0

Let $$\langle x'|p'\rangle = N \exp(\frac{ip'x'}{\hbar})$$ be the overlap between position and momentum space, where $N$ is a normalization constant to be determined.

We can then compute $N$ by $$ \langle x'|x''\rangle = \int \mathrm{d}p'\langle x'|p'\rangle \langle p'|x''\rangle\\ \rightarrow \delta(x'-x'')=|N|^2 \int \mathrm{d}p'\exp(\frac{ip'(x'-x'')}{\hbar}) \\ =2\pi\hbar |N|^2 \delta(x'-x'') $$

see Sakurai - "Modern Quantum Mechanics" 2nd, Pearson, p. 54.

How was the factor $2\pi\hbar$ obtained? The intermediate step should look something like this: $$ \rightarrow \delta(x'-x'')=|N|^2 \int \mathrm{d}p'\exp(\frac{ip'(x'-x'')}{\hbar}) \\ =|N|^2 \int \mathrm{d}p'\delta(x'-x'') $$

But all I can think of is that one can say $\int \mathrm{d}p'=\hbar k$ where $k$ is the wavenumber.

What am I missing?

Qmechanic
  • 201,751
EpsilonDelta
  • 370

2 Answers2

2

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 $p=\hbar k$:

$$\int dp\ e^{ipx/\hbar} = \hbar \int dk\ e^{ikx} = 2\pi\hbar$$

Javier
  • 28,106
  • Thanks! You silently grabbed $\mathcal{F}(1)= (2\pi)^n\delta$ from the Theory of Distributions, did you? – EpsilonDelta May 09 '15 at 14:51
  • @EpsilonDelta you don't need fancy theory to show that. You can show that $\int \exp(-i (k-k') x)dx = 2\pi \delta(k-k')$ by just integrating the left hand side against any function of $k$ and messing around. – DanielSank May 09 '15 at 15:52
1

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.

Qbit
  • 11