I have a book saying,

$\int \delta(x-x')\psi(x)dx = \psi(x')$ where $\psi(x) = \langle x\lvert\psi\rangle$, so our definition of delta function would be $\langle x'\lvert x\rangle = \delta(x-x')$.

However I could find some documents (example; refer to 3. Position Space) saying,

$$\delta(x'-x'') = \langle x'\lvert x''\rangle$$

which corresponds to $\delta(x-x') = \langle x\lvert x'\rangle$.

So the result should be

$$\delta(x-x') = \langle x\lvert x'\rangle (=) \langle x'\lvert x\rangle \tag{1}$$

I think neither of them is an error, because my book uses the definition many times and I have found many documents explaining as $\delta(x-x') = \langle x\lvert x'\rangle$. Is (1) correct?


2 Answers 2


As DJBunk mentions the delta function is symmetric $$\delta(x)=\delta(-x)$$ so you certainly have $$\delta(x-x')=\delta(x'-x).$$

But you should also know that in general we have $$\langle a | b \rangle = \langle b | a \rangle^* $$ and since in this case the inner product is real, you will also have $$\langle x | x' \rangle = \langle x' | x \rangle .$$ So it doesn't matter which way you write the delta function or the inner product.

  • 1
    $\begingroup$ Keep in mind, though, that the delta-function symmetry only holds in the distributional sense. That is, $\delta(-x) = \delta (x) $ simply and only means that $\int_{-\infty}^\infty f(x)\delta(-x)dx=f(0)=\int_{-\infty}^\infty f(x)\delta(x)dx$, which you can prove by changing variables to $y=-x$. $\endgroup$ Commented Aug 31, 2012 at 20:46

Either way is fine:

$\delta(x-x')= \langle x| x' \rangle= \delta(x' - x) =\langle x'| x \rangle$.

You can see this either from the fact that in the limiting definition

$\delta (x-x') = \lim_{\sigma \rightarrow \infty} \frac{1}{\sqrt{2 \pi} \sigma} e^{-\frac{(x-x')^2}{2 \sigma^2}}$

is symmetric in $x$ and $x'$

or that either form does the same job under an integral:

$ \int^\infty_{-\infty} dx \delta(x-x') = \int^\infty_{-\infty} dx \delta(x' - x) = 1 $

and as Mistake Ink pointed out:

$ \int^\infty_{-\infty} dx f(x' )\delta(x-x') = \int^\infty_{-\infty} dx f(x' ) \delta(x' - x) = f(x) $

  • $\begingroup$ You should really put a test function under the integrals e.g. $f(x)$ and then see that they both give $f(x')$ to prove what you are trying to say. $\endgroup$ Commented Aug 31, 2012 at 16:28

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