How do we normalize a Dirac delta function position space wave function?

Question

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?

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
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 $\sqrt 2$ — Phoenix87, Dec 27 '14 at 21:43

score 2 · Accepted Answer · edited Apr 13 '17 at 12:39

2

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.

edited Apr 13 '17 at 12:39

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answered Dec 27 '14 at 20:41

glS

14,271

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
Could you point out what you mean explicitly? I just can't get it. – sbp Dec 28 '14 at 17:55
$ 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
Okay now I get it, you are considering $|a \rangle$ and $|-a \rangle$ as two distinct position base kets. Ofcourse they are orthogonal. Is this the reason? – sbp Dec 28 '14 at 19:39
yes, as long as, as I said, $a \neq 0 $ – glS Dec 29 '14 at 11:17
Any input as to how I may go about computing $\Delta x$? – sbp Dec 29 '14 at 17:13

How do we normalize a Dirac delta function position space wave function?

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