# Normalizing the Momentum Eigenfunctions

1. Jun 24, 2008

### Domnu

I know that the momentum eigenfunctions are of the form $$\phi = Ce^{ikx}$$, but how would we normalize them? We just get

$$\int_{-\infty}^{\infty} C^2 dx = 1$$

which means that $$C$$ is infintesimally small...

2. Jun 24, 2008

### CompuChip

Indeed, so these wave-functions are non-normalizable. You either choose C = 0 and you have a trivial wave function, or you have a diverging integral.

3. Jun 24, 2008

### Domnu

Well, see the thing is that there's this formula in my book which I have absolutely no idea as to how they got... here it is... could someone show a proof of this?

$$\int_{-\infty}^{\infty} e^{ix(k'-k)} dx = 2\pi \cdot \delta(k' - k)$$

It makes absolutely no sense to me... if we have that $$k' \neq k$$, then

$$\int_{-\infty}^{\infty} e^{ix(k'-k)} dx = \int_{-\infty}^{\infty} \cos [(k'-k) x] dx$$

since sin is an odd function. But even this is a large step to claim... it's like saying that 1 - 1 + 1 - 1 + 1 - 1 + ... = 0. But can we just say that cos is a sideways shift of sin, so we can just say that it evaluates to zero as well? Or do the limits mean that we take a value that approaches infinity and it always equals zero? I think I understand why the k'=k case works, however...delta(0) = infinity, which is also the lhs. Lastly, why does the 2 pi have to be on the RHS?

Last edited: Jun 24, 2008
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?