Shankar’s PQM.
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Algebra method for eigen states
The Hamiltonian of a harmonic oscillator is
1 2 1 2
H
P kX
2m
2
Y mk 2 1 4 X
Defining dimensionless coordinate
, so
1
4
Q 1 2 mk P
2
H
Y Q 2 , where k m
2
position and momentum are now on equal foot.
y q
Classical dynamics:
q y
q
y
Cyclic trajectory in the phase space.
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Quantum mechanics: X , P i or Y , Q i
Let us “rotate” the coordinates by an imaginary angle so that the
cyclic rotation in the phase space is automatically taken into account
by the transformation
Y iQ
Y iQ
†
a
and a
2
2
a, a 1
†
Note: Do you still remember how
to write a circularly polarized light
(whose electric field rotates) in
terms of linear polarization?
a and a†can be viewed as "position" and "momentum" in the
coordinates of the phase space rotated by an imaginary angle.
The Hamiltonian becomes
1
† 1
†
†
H a a aa a a
2
2
How to get the eigen states and the eigen values?
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A few general properties to be used:
1. The energy spectrum must be lower bounded, for
H
2
Y 2 Q2 0
So there must be a ground state H 0 E0 0
2. There should be no continuum state, i.e., the eigenstate
wavefunction should be normalized to one.
1
Otherwise the energy of the state cannot be finite due to
the infinite potential diverging at the remote positions.
3. Theorem: In one-dimension space, a discrete state cannot be
degenerate (see Shankar, PQM page 176 for a proof).
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If there is an eigenstate H E E E
H , a a
Ha E aH
a E E a E
i.e, a E is also an eigen state with energy E
H , a† a †
Ha† E a† H a† E E a † E
i.e, a† E is also an eigen state with energy E
Repeating the process, we get a series of eigen states
, aa E , aa E , a E , E , a† E , a†a† E ,
Their energies form an equally space ladder
, E 2 , E , E, E , E 2 ,
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The energy ladder has to be lower bounded, so we must have
a 0 0
So the ground state energy is given by
1
1
1
H 0 a † a 0 0 i.e., E0
2
2
2
All the other eigen states can be obtained by
n Cn a
† n
0 , where Cn is a normalization factor
H n En n
En E0 n n 1 2
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Using a, a† aa† a† a 1 and a 0 0 :
aa† 0 1 a† a 0
aaa†a† 0 2! 0
aaaa†a†a† 0 3! 0
n n Cn
2
0 a a
n
† n
0 1 Cn 1
n!
1
† n
n
a 0
n!
a n1 n n 1 n and a† n1 n n n 1
a n n n 1
matrix forms in the eigen state basis:
a n n 1 n 1
†
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0
0
a 0
0
1
0
0
0
0
2
0
0
0
0
3
0
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The dimensionless position and momentum operators are
a a†
a a†
Y
and Q
2
i 2
a
Y iQ
Y iQ
and a†
2
2
matrix forms in the eigen state basis:
1
Y
2
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0
1
1
0
0
0
2
0
0
2
0
3
0
0
3
0
0 i 1
0
0
i 2
i 1
1
Q
0
i 2
0
2
0
i 3
0
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0
0
i 3
0
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Wavefunctions
Let us first consider the ground state:
Y iQ
i.e.,
0 0
2
a 0 0
In the dimensional coordinates:
y
Y iQ
1
d
0
y
0 y 0
dy
2
2
d
0 y y 0 y
dy
1
2
y
exp
y
2
The solution is 0
14
A nice property of Gaussian function is that its Fourier transformation is
also a Gaussian function. The wavefunction in the q-representation
would have the same form (remember Y and Q are inter-exchangeable.
So the ground state wavefunction in the real space is:
14
mk
0 x 2 exp mk
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2
2
x 2
A wavepacket centered at
the potential minimum.
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Now consider the excited states:
n
1
a
n!
† n
Y iQ
a
2
†
0
Thus the wavefunction in real space is
n y
1
2n
n
d
1
2
y
exp
y
2
H n y exp y 2 2
dy
n!
2n n !
where H n y is the nth order Hermite polynomial.
The above equation actually defines
the generation of Hermite polynomials.
The wavefunction in the momentum-representation is
n q
1
2n
d
i H q exp q 2 2
2
i
iq
exp
q
2
n
n
n ! dq
2 n!
n
n
defining a nice property of the F.T. of Hermite polynomials.
