Timeline for Why $\langle x|p \rangle$ is momentum eigenstate of which eigenvalue is $p$ at position basis
Current License: CC BY-SA 4.0
7 events
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| Sep 16, 2023 at 13:59 | comment | added | Himanshu | (1) The form of the matrix element of $P$ in position basis is sometimes considered as a Postulate of QM. See R. Shankar Chapter 4. But you can derive them assuming the commutation relation between $X$ and $P$. (2) In $\langle x|p\rangle $, $x$ and $p$ are assumed to be continuous variables; therefore, the inner product is written as a function. | |
| Sep 6, 2023 at 1:42 | comment | added | Plantation | @ Young Kindaichi : Can I ask more? 1) Why $\langle x | P | x' \rangle = -i \hbar \frac{\partial}{\partial x} \langle x| x' \rangle = -i \hbar \frac{\partial}{\partial x} \delta(x-x')$? Can the momentum operator $P$ be pulled outside from braket ? Why? 2) Just looking, the notation $\langle x | p \rangle$ seems to mean inner product of position bra vector $\langle x | $ and momentum ket vector $| p \rangle$. And I think that inner product is number. And how can we set $\langle x | p \rangle$ as a function $\psi_{p}(x)$? | |
| Sep 5, 2023 at 19:25 | comment | added | Himanshu | @Plantation Please look into the properties of the Dirac delta function. In particular, try integration by part. | |
| Aug 26, 2023 at 8:18 | comment | added | Plantation | Can I ask question? In the note, why the second and third equalities are true? | |
| Apr 12, 2021 at 13:02 | vote | accept | XX X | ||
| Apr 12, 2021 at 13:02 | vote | accept | XX X | ||
| Apr 12, 2021 at 13:02 | |||||
| Apr 9, 2021 at 8:15 | history | answered | Himanshu | CC BY-SA 4.0 |