How Does Particle Behavior Change in a Shifted Infinite Square Well?

[eku_girl83]

Any help is much appreciated!

 

[nrqed]

Suppose there is a particle in a box (infinite square well) in the ground state. The wall begins at x=L and ends at x=3L. I need to find the probability distribution for the particle.

I know how to work this problem for a well with boundaries at x=0 and x=L. In the new problem, since the width of the box is 3L - L = 2L, can I simply say that the box dimenstions are x=0 to x=2L?

No, because the whole idea here is to have you work with different boundary conditions! If you understand well the case from 0 to L, you should be able to do it from L to 3L.

This preserves the width and makes it much easier to use my boundary conditions to determine k (U(x) = A cos kx +B sin kx), since x = 0 makes the B sin kx term disappear.

Of course but there would be no point in asking this question if you could simply do this. The idea here is that someone has already made the choice to fix the origin at a distance L to the left of the left side of the well, You have to work with that. You then must give the wavefunctions with that choice of origin. It is not as simple as having the box between 0 and L but that`s the whole idea of the question!

There are two ways to proceed (I am not sure if your prof has a preference). You may start from scratch and impose that function is zero at L and at 3L and work out the conditions on A and B. And then you normalize. OR you may start from the already known solutions for 0 to L and shift the origin in these solutions to -L. Using trig identities you will get that the eiegnfunctions are linear combinations of sine and cosine functions instead of being pure sine.

Even better, you sould do it both ways and check that you get the same thing.

Also, how do I determine the ground state psi (x,0)? Is it Ax for O<x<L and A(L-x) for L<x<2L?

Any help is much appreciated!

? As you said above, the solutions are of the form A cos(kx) + Bsin(kx)! There is no way to get the function you just wrote as an eigenstate! You must be confusing with another problem!

For the 0 to L well, the ground state is what? A single sine function. In your new well, it will be alinear combination of a sin and a cos function!

Patrick