What is $langle x_1 |hat V(hat x)| x_2 rangle$?

So I’ve become rusty in Quantum Mechanics. What is $langle x_1 |V(hat x)| x_2 rangle$? Where $V$ is the potential and $|x rangle$ is the postion eigenket?
$$ langle x_1 |hat V(hat x)| x_2 rangle = ? $$

Is it with $V(x_1) delta(x_1 -x_2)$ but since the position operator can also act on a bra is it $V(x_2) delta(x_1 -x_2)$ or some combination of $x_1$, $x_2$ like $V(x_1,x_2) delta(x_1 -x_2)$? If it does not make any difference can you prove so? ( I was doing a calculation where it seemed to make a difference).

Thanks

Calculation where is matters

The following is a snippet of what I was doing.

Let the Hamiltonian $hat H$ of the particle be:

begin{equation}
hat H = hat T + hat V
end{equation}

where $hat T$ is the kinetic energy and $hat V$ is the potential energy. Now, to find the velocity:

begin{equation}
hat v = frac{-i}{hbar}[hat H , hat x] = frac{-i}{hbar} [ hat T , hat x] = hat T'(hat p)
end{equation}

where $hat T'(hat p)$ is the velocity which is a function of momentum. Now, if we further assume $T'(p)$ is of degree $1$ and find the acceleration $hat a$:

begin{equation}
hat a = frac{-i}{hbar} [ hat H , hat T'(hat p) ] =frac{-i}{hbar} [hat V,hat T'( hat p )] = hat a (hat x)
end{equation}

We know that acceleration must be a function of $hat x$ since we have already assumed $T'(p)$ is of degree $1$ and find the acceleration $hat a$.
Hence,

begin{equation}
[hat a, hat x] = 0
end{equation}

Now, multiplying $| x rangle$ on the $hat a$ equation:

begin{equation}
langle x ‘ | hat a | x rangle = langle x ‘ | [hat V,hat T'( hat p )] | x rangle
end{equation}

Using the eigenvalue equation with eigenvalue $a$ (which represents the acceleration at a position $| x rangle $ ):

begin{equation}
a langle x ‘ | x rangle= a delta( x’ – x) = frac{-i}{hbar} langle x ‘ | (hat V hat T'( hat p ) – hat T'( hat p ) hat V) | x rangle = frac{-i}{hbar} (V(x’) – V(x)) langle x’ | hat T'( hat p ) | x rangle
end{equation}

Dividing both:

begin{equation}
frac{i hbar a}{(V(x’) – V(x)) } delta( x’ – x) = langle x’ | hat T'( hat p ) | x rangle
end{equation}

Let us consider $langle x’ | hat T'( hat p ) | psi rangle$:

begin{equation}
langle x’ | hat T'( hat p ) | psi rangle = int_{-infty}^infty langle x’ | hat T'( hat p ) | x rangle langle x| psi rangle dx = int_{- infty}^infty frac{i hbar a}{(V(x’) – V(x)) } delta( x’ – x) psi(x)dx
end{equation}

Or:

begin{equation}
langle x’ | hat v | psi rangle = int_{- infty}^infty frac{i hbar a}{(V(x’) – V(x)) } delta( x’ – x) psi(x) dx
end{equation}

Due to the quantization condition of position and momentum:

begin{equation}
frac{- i hbar}{m }frac{partial }{ partial x’} psi (x’) = int_{- infty}^infty frac{ a i hbar}{(V(x’) – V(x)) } delta( x’ – x) psi(x) dx
end{equation}

OR:

begin{equation}
frac{partial }{ partial x’} psi (x’) = int_{- infty}^infty frac{ – m a }{(V(x’) – V(x)) } delta( x’ – x) psi(x) dx
end{equation}

Let us now try $V(x) = – G frac{m M}{x}$ (point particle potential):

begin{equation}
frac{partial }{ partial x ‘} psi (x’) = int_{- infty}^infty frac{- a}{(-frac{GM}{x’} + frac{GM}{x}) } delta( x’ – x) psi(x) dx
end{equation}

Simplifying:

begin{equation}
GM frac{partial }{ partial x ‘} psi (x’) = int_{- infty}^infty frac{ a x x’}{(x’ – x) } delta( x’ – x) psi(x) dx
end{equation}

Again Taylor expanding around $psi (x)$ around $x’$:

begin{equation}
GM frac{partial }{ partial x ‘} psi (x’) = int_{- infty}^infty frac{ a x x’}{(x’ – x) } delta( x’ – x) (psi(x’) + (x-x’) partial_{x’} psi(x’) + dots) dx
end{equation}

Plugging in the ansatz $a = – frac{GM}{x x’}$ we get:

begin{equation}
GM frac{partial }{ partial x ‘} psi (x’) = int_{- infty}^infty – frac{ GM}{(x’ – x) } delta( x’ – x) (psi(x’) + (x-x’) partial_{x’} psi(x’) + dots) dx
end{equation}

Note the term $int_{- infty}^infty frac{ GM}{(x’ – x) } delta( x’ – x) psi(x’) $ goes to $0$ since is antisymmetric.

begin{equation}
GM frac{partial }{ partial x ‘} psi (x’) = GM frac{partial }{ partial x ‘} psi (x’)
end{equation}

Hence, we get a consist solution which agrees with classical calculations.
Note: $a = – GM{x^{-2}}$ or $a = – GM{x’^{-2}}$ both will give wrong answers.