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I can understand the position operator in Dirac notation: $$\langle x'|\hat x|x''\rangle = \langle x'| x''|x''\rangle = x''\langle x'|x''\rangle = x'' \delta(x'-x'').$$ $\hat x$ is the position operator and the above equation is given by the eigenequation $$\hat x|x''\rangle=x''|x''\rangle.$$ But how to calculate the potential operator $\langle x'|V(\hat x)|x''\rangle$? In one quantum mechanics textbook it says $$\langle x'|V(\hat x)|x''\rangle=V(x'')δ(x'-x'').$$ But how to prove it directly?

(The $x'$ and $x''$ are the ordinal numbers of column and row of the matrix form of the operator in position representation)

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I've got an idea but I don't know if it's right. At first it's easy to prove that a polynomial of operator x could be taken out of the braket: $\langle x'|\hat{x}^{n}|x''\rangle=(x'')^{n}\,\delta(x'-x'')$. And the potential V(x) could be expanded in terms of Taylor series. So it's obviously correct to take V(x) out of the braket directly.

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  • $\begingroup$ It is indeed. More generally, one defines any multiplicative operator $\hat{V}$ as $\hat{V}|x\rangle =V(x)|x\rangle $, where $V(x)$ is a function. $\endgroup$
    – Giorgio Comitini
    Commented Apr 13, 2020 at 14:04
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You want to have the property that acting $V$ on a vector $\psi$ just multiplies, so that in the x basis, the components of the new vector are $V(x)\psi(x)$:

$\langle x|V|\psi\rangle =V(x)\psi(x)$

Include an identity via $\int |x'\rangle \langle x'|dx'$ in the LHS:

$\int \langle x|V|x'\rangle \langle x'|\psi\rangle dx'=V(x)\psi(x)$

To satisfy this, you need $\langle x | V|x'\rangle =\delta(x-x')V(x')$. It doesn't matter whether you put the argument of V as $x$ or as $x'$ since the delta function is symmetric in those arguments. Then, noting that $\langle x'|\psi\rangle =\psi(x')$, we get

$\int V(x')\delta(x-x') \psi(x') dx'=V(x)\psi(x)$

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  • $\begingroup$ sorry I still don't understand. it seems that you just use the equation ⟨x|V|x′⟩=δ(x−x′)V(x′)，which is needed to be proved, as a condition to derive. $\endgroup$
    – Afish
    Commented Apr 13, 2020 at 13:15
  • $\begingroup$ I understand your concern, it depends whether you can trust that this is the only formula for $\langle x | V | x'\rangle$ which lets you satisfy that equation. That uniqueness was the point. For me that is intuitive - that putting any other function in that integral wouldn't just give you back just $V(x)$, the delta function is the unique function to do that. But if you want a rigorous proof, yes, that isn't in my answer. It's possible you can modify $V(x)$ on sets of points with measure 0, but such modifications are kind of pathological - physical functions tend to be continuous. $\endgroup$
    – don't train ai on me
    Commented Apr 13, 2020 at 14:32
  • $\begingroup$ I think complete rigor might also be difficult with the taylor series approach for any $V(x)$ which has a finite radius of convergence of its taylor series $\endgroup$
    – don't train ai on me
    Commented Apr 13, 2020 at 14:33
  • $\begingroup$ It might help in your intuition that "diagonalizing a matrix" in operator language is the same thing as "finding a basis in which is it just a multiplication operator". And a diagonal matrix is proportional to the identity matrix with components $\delta _{ij}$ in the same way that a diagonal operator is proportional to the identity operator $\delta(x-x')$ $\endgroup$
    – don't train ai on me
    Commented Apr 13, 2020 at 14:47

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