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.
2016/3/16
Chang-Kui Duan, Institute of Modern Physics, CUPT
1
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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2
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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3
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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5
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   n1 n n  1 n and a†   n1 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
2
0
0
0
0
0
0
3
0








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7
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


Chang-Kui Duan, Institute of Modern Physics, CUPT
0
0
i 3
0









8
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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9
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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10
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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11
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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12
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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13
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  eCC
*

2

n 0
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

Cn
n
 0 
n!
n 0

Cn
n
n!
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14
The state is normalized (of course) as can be checked directly:
C C  eCC
*

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  eCC
*

2

n 0
n

*
C
Cn
n n 1  C C
a n  eCC 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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15
The expectation value of the boson number is
n  C aaC C
†
2
Boson number distribution (Poisson distribution)
2n
n
2
C
n n
C
p n  n C C n 
e

e
n!
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
eiqy
 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  ei
2Cr Q
0 , i.e., C is real
The wave function in the momentum representation is
 C  q   ei
 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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19
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  2C
i 2
a  a†
Y  CY C  C
C  2C
2
The center of the wavepacket in the phase space is at
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2C
22
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
2C

2C

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 , m 0

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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27
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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28
Time evolution of a coherent state
If we have an initial coherent state:
C  eCC
*

2

n 0
Cn
n
n!
The time evolution is simply
C  t   eCC
*

2

n 0
C n int it 2
e
n
n!
C  t   eit 2 Ceit
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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29
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 idt C  t   i  dt   C t 
 idtC  t   i  dt
d
The equation of motion is:
C  t   iC  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!
Chang-Kui Duan, Institute of Modern Physics, CUPT
32
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.
2016/3/16
Chang-Kui Duan, Institute of Modern Physics, CUPT
33