What is the Time Dependent of a Particle in an Infinite Square Well Potential?

[Fourier mn]

Homework Statement

consider a particle of mass m in the ground state of an infinite square well potential width L/2. What is the probability of detecting the particle at x=L/4 in a range of \Deltax=0.01L (d not integrate)? Assume that the particle is in the normalized state \Psi(x,0)=c₁\Psi₁+c₂\Psi₂, what is the time dependent?

Homework Equations

The Attempt at a Solution

Help, i don't know from where to start even...

 

[Pacopag]

If you WERE allowed to integrate to get the probability, would you know what to do?

 

[Fourier mn]

i would integrate the wavefunction (multiplied by its complex part) to find <x> and <x^2>, and then use the uncertainty. isn't <x> always zero though?

 

[Pacopag]

Close, but not quite. If you wanted to know the probability of finding the particle between x and \Delta x you would get it from
P = \int_x^{x+\Delta x} \Psi^* \Psi dx.
Now think about what a definite integral "means" and find an approximation to the integral if \Delta x is really small. Maybe drawing a picture would help. You do not have to integrate.

 

[Fourier mn]

So I need from 0 to L/4 in increments of 0.01L, if delta x is really small isn't just going to be the wavefunction multiplied by itself with x=L/4. I mean i know that a definite integral is the area under the curve, maybe just the function where x=L/4 multiplied by 0.01L?

 

[Pacopag]

You're very close. But your not interested in the particle between 0 and L/4. You are interested in a small interval AROUND L/4. Catch my drift?

 

[Fourier mn]

delta and expsilon? ===>limit?

 

[Pacopag]

Wait!. I should've read your reply to the end. Your last statement
"maybe just the function where x=L/4 multiplied by 0.01L?"
is 99 percent right. Just use the probability AMPLITUDE instead of the wavefunction itself.

 

[Fourier mn]

ohhh...thanks. I've already started taking the limit...lol. I'll do that instead

 

[Fourier mn]

so for the first part i got 0.04 (4%), and the second part is just an infinite series.
now, I'm asked to find the expectation values of energy <E>. Is that just--
<E>=\int\Psi(x,t)*\Psi(x,t)Hdx?

 

[Pacopag]

yep!

 

[Fourier mn]

actually I got 8% on the first part. for the second part i got <E>=E.
Another question- "now that the wave function returned to the ground state. at t=0 the well suddenly changes to an infinite square well of width L without affecting the wave function. find the probability that a measurement of energy right after the expansion will yield E=(h bar*Pi)^2/(2mL^2)?


Isnt it a stationary state===> thus the probability never changes and stays 1?

 

[Pacopag]

It is in a stationary state for the OLD well. The NEW well has different stationary states. You have to write your state as a linear combo of the NEW stationary states.

 

[Fourier mn]

but it states in the question that the width changes without affecting the wave function

 

[Pacopag]

Right. But when the width changes, you have NEW eigenfunctions (i.e. stationary states). So your old state is no longer one of the stationary states for the new problem.

 

[Fourier mn]

<E>=\SigmaC^2_nE_n
where \SigmaC^2_n=1
P₁=C^2₁=1, thus, the probability of E didnt change b/c the expectation value of E didnt change.

 

[Pacopag]

You can't tell much about the probability from the expectation value. They aren't really related.

 

[Pacopag]

Ok. You are partly right. It seems to me that the state IS still a stationary state, but now it is NOT the ground state anymore. The state corresponds to an eigenvalue that is not the one for n=1.

 

[Pacopag]

I don't know why the heck my last reply didn't show up.
It turns out that your state is actually a stationary state (even after the expansion), but the catch is that it is not the "ground" state for the new well. You diggin' what I'm buryin'?

 

[Fourier mn]

lol...got it. Thanks. Why your earlier replies didnt show up? weird

 

[Pacopag]

Actually, I just didn't know to go to page 2. All my replies are there :)
You got it! That's great. Good job.