Friday, December 4, 2015

quantum mechanics - Why is $langle x| x' rangle=delta(x-x')$?



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.



Answer



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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