# KE operator and eigenfunctions

1. May 20, 2015

### dyn

I have just done a question and then looked at the solution which I don't get. The question gives a wavefunction as u = x - iy. It then asks if this function is an eigenfunction of the kinetic energy operator in 3-D. Applying this operator to u gives zero. I took this to mean that u is an eigenfunction with eigenvalue zero but the solution says that u is not an eigenfunction. Am I right or what is wrong with my argument ? Thanks

2. May 20, 2015

### Staff: Mentor

It is zero - so is an eigenfunction with eigenvalue 0.

Its not physically realisable of course - but formally it is true.

Thanks
Bill

Last edited: May 20, 2015
3. May 20, 2015

### dyn

The kinetic energy operator contains a Laplacian which involves 2nd derivatives in x , y , z so the function u becomes zero. The solution agreed with me that the KE operator acting on the function gives zero but I took that to mean u is an eigenfunction with eigenvalue zero but the solution says it is not an eigenfunction.

4. May 20, 2015

### Staff: Mentor

Well obviously it is - there is an error.

What it may be alluding to is its not physically realisable since its non zero at infinity. But then again so are the normal solutions.

Thanks
Bill

5. May 20, 2015

### Staff: Mentor

Just to expand a bit further what's going on is you have run into a pathological case with degeneracy issues due to an eigenvalue of zero.

If you solve the free SE via separation of variables you end up with the multiple of wave solutions. But things get tricky when those waves have infinite wavelength as in your example. Work through the detail - I just did and its interesting.

Thanks
Bill

Last edited: May 20, 2015
6. May 20, 2015

### fzero

I suppose the most straightforward argument that that is not an eigenvector is simply that it is not orthogonal to any of the eigenvectors with nonzero eigenvalue. Since the operator is Hermitian, it is not a proper eigenfunction (pun intended!)

7. May 21, 2015

### stevendaryl

Staff Emeritus
In QM, the usual meaning of "eigenfunction" is a square-integrable function that has a definite value for some operator. But $\psi = x - iy$ isn't square-integrable, so it's not a legitimate wave function. In many instances in quantum mechanics, it's the constraint that the wavefunction be square-integrable that forces its eigenvalues to be discrete. So usually when people talk about the eigenvalues of an operator $O$, they mean those values $\lambda$ such that there is a square-integrable function $\psi$ with $O \psi = \lambda \psi$.

8. May 21, 2015

### Staff: Mentor

I am pretty sure that's the reason they wanted it excluded as an eigenfunction - it blows up to infinity at infinity. My concern is I have zero doubt wherever the OP got it from would have the usual wave solution to the free particle SE. That, while not blowing up to infinity, is nonetheless bounded there so is not valid either. This is the type of thing that can easily confuse a beginning student that actually thinks.

In fact it can confuse more advanced students until they come to grips with Rigged Hilbert Spaces.

Thanks
Bill

9. May 21, 2015

### dyn

the question was from an introductory QM course. At that level Rigged Hilbert Spaces are unheard of. So at the introductory level is there a definite answer of eigenfunction or not ?

10. May 21, 2015

### Staff: Mentor

No it isn't because it blows up at infinity so isn't normalisable - but like I said its inconsistent.

Thanks
Bill

11. May 21, 2015

### dyn

Sorry to be a pain here but the mathematical definition of eigenvectors doesn't put any requirements on whether they can be normailised or not. So if it is given as a wavefunction surely it is an eigenfunction with eigenvalue zero. Another part of the question asked whether Aexp(ikx) + Bexp(-ikx) was an eigenfunction of the kinetic energy operator and gave the answer that it was but this function is also not square-integrable

12. May 21, 2015

### Staff: Mentor

They are part of a normed vector space - that vector has infinite norm.

That function you gave has exactly the same issue but is considered valid - but it really isn't - that's the part that makes thinking students pull their hair out.

I could explain what's going on but it will involve a long sojourn into distribution theory:
https://terrytao.wordpress.com/2009/04/19/245c-notes-3-distributions/#more-2072

Really your teacher needs to explain it - post here with what he/she says - it should prove interesting.

Thanks
Bill

Last edited: May 21, 2015
13. May 21, 2015

### dyn

I don't have a teacher. Just self-studying. Obviously the QM case is never straight forward - no surprise there but I have a last question. If it was just a maths question regarding eigenvalues and I had a function similar to the KE operator and the function x-iy would it be an eigenfunction with eigenvalue zero ? Whenever I have done eigenfunction questions I have just had to check if it satisfies the eigenvalue equation. Is this not enough ? Do I now need to check normed vector spaces ?

14. May 21, 2015

### Staff: Mentor

You need to specify a space. To really understand this stuff you need to study linear algebra first, then Hilbert Spaces. For now my advice is simply to persevere and move on.

Thanks
Bill

15. May 21, 2015

### Avodyne

One way to deal with this kind of issue, at least in part, and without going to the full machinery of rigged Hilbert spaces, is to put the particle in a box with some boundary conditions, like the wave function must vanish on the walls, or periodic boundary conditions. Then require wave functions to be square integrable. Then x-iy is not an allowed wave function, because it doesn't satisfy the boundary conditions.

Whatever source the OP had that asked the original question is badly written and should be dropped.

16. May 21, 2015

### Staff: Mentor

Agreed.

Students should not be confronted with tricky issues like this to start with.

Thanks
Bill

17. May 21, 2015

### dyn

Thanks guys. I'm going to forget I ever saw that question.