Double occupied state in SU(2) slave-boson

Question

The SU(2) generalization of the U(1) slave-boson has been introduced in PRL 76, 503 (1996), and PRB 86, 085145 (2012). (A generic recipe for constructing the SU(2) slave-particle framework has been discussed Here.) In this framework, the electronic annihilation operator with spin $\sigma$ can be expressed as $$c_{\sigma}=\frac{1}{\sqrt{2}} (b^{\dagger}_{1} f_{\sigma} -b^{\dagger}_{2}f^{\dagger}_{\overline{\sigma}}),$$ where $b~(b^{\dagger})$ is a slave-boson annihilation (creation) operator, and $f^{\dagger}_{i}(f_{i})$ creates (annihilates) a fermion with spin $i$. (The Nambu-like version of the above expression is presented in this link.) In the SU(2) slave-boson language, it is claimed that "the double occupied state is automatically ruled out". As a result, the one-site electronic states will be translated into the SU(2) slave-boson language as $$\begin{align} |0 \rangle_{c} &= \frac{1}{\sqrt{2}}(b^{\dagger}_{1} + b^{\dagger}_{2} f^{\dagger}_{\uparrow} f^{\dagger}_{\downarrow}) |0\rangle_{sb},\\ c^{\dagger}_{\uparrow}|0 \rangle_{c} &= f^{\dagger}_{\uparrow} |0\rangle_{sb},\\ c^{\dagger}_{\downarrow}|0 \rangle_{c} &= f^{\dagger}_{\downarrow} |0\rangle_{sb},\\ c^{\dagger}_{\uparrow}c^{\dagger}_{\downarrow}|0 \rangle_{c} &= \frac{1}{\sqrt{2}} ( b_{1} f^{\dagger}_{\uparrow}f^{\dagger}_{\downarrow} -b_{2}) |0\rangle_{sb}=0 \end{align}$$ This result is a consequence of two points:

1. The SU(2) nature of this representation which is under the constraint of $$b^{\dagger}_{1}b_{1} -b^{\dagger}_{2}b_{2}+\sum\limits_{\sigma \in \{\uparrow, \downarrow\}} f^{\dagger}_{\sigma}f_{\sigma}=1.$$
2. Exploiting, at most, only one species of each auxiliary particles per state. In other words, enforcing $$\forall \alpha \in \{1,2\} \rightarrow (b^{\dagger}_{\alpha}b_{\alpha} )^{2} =b^{\dagger}_{\alpha}b_{\alpha}, \qquad \forall \sigma \in \{\uparrow,\downarrow\} \rightarrow (f^{\dagger}_{\sigma}f_{\sigma} )^{2} =f^{\dagger}_{\sigma}f_{\sigma}.$$

The question is that what is the necessity of applying the second implicit constraint? In addition, why we can not define a background charge for the vacuum of slave-boson such that the double-occupied state can survive?

This post imported from StackExchange Physics at 2017-07-02 10:53 (UTC), posted by SE-user Shasa

Comments

I guess you of course understand the meaning of physical constraint, so are you asking what's the result in practice of imposing such constraint?