Coherent States of the Simple Harmonic Oscillator

Michael Fowler, 1/24/06

What is the Wave Function of a Swinging Pendulum?

Consider a macroscopic simple harmonic oscillator, and to keep things simple assume there are no interactions with the rest of the universe.  We know how to describe the motion using classical mechanics: for a given initial position and momentum, classical mechanics correctly predicts the future path, as confirmed by experiments with real (admittedly not perfect) systems.  But from the Hamiltonian we could also write down Schrödinger’s equation, and from that predict the future behavior of the system.  Since we already know the answer from classical mechanics and experiment, quantum mechanics must give us the same result in the limiting case of a large system.

 

It is a worthwhile exercise to see just how this happens.  Evidently, we cannot simply follow the classical method of specifying the initial position and momentum—the uncertainty principle won’t allow it.  What we can do, though, is to take an initial state in which the position and momentum are specified as precisely as possible. Such a state is called a minimum uncertainty state (the details can be found in my earlier lecture on the Generalized Uncertainty Principle).

 

In fact, the ground state of a simple harmonic oscillator is a minimum uncertainty state. This is not too surprising—it’s just a localized wave packet centered at the origin.  The system is as close to rest as possible, having only zero-point motion.  What is surprising is that there are excited states of the pendulum in which this ground state wave packet swings backwards and forwards indefinitely, a quantum realization of the classical system, and the wave packet is always one of minimum uncertainty.  Recall that this doesn’t happen for a free particle on a line—in that case, an initial minimal uncertainty wave packet spreads out because the different momentum components move at different speeds.  But for the oscillator, the potential somehow keeps the wave packet together, a minimum uncertainty wave packet at all times. These remarkable quasi-classical states are called coherent states, and were discovered by Schrodinger himself.  They are important in many quasi-classical contexts, including laser radiation.

 

Our task here is to construct and analyze these coherent states and to find how they relate to the usual energy eigenstates of the oscillator.

Classical Mechanics of the Simple Harmonic Oscillator

To define the notation, let us briefly recap the dynamics of the classical oscillator: the constant energy is

 

or

.

 

The classical motion is most simply described in phase space, a two-dimensional plot in the variables .  In this space, the point  corresponding to the position and momentum of the oscillator at an instant of time moves as time progresses at constant angular speed w in a clockwise direction around the circle of radius  centered at the origin. 

 

This motion is elegantly described by regarding the two-dimensional phase space as a complex plane, and defining the dimensionless complex variable

 

.

 

The time evolution in phase space is simply .

 

The particular choice of (quantum!) scaling factor in defining z amounts to defining the unit of energy as , the natural unit for the oscillator:  it is easy to check that if the classical energy  then the dimensionless  is simply the number  (which is of course very large, so the ½ is insignificant).

Minimum Uncertainty Wave Packets

We established in the lecture on the Generalized Uncertainty Principle that any minimum uncertainty one-dimensional wave function (so ) for a particle must satisfy the linear differential equation (here )

 

 

where , l are constants, and l is pure imaginary.  The equation is easy to solve: any minimum uncertainly one-dimensional wave function is a Gaussian wave packet, having expectation value of momentum , centered at  and having width .   (Dx is defined for a state  by .)

 

That is to say, the minimum uncertainly solution is:

 

 

with C the normalization constant.

 

In fact, the simple harmonic oscillator ground state  is just such a minimum uncertainty state, with

 

.

 

Furthermore, it is easy to see that the displaced ground state    (writing the normalization constant ) must also be a minimum uncertainty state, with the same .  Of course, in contrast to the ground state, this displaced state is no longer an eigenstate of the Hamiltonian, and any such initial state will change with time.

 

(Both these states have the same spread in x-space , and the same spread in p-space, the only difference in the p direction being a phase factor for the displaced state.)

 

 

What about the higher eigenstates of the oscillator Hamiltonian? They are not minimally uncertain states—for the nth state, , as is easily checked using .  So, if we construct a minimally uncertain higher energy state, it will not be an eigenstate of the Hamiltonian.

 

Exercise: prove  for the nth energy eigenstate. (Hint: use creation and annihilation operators.)

Time Development of a Coherent State: the Role of the Annihilation Operator

In this section, we shall establish a remarkable connection between minimally uncertain oscillator states and the annihilation operator, then use properties of that oscillator to find the time-development of the minimally uncertain states.

 

Suppose that at t = 0 the oscillator wave function is the minimum uncertainty state

 

 

centered at  in phase space (as defined above for the classical oscillator), and with to give it the same spatial extent as the ground state.

 

From the preceding section, this satisfies the minimum uncertainty equation

 

 

Rearranging this equation shows it in a different light:

 

.

 

This is an eigenvalue equation!  The wave packet  is an eigenstate of the operator  with eigenvalue .  It is not, of course, an eigenstates of either or taken individually.

 

Furthermore, the operator  is just a constant times the annihilation operator --recall

 

 

Therefore, this minimally uncertain initial wave packet  is an eigenstate of the annihilation operator , with eigenvalue .  By the way, it’s ok for  to have complex eigenvalues, because  is not a hermitian operator.

 

We can now make the connection with the complex plane representation of the classical operator: the eigenvalue  is precisely the dimensionless complex parameter!

 

This means that if we have a minimal uncertainty oscillator wave packet

 

having the same spatial extent as the ground state, centered at  in phase space, and we write

Then

                                                                             

.

 

That is to say, all eigenstates of the annihilation operatorare minimal uncertainty wave packets having the same spatial width as the oscillator ground state.

 

Turning now to the time development of the state, it is convenient to use the ket notation

 

so as usual

 

.

