Timeline for Expected value of $\hat{L}_x$, $\hat{L}_y$, $\hat{L}_z$ real or complex?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 17, 2022 at 12:06 | vote | accept | peter mafai | ||
| Jan 17, 2022 at 12:06 | |||||
| Jan 16, 2022 at 13:18 | comment | added | Bobsn | Ah I understand! No it's the other way around: $\hat{L}_{\pm}$ are not hermitian, because $\hat{L}_{y}$ is. $\hat{L}_{\pm}$ are abstract operators and are immeasurable, since they do not represent observables, while all angular momentum components do correspond to physical observables. For the ladder operators the following conjugation condition holds: \begin{align} \hat{L}_{+}^\dagger &= \hat{L}_{-} \\ \hat{L}_{-}^\dagger &= \hat{L}_{+}\,. \end{align} | |
| Jan 16, 2022 at 12:33 | comment | added | peter mafai | ahh. okay. i took the easy route. i said "because $L_\pm=L_x+iL_y$ is real it means that $L_y=0$". Okay but does that mean then that $L_y$ is an anti hermitian operator? And if yes: Is it not measureable ? or what does it physically mean that an operator is anti hermitian? | |
| Jan 16, 2022 at 12:03 | comment | added | Bobsn | And $\hat{L}_x = \frac{1}{2}(\hat{L}_{+} + \hat{L}_{-})$. So you forgot a factor of $\frac{1}{2}$ in your calculation, I think. | |
| Jan 16, 2022 at 12:01 | comment | added | Bobsn | I am a little confused that you get the zero operator for \begin{align} \hat{L}_y = -\frac{i}{2}(\hat{L}_+ - \hat{L}_-) \end{align} using your results you should get \begin{align} \hat{L}_y = -\frac{i}{2} \hbar \begin{pmatrix} 0 & \sqrt{2} & 0 \\ -\sqrt{2} & 0 & \sqrt{2} \\ 0 & -\sqrt{2} & 0 \end{pmatrix} \end{align} Moreover, I think you should reverse the order of rows in your matrices, since your first matrix element should e.g. look like $\langle Y_{1, -1} \vert \hat{L}_j \vert Y_{1,-1} \rangle$. But that may be just a choice of convention as long as you do it consistently. | |
| Jan 16, 2022 at 11:49 | comment | added | peter mafai | But then $[L_x,Ly]=L_xL_y-L_yL_x=0\neq i\hbar L_z$ since $$ L_z=\left(\begin{eqnarray} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1 \end{eqnarray}\right)\hbar $$ | |
| Jan 16, 2022 at 11:45 | comment | added | peter mafai | So as a result for the ladder operators i get $$ L_+=\left(\begin{eqnarray} 0 & \sqrt2 & 0\\ 0 & 0 & \sqrt2\\ 0 & 0 & 0 \end{eqnarray}\right)\hbar $$ $$ L_-=\left(\begin{eqnarray} 0 & 0 & 0\\ \sqrt2 & 0 & 0\\ 0 & \sqrt2 & 0 \end{eqnarray}\right)\hbar $$ Then with $L_\pm=L_x\pm iL_y$ i get $$ L_y=\left(\begin{eqnarray} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{eqnarray}\right) $$ and $$ L_x=\left(\begin{eqnarray} 0 & \sqrt2 & 0\\ \sqrt2 & 0 & \sqrt2\\ 0 & \sqrt2 & 0 \end{eqnarray}\right)\hbar $$ is that right ? | |
| Jan 16, 2022 at 11:26 | history | edited | Bobsn | CC BY-SA 4.0 |
added 6 characters in body
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| S Jan 16, 2022 at 11:17 | review | First answers | |||
| Jan 16, 2022 at 13:00 | |||||
| S Jan 16, 2022 at 11:17 | history | answered | Bobsn | CC BY-SA 4.0 |