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Regarded as boson
Now that all the eigen states of a harmonic oscillator are equally spaced., we can
take the rising from one state to the next one as the addition of one particle with
the same energy to a mode. The ground state contains no particle and hence is the
vacuum state. Simple one mode can have an arbitrary number of particles, this
particle is a boson.
0
Vacuum state
n
The state with n bosons, called a Fock state
a
The operator annihilating one boson
a†
The operator creating one boson
n a†a
2
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The boson particle number operator
The energy of the boson
The energy of the vacuum
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Coherent state
A coherent state of a harmonic oscillator is defined as
C exp Ca† C a 0 where C is a complex number.
In terms of the position and momentum operators, it is
C e
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i 2 CiY Cr Q
0 where Cr C and Ci C.
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Coherent state in Fock state basis
C exp Ca C a 0 C n
†
*
n 0
n
To derive the wave function of the coherent state in the Fock
state basis, we use the Baker-Hausdorff theorem
e A B e
1
A, B A B
2
e e if A, A, B A, B , B 0
Ca† , C *a CC is a c-number,
so the condition for the theorem is satisfied.
C e
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CC 2 Ca† C a
e e
0
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Coherent state in
Fock
state
basis
2
*
e
C *a
C
1 C a
*
2!
but a 0 0 so e
C *a
a
2
0 0
C exp CC 2 exp Ca† exp C a 0
exp CC 2 exp Ca† 0
eCa
†
2
C
0 1 Ca†
a†2
2!
The expansion is C eCC
*
2
n 0
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Cn
n
0
n!
n 0
Cn
n
n!
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The state is normalized (of course) as can be checked directly:
C C eCC
*
n *n
*
C *m C n
CC
CC
m n e
1
n!
m! n!
n ,m 0
n 0
Coherent state is an eigen state of the annihilation operator: Removing
one boson does NOT change a coherent state!
a C eCC
*
2
n 0
n
*
C
Cn
n n 1 C C
a n eCC 2
n!
n!
n 0
Like x being an eigen state of the position operator X , a coherent
state can be viewed as an eigen state of the "position" operator a in
the phase-space coordinates rotated by an imaginery angle.
However, adding one boson changes the state
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a C C
†
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The expectation value of the boson number is
n C aaC C
†
2
Boson number distribution (Poisson distribution)
p n n C C n
C
2n
n!
e
C
2
n
n
n!
e
n
p n
n
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To understand the nature of the coherent state, let us consider first a
few special case:
1. C ei
2CiY
0 , i.e., C is pure imaginery
The real space wave function of this state is
C y ei
0 y
2Ci y
Or in the momentum representation
eiqy
C q C y
dy 0 q 2Ci
2
The state is the ground state with the
momentum distribution shifted by
q
2Ci
y
2Ci
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2. C ei
2Cr Q
0 , i.e., C is real
The wave function in the momentum representation is
C q ei
0 q
2Ci q
The real space wave function of this state is
eiqy
C y C q
dq 0 y 2Cr
2
The state is the ground state with the
real-space distribution shifted by
2Cr
q
y
2Cr
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3. For an arbitary complex number C
C e
i 2 CiY Cr Q
0
Baker-Hausdorff theorem
e A B e
e
1
A, B A B
2
e e if A, A, B A, B , B 0
i 2 CiY Cr Q
e
i C
2
e
i 2CiY i 2Cr Q
e
The state is shifted from the ground
state in the phase space y, q by
q
2C
2 Cr , Ci
y
For more, do the homework.
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Quantum fluctuation
According to Heisenberg principle
Y , Q i
Y Q 1 2
a a†
a a†
Y
and Q
2
i 2
Fock states
Q n Q n 0 n Y n n Y n
†
†
† †
aa
a
a
aa
a
a
1
2
2
Y nY n n
n n
2
2
†
†
† †
aa a a aa a a
1
2
2
Q nQ n n
n n
2
2
Y Q n 1 2 1 2
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In particular, for the vacuum state
Q 0 Q 0 0 0 Y 0 0 Y 0
†
†
† †
aa
a
a
aa
a
a
1
2
2
Y 0Y 0 0
0
2
2
†
†
† †
aa
a
a
aa
a
a
1
Q2 0 Q2 0 0
0
2
2
Y Q 1 2
The vacuum state has minimum quantum fluctuation
(the most classical state)
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Quantum fluctuation: Coherent state
For a coherent state
a a†
Q CQC C
C 2C
i 2
a a†
Y CY C C
C 2C
2
The center of the wavepacket in the phase space is at
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2C
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The variance can be obtained by calculating
Q
Y
aa† a† a aa a †a †
1
C
C
2
2
aa† a† a aa a† a†
1
C
C
2
2
2
2
2C
2C
2
2
Y Q 1 2
A coherent state has minimum quantum fluctuation.
