# How do we normalize a delta function position space wave function? [duplicate]

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?

• $\int_I\delta(x-a)\text dx = 1$ whenever $I$ contains $a$, otherwise is 0. So the integral over $\mathbb R$ is bound to contain any point – Phoenix87 Dec 27 '14 at 20:04
• Yup, I do get it, but I need to compute something like $\int \delta (x-a)^2 dx$. So how to do this? – sbp Dec 27 '14 at 20:08
• How to integrate the square of a delta function? – glS Dec 27 '14 at 20:12
• Thanks for the insight @glance, but I do have one question, $\delta(\lambda-\lambda) = \delta(0)$, but what does that mean? – sbp Dec 27 '14 at 20:17
• I do have an idea, I mean $\delta(x) = \frac{1}{2 \pi} \int \exp(\imath k x) dx$, using Fourier's theorem. But putting $x=0$, will get me nowhere! So how to get the value of $\delta(0)$? – sbp Dec 27 '14 at 20:23

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.
• I have a couple of questions. First $|x \rangle$ is the base ket I am using. Now $|a \rangle$ and $|-a \rangle$ are my two states. Now while computing $\langle\psi | \psi \rangle$, I get two cross terms, viz. $\langle a|-a \rangle$ and $\langle -a|a \rangle$. Are they orthogonal to each other? If yes, how? – sbp Dec 28 '14 at 17:33
• Do you know the properties of a dirac delta function? If yes, the third equation I wrote answers your question (remembering $a\neq 0$). If not, wikipedia is your friend! – glS Dec 28 '14 at 17:37
• $x \neq 0 \Longrightarrow \delta(x)=0$ – glS Dec 28 '14 at 17:58
• Any pointers on how I can go about computing, $\Delta x$? – sbp Dec 28 '14 at 18:44