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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.

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marked as duplicate by FraSchelle, Wolpertinger, Kyle Kanos, Qmechanic quantum-mechanics May 4 '17 at 5:48

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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

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

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

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  • $\begingroup$ The argument is not conclusive. This way you can only conclude that $\langle x| x'\rangle - \delta(x,x')$ produces $0$ when smeared with any function, not $\langle x| x'\rangle = \delta(x,x')$... $\endgroup$ – Valter Moretti May 2 '17 at 14:18
  • $\begingroup$ @ValterMoretti: This point interests me--could you elaborate? $\endgroup$ – daniel May 2 '17 at 14:48
  • $\begingroup$ There is not much to elaborate: from $\psi(x) = \int dx' \psi'(x) \langle x|x'\rangle$ and $\psi(x) = \int dx' \psi'(x) \delta(x-x')$ you get $0 = \int \psi (x') (\langle x|x'\rangle - \delta(x-x'))$ for every $\psi$. Does it imply $\langle x|x'\rangle = \delta(x-x')$? $\endgroup$ – Valter Moretti May 2 '17 at 14:54
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    $\begingroup$ Without a precise definition of "delta function", a choice of the relevant class of functions and so on, it is difficult to develop any rigorous reasoning. What we can say is just that the formal manipulations of the symbols suggest that $\langle x|x'\rangle = \delta(x-x')$. $\endgroup$ – Valter Moretti May 2 '17 at 14:57
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    $\begingroup$ To make conclusive this argument at this level of rigour it is enough to assume some sloppy axiom like this "$\int f(x) g(x) dx=0$ for all $g$ implies $f(x)=0$ for all $x$"... $\endgroup$ – Valter Moretti May 2 '17 at 15:04
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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 $\left|x\right\rangle\left\langle x|x'\right\rangle=\delta\left( x-x'\right)$.

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