 

We shall show that  remains an eigenstate of the annihilation operator for all times t:  it therefore continues to be a minimum uncertainty wave packet!

 

The key point in establishing this is that the annihilation operator itself has a simple time development in the Heisenberg representation,

 

.

 

To prove this, consider the matrix elements of between any two eigenstates of the Hamiltonian

 

so

.

 

Since the only nonzero matrix elements of the annihilation operator  are for , and the energy eigenstates form a complete set, this simple time dependence is true as an operator equation

  

.

 

It is now easy to prove that  is always an eigenstate of :

 

 

Therefore the annihilation operator, which at t = 0 had the eigenvalue , corresponding to a minimal wave packet centered at  in phase space, evolves in time t to another minimal packet (because it’s still an eigenstate of the annihilation operator), and writing

 

,

 

the new eigenvalue

 

 

or more succinctly

.

 

Therefore, the center of the minimal wave packet in phase space follows the classical path in time.  This is made explicit by equating real and imaginary parts:

 

So we’ve found Schrödinger’s “best possible” quantum description of a classical oscillator.

The Translation Operator

It’s worth repeating the exercise for the simple case of the oscillator initially at rest a distance  from the center.  This gives a neat tie-in with the translation operator.

 

Let us then take the initial state to be  where  is the ground state wave function—so we’ve moved the packet to the right by.

 

From the Taylor series expansion (taking  to be the variable!)

 

Evidently the translation operator   shifts the wave function a distance to the right.

 

Since , the translation operator can also be written as, and from this it can be expressed in terms of , since

,   so ,

 

 being Hermitian, and       

.

 

Therefore the displaced ground state wave function can be written

 

for real , since  is zero for this initial state (the wave function is real).

 

How do we generalize this translation operator to an arbitrary state, with nonzero ? Thinking in terms of the complex parameter space z, we need to be able to move in both the x and the p directions, using both  and .  This is slightly tricky since these operators do not commute, but their commutator is just a number, so (using a theorem from the Appendix) this will only affect the overall normalization. 

 

Furthermore, both and  are combinations of , so the generalization of from real  to complex  z, to be unitary, must have an antihermitian combination of in the exponent—a unitary operator has the form , where H is Hermitian, so iH is antihermitian.. 

 

We are led to the conclusion that

 

,

 

conveniently labeling the coherent state using the complex parameter z of its center in phase space.  Since the generalized translation operator is unitary, the new state is correctly normalized.

 

This equation suggests the possibility of representing the displaced state  in the standard energy basis .  Expanding the exponential yields an infinite series of terms .  These could then be evaluated—in principle.  It turns out not to be such a good idea—since  don’t commute, the higher order terms become intractable.

 

Fortunately, the problem is easily solved using the operator identity , true for operators A and B provided  commutes with A and B.   (See the Appendix for a proof.) 

Using this identity with

 

 

where we have used  since .

 

It is now straightforward to expand the exponential:

 

 

and recalling that the normalized energy eigenstates are

 

we find

 

 

Exercise: Check that this state is correctly normalized, and is an eigenstates of .

Time Development of an Eigenstate of a Using the Energy Basis

Now that we have expressed the eigenstate  as a sum over the eigenstates  of the Hamiltonian, finding its time development in this representation is straightforward: since ,

 

 

which can be written

 

equivalent to the result  derived earlier.

Some Properties of the Set of Eigenstates of a

In quantum mechanics, any physical variable is represented by a Hermitian operator. The eigenvalues are real, the eigenstates are orthogonal (or can be chosen to be so for degenerate states) and the eigenstates for a complete set, spanning the space, so any vector in the space can be represented in a unique way as a sum over these states.

 

The operator  is not Hermitian.  Its eigenvalues are all the numbers in the complex plane.  The eigenstates belonging to different eigenvalues are never orthogonal, as is immediately obvious on considering the ground state and a displaced ground state.  The overlap does of course decrease rapidly for states far away in phase space.

 

The state overlap can be computed using :

 

 

And we can then switch the operators using the theorem from the Appendix , leaving  and

 

.

Finally, using,  we can construct a unit operator using the ,

 

Where the integral is over the whole complex plane  (this x is not, of course, the original position x).  Therefore, the  span the whole space.

Appendix: Some Exponential Operator Algebra

Suppose that the commutator of two operators A, B 

 

where c commutes with A and B, usually it’s just a number, for instance 1 or .

Then

 

That is to say, the commutator of A with elB is proportional to elB itself. 

 

That is reminiscent of the simple harmonic oscillator commutation relation  which led directly to the ladder of eigenvalues of H separated by .  Will there be a similar “ladder” of eigenstates of A in general?

 

Assuming A (which is a general operator) has an eigenstate  with eigenvalue a,

 

 

Applying to the eigenstate :

 

 

Therefore, unless it is identically zero, is also an eigenstate of A, with eigenvalue a + lc. We conclude that instead of a ladder of eigenstates, we can apparently generate a whole continuum of eigenstates, since l can be set arbitrarily! 

 

To find more operator identities, premultiply  by e-lB to find:

 

 

This identity is only true for operators A, B whose commutator c is a number. (Well, c could be an operator, provided it still commutes with both A and B).

 

Our next task is to establish the following very handy identity, which is also only true if [A,B] commutes with A and B:

 

The proof (due to Glauber, given in Messiah) is as follows:

 

Take

 

It is easy to check that the solution to this first-order differential equation equal to one at x = 0 is

 

 

so taking x = 1 gives the required identity,

 

It also follows that  provided—as always—that [A, B] commutes with A and B.