Unravelling the Mystery of Infinite Potential Solutions

[ehrenfest]

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?

Homework Equations

The Attempt at a Solution

 

[Dick]

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

 

[ehrenfest]

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

 

[Dick]

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

 

[ehrenfest]

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]

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.

 

[ehrenfest]

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?

 

[Dick]

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

 

[ehrenfest]

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

 

[Dick]

The derivatives at a are continuous.

 

[ehrenfest]

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.

 

[ehrenfest]

Yes. I agree with that.

 

[Avodyne]

So, that's the solution. If you don't use normalization, then you don't determine the values of all three coefficients, but only their ratios; the 3rd equation then determines the allowed eigenvalues.

 

[ehrenfest]

Thanks. I see.