- #1
hbweb500
- 41
- 1
I am trying to see why exactly the momentum eignenstates for a free particle are orthogonal. Simply enough, one gets:
[tex]\int_{-\infty}^{\infty} e^{i (k-k_0) x} dx = \delta(k-k0)[/tex]
I can see why, if k=k0, this integral goes to zero. But if they differ, I don't see why it goes to zero. You have:
[tex] \int_{-\infty}^{\infty} e^{i(k-k0)x} dx = \int_{-\infty}^{\infty}( \cos [(k-k0)x] + i \sin [(k-k0)x]) dx [/tex]
Now the sine vanishes by symmetry, but what about Cos[x]? I would imagine this integral diverges, but it must go to zero for these to be orthogonal...
I am recalling a bit from complex analysis that might be useful, but for now I am in the dark. Why is this integral the dirac delta function.
[tex]\int_{-\infty}^{\infty} e^{i (k-k_0) x} dx = \delta(k-k0)[/tex]
I can see why, if k=k0, this integral goes to zero. But if they differ, I don't see why it goes to zero. You have:
[tex] \int_{-\infty}^{\infty} e^{i(k-k0)x} dx = \int_{-\infty}^{\infty}( \cos [(k-k0)x] + i \sin [(k-k0)x]) dx [/tex]
Now the sine vanishes by symmetry, but what about Cos[x]? I would imagine this integral diverges, but it must go to zero for these to be orthogonal...
I am recalling a bit from complex analysis that might be useful, but for now I am in the dark. Why is this integral the dirac delta function.