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is exp (-kx) an eigenfunction?

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In summary, eigenfunctions are mathematical entities that can represent physical entities such as momentum, position, and energy. The function exp(-kx) could potentially be an eigenfunction of a mathematical operator, but may not be a physical eigenfunction due to its behavior at negative infinity. However, if the function is defined on a semi-infinite region, it could be a valid eigenfunction. In general, the relationship that defines eigenfunctions and eigenvalues of an operator is that the function will be mapped to another function which is the same function multiplied by the eigenvalue. In the case of d/dx exp(-kx) = -k.exp(-kx), the eigenvalue is -k and the function remains exp(-kx). Therefore

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is exp (-kx) an eigenfunction?

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eigenfunctions have to be eigenfunctions

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).

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That's like asking "Is 23 an answer?" - it might be, but we have to know what the question is first.Amy B said:is exp (-kx) an eigenfunction?

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.

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

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Amy B said: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.

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jtbell said:definesthe 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.

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Amy B said: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##.

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