I have a position space wavefunction $$\psi(x) = \delta(x-a) + \delta(x+a).$$ Now the question states to compute the following:
- The Fourier transform of $\psi(x)$. (Which invariably is the momentum space wave function $\phi(p)$)
- Compute $\Delta p \Delta x$
Now to compute $\phi(p)$ one first need to normalize $\psi(x)$, but while doing so, one need to compute, $$|N|^2\int\limits_{-\infty}^{+\infty} (\delta(x-a) + \delta(x+a))^2 dx =1,$$ $N$ being the normalization constant. But my question is how to go about computing this integral?
The normalization integral is now done, any help, in solving $\Delta x$, or $\Delta p$? I can't see through this problem?
Answer
The correct normalization factor is $$ N = \frac{1}{\sqrt{2}}.$$ To see this, note that you can write your wave-function in ket notation as $$\psi(x) = \langle x | a \rangle + \langle x | -a \rangle \equiv \langle x | \psi \rangle, $$ where we have used the usual basis for the (one dimensional) position representation, with normalization $$ \langle x | y \rangle = \delta(x-y),$$ and we have defined the state $|\psi\rangle$ as the state corresponding to your wave function: $$| \psi \rangle = | a \rangle + |-a \rangle.$$
Your question now translates into: what is $\langle \psi | \psi \rangle$? The answer is readily obtained: this is a sum of two orthogonal states (assuming $a\neq0$), hence $$ \langle \psi | \psi \rangle = \langle a | a \rangle + \langle -a | -a \rangle = 2$$ and $$ | \psi \rangle = \frac{1}{\sqrt{2}} ( |a\rangle + |-a\rangle),$$ $$ \psi(x) = \frac{1}{\sqrt{2}} ( \delta(x-a)+ \delta(x+a)).$$
Note that if you try to compute $\langle a | a \rangle$ as an integral you get $\infty$: $$ \langle a | a \rangle = \int dx \delta(x-a)^2 \approx \delta(0) \approx \infty $$ where the $\approx$ is because we are being mathematically sloppy here, see this discussion of Qmechanic about it. This is due to the use of a continuous base. You can always think of it as the limiting case of a discrete base, taking case this way of the mathematical difficulties.
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