How to derive momentum operator in position basis via its definition?

[LightPhoton]

Let the representation of a Hermitian Operator in some basis $x$ be
$$\hat A\equiv A(x)$$
and let
$$\psi(x)=\langle x\vert\psi\rangle$$

Then we define

$$A(x)\,\psi(x)=\langle x\vert\hat A\vert\psi\rangle$$

This is the Wikipedia article that mentions this.

From here how do we derive the momentum operator (or any other) on a position basis?

 

[PeterDonis]

From here how do we derive the momentum operator (or any other) on a position basis?

You don't. The information given is not enough to specify any particular operator. All it is telling you is general properties that any operator will satisfy in the position basis.

 

[PeterDonis]

how do we derive the momentum operator

Generally representations for specific operators are derived by looking at the physical properties that the operators are expected to satisfy as observables. See, for example, Chapters 3 (particularly section 3.4) and 4 (particularly section 4.1) of Ballentine.

 

[pines-demon]

Let the representation of a Hermitian Operator in some basis $x$ be
$$\hat A\equiv A(x)$$
and let
$$\psi(x)=\langle x\vert\psi\rangle$$

Then we define

$$A(x)\,\psi(x)=\langle x\vert\hat A\vert\psi\rangle$$

You are missing a key ingredients here, you are missing the
$$\langle x|\hat{A}|\psi\rangle=\int\mathrm dx' \langle x|\hat{A}|x'\rangle\langle x'|\psi\rangle=\int\mathrm dx' \langle x|\hat{A}|x'\rangle\psi(x')$$
If $\hat{A}$ has a nice way to represent it in position space then you can write $\hat{A}|x\rangle=A(x)|x\rangle$ so you get
$$\langle x|\hat{A}|\psi\rangle=\int\mathrm dx' \langle x|x'\rangle A(x')\psi(x')=\int \mathrm d x' \delta(x-x')A(x')\psi(x')=A(x)\psi(x)$$

Edit: this is important if you want to derive the momentum operator. The crucial step consist in introducing a momentum base between $\hat{A}$ and $|\psi\rangle$ but then you would also need to know what is $\langle p|x\rangle$.

 

[LightPhoton]

You are missing a key ingredients here, you are missing the
$$\langle x|\hat{A}|\psi\rangle=\int\mathrm dx' \langle x|\hat{A}|x'\rangle\langle x'|\psi\rangle=\int\mathrm dx' \langle x|\hat{A}|x'\rangle\psi(x')$$
If $\hat{A}$ has a nice way to represent it in position space then you can write $\hat{A}|x\rangle=A(x)|x\rangle$ so you get
$$\langle x|\hat{A}|\psi\rangle=\int\mathrm dx' \langle x|x'\rangle A(x')\psi(x')=\int \mathrm d x' \delta(x-x')A(x')\psi(x')=A(x)\psi(x)$$

Edit: this is important if you want to derive the momentum operator. The crucial step consist in introducing a momentum base between $\hat{A}$ and $|\psi\rangle$ but then you would also need to know what is $\langle p|x\rangle$.

Thanks, really appreciate the derivation!

 

[LightPhoton]

$\hat{A}|x\rangle=A(x)|x\rangle$

Actually, now that I think about it, what does this equation even mean? We are acting an operator in position representation (let's say $\hat p=-i\hbar\partial_x$) on a ket which is in Hilbert space.

 

[pines-demon]

Actually, now that I think about it, what does this equation even mean? We are acting an operator in position representation (let's say $\hat p=-i\hbar\partial_x$) on a ket which is in Hilbert space.

Ah you see that's why I am saying that it makes sense only if the operator behaves well with the basis. It is not so clear what $\hat{P}|x\rangle$ does or if it even makes sense but in momentum basis it does the following $\hat{P}|p\rangle=p|p\rangle$