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What would be a good basis for a modified Hamiltonian that reads:

$$ H_1 = \frac{1}{2}(L_+S_- + L_-S_+) + c_1 L_x + c_2 S_x,$$

from a symmtry point of view?

The Hamiltonian itself is not difficult to diagonalize, but I want to know if there is any hidden symmetry that I am not aware of in this situation. Or what is an invariant subspace of this Hamiltonian, if it exists.

(As opposed to the normal spin-orbit coupled Hamiltonian:

$$ H_2 = L_z S_z + \frac{1}{2}(L_+S_- + L_-S_+) + c_1 L_x + c_2 S_x,$$

in which, the eigenkets of $L^2$, $S^2$, $J_x$, where $J= L+S$, forms a good basis, since $J_x$ are good quantum numbers for this problem.)

Examples:

Let's plot the eigenvalues of these two Hamiltonian for information.

(Left Panel) $$ H_1(t) = (1-t) \frac{1}{2}(L_+S_- + L_-S_+) + t L_x$$ (Right Panel) $$ H_2(t) = (1-t) (L_zS_z+\frac{1}{2}(L_+S_- + L_-S_+)) + t L_x $$

L=2, S=1/2.

enter image description here

L=3, S=1/2.

enter image description here

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  • $\begingroup$ If all you're looking for is hidden symmetries, then those can be ruled out easily by looking at the eigenspace degeneracies. If all the eigenspaces are non-degenerate, there can't be any nontrivial hidden symmetries. It's only a problem if you do have degenerate subspaces - does your numerical evidence point in that direction? $\endgroup$ – Emilio Pisanty Nov 14 '18 at 18:10
  • $\begingroup$ @EmilioPisanty, that's a good point. I added the numerical evidence here. There seems to be a different symmetry that plays a role. Do you know what this symmetry should be? $\endgroup$ – akpc Nov 14 '18 at 19:28
  • $\begingroup$ I only see (occasional) accidental degeneracies there. $\endgroup$ – Emilio Pisanty Nov 14 '18 at 19:36
  • $\begingroup$ @EmilioPisanty The question is about the invariant subspace. Maybe I should say it that way. also. In the L=3, S=1/2 (left), the E=0 line is degenerate. $\endgroup$ – akpc Nov 14 '18 at 20:02
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The new symmetry here is parity. Compared to $H_2$, $H_1$ is no longer symmetric under rotation. That is, the symmetric has been reduced to $\mathbb{Z}(2)$.

Therefore the invariant subspace here is the parity subspace.

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