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I've tried to find any solution or proof for $$\langle x| x' \rangle=\delta(x-x'),$$ but I only came to this post: Wave function and Dirac bra-ket notation

So I got the information, that the vector $|x\rangle$ form a dirac-normalized basis for the Hilbert Space.

I know that the dirac-delta distribution is defined like this: $$\delta(x-x') = \begin{cases} 0 &\mbox{if } x\neq x' \\ \infty & \mbox{if } x=x' \end{cases},$$ this means that my x' is a point on my x axis where I have my infinite high peak. And also $$\int_{-\infty}^{\infty}dx\cdot \delta(x-x')=1.$$

But how actually correlate this with the scalar product of vectors x, x' in the Hilbert Space that form a so-called 'diracl-normailzed" basis of it?

Can you give me some tips on this please? Or maybe you actually know a link, where this is explained.

Qmechanic
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physics
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2 Answers2

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Isn't it just from the sifting property?

$$f(x) = \int\mathrm{d}x'\;f(x')\,\delta(x - x')$$

That is, if you accept the above and if you accept that

$$|\psi\rangle = \int \mathrm{d}x'\,\psi(x')\,|x'\rangle$$

then

$$\psi(x) = \langle x|\psi\rangle = \langle x| \int \mathrm{d}x'\,\psi(x') \,|x'\rangle = \int \mathrm{d}x'\,\psi(x')\,\langle x|x'\rangle$$

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

Alfred Centauri
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First, convince yourself that any finite-dimensional vector space, which you can easily think about in notation more familiar than bra-ket, $\sum_i \left|i\right\rangle\left\langle i\right|$ is the identity operator, where the sum is over elements $\left|i\right\rangle$ of some orthonormal basis. Indeed, $$\sum_i \left|i\right\rangle\left\langle i|j\right\rangle=\sum_i \left|i\right\rangle\delta_{ij}=\left|j\right\rangle$$proves the putative identity acts as expected on basis elements, and the general case then follows from linearity. All we needed was the orthonormality condition $\left\langle i|j\right\rangle=\delta_{ij}$.

Next we'll go to a space that is not only infinite-dimensional, but has a continuous spectrum of basis element labels. Our identity operator is now an integral instead of a sum, $\int dx\left|x\right\rangle\left\langle x\right|$. We want$$\left|x'\right\rangle=\int dx\left|x\right\rangle\left\langle x|x'\right\rangle,$$which is clearly true iff $\left|x\right\rangle\left\langle x|x'\right\rangle=\delta\left( x-x'\right)$.

J.G.
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