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

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

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  • $\begingroup$ 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. $\endgroup$ – jacob1729 Jun 2 at 19:52
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No, it's only true for operators that are diagonal in the position basis. (This follows trivially: your equation is in fact the definition of "being diagonal in the position basis".)

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 $M(x)$.

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  • $\begingroup$ I edited my question a bit , can u clear my concept now please $\endgroup$ – Robin Raj Jun 2 at 19:33
  • $\begingroup$ do you see my edits@tparker $\endgroup$ – Robin Raj Jun 3 at 4:01
  • $\begingroup$ I asked same question in Mathematics stalk exchange they answered differently , its link is math.stackexchange.com/questions/3249163/… @tparker $\endgroup$ – Robin Raj Jun 4 at 3:51

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