# Why are there no potential operators that are non-diagonal in position basis?

The potential energy operator in the Hamilton operator can be expressed in the following way \begin{aligned} \hat V &=\int dx\int dx'|x \rangle\langle x|\hat V|x'\rangle \langle x'| \\ &=\int dx\int dx'|x \rangle V(x,x') \langle x'| \, .\\ \end{aligned} Why is it legit to assume that the potential has the follwing form $$V(x,x') = \delta(x-x')\tilde V(x)$$ effectively removing the dependency on the second variable leading to the common form $$\hat V =\int dx \ |x\rangle \tilde V(x) \langle x| \, .$$ Are there potentials that are non diagonal in position basis or are all physically meaningful potentials diagonal ?

I am mostly wondering because I started with a general expression of the form $$\langle \Psi|\hat H|\Psi\rangle =\int\int dx dx' \langle \Psi|x\rangle \langle x|\hat H |x'\rangle \langle x'|\Psi\rangle \, ,$$ and then inserting $$\hat H = \hat T + \hat V$$ leads then to terms as described above for the potential, but all Hamiltonians I have seen so far in position basis are assumed to have only a $$V(x)$$ term implying the diagonal form.

• Possible search term: "velocity dependent potential"? – dmckee Jun 18 at 19:53
• A potential that depends on more than one position is basically a particle/field interacting with itself, which can happen especially if you use math tricks to get rid of the part of the Hamiltonian representing other particles/fields. – DanielSank Jun 18 at 20:30
• You could have nonlocal interactions. So far there is no experimental evidence for such interactions. Nonlocal interactions can give you off-diagonal terms. (I do not think that velocity-dependent terms do that per se @dmckee.) – user178876 Jun 18 at 20:48