- #1

quasar987

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[tex]\int_{-\infty}^{\infty} e^{i(p-p')x/\hbar}dx = 2\pi \hbar \delta(p-p')[/tex]

Where does that come from? I mean, set i(p-p')/h = K. Then the solution is

[tex]\frac{\hbar e^{i(p-p')x/\hbar}}{i(p-p')}[/tex]

and evaluate at infinity, it doesn't exists as the limit of cos and sin at infinity do not exist.

There must be something in the fact that the integral is complex but I haven't studies complex analysis yet so go easy on me plz.