# What's the difference and relations between $SU(N)$ Schwinger boson and $CP(N!-!1)$ non-linear sigma model?

There are two ways when dealing with spin system(Heisenberg model): non-linear $$sigma$$ model and Schwinger boson.

# Non-linear $$sigma$$ model

When taking large $$S$$ limit, the quantum fluctuation of spin will be suppressed, which is so called "semi-classical" approximation. This means the start point is the classical configuration. If the correlation length of spin is not too short, Haldane has proven that we can map the Heisenberg model to non-linear $$sigma$$ model, which is characterized by unit vector $$boldsymbol{n}$$:
$$Z=int_{Lambda} mathcal{D}[n] e^{i Gamma[n]} exp left[-frac{1}{f} int d^{D} xleft|partial_{x} boldsymbol{n}right|^{2}right]$$
If $$S$$ is not so large, i.e. $$f$$ is small, which means "strong coupling", we can use $$CP(1)$$ representation to re-write the non-linear $$sigma$$ model:
$$boldsymbol{n}(x)=z^{dagger}(x) sigma z(x)=left(z_{1}^{*}(x) z_{2}^{*}(x)right) sigmaleft(begin{array}{c}z_{1}(x) \ z_{2}(x)end{array}right)$$
where $$sigma$$ here is the Pauli matrix. As this answer says, this representation is actually the standard formalism of fractionalization: replace the physical degree of freedom (the spin $$boldsymbol{n}$$) as fractionalized degree of freedom (the spinon $$z$$) with gauge structures. Then, we can discuss something about gauge field, e.g. confine phase v.s. deconfine phase. For example, when frustration is strong like in triangular lattice, there may exist deconfine spinon (Ref. 1). Also, to deal with it more elegantly, sometimes we also prefer to using large-$$N$$ expansion, i.e. $$CP(1)mapsto CP(N-1)$$.(Ref.2)

# Schwinger boson

Similarly, we are also familiar with another standard formalism of fractionalization: Schwinger boson, but here we fractionalize spin operator $$hat S$$ directly rather then unit vector $$boldsymbol{n}$$ above:
$$boldsymbol{S}_{i}=frac{1}{2} b_{i}^{dagger} boldsymbol{sigma} b_{i}=frac{1}{2} b_{i, alpha}^{dagger} boldsymbol{sigma}_{alpha beta} b_{i, beta}=left(begin{array}{cc}b_{i uparrow}^{dagger} & b_{i downarrow}^{dagger}end{array}right)(boldsymbol{sigma})left(begin{array}{c}b_{i uparrow} \ b_{i downarrow}end{array}right)$$
and we can use Schwinger boson $$b$$ re-write the Heisenberg model,resulting in fractionalized degree of freedom (the spinon $$b$$) with gauge structures. Similarly, we can also discuss the something about gauge structure. And we can also generalized to $$SU(N)$$ expansion.

# Question

1. "Large-S" v.s. "Large-N": Both "large-S"(non-linear $$sigma$$ model and spin wave) and "large-N" ($$CP(N-1)$$ and $$SU(2N)$$ Schwinger boson) suppress quantum fluctuation, so that I am wonder the difference between them. (I have this confusion since the spin in practice often has small $$S$$ and small $$N$$, thus, I am wonder that which kind of expansion is better.)

2. "Large-N" v.s. "Large-N": Both $$CP(N-1)$$ and $$SU(2N)$$ can be considered as large-N expansion, but I don’t know the difference and relation between them. Does they give different physical effect?

### Reference

1. Ch19.3.4, Subir Sachdev, Quantum Phase Transition
2. Auerbach, Interacting electrons and Quantum Magnetism

Physics Asked on November 11, 2021

1. Tuning $$S$$ is equivalent to tuning the representation of SU(2), i.e. one considers spin-1/2 [spin operators represented by $$2 times 2$$ matrices], spin-1 [spin operators represented by $$3 times 3$$ matrices], spin-3/2 [spin operators represented by $$4 times 4$$ matrices] , etc. Tuning $$N$$ takes us away from SU(2) to SU(N) spins. Mathematically, a key difference between the two semi classical limits, therefore, arises in terms of the number of generators: while spin-S has 3 generators (irrespective of the value of S), SU(N) has $$N^2-1$$ generators.

2. CP(N-1) is a specific kind of SU(N) generalization in that it is the coset space of a special pattern of SU(N) symmetry breaking where SU(N) $$to$$ U(N-1) [this generalizes the more familiar pattern of (internal) symmetry breaking that one encounters in magnetism which corresponds to the case N=2]. Therefore, they are inequivalent large $$N$$ limits.

Answered by vik on November 11, 2021

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