Operators and the position representation

[tom.fay]

I have a question about the formalism of quantum mechanics. For some operator A...

$\langle x |A|\psi\rangle = A\langle x | \psi \rangle$

Can this be derived by sticking identity operators in or is it more a definition/postulate.

Thanks.

 

[The_Duck]

Your right-hand side doesn't really make sense; operators act on states but on the right-hand side A is trying to act on a number. I think what you want is

$\langle x | A | \psi \rangle = \int dx' \langle x | A | x' \rangle \langle x' | \psi \rangle$

The left hand side is the position-space wave function of the state $A | \psi \rangle$ which the right-hand side gives in terms of the position-space wave function of the state $| \psi \rangle$ and the position-space matrix elements of the operator A.

For example, you can derive the position-space matrix elements of the momentum operator $p$ from the canonical commutation relation $[x, p] = i\hbar$. You find

$\langle x | p | x' \rangle = i \hbar \delta ' (x' - x)$

Where $\delta'(x)$ is the derivative of the Dirac delta function. Plugging this into the above formula you get

$\langle x | p | \psi \rangle = \int dx' i \hbar \delta ' (x' - x) \langle x' | psi \rangle = -i \hbar \int dx' \delta(x' - x) \frac{d}{dx'} \langle x' | \psi \rangle = -i \hbar \frac{d}{dx} \langle x | \psi \rangle$

which tells you that the position space wave function of $p | \psi \rangle$ is -i hbar times the derivative of the position space wave function of $| \psi \rangle$

 

[tom.fay]

Aaah that makes a lot more sense. Thank you.