Eigenfunctions and eigenvalues

[Amy B]

is exp (-kx) an eigenfunction?

 

[secur]

Well, it could be, but ...

eigenfunctions have to be eigenfunctions of something. Namely, a matrix, or an operator (which is like a matrix, but often infinite-dimensional). They're mathematical entities which can represent physical entities. Your function could certainly be an eigenfunction of a mathematical operator, but it probably wouldn't be a physical eigenfunction. (Since this is the quantum physics forum I assume physical QM-related eigenfunctions are the main interest.)

Physical observables like momentum, position, energy are represented by operators in a Hilbert Space, and they have eigenfunctions - also called eigenvectors or eigenstates.

The thing is, normally k is used for a real number, and I assume it is here. The reason your function is unlikely to be a physical eigenfunction is that it blows up at negative infinity (supposing k is positive). If your function were only defined on the positive real number axis it could be a physical eigenfunction, because it goes to 0 and is square integrable.

In fact it can appear in tunneling problems; is that where you got it?

More likely you're probably missing an "i": it should be exp(-ikx). If k is square root (2mE) / hbar, for instance, then this could be a typical eigenfunction for the energy of a free particle (non-relativistic, time-independent).

 

[Nugatory]

is exp (-kx) an eigenfunction?

That's like asking "Is 23 an answer?" - it might be, but we have to know what the question is first.

An eigenfunction is a solution to an equation (of a particular type) so if you give me a function and ask if it's an eigenfunction, I have to answer that it depends on what equation you had in mind.

 

[Amy B]

oops, I forgot to mention that the operator is d/dx

 

[jtbell]

OK, so what is the general relationship that defines the eigenfunctions and eigenvalues of an operator? If function f is an eigenfunction of operator O, with eigenvalue E, what relationship has to be true?

 

[stevendaryl]

is exp (-kx) an eigenfunction?

In quantum mechanics, the set of possible wave functions depends on the boundary conditions. With the usual one-dimensional case in which space is the real numbers -\infty < x < +\infty, the boundary condition is that \int_{-\infty}^{+\infty} |\psi(x)|^2 dx < \infty. In the case of e^{-kx}, this integral is not defined (because that function goes to infinity as x \rightarrow -\infty.

On the other hand, if space (or the portion of space that is occupied by the particle) is a semi-infinite region 0 \leq x < \infty, then that function is perfectly acceptable.

 

[Amy B]

OK, so what is the general relationship that defines the eigenfunctions and eigenvalues of an operator? If function f is an eigenfunction of operator O, with eigenvalue E, what relationship has to be true?

if the operator O acts on the function F, then the function will be mapped to another function which is the same function f times an eigenvalue E.
so if d/dx exp(-kx) = -k.exp(-kx) the eigenvalue E is -k and the function f remains exp(-kx), so I think this should be an eigenfunction, but just want to be sure.

 

[jtbell]

so if d/dx exp(-kx) = -k.exp(-kx) the eigenvalue E is -k and the function f remains exp(-kx)

Correct!

To re-phrase it slightly, $e^{-kx}$ is an eigenfunction of $\frac{d}{dx}$, with eigenvalue $-k$.