Building eigenfunctions from eigenkets

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?

• 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). Commented Dec 28, 2020 at 4:20

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 $$x$$.

Note:

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 $$\langle x| Â | a_k \rangle = \langle x| (Â a_k) \rangle:= (Â u_k) (x)$$

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