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Given

$$ | \phi \rangle = A | \psi \rangle $$

I'm trying to show that

$$ \phi (x) = \int_{-\infty}^\infty A(x,x')\psi(x')dx' $$

I've used the relation between the states and the wavefunction to obtain

$$ | \phi \rangle = \int_{-\infty}^\infty \langle x | A| \psi \rangle | x \rangle dx $$

By using the identity relation I get

$$ | \phi \rangle = \int_{-\infty}^\infty\int_{-\infty}^\infty \langle x | A| x' \rangle\langle x' | \psi \rangle | x \rangle dx' dx = \int_{-\infty}^\infty\int_{-\infty}^\infty A(x,x')\psi(x') | x \rangle dx' dx $$

It seems the step I'm missing is multiplying by a $\langle x |$ on both sides of the equation, but I'm not sure I see how it's legal to bring the $\langle x |$ into the integral

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  • $\begingroup$ It may be useful to point out that not every operator admits such a representation with a functional kernel $A(x,x')$. If the operator is Hilbert-Schmidt, then the kernel is an $L^2$ function (in both variables). If else, it may be a distribution (think e.g. of the identity operator, whose integral kernel is a delta). $\endgroup$ – yuggib May 3 '17 at 7:13
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$\vert \phi \rangle = A\vert \psi \rangle$

$\phi(x) = \langle x \vert \phi \rangle$

Inserting completeness relation in the first equation, we get:

$\int dx' \vert x' \rangle \langle x' \vert \phi \rangle = \int dx' \vert x' \rangle \langle x' \vert A \vert \psi\rangle$

Inserting completeness relation again the the RHS of the above equation, we get:

$\int dx' \vert x' \rangle \langle x' \vert \phi \rangle = \int dx' dx'' \vert x' \rangle \langle x' \vert A \vert x'' \rangle \langle x'' \vert \psi \rangle$, which is just:

$\int dx' \vert x' \rangle \phi(x') = \int dx' dx'' \vert x' \rangle \langle x' \vert A \vert x'' \rangle \psi(x'')$

Now act $\langle x''' \vert$ on both sides of the equation from the left. It is legal to do so, and since it is a bra, it will act on a ket, which is nothing but $\vert x' \rangle$:

$\int dx' \langle x'''\vert x' \rangle \phi(x') = \int dx' dx'' \langle x'''\vert x' \rangle \langle x' \vert A \vert x'' \rangle \psi(x'')$, which gives:

$\int dx' \delta(x''' - x') \phi(x') = \int dx' dx'' \delta(x''' - x') \langle x' \vert A \vert x'' \rangle \psi(x'')$

$\Rightarrow \phi(x''') = \int dx'' A(x''', x'') \psi(x'')$

Now just relabel x''' $\rightarrow$ x, and $x'' \rightarrow x'$ to get the final expression:

$ \phi(x) = \int dx' A(x, x') \psi(x')$

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  • $\begingroup$ You can make the calculation much easier if in the first step you don't insert the completeness relation $\int dx' |x' \rangle\!\langle x'|$, but you just act with $\langle x' |$ from the left. Then you won't need the whole ${{x'}'}'$ step. (Also, you could use the correct labels $x$ and $x'$ in the right places from the start.) $\endgroup$ – Noiralef May 3 '17 at 7:38
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    $\begingroup$ Absolutely. The derivation can be done in one line. The OP had already used the completeness relation in their definition of $\vert \phi \rangle$, so I did the same. They were confused about the legality of a bra acting on an integral. I just wanted to showcase the cute tricks of the Dirac delta function. @user201756 - In the first step, you can omit the insertion of the completeness relation on the LHS. Just act on it by a bra, like I did anyway later. $\endgroup$ – Avantgarde May 3 '17 at 7:55

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