# Eigenfunction of the momentum "operator"

1. Mar 6, 2006

### 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?

2. Mar 6, 2006

### 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

3. Mar 6, 2006

### 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.

Last edited: Mar 6, 2006
4. Mar 6, 2006

### AKG

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

5. Mar 6, 2006

### 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 dont understand is how to satisfy: $$-i \hbar \frac{\partial}{\partial x}$$

6. Mar 6, 2006

### 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) = ea2x 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?

7. Mar 7, 2006

### UrbanXrisis

ok, so the eigenvalue is really -iha² for ea2x?

as for cos(x) the function changes to sin so it cant be an eigenfunction for -ih(d/dx)?