Eigenfunctions and eigenvalues

  • #1
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Main Question or Discussion Point

is exp (-kx) an eigenfunction?
 

Answers and Replies

  • #2
381
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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).
 
  • #3
Nugatory
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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.
 
  • #4
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oops, I forgot to mention that the operator is d/dx
 
  • #5
jtbell
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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?
 
  • #6
stevendaryl
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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 [itex]-\infty < x < +\infty[/itex], the boundary condition is that [itex]\int_{-\infty}^{+\infty} |\psi(x)|^2 dx < \infty[/itex]. In the case of [itex]e^{-kx}[/itex], this integral is not defined (because that function goes to infinity as [itex]x \rightarrow -\infty[/itex].

On the other hand, if space (or the portion of space that is occupied by the particle) is a semi-infinite region [itex]0 \leq x < \infty[/itex], then that function is perfectly acceptable.
 
  • #7
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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.
 
  • #8
jtbell
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so if d/dx exp(-kx) = -k.exp(-kx) the eigenvalue E is -k and the function f remains exp(-kx)
Correct! :woot:

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

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