Eigenfunction of the momentum "operator"

[UrbanXrisis]

which of the following functions is an eigenfunction of the momentum "operator"

-i \hbar \frac{\partial}{\partial x}:

f_1 =cos(kx- \omega t)
f_2 =e^{a^2x}
f_3 =e^{-(\omega t+kx)}

for this question, I'm not sure what they are looking for...

for f1
i \hbar k sin(k x -\omega t)
for f2
-i \hbar a^2 e^{a^2x}
for f3
- \hbar ke^{-(\omega t+kx)}

the eigenvalues for f1 is -k^2
the eigenvalues for f2 is a^2
the eigenvalues for f3 is -ik

how do I find the correct eigenfunction?

 

[xman]

careful with your assertions, remember an eigen function says,
H \mid v \rangle = \lambda \mid v \rangle
in other words when you act on a vector, you get the function back times the eigenvalue right...check your calculations...hope this helps, sincerely,x

 

[Physics Monkey]

Hi UrbanXrisis,

Typically you say that f is an eigenfunction of an operator like \mathcal{O} = - i \hbar \frac{\partial}{\partial x} if \mathcal{O} f = \lambda f where \lambda is just some number called the eigenvalue. In other words, if when taking the derivative of f you get the same function back up to an overall constant, then f is an eigenfunction of the momentum operator. There are issues associated with the ability to normalize such functions, but I don't think that's the point of this problem.

 

[AKG]

Do you know what it means for a function f to be an eigenfunction of an operator T?

 

[UrbanXrisis]

ok, so taking f_2 =e^{a^2x} as an example, the eigenfunction is e^{a^2x} and the eigenvalue is a^2 because if I took the derivative of the function, it would keep get the eigenfunction multiplied by a^2 over and over again. what I don't understand is how to satisfy: -i \hbar \frac{\partial}{\partial x}

 

[AKG]

No, first look up the definition of an eigenfunction. There's no "satisfying" -ih(d/dx). -ih(d/dx) is an operator, you can apply it to any function. Consider the function f(x) = x². There's no "satisfying" f. You can plug in any number for x, and you get another number x². It's the same idea with operators. You plug in your function f into your operator, and you get another function.

Plug the function f(x) = e^a2x into -ih(d/dx), and you get:

-ih(d/dx)(exp(a²x)) = -iha²exp(a²x) which is just another function, agree? It's g(x) = -iha²exp(a²x). Now the question is, is g(x) a scalar multiple of f(x)? Yes it is. What is that multiple (it's not a²).

Try a new example, f(x) = cos(x). Plug it in:

-ih(d/dx)(cos(x)) = ihsin(x)

Again, this is just another function, g(x) = ihsin(x). Is this a scalar multiple of f(x)? That is, is there some complex constant z such that

g(x) = zf(x) ?
ihsin(x) = zcos(x) ?
sin(x) = (z/ih)cos(x) ?

No! Do you know why?

 

[UrbanXrisis]

ok, so the eigenvalue is really -iha² for e^a2x?

as for cos(x) the function changes to sin so it can't be an eigenfunction for -ih(d/dx)?