SU(2) operators to SU(N) generators for Heisenberg XXX

  • #1
Maybe_Memorie
353
0
A paper I'm reading says

"Our starting point is the [itex]SU(N)[/itex] generalization of the quantum Heisenberg model:
[tex]H=-J\sum_{\langle i,j \rangle}H_{ij}=\frac{J}{N}\sum_{\langle i,j \rangle}\sum_{\alpha , \beta =1}^N J_{\beta}^{\alpha}(i)J_{\alpha}^{\beta}(j)
[/tex]
The [itex]J_{\beta}^{\alpha}[/itex] are the generators of the [itex]SU(N)[/itex] algebra and satisfy the usual commutation relations.

** The [itex]SU(N)[/itex] Heisenberg model can alternatively be written as an [itex]SU(2)[/itex] system with spin [itex]S=(N-1)/2[/itex] moments interacting via higher-order exchange processes.

An exact mapping connects the conventional [itex]SU(2)[/itex] spin operators to the [itex]SU(N)[/itex] generators as follows:
[tex]STUFF
[/tex]

The Hamiltonian can then be expressed in terms of
[tex]STUFF
[/tex]"

This is the paper http://arxiv.org/pdf/0812.3657.pdf. The stuff in question is on page 2.

Sorry I didn't LaTeX the full thing but I'm using a foreign keyboard and it would've taken ages.

My questions... How is ** arrived at? Presently my Lie algebra knowledge is very lacking but i'm working on it. This paper is about a square lattice. Can the result still be generalised for a 1-dim spin chain such as the Heisenberg XXX model with [itex]SU(N)[/itex]?

So essentially my real question is can I express SU(N) symmetry in terms of SU(2) symmetry with higher spin for the 1-dim spin chain, and if so how is the result arrived at?

Many thanks.
 

Answers and Replies

  • #2
18,823
8,987
I'm sorry you are not generating any responses at the moment. Is there any additional information you can share with us? Any new findings?
 

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