Any help is much appreciated!
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.
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.
??? 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!
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