# Simple Question about Dirac Notation

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!

• Are you sure you want to obtain $\psi(x)$ rather than $\psi(x')$?
– TLDR
Jun 5, 2016 at 2:17
• Sorry, my mistake! Have fixed it, Jun 5, 2016 at 2:20
• 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? Jun 5, 2016 at 2:22
• 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! Jun 5, 2016 at 2:45
• 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 $\psi(x)$. Jun 5, 2016 at 2:48

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.
• 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! Jun 5, 2016 at 2:48
• @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. Jun 5, 2016 at 2:51