Michael Fowler, 10/14/07
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.
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
in a clockwise
direction around the circle of radius
centered at the
origin.
(Note: phase space
is usually defined in terms of the variables
, but in describing the simple harmonic oscillator, the pair
are more convenient.)
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).
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
are constants, and
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
. (
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
, with
, and writing the normalization constant
, must also be a minimum uncertainty state, with the same
. (It satisfies the
necessary differential equation.) Of
course, in contrast to the ground state, this displaced state is no longer an
eigenstate of the Hamiltonian, and will therefore change with time.
(Both these states,
and
, 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.)
Notation: We’ll write
![]()
We restrict our attention here to those minimum uncertainty states having the same spatial width as the oscillator ground state—these are what we need, and these are the ones we’ll show to be eigenstates of the annihilation operator. (Actually, more general minimum uncertainty states, known as squeezed states, are also interesting and important, but we’ll not consider them here.)
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 (and multiplying by –i) 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 parameter
labeling the position of the classical operator in phase
space in natural dimensionless units!
That is to say, a minimum uncertainty oscillator wave packet
![]()
centered at
in phase space and
having the same spatial extent as the ground state, is an eigenstate of the
annihilation operator
.
with eigenvalue the position of its center in phase space, that is,
![]()
Turning now to the time development of the state, it is convenient to use the ket notation
![]()
with
denoting a minimum
uncertainly wave packet (with the same spatial width as the ground state)
having those expectation values of position and momentum.
The time development of the ket, as usual, is given by
.
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! (And, of course, with constant
spatial extent.)
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 of ![]()

Therefore, the center of the 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.
We have chosen to work with the original position and momentum variables, and the complex parameter expressed as a function of those variables, throughout. We could have used the dimensionless variables introduced in the lecture on the simple harmonic oscillator,
,
This would of course
also give
a more compact
representation, but one more thing to remember.
It’s also common to denote the eigenstates of
by
very elegant, but
we’ve used z to keep reminding
ourselves that this eigenvalue, unlike most of those encountered in quantum
mechanics, is a complex number. Finally, some use the dimensionless variables
differing from
The eigenvalue
equation for the annihilation operator is very neat in this notation:
We’ve avoided it,
though, because our recommended textbook, Shankar, uses X, P for the ordinary
position and momentum operators.
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 (defined below).
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
.
Now do a
to be the variable!):

It’s clear from this that 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
![]()
(
being Hermitian) so
.
Therefore the displaced ground state wave function can be written

for real
, since
is zero for this initial
state (the wave function is real).
In the ket notation, we have established that the minimal uncertainty state centered at x0, and having zero expectation value for the momentum, is
![]()
But it’s not exactly obvious that this is an eigenstate of
with eigenvalue z0! (As it must be.)
It’s worth seeing how to prove that just from the properties of the operators—but to do that, we need a couple of theorems concerning exponentials of operators given in the Appendix.
First, if the commutator [A,B] commutes with A and B, then
This result
simplifies the right hand side of the above equation, for

where we have used ![]()
This is simpler, but it’s still not obvious that we have an
eigenstate of
: we need the commutator
.
The second
theorem we need is: if the commutator of two operators
itself
commutes with A and B, then
![]()
(This is easily proved by expanding the exponential—see the Appendix.)
Applying this to our case,
![]()
It follows
immediately that
is indeed an
eigenstate of
with eigenvalue
. (It must also be
correctly normalized because the translation
is a unitary operation
for real z0.)
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 the theorem proved in the Appendix) this will only affect the
overall normalization.
Furthermore, both
and
are combinations of
, so for the generalization of
from real
to complex z
to be unitary, it 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 this generalized translation operator is unitary, the new state is automatically correctly normalized.
The equation above suggests the possibility of representing
the displaced state
in the standard energy
basis
. We can simplify with
the same trick used for the spatial displacement case in the last section, that
is, the theorem
where now
:
![]()
using
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 eigenstate of
.
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.
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
, then since
we’re left with
![]()
from which
.
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, recall for the wave
function just displaced along the axis
). Therefore, the
span the whole space.
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
is proportional to