Is it possible to find out the generic eigenvalue spectrum of the non-Hermitian operator $L_x+iL_y$, without using any representation?
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2 Answers
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I am assuming that $(L_x,L_y,L_z)$ satisfy the usual angular momentum algebra. Then, one knows that all states can be labelled by the eigenvalues of $(L^2,L_z)$. Pick such a state and label it as $|j,m\rangle$ where the $L^2$ eigenvalue is $j(j+1)$ and the $L_z$ eigenvalue is $m$. Note that we have not made any assumption about the allowed values of $(j,m)$. The action of $L_\pm$ produces states $|j,m\pm1\rangle$. This can be continued to get an infinite set of states. Several possibilities occur:
- The process of raising using $L_+$ terminates at some state -- call it the highest weight state (the norm of $L_+$ on the highest weight state is zero) .
- The process of lowering using $L_-$ terminates at some state -- call it the lowest weight(the norm of $L_-$ on the lowest weight state is zero).
Recall that the groups $SU(2)$ and $SL(2,\mathbb{R})$ arise from the same Lie algebra (over $\mathbb{C}$). So the above construction holds for both of them. Unitarity (each state in the irrep has positive norm) decides on the allowed values of $(j,m)$. For $SU(2)$, all unitary irreps have an highest and lowest weight state. However, for $SL(2,\mathbb{R})$, the unitary irreps are different. In particular, they are infinite dimensional. For instance, one could have a highest weight state but no lowest weight state. You can look up unitary irreps of $SL(2,\mathbb{R})$ to see all the cases. See here for example: https://en.wikipedia.org/wiki/Representation_theory_of_SL2%28R%29
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3$\begingroup$ 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. $\endgroup$ Commented Feb 18, 2014 at 2:25
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$\begingroup$ 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 $L_+$? $\endgroup$– sureshCommented Feb 18, 2014 at 2:33
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1$\begingroup$ 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. $\endgroup$ Commented Feb 18, 2014 at 2:36
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$\begingroup$ Then, the only possible eigenvalue is zero and any highest weight state is an eigenvector. $\endgroup$– sureshCommented Feb 18, 2014 at 2:43
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$\begingroup$ @joshphysics- Thanks. But how did you conclude that all the eigenvalues will be zero? Is it trivial to guess? $\endgroup$– SRSCommented Feb 18, 2014 at 5:01
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I'm not sure this is a complete answer, but here's my stab at it:
Looking for the eigenstates and eigenvalues of $L_-$.
These might be something analogous to to the coherent states of quantum optics: wiki
Where $L_-$ takes the role of the annihilation operator, and the Fock states become the eigenstates of $L_z$. I'm still not sure what role the total angular momentum plays here, or the physical interpretation of such a sate. You could construct such something close to such a sate from a superposition of $m$ values for a very high $n$ Rydberg state.
Still think I'm missing something though
