Tag Info

New answers tagged


You can express the dirac-delta-function as: $$\delta(x-x')=\frac{1}{2 \pi}\int dp e^{i p (x-x')}$$ (simply fourier-transform the dirac-function) compare it with your expression and you get the factor. p.s Your last line from intermediate step is wrong.


You're close, but you seem to be saying that $\exp(ikx) = \delta(x)$, which is not true, and you're missing a $2\pi$. The correct identity is $$\int dk\ e^{ikx} = 2\pi \delta(x)$$ Therefore, with a change of variables $p=\hbar k$: $$\int dp\ e^{ipx/\hbar} = \hbar \int dk\ e^{ikx} = 2\pi\hbar$$

Top 50 recent answers are included