Suppose I have a wavefunction into which I insert the completeness relation of some discrete basis as

$$ \psi_\alpha(x)=\langle x|\alpha\rangle=\sum_k \langle x|a_k\rangle\langle a_k|\alpha\rangle=\sum_k u_k(x) c_k~~. $$

The $u_k(x)$ are called the eigenfunctions of operator $\hat A$ if

$$ \hat A|a_k\rangle=a_k|a_k\rangle~~. $$

To show that the $u_k(x)$ are indeed eigenfunctions, I multiply the eigenket equation from the left with $\langle x|$ to get

$$ \langle x|\hat A|a_k\rangle=\langle x|a_k|a_k\rangle=a_k\langle x|a_k\rangle=a_k u_k(x)~~. $$

To show that $u_k(x)$ is properly an eigenfucntion, I need to bring $\langle x|$ to the right of $\hat A$ in the leftmost expression as

$$\langle x|\hat A|a_k\rangle=\hat A u_k(x)~~.$$

How do I know that I can do that?

  • 3
    $\begingroup$ I believe that for full clarity, you shouldn't write something like $\hat A u_k(x)$. $\hat A$ is an operator acting on the Hilbert space, but $u_k(x)$ is not an element of the Hilbert space (at least, not the one $\hat A$ acts on). $\endgroup$
    – anonymous
    Dec 28, 2020 at 4:20

2 Answers 2


You cannot do that because it is not correct!

What you can do instead is expressing the operator $\hat{A}$ in the complete basis of the position variable $x$: $$\langle x|\hat A|a_k\rangle= \int dx^\prime \langle x|\hat A|x^\prime\rangle \langle x^\prime|a_k\rangle = \int dx^\prime A(x,x^\prime) u_k(x^\prime)$$ where $A(x,x^\prime) := \langle x|\hat A|x^\prime\rangle$.

Combining this, with your derivation gives $$\int dx^\prime A(x,x^\prime) u_k(x^\prime) = a_k u_k(x)$$ which shows that $u_k(x)$ is an eigenfunction of the functional operator $\int dx^\prime A(x,x^\prime)$. This is really nothing but your original assumption $\hat A|a_k\rangle=a_k|a_k\rangle$ expressed in the basis of the position variable $x$.


The integration over $x^\prime$ may seem unfamiliar, but it is necessary for the general case. In the special case of local operators, the integration actually disappears because of an implicit delta function in $A(x,x^\prime)$. Take the case of momentum operator, for example, where $$ p(x,x^\prime) := \langle x|\hat p|x^\prime\rangle = -i\hbar \delta(x-x^\prime) {\frac {\partial }{\partial x}}$$ and so $$ \int dx^\prime p(x,x^\prime) = i\hbar {\frac {\partial }{\partial x}}$$


Since $Â$ is an operator, $Â| a_k \rangle = |(Â a_k) \rangle$ is a ket, so we have $\langle x| Â | a_k \rangle = \langle x| (Â a_k) \rangle:= (Â u_k) (x) $

  • $\begingroup$ What steps did you do where you have the $:=$ sign? How did you move $\langle x|$ past $\hat A$? $\endgroup$ Dec 28, 2020 at 3:01

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