Can I define $\hat{x}$ as $x\delta(x'-x)$?

Question

Operators can be thought of as matrices. Since matrices have two indices and involve summing:

$A^i_j \psi^j\equiv A^i_0\psi^0 + A^i_1\psi^1 + A^i_2\psi^2 \dots$

and a summation between 2 vectors turns into an integral in the continuum:

$\psi_i \psi^i \to \int dx \psi^*(x) \psi(x)$

The usual definition of the position operator:

$$\hat{x}\psi(x) = x\psi(x)$$

involves neither summing, nor has two indices.

I discovered this definition $\hat{x}\equiv \delta(x'-x)x$

such that:

$$\hat{x}\psi(x) = \int dx \delta(x'-x)x\psi(x) = x'\psi(x')$$

which involves summing and has two indices $x, x'$, and it reduces to the original definition. Does this appear anywhere? How would one define momentum operator then? I know it's needlessly complicated, but it helps me grasp some other concepts.

The integration has to be over $x'$, the result should be $x\psi(x)$, not $x'\psi(x')$ since $x'$ is auxiliary integration variable. — Ján Lalinský, Aug 14 '22 at 16:04
I.e. $\hat{x}\psi = \int dx' x'\delta(x'-x)\psi(x') = x\psi(x)$. — Ján Lalinský, Aug 14 '22 at 16:05
It doesn't matter it's a variable name, but I believe mine is accurate because similar to how vectors change indices: $\psi^i \to A^j_i \psi^i = \psi'^j$. — Habouz, Aug 14 '22 at 16:15
You should have an $x'$ rather than an $x$ on the LHS: $(\hat x \psi)(x')$. You are integrating over $x$, so the output of the integral cannot depend on $x$. — mike stone, Aug 14 '22 at 16:50
@Habouz it does matter, you wrote an expression that is not true except when $x=x'$. One way to fix this error is as I have described, the other one is as mike has described. — Ján Lalinský, Aug 14 '22 at 18:24
Related : My answer here Hermiticity of Momentum Operator (matrix) Represented in Position Basis. See equations (12)-(15). — Frobenius, Aug 14 '22 at 18:45

Can I define $\hat{x}$ as $x\delta(x'-x)$?

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