Maybe_Memorie
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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.
"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.