Timeline for Expectation value of $ Y \otimes I \otimes I $ for a charged particle in a magnetic field
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14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Dec 20, 2021 at 23:12 | history | bounty ended | anon.jpg | ||
| S Dec 20, 2021 at 23:12 | history | notice removed | anon.jpg | ||
| Dec 19, 2021 at 20:59 | vote | accept | anon.jpg | ||
| Dec 19, 2021 at 20:54 | answer | added | d_b | timeline score: 1 | |
| Dec 19, 2021 at 20:16 | comment | added | anon.jpg | @d_b I think my confusion either came from that or from some misconception that they commuted because they were in different spaces... even though $ d/dy $ clearly commutes with $ d/dx $ anyway. Wow, now a lot has clicked. Now the construction of $ L_i $ makes more sense. if you write this as an answer I'll accept it and give u your reputation. thanks chief!!! | |
| Dec 19, 2021 at 20:13 | comment | added | d_b | I think you are mixing up the Euclidean unit vectors for the different components of $\vec{r}$, $\vec{p}$, etc. with tensor factors. | |
| Dec 19, 2021 at 20:11 | comment | added | d_b | That is incorrect. $\hat{p}_x$ is just $\hat{p}_x$, not tensored with anything. The hamiltonian is also not a tensor product, it is just an ordinary sum $(\hat{p}_x - e\hat{A}_x/c)^2 + (\hat{p}_y - e \hat{A}_y /c)^2 + \ldots$ | |
| Dec 19, 2021 at 20:05 | comment | added | anon.jpg | @d_b thank you for your response. Can you expand on this? For example, $ P_x $ is really $ P_x \otimes I \otimes I $. I also know that the entire Hamiltonian cannot be written as something like $ (P_x - \frac{q}{c} A_x)^2 \otimes (P_y - \frac{q}{c} A_y)^2 .... $ and so on, if this is what you mean. | |
| Dec 19, 2021 at 20:00 | comment | added | d_b | The different coordinates in the position basis don't correspond to different tensor product factors. That is, $\mathcal{H} \neq \mathcal{H}_x \otimes \mathcal{H}_y \otimes \mathcal{H}_z$. There is just one Hilbert space, and the states $|x, y, z\rangle$ form a basis. So there is no $\psi(x)$, $\psi(y)$, etc., just $\psi(x, y, z) = \langle x, y, z| \psi \rangle$ | |
| S Dec 19, 2021 at 19:44 | history | bounty started | anon.jpg | ||
| S Dec 19, 2021 at 19:44 | history | notice added | anon.jpg | Draw attention | |
| Dec 17, 2021 at 6:09 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
edited tags; edited title
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| Dec 17, 2021 at 5:35 | comment | added | anon.jpg | Ok, I already know I'm at least slightly off because even in going to the position basis, I assume that $ \langle x | Y = y \langle x | $. | |
| Dec 17, 2021 at 5:20 | history | asked | anon.jpg | CC BY-SA 4.0 |