Timeline for Eigenvalue spectrum of $L_x+iL_y$
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 18, 2014 at 7:02 | comment | added | suresh | Use the commutator $[L_z,L_+]$ to show that $L_+$ maps any state to an orthogonal state. So it has no diagonal entries and its matrix representation will be upper triangular with zero on the diagonals. | |
| Feb 18, 2014 at 5:01 | comment | added | SRS | @joshphysics- Thanks. But how did you conclude that all the eigenvalues will be zero? Is it trivial to guess? | |
| Feb 18, 2014 at 2:43 | comment | added | suresh | Then, the only possible eigenvalue is zero and any highest weight state is an eigenvector. | |
| Feb 18, 2014 at 2:36 | comment | added | joshphysics | I think he's simply asking "given a representation $\rho$ of $\mathfrak{sl}(2,\mathbb C)$ on a vector space $V$, what are the eigenvalues of $\rho(L_+)$?" (but I could be wrong). In this case, I'm fairly certain that all of the eigenvalues are zero. | |
| Feb 18, 2014 at 2:33 | comment | added | suresh | Yeah, I am answering something totally different. Thanks.:-) @roopam: What is the meaning of a non-hermitian operator if there is no vector/Hilbert space on which it can act? What defines $L_+$? | |
| Feb 18, 2014 at 2:25 | comment | added | joshphysics | I think the OP is asking "What, if any, are the eigenvalues of L_+?" So, it's not clear to me how this answers the question. | |
| Feb 18, 2014 at 1:39 | history | edited | suresh | CC BY-SA 3.0 |
added 88 characters in body
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| Feb 18, 2014 at 0:52 | history | answered | suresh | CC BY-SA 3.0 |