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

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

  • $\begingroup$ 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

$\vert \phi \rangle = A\vert \psi \rangle$

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

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')$

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.