# How do operators on kets and wavefunctions correspond?

Let $$\hat{A}$$ be an operator on Hilbert space vectors. How does one show that there always exists a corresponding operator $$\hat{a}$$ on wave functions? i.e. $$\exists \hat{a}:L^2\rightarrow L^2$$ s.t. $$\left=\hat{a}\left$$ for square integral functions $$L^2$$. Is it by showing that this holds for position and momentum operators, and then have the result follow from any observable having to be a function of these?

• I dont understand why this is closed. Cant vote to open though. But what prevents you from constructing the rhs by definition? Even if you want a to be linear, just check for linearity and you are done. Commented Jul 19 at 11:54
• Is this always true? Any restrictions on the rank of $A$? Commented Jul 19 at 12:32
• @MariusLadegårdMeyer In the language of physicists: If you associate to the ket $|\psi\rangle$ the wave function $\psi$, then you can associate to the ket/state $A|\psi\rangle=|\tilde \psi\rangle$ the wave function $\tilde \psi$, which implicitly defines an operator which maps $\psi\mapsto \tilde \psi$. Commented Jul 19 at 12:45
• You cannot extract the integral operator out of $\left<x|\hat{A}|\psi\right>= \int \!\! dy \langle x|\hat A|y\rangle \langle y|\psi\rangle$ ? Commented Jul 19 at 13:25
• Cosmas is giving the full answer, but I think it is important to hammer the answer down. What you are supposed to do, is to insert the resolution of the identity operator in the following way, and then discover that $a=\left<x\right|\hat A\left|y\right>$ always exists in the following: $$\left<x\right|\hat A\left|\psi\right>=\left<x\right|\hat A\hat{\mathbb I}\left|\psi\right>=\int\left<x\right|\hat A\left|y\right>\mathrm dy\left<y|\psi\right>$$ Commented Jul 19 at 13:46

$$\left = \left\\ = \int \!\! dy ~\langle x|\hat A|y\rangle \langle y|\psi\rangle =\hat{a} \left.$$
Many write the position representation (chosen by your leftmost bra, $$\langle x|$$) of the operator as $$\hat a _x$$ to underscore this, so a routine integral operator, $$\hat a _x \psi (x) \equiv \int \!\! dy \langle x|\hat A|y\rangle ~~~\psi (y).$$ For example, for $$\hat A=\hat p$$, you get $$\hat a_x = -i\hbar\partial_x$$.
Conversely, you may, of course, express $$\hat A$$ in terms of $$\hat a_x$$, $$\hat A= \int\!\! dx ~|x\rangle \hat a_x \langle x| ~.$$