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I have seen multiple papers talking about the manifold, $M$ of the Neel order for an $SU(N)$ magnet is

$$M~=~\frac{U(N)}{U(m)\times U(N-m)}.$$

So for instance, a $SU(2)$ magnet has manifold

$$M ~=~ \frac{U(2)}{U(1)\times U(1)} ~\sim~ S^2.$$

Could someone explain to me physically how I could see why this is the appropriate manifold?

References:

  1. Subir Sachdev, NATURE OF THE DISORDERED PHASE OF LOW-DIMENSIONAL QUANTUM ANTIFERROMAGNETS, lectures (1990) page 13. The pdf file is available here.
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This is slightly too localized. Where did you find this information, and what are you trying to apply it to? What should it be "appropriate" for? –  Emilio Pisanty May 12 '13 at 21:11
    
For example, qpt.physics.harvard.edu/c6.pdf pg 13. –  Talon May 12 '13 at 22:43
    
What is $m$?$ $ –  leongz May 13 '13 at 0:13
    
It seems to be an arbitrarily chosen value of the number of fermions placed on a site (the remaining N-m terms go onto the other (conjugate?) site). And these states then transform according to a totally antisymmetric representation of SU(N). It is also related to the number of rows a Young tableau has. It is precisely this physical interpretation that I'm unable to understand and would appreciate any explanation. –  Talon May 13 '13 at 0:23

1 Answer 1

I am going to try and reinterpret some of that paper, see if we can get some kind of answer started. Please comment and contribute, I think this is an interesting physical system.

It would seem that they are using the fact that there are both boson (symmetric) and fermion (antisymmetric) representations of $SU(N)$ to generalize the usual $SU(2)$ magnet. The Schwinger representation of $SU(2)$ gives a state of spin $S=n_c/2$ with $n_c$ bosons at each site. If you allow $N$ bosons (rather then 1 with up and down states), the state transform under an $SU(N)$ representation. These are symmetric, with a Young's tableau of a single row with $n_c$ boxes.

This same representation can be described by fermionic operators. These are totally antisymmetric so they are a Young's tableau with a column of $m$ boxes (for $m$ fermions.)

For full generality, you want to describe a representation of arbitrary symmetry; a Young's tableau with arbitrary rows and columns. The author says specifically they are interested in the case when the two sublattices have conjugate representations. This is to match the results with the usual $SU(2)$ case.

The results are that in the semiclassical limit (bosons $n_c\to \infty$ with constant $N$ and fermions $m$) the expectation value of the spin of the coherent states has an $U(N)/((U(m)\times U(N-m))$ symmetry. So the fermion content determines the symmetry of the resulting classical system. The integer $N$ is something like a generalized spin quantum number where normally $N=2$ for up and down.

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