# Qm finite well

## Homework Statement

Say I have a potential that is infinite from - infinity to 0, -V_0 from 0 to a, and 0 from a to infinity.

Then I have three unknowns, two for the plane wave solution in the finite well and one for the exponential outside the well.

But I also have four equations. Two wavefunction continuity boundary conditions and a wavefunction derivative continuity boundary conditions. I also have the normalization equation!

How is this possible? How can the solutions depend on which 3 of the four equations I select to use?

## The Attempt at a Solution

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Dick
Homework Helper
You already used the normalizability constraint in eliminating the exp(+ax) solution outside the well. Otherwise you would have four equations in four unknowns.

But I never used the constraint that square integral over all space of the piecewise wavefunction must be 1.

Dick
Homework Helper
You used that it should be finite and then arbitrarily threw away a free constant.

But if I use the 3 boundary conditions to determine the 3 remaining constants then I could get a wavefunction with no adjustable parameters left and with a square integral over all of space not equal to 1, right?

Dick
Homework Helper
No. If you've done this problem, and I know you have, you should know that the remaining three constants are only determined up to a multiplicative constant. If you want to continue bickering about this just do it with four equations and four unknowns. The same thing will happen.

I do not want to bicker. I just want to understand the problem.

I found that it is not even possible to implement the continuity of the derivatives at a which makes some sense considering the shape of the functions. So I was wrong, there are only a total of three constraints left and all is well.

I am not sure why it makes sense that the derivatives would not be continuous since there are no delta functions at boundaries, though?

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Dick
Homework Helper
The derivative at zero is not continuous because there is an infinite potential at x=0. That's worse than a delta function.

Of course. I am talking about the derivatives at a.

Dick
Homework Helper
The derivatives at a are continuous.

So, maybe I did something wrong. I still do not understand what you were saying about the "the remaining three constants are only determined up to a multiplicative constant".

We have three equations and three unknowns, we should be able to uniqually determine each of the unknowns if there is a solution.

Avodyne
The point is that your four equations will determine the three coefficients AND the set of allowed energy eigenvalues.

Yes. I agree with that.

Avodyne