Re: The Schrödinger Equation, potential box problem

Greg Hansen wrote:
Carl wrote:

Hello,

While trying to solve a physics problem relating to the Schrödinger
equation, we found something that seems unlogical. The problem given is
a real "standard" problem about separable time- and space-dependent
parts of the solution to the Schrödinger Equation.

The following is given:
An electron in the potential box
V(X)={inf for x<0, 0 for 0<=x<=L, inf for x>L}
is at t=0 described by the wave function
psi(x, 0) = 0.8psi_1(x)+0.6psi_2(x)
where the orthonormal eigenfunctions
psi_n(x) = sqrt(2/L)sin(n*pi*x/L)
and the corresponding eigenvalues
E_n = (n*pi*h~/L)^2/(2m)=n^2*E_1, n=1, 2, ...

It is asked for:
a) Make sure psi is normalized.
b) Sketch the probability of finding the electron as a function of x at
t=0, that is, P(x,0).
c) Sketch the corresponding graph at time t_1=pi*h~/(3E_1)

So, to our problem: a) is straight-forward. b) gives a sketch of an
unsymmetric function, heavy on the left side, with a peak at around
x=0.3L.

Our problems start when trying to solve c). The reason is that our
calculations and reasonings give that P(x, t_1) would be the same as
P(x, 0). Because a number of the following questions regard the
differences of P(x, 0) and P(x, t_1) and the corresponding expectation
values and so on, we feel that we must be wrong.

Well, stop for a moment and think about what it *should* be. Quantum mechanics has waves. What would you expect an asymmetric wave in a box to do? It's heavy on the left side. Then it will slosh over to the right side, and back to the left side, and at a particular time it will be identical to what it was as t=0. It will slosh back and forth.

Sorry, I don't really get your point here. If the probability function
is heavy on the left side, then the probability of finding the particle
is bigger in the left half of the box than in the right half. How could
that be? What could be the cause of that? I argue that since there is
no potential (for example an electric field) I can't see why it could
be more probable that the particle is found on the left side than on
the right.

However, for any potential function V(x) that is not time-dependent
(V(x)=0 for all x within [0,L] clearly is time-independent), psi shall
be a time-space-dependance separable function. In those cases, denoting
the time-dependent function as phi(t), phi(t) can be written as
phi(t)=exp(-iEt/h~).
That would mean that for P(x,t),
P(x,t)=psi(x,t)*complex-conjugate(psi(x,t))=psi(x)^2 since the phi's
will multiply to unity for all t. That would mean that P(x,t) is the
same for all t.

What are we doing wrong? We just cannot believe that P(x, t) is
independent of t. We are also annoyed by the fact that P(x,t) is
unsymmetric over x (we would expect symmetry around L/2, especially
since there is no potential). If there would be a time-dependance, and
a phase velocity for P(x, t), we could still achieve symmetry if
calculating the average over time.

Your wavefunction is not an eigenfunction of energy. The time evolution operator is U=exp(-iHt/hbar), so

psi(t) = U psi(0)

= U psi_1 + U psi_2 {and psi1, psi2 are energy eigenfunctions}

= exp(-i w1 t) psi_1 + exp(-i w2 t) psi_2

Find the complex conjugate of that, and the time dependence won't go away.

At first: sorry that I lack some basic understanding of these topics.

I still don't rellay get it, but seeing w1 and w2 (that is, separate
angular velocities) in what is called the U operator by you gives me a
clue.

I multiply
psi(x)=0.8psi_1(x) + 0.6psi_2(x) with exp(-iEt/h~) to get the time
dependent function, that is,
psi(x,t)=( 0.8psi_1(x) + 0.6psi_2(x) )*exp(-iEt/h~)
Is this my error? Do I have to multiply psi_1 and psi_2 separately,
substituting E in exp(-iEt/h~) to E_1 and E_2? If that is the case, it
is clear that P(x,t) will be dependet of both x and t.

.

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