Particle in an infinite square well

[KaiserBrandon]

Homework Statement

Consider a point particle of mass m contained between two impenetrable walls at +/- 2a. The potential V(x) between the walls is zero. Assume that at time t=0 the state of the particle is described by the wave function
$$\Psi(x) = A\frac{1+cos(\frac{2*\pi*x}{a})}{2} for |x|\leq a/2$$
=0 otherwise
a)normalize the wave function
b) Using configuration space, calculate the expectation value of the hamiltonian.
c) If a precise measurement of the energy of that particle were made, what would be the possible outcomes?
d)calculate the probability of finding the particle in the ground state of this well.

Homework Equations

the postulates of quantum mechanics and the mathematics of quantum mechanics.

The Attempt at a Solution

a)Normalizing the wave-function, I got
$$A=\sqrt{\frac{8}{3a}}$$
b) I calculated the expectation value to be
$$<H> = \frac{2\hbar^{2}\pi^{2}}{3ma^{2}}$$

c) Now this is the part I'm confused about. According to postulate 4, any precise measurement of the energy has a possible result of one of the eigenvalues of the operator H, correct? so then if we have,
$$H\Psi=E\Psi$$
where
$$H=\frac{-\hbar^{2}}{2m}\frac{d^{2}}{dx^{2}}$$
since V=0 inside the well, and E is the eigenvalue.
So then the eigenvalue I get is
$$E=\frac{\hbar^{2}*(2 \pi)^{2}*cos(\frac{2\pi x}{a})}{4ma^{2}*(1+cos(\frac{2 \pi x}{a})}$$
but this would mean that E approaches infinity as x approaches +/- a/2.

 

[elkement]

The energy eigenvalues should not depend on x.

Note that the wave function is given at t=0. Start from a time-dependent wave function by multipliying with an exponential function depending on Omega t. Plugging into the time-dependent Schrödinger equation should give you the dispersion relation (energy as function of wave vector).

I would follow the steps described here for the infinite square well potential:
http://en.wikipedia.org/wiki/Particle_in_a_box

 

[KaiserBrandon]

ok, so doing that I just get
$$E_{n}=\hbar \omega_{n}$$
For this question though, it seems like the prof keeps specifying that we are just considering the initial wave function at time t=0. For the next part, we're supposed to consider if the well is doubled in width, and find the energies under these circumstances. When I asked him, he said that the calculations will be different for the wider well, even if you are analyzing the exact same wavefunction at time t=0. This just doesn't seem to make any sense to me whatsoever.

 

[ideasrule]

I don't know how you got your answers, but the standard way of solving this type of problem is:

1) Use the time-independent Schrodinger's equation to find the stationary states. Normalize them (this should be easy for the infinite square well).

2) Apply boundary conditions. Here, you know that the wavefunction has to be 0 at both ends of the well. These boundary conditions give you the possible energies corresponding to the stationary states.

3) Represent the initial wavefunction as a sum of the normalized stationary states. Suppose the coefficient of the nth stationary state is c_n. When you make a measurement on the particle, there is a probability of p=|c_n|^2 that the wavefunction would collapse into the nth stationary state. Hence, there is a p=|c_n|^2 chance that you'll measure E_n as your energy.

For your question about doubling the well: the answers will indeed be different. For the expanded well, you know that the wavefunction is 0 from x=a to x=2a. For the initial well, the region from x=a to x=2a had V=infinity, so for the purposes of finding the wavefunction inside the well it was irrelevant.

 

[KaiserBrandon]

ok, so just treating this as an infinite square well problem, I get

$$E_{n}=\frac{\hbar^{2}}{2m}\frac{(n-1/2)\pi}{2a}$$

and

$$\Psi_{n}=A_{n}cos(\frac{(n-1/2)\pi}{2a}x) A_{n}=\sqrt{\frac{1}{2a}}$$

now when you say represent the initial wave function as a superposition of the stationary states, are you saying we represent the function at t=0 that is given in the question as a superposition of the states?

 

[ideasrule]

ok, so just treating this as an infinite square well problem, I get

$$E_{n}=\frac{\hbar^{2}}{2m}\frac{(n-1/2)\pi}{2a}$$

and

$$\Psi_{n}=A_{n}cos(\frac{(n-1/2)\pi}{2a}x) A_{n}=\sqrt{\frac{1}{2a}}$$

No, those aren't correct. This question is much easier to solve algebraically if you shift the reference frame so that the boundaries of the well are x=0 and x=4a. That way, you can use the well-known solution of the infinite square well. The possible energies remain unchanged (it would be quite surprising if they didn't, since all we did was redefine the coordinates). The eigenfunctions are simply shifted by 2a.

now when you say represent the initial wave function as a superposition of the stationary states, are you saying we represent the function at t=0 that is given in the question as a superposition of the states?

I'm saying represent the initial wave function as such. Of course, the initial wave function is just the function given, plus 0 for the two "sides" (x<-a/2 and x>a/2).