6210/Spring Physics 2007/Lecture 18
Lecture 18
Relevant sections in text: 2.3
Stationary states and classical mechanics
Here we use the oscillator to illustrate a very key point about the relation between
classical and quantum mechanics.
The stationary states we just studied do not provide any explicitly dynamical behavior.
This is not a specic feature of the oscillator, but a general feature of stationary states in
quantum mechanics. This is, at rst sight, a little weird compared to classical mechanics.
Think about it: the probability distributions for position and momentum are time independent in any state of denite energy. In classical mechanics the position and momentum
(and the energy) can, at each moment of time, be determined with certainty; the values
for the position and momentum oscillate in time in every state but the ground state.*
In light of the incompatibility of the position, momentum and energy observables in the
quantum description, one cannot directly compare the classical and quantum predictions
in the excited states. The quantum predictions are purely statistical, involving repeated
state preparations states specied only by their energy and measurements of various
observables. If we want to compare the quantum and classical descriptions we need to ask
the right questions of classical mechanics this means statistical questions. Let us pause
for a moment to expand on this.
In classical mechanics every solution of the equations of motion is a state of denite
energy (assuming energy is conserved). Fix an energy E . Let us ask the same question
that we ask in quantum theory: Given the energy what is the probability for nding
the particle at various locations? To answer this question we dene the probability for
nding the classical particle to be in the interval [x, x + dx] to be proportional to the time
spent in that region. (The proportionality constant is used to normalize the probability
distribution.) For a classical oscillator at the point x, a displacement dx takes place in the
time dt where (exercise)
dx
dt =
.
2E 2 x2
m
Here it is understood that x lies between the two classical turning points, where
1
E = m 2 x2 .
2
* In the ground state, the classical motion is of course trivial the position (relative to
the equilibrium position) and the momentum vanish. In the quantum ground state the
position and momentum have Gaussian probability distributions centered at the vanishing
values. For macroscopic values of m and the widths of these distributions are negligible.
1
Physics 6210/Spring 2007/Lecture 18
Up to a constant normalization factor, this denes the probability density P (x) for the
classical oscillator:
1
P (x) = (const.)
.
2E 2 x2
m
(The probability density vanishes outside the turning points.) For E = 0, the resulting
probability density is strongly peaked near the classical turning points of the motion and
relatively small and at near the equilibrium position. This just reects the fact that,
because the oscillator moves slowly near the turning points and rapidly at the equilibrium
position, it is more likely to nd the particle near the turning points. In general, a quantum
oscillator does not have anything like this classical probability distribution. First of all, it
oscillates in a non-trivial way because of the Hermite polynomials. Furthermore, while the
probability density is exponentially damped outside of the turning points, the probability
distribution is not identically zero there. However, it can be shown that the quantum
probability distribution does approximate the classical distribution in the limit of large
energy. Here, large means suciently macroscopic, i.e., En >> so that n >> 1.
h
2 for a large n. The result is a very rapidly oscillating
To see this one simply computes |un |
wavefunction; the oscillations occur about an average curve which approaches the classical
probability distribution as n gets larger and larger. For any nite interval of x, a large
enough n will result in a probability for nding the particle in that interval to agree with
the classical prediction to any desired accuracy. To see this quickly, simply ask your favorite
computer math software plot to the graphs!
We have seen that the probability distribution for position predicted by quantum
mechanics approaches that predicted by classical statistical mechanics in the limit of large
quantum numbers. This result is satisfying, but is far from the complete story of the
relation of the classical and quantum features of the oscillator. A much better way to model
the classical oscillator does not use stationary states but instead minimum uncertainty
coherent states (see the exercises at the end of chapter 2). Recall that the excited
states are not minimum uncertainty states in position and momentum. The coherent
states are minimum uncertainty states; for macroscopic mass and frequency the oscillator
has a small enough uncertainty in energy, position and momentum and a suitable time
dependence to model a classical oscillator. In particular, because the coherent states
arent stationary (except for the ground state), the position and momentum probability
distributions can mimic their classical counterparts sinusoidal oscillations. We will not
explore the coherent states here. We do note that the ground state of the oscillator is one
such state. For macroscopic values of m and the width of the ground state Gaussian
h
position probability distribution controlled by x0 = m is truly negligible compared to
macroscopic length scales, so you can see at least in this state that classical behavior is
recovered.
2
Physics 6210/Spring 2007/Lecture 18
Oscillator Dynamics
To explicitly see the oscillatory behavior of an oscillator in quantum mechanics one
needs to use non-stationary states, that is, superpositions of energy eigenstates, as the
initial state. Let us examine this feature using the Heisenberg picture. The principal job is
to get the basic observables at time t in terms of their representation at some xed initial
time, say, t0 = 0. We have
X (0) X, P (0) P.
We need to compute X (t) and P (t). We can do this directly using
i
i
i
X (t) = e h Ht X (0)e h Ht ,
i
P (t) = e h Ht P (0)e h Ht ,
and
P 2 (t) 1
P 2 (0) 1
+ m 2 X 2 (t) = H (0) =
+ m 2 X 2 (0).
2m
2
2m
2
To evaluate X (t) and P (t) you can either expand the exponentials in power series and, by
studying the general term, try to deduce a closed form expression for the result. There are
other tricks as well for manipulating the similarity transformation. However, a somewhat
easier way to get expressions for X (t) and P (t) is to directly solve the Heisenberg equations.
Using (exercise)
[X (t), P (t)] = i I,
h
H = H (t) =
we have
i
h
d
P (t)
X (t) = [X (t), H ] = i
h
,
dt
m
d
P (t) = [P (t), H ] = i m 2 X (t),
h
dt
So that the Heisenberg equations can be written as
i
h
P (t)
d
d
X (t) = [X (t), H ] =
,
P (t) = [P (t), H ] = m 2 X (t).
dt
m
dt
These are mathematically the same as the classical Hamilton equations of motion, and are
as easily solved. Taking account the initial conditions at t = 0, we have
X (t) = (cos t)X (0) + (
1
sin t)P (0),
m
P (t) = (cos t)P (0) (m sin t)X (0).
In the Heisenberg picture, the time dependence in physical predictions (i.e., probability
distributions) is obtained using these observables and a xed state vector. As an example,
let us compute the statistical mean of the position and momentum at time t. Taking
diagonal matrix elements of the Heisenberg operators, we get
X (t) = (cos t) X (0) + (
3
1
sin t) P (0),
m
Physics 6210/Spring 2007/Lecture 18
P (t) = (cos t) P (0) (m sin t) X (0).
From the above equations you can see that we get the usual sort of behavior expected of a
harmonic oscillator with mass m and frequency . Indeed, we see that the expectation values behave exactly as do the classical position and momentum, in accord with Ehrenfests
theorem.
Of course, if the system is in a stationary state |n then expectation values do not
change in time. This seems paradoxical in light of our result shown above. The resolution
of this paradox in the above example is that, for stationary states, the initial expectation
values vanish, e.g.,
X (t) = X (0) = 0,
P (t) = P (0) = 0.
As an exercise see if you can see how the expectation values of, say, X 2 manages to be
time independent in a stationary state.
4