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?

  • 2
    $\begingroup$ How to integrate the square of a delta function? $\endgroup$
    – glS
    Dec 27, 2014 at 20:12
  • $\begingroup$ Thanks for the insight @glance, but I do have one question, $\delta(\lambda-\lambda) = \delta(0)$, but what does that mean? $\endgroup$
    – sbp
    Dec 27, 2014 at 20:17
  • $\begingroup$ 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 $\delta(0)$? $\endgroup$
    – sbp
    Dec 27, 2014 at 20:23
  • $\begingroup$ as people have said, $\delta^2$ makes no sense. so the "best" way to deal with this is to observe that the two deltas are independent so you can treat them as independent vectors, hence the normalisation factor of $\sqrt 2$ $\endgroup$
    – Phoenix87
    Dec 27, 2014 at 21:43

1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$
    – sbp
    Dec 28, 2014 at 17:33
  • $\begingroup$ 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! $\endgroup$
    – glS
    Dec 28, 2014 at 17:37
  • $\begingroup$ Could you point out what you mean explicitly? I just can't get it. $\endgroup$
    – sbp
    Dec 28, 2014 at 17:55
  • $\begingroup$ $ x \neq 0 \Longrightarrow \delta(x)=0$ $\endgroup$
    – glS
    Dec 28, 2014 at 17:58
  • $\begingroup$ Any pointers on how I can go about computing, $\Delta x$? $\endgroup$
    – sbp
    Dec 28, 2014 at 18:44

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