This justify our viewing a coherent state as a wavepacket centered
at a point in the phase space and a most classical state.
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Coherent states as a basis
The coherent states for all complex numbers form a complete basis
Completeness condition:
1
C C d 2C 1
1
C
e
C
2
C d C
2
C C d C e
m
*n
2
1
2
m
ne
C
2
n ,m0
n m i m n
2 n,m e
e
2
C m C *n 2
d C
n !m !
d d
2n d
n ! n,m
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The Hilbert space has been expanded by a discrete set of states (the
Fock states). But the complex numbers form a continuum. So the
coherent states must be over complete, i.e., more than enough, since
they are not orthogonal:
C C
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2
exp C C
2
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The most classical quantum system
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We have seen that a coherent state can be viewed as a shift in the
phase space from the vacuum state.
The vacuum state, of course, is also a coherent state.
What if we do the shift from a coherent state other than the vacuum?
C , C exp C a† C *a C ?
U exp C a† C a exp Ca † C a ?
Repeatedly using Baker-Hausdorff theorem:
C
2
U e
2
C
e
e
2 C
2
2 C
2
2 C a† C a Ca† C a
e e e e
2 CC C a† Ca† C a C a
e
e e e e
CC 2 C C 2 C C
e
2
2
e
C C a†
e
C C a
C , C exp C C * 2 C C * 2 C C
It is still a coherent state, up to a trivial phase factor.
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exp C a† C *a C exp C C * 2 C C * 2 C C
A shift in the phase space from a coherent state is still a coherent state,
the total shift from the vacuum is just the sum of the two shifts.
Thus we can define a shift operator
DC exp Ca† C *a
C DC 0
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Time evolution of a coherent state
If we have an initial coherent state:
C eCC
*
2
n 0
Cn
n
n!
The time evolution is simply
C t eCC
*
2
n 0
C n int it 2
e
n
n!
C t eit 2 Ceit
q
2C t
It is a coherent state with its shift from
the vacuum rotating in the phase space
like a classical oscillator.
y
2C
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Harmonic oscillator driven by a force
Let us consider the motion of a harmonic oscillator starting from
the ground state
1 2 1 2
H
p kx exE t
2m
2
In the form of boson operators:
H a†a 1 2 a† a t
Suppose the state at a certain time is
The Schroedinger equation is
t
i t t H t t
Consider an infinitesimal time increase
t dt t iHdt t exp iHdt t
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exp iHdt 1 iHdt 1 i a† a 1 2 dt i a † a t dt
1 i a†a 1 2 dt 1 i a† a t dt
exp i a† a 1 2 dt exp i a† a t dt
The first term is a free evolution of the state.
U 0 t exp i a†a 1 2 dt
The second term is the shift operator which shift a state in the
phase space along the position axis.
U1 t exp i a† a t dt
The evolution from the initial state in a finite time is
t U 0 t U1 t U 0 t dt U1 t dt U 0 dt U1 dt U 0 0 U1 0 0
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So, if the initial is a coherent state, say, the vacuum state, the state
after a finite time of evolution is still a coherent state
C 0 C t
And we have the shift for an infinitesimal time increase to be
dC t C t dt C t e idt C t i dt C t
idtC t i dt
d
The equation of motion is:
C t iC t i
dt
i.e., Cr t Ci t i
Ci t Ci t r
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The same as the classical eqns.
for position and momentum!
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Conclusion: A harmonic oscillator driven by a classical force from
the ground state is always in a coherent state.
We have seen that the coherent state follows basically the
equations for the classical eqns for position and momentum. It
could be taken as a reproduction of the classical dynamics from
quantum mechanics. The coherent state could be understood as
classical particle, though it is quite a wavepacket (it is just so
small that we had not enough resolution to tell it from a particle).
So, if we have only classical forces and harmonic oscillators, there
is no way to obtain a “real” quantum state from the vacuum or the
ground state. That is why we call harmonic oscillators the most
classical quantum systems.
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