# Position representation of an operator

$$\langle\ x\rvert M\lvert\ x'\rangle=M(x)\langle\ x\lvert\ x'\rangle=M(x)\delta(x-x')$$

I know this is true for if $$M$$ is a momentum operator or position operator, is this is true for a general operator $$M$$?

$$\langle\ x'\rvert M\lvert\psi\rangle=\langle\ x'\lvert\alpha\rangle=\alpha(x')$$

this is equivalent to

$$\int dx \langle\ x'\rvert M\lvert\ x\rangle\langle\ x\lvert \psi\rangle$$

If equation 1 is true we can write this as

$$\int dx \langle\ x'\rvert M\lvert\ x\rangle\langle\ x\lvert \psi\rangle=\int dx M(x)\delta(x-x') \psi(x)=M(x') \psi(x')=\alpha(x')$$

equation one is not true imples equation 2 is also not true, but actually equation 2 is true in fact this implies equation one is also true

hear by M(x') I mean operator operates on position representation of wavefunction , please clarify me where I am wrong, please help me to clear my concept

• If by $\lvert \alpha \rangle$ you mean $M\lvert \psi \rangle$ then it is not true generally that $M(x)\psi(x)=\alpha(x)$. It might help to think about the more familiar finite dimensional case where $M$ becomes a matrix and $\psi$ a column vector. – jacob1729 Jun 2 at 19:52

Also, your equation does not hold for the momentum operator, whose position-basis matrix elements are $$\langle x | \hat{P} | x' \rangle = i \hbar \delta'(x - x')$$, not $$M(x) \delta(x - x')$$ for any function $$M(x)$$.