Physical Chemistry/Quantum Mechanics Eigenvalues

[trf5]

Homework Statement

Indicate which of the following expressions yield eigenvalue equations and identify the eigenvalue.

a) d/dx (sin(∏x/2))
b) -i*hbar * ∂/∂x (sin(∏x/2))
c) ∂/∂x (e-x^2)

The Attempt at a Solution

I know that if the wave equation yields an eigenvalue equation, it will give me the wave equation multiplied by the eigenvalue back. I calculated the derivatives:

a) 1/2∏cos(∏x/2)
b)(-i*hbar∏/2)(cos(∏x/2))
c) -2x

Because none of these give me back the original wave function, does that mean none of these are eigenvalue equations? I don't think my professor would give us a problem that asks to calculate the eigenvalues, where there are no eigenvalues. In parts A and B, the ∏x/2 is in both the question and the answer, but the sin changes to cosine when you take the derivative, so does that mean it is not an eigenfunction? Any help would be appreciated.

 

[LCKurtz]

That seems like an unusual way to phrase the question. Are you really asking whether any of those functions satisfy $y''+\lambda y = 0$? Note the second derivative, not just the first.

 

[trf5]

I don't think so. I have taken a differential equations course and I have not seen eigenvalues in the context he is asking for before. From my notes, using the schroedinger wave equation (time independent), you can use the Hamiltonian operator on a wave function, and if the wave exists, after using the Hamiltonian, you get the energy of the wave multiplied by the wave function. Therefore, I should get the wave function back in my answers and what he is calling the "eigenvalue" is the energy of the system. Is that any clearer?

Edit: Oops, I just realized I posted in the math section.

 

[LCKurtz]

Well, I don't know about that physics stuff, so it may in fact be clearer, but not to me. But sines and cosines do give multiples of themselves back on differentiating twice. For example, this shows up solving the one dimensional wave equation for a vibrating string with fixed ends, where they are the eigenfunctions. That's why I speculated as I did in post #2. But I will leave it to one of the resident physicists to actually answer your question.

 

[Dick]

There's no way to say whether a function is an eigenfunction until you say what operator they are supposed to be an eigenfunction of. Every function is eigenfunction of the identity operator. What hamiltonian are you talking about? If the operator part is supposed to be the d/dx and the rest of the expression the eigenvector then I'd agree none of them are eigenvectors. Pretty badly phrased question.