Timeline for What exactly is the definition of the representation of an operator in position or momentum space?
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| Sep 18, 2023 at 11:06 | comment | added | Abdul Qadeer | And thanks for the second comment, I think that clears up what position representation means. | |
| Sep 18, 2023 at 11:02 | comment | added | Abdul Qadeer | @TobiasFünke in regards to the first comment, what I meant to say is that if $V$ is some arbitrary operator, not necessarily represented as a function of the position operator $x$, and we wanted to find out what it looks like in position space, eqn. (1) was what first came to mind (as in ij-th matrix elements of the matrix V in position states), but I see that it's not right because elements of an operator are not the operator itself. In regards to $\langle x| V(X) |x' \rangle = V(x) \langle x|x' \rangle$, could we say that $\langle x| V(P) |x' \rangle = V(-ih d/dx) \langle x|x' \rangle$? | |
| Sep 18, 2023 at 9:53 | comment | added | Tobias Fünke | And yes, roughly speaking: This induces what colloquially is called the position representation. For an operator $V$ on an abstract space $H$, you can define an operator $\tilde V$ on $L^2(\mathbb R)$ via $\langle x|V|\psi\rangle=\int \mathrm dx\, V(x,x^\prime)\psi(x^\prime):=(\tilde V\psi)(x)$ with $V(x,x^\prime):= \langle x|V|x^\prime\rangle$. So if $V$ maps $|\psi\rangle$ to $|\phi\rangle$, then $\tilde V$ maps $\psi \in L^2$ to $\phi\in L^2$. So any vector in $H$ is in a one-to-one correspondence with a function in $L^2$, and likewise for operators on those spaces (discarding rigor). | |
| Sep 18, 2023 at 9:49 | comment | added | Tobias Fünke | Eq. $(1)$ does not make sense and your notation seems also confused. If you have a position-operator (X) dependent potential $V(X)$, then $\langle x|V(X)|\psi\rangle=V(x)\langle x| \psi\rangle =V(x) \psi(x)$, by the very definition of functions of operators and the convention/definition $\psi(x):=\langle x|\psi\rangle$. If $|\psi\rangle=|x^\prime\rangle$, then you find $\langle x|V(X)|x^\prime\rangle=V(x) \langle x|x^\prime\rangle = V(x)\delta(x-x^\prime)$. Does that clear up a bit the confusion? | |
| Sep 18, 2023 at 9:20 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
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| Sep 18, 2023 at 9:05 | history | asked | Abdul Qadeer | CC BY-SA 4.0 |