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Hello I am doing introductory QM and I am getting myself hopefully confused with some Dirac notation. We have that \begin{align*} \langle x' | \psi \rangle &= \langle x' | \hat{I} |\psi \rangle \\ &= \int \langle x' | x \rangle \langle x|\psi\rangle dx \\ &=\int \delta(x'-x)\langle x | \psi \rangle dx \\ \end{align*} then I am unsure on how to proceed to the next step. I simply want to obtain $\psi(x')$. Any help would be much appreciated!

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    $\begingroup$ Are you sure you want to obtain $\psi(x)$ rather than $\psi(x')$? $\endgroup$
    – TLDR
    Commented Jun 5, 2016 at 2:17
  • $\begingroup$ Sorry, my mistake! Have fixed it, $\endgroup$
    – Cococabana
    Commented Jun 5, 2016 at 2:20
  • $\begingroup$ Obtain $\psi(x)$ from what? What's your starting point? I guess I don't understand the question. $\psi(x') = \left<x'\,|\,\psi\right>$, right? $\endgroup$
    – garyp
    Commented Jun 5, 2016 at 2:22
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    $\begingroup$ I think you're going in circles because you don't have a definition of what $\psi(x)$ is. $\psi(x)$ is, as the notation implies, a function of $x$. You find the value of the function by doing $\langle x|\psi\rangle$. So $\psi(x)=\langle x|\psi\rangle$. There is nothing to prove; this is just a definition. And without this definition, you'll never be able to "prove" the equality, because you won't even know what the object you're supposed to be proving things about is! $\endgroup$ Commented Jun 5, 2016 at 2:45
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    $\begingroup$ Another way to say this that you might like better: we want to represent $|\psi\rangle$ in the $|x\rangle$ basis. So $|\psi\rangle=\int \psi(x)|x\rangle dx$ for some coefficients we've named $\psi(x)$. From this definition of $\psi(x)$ you should be able to prove $\langle x|\psi\rangle=\psi(x)$. Again though, this includes a definition of $\psi(x)$. $\endgroup$ Commented Jun 5, 2016 at 2:48

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Well, by definition, $\psi(x)=\langle x | \psi\rangle$. So you have

$$ \begin{array}{rcl} \langle x'|\psi\rangle &=& \int\delta(x-x')\langle x | \psi\rangle dx\\ &=& \int \delta(x-x')\psi(x)dx\\ &=& \psi(x')\\ \end{array} $$ where in the second step I used the definition of $\psi(x)$, and in the third I used the defining property of the delta function.

Of course, you could have gotten the result in step one just by using the definition of $\psi(x')$, without ever introducing the integral. But I'll assume you wanted to do it a hard way for some reason.

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  • $\begingroup$ Ahh so then that means that $\langle x' | x'' \rangle = \int \delta(x-x') \delta (x-x'') dx = \delta (x'-x'')$. This makes a lot of sense, thankyou! $\endgroup$
    – Cococabana
    Commented Jun 5, 2016 at 2:48
  • $\begingroup$ @Cococabana Right, it's all self-consistent. Although of course, that's not how you PROVE $\langle x'|x''\rangle=\delta(x'-x'')$; that's again just a definition. But performing those manipulations shows the definition is well-defined. $\endgroup$ Commented Jun 5, 2016 at 2:51

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