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This is a very good question. The same operator algebra does not imply that $H(J,h)$ and $H(h,J)$ have the same spectrum. As has been mentioned in Dominic's answer, even the ground state degeneracy is different under the interchange of $J$ and $h$ ($J\gg h$: symmetry-broken two-fold degeneracy, and $J\ll h$ unique ground state), therefore it is impossible to ...


1

1) In general, an algebra can have many representations. In this case, however, if you assume that there is a unique joint +1 eigenstate of the $\sigma_i$'s, that determines the representation uniquely. [All the other states can be found from this state by applying products of $\sigma_i^x$to it. And from the anti-commtation of $\sigma_i^x$ and ...



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