One of the $SU(2)$ slave-boson decompositions has been introduced by X.-G. Wen and P. A. Lee in PRL, 76, 503 (1996). (A generic recipe for constructing the SU(2) slave-particle framework has been discussed Here.) It has been suggested that the electronic annihilation operators can be expressed as

$$ \big(c_{\uparrow}\quad c_{\downarrow} \big)= \frac{1}{\sqrt{2}} \big( b^{\dagger}_{1} \quad b^{\dagger}_{2} \big) \begin{pmatrix} f_{\uparrow} & f_{\downarrow} \\ f_{\downarrow}^{\dagger} & -f^{\dagger}_{\uparrow} \end{pmatrix} , $$ where $c_{\sigma}$ is electronic annihilation operator with spin $\sigma$, $b~(b^{\dagger})$ is a slave-boson annihilation (creation) operator, and $f^{\dagger}_{i}(f_{i})$ creates (annihilates) a fermion with spin $i$. It has been shown that the above representation preserves the $SU(2)$ symmetry if slave-particles satisfy $$ b^{\dagger}_{1}b_{1} -b^{\dagger}_{2}b_{2}+\sum\limits_{\sigma \in \{\uparrow, \downarrow\}} f^{\dagger}_{\sigma}f_{\sigma}=1. $$

To assess the anti-commutation relation of electrons I have followed the common approach in slave-particle methods, namely employing the constraint to enforce the Fermionic statistics.

$$ \begin{align} &\qquad \qquad \qquad \{ c_{\sigma}, c^{\dagger}_{\sigma} \}=1,\\ &2c^{\dagger}_{\sigma} c_{\sigma}= f^{\dagger}_{\sigma} b_{1} b^{\dagger}_{1} f_{\sigma} + \sigma^{2} f_{\overline{\sigma}}b_{2}b^{\dagger}_{2} f^{\dagger}_{\overline{\sigma}} +\sigma f^{\dagger}_{\sigma} b_{1} b_{2}^{\dagger}f_{\overline{\sigma}}^{\dagger} +\sigma f_{\overline{\sigma}} b_{2} b_{1}^{\dagger}f_{\sigma},\\ &2c_{\sigma}c^{\dagger}_{\sigma}= b^{\dagger}_{1} f_{\sigma} f^{\dagger}_{\sigma} b_{1} +\sigma^{2} b^{\dagger}_{2} f^{\dagger}_{\overline{\sigma}} f_{\overline{\sigma}} b_{2} +\sigma b^{\dagger}_{1}f_{\sigma} f_{\overline{\sigma}}b_{2} +\sigma b_{2}^{\dagger}f_{\overline{\sigma}}^{\dagger} f_{\sigma}^{\dagger}b_{1} \end{align} $$ It is straight forward to show that the summation of last two terms of $c^{\dagger}_{\sigma}c_{\sigma}$ and $c_{\sigma} c^{\dagger}_{\sigma}$ are zero. Following the same strategy, I can show that $$ \begin{align} 2c^{\dagger}_{\sigma} c_{\sigma} &= f^{\dagger}_{\sigma} b_{1} b^{\dagger}_{1} f_{\sigma} + \sigma^{2} f_{\overline{\sigma}}b_{2}b^{\dagger}_{2} f^{\dagger}_{\overline{\sigma}} \\ &= f^{\dagger}_{\sigma} (1+ b^{\dagger}_{1}b_{1} ) f_{\sigma} + \sigma^{2} f_{\overline{\sigma}} (1+ b^{\dagger}_{2} b_{2} ) f^{\dagger}_{\overline{\sigma}} \\ &= f^{\dagger}_{\sigma}f_{\sigma} +f^{\dagger}_{\sigma} b^{\dagger}_{1}b_{1}f_{\sigma} + \sigma^{2} f_{\overline{\sigma}}f^{\dagger}_{\overline{\sigma}} + \sigma^{2}f_{\overline{\sigma}}b^{\dagger}_{2} b_{2}f^{\dagger}_{\overline{\sigma}} \\ &= f^{\dagger}_{\sigma}f_{\sigma} + b^{\dagger}_{1} ( 1- f_{\sigma}f^{\dagger}_{\sigma} )b_{1} + \sigma^{2} f_{\overline{\sigma}}f^{\dagger}_{\overline{\sigma}} + \sigma^{2} b^{\dagger}_{2} (1- f^{\dagger}_{\overline{\sigma}} f_{\overline{\sigma}}) )b_{2} \\ &= f^{\dagger}_{\sigma}f_{\sigma} + \sigma^{2} f_{\overline{\sigma}}f^{\dagger}_{\overline{\sigma}} +b^{\dagger}_{1} b_{1} +b^{\dagger}_{2}b_{2} -b^{\dagger}_{1} f_{\sigma}f^{\dagger}_{\sigma} b_{1} -\sigma^{2} b^{\dagger}_{2}f^{\dagger}_{\overline{\sigma}} f_{\overline{\sigma}}b_{2}. \end{align} $$ By setting $\sigma^{2}=1$, and cancelling the last two terms of $2c^{\dagger}_{\sigma}c_{\sigma}$ by the first two terms of $2 c_{\sigma} c^{\dagger}_{\sigma}$, I would obtain $$ \begin{align} 2c^{\dagger}_{\sigma}c_{\sigma} + 2c_{\sigma} c^{\dagger}_{\sigma} &= f^{\dagger}_{\sigma}f_{\sigma} + f_{\overline{\sigma}}f^{\dagger}_{\overline{\sigma}} +b^{\dagger}_{1} b_{1} +b^{\dagger}_{2}b_{2} \\ &= f^{\dagger}_{\sigma}f_{\sigma} + 1- f^{\dagger}_{\overline{\sigma}}f_{\overline{\sigma}} +b^{\dagger}_{1} b_{1} +b^{\dagger}_{2}b_{2} \end{align} $$ which is not the same as the constraint. Where did I make a mistake?

  • 4
    $\begingroup$ Please do not make trivial edits to your question solely to bump it on the front page. $\endgroup$
    – ACuriousMind
    Commented Jul 3, 2017 at 15:44
  • $\begingroup$ I'm not familiar with this particular formulation of slave boson. However, why would you need two bosons? $\endgroup$
    – Roger V.
    Commented Apr 19, 2020 at 11:52

1 Answer 1


After trying it a couple of times, I get the same as you at the equation for $2c^{\dagger}_{\sigma}c_{\sigma}$, but then I don't understand where the last two terms went. Basically, I get $$ \begin{array}{ccl} \{c_{\sigma},c^{\dagger}_{\sigma}\} & = & f^{\dagger}_{\sigma}f_{\sigma} b_1b_1^{\dagger} + f_{\bar{\sigma}}f^{\dagger}_{\bar{\sigma}} b_2b_2^{\dagger} \\ & = & f^{\dagger}_{\sigma} f_{\sigma} + 1 - f^{\dagger}_{\bar{\sigma}} f_{\bar{\sigma}} +b^{\dagger}_2 b_2 + b^{\dagger}_1 b_1 f^{\dagger}_{\sigma} f_{\sigma} - b^{\dagger}_2 b_2 f^{\dagger}_{\bar{\sigma}} f_{\bar{\sigma}}. \end{array} $$ By summing on sigma, the $ff$ terms disappear and I get $$ \begin{array} a\sum_{\sigma} \{c_{\sigma}, c^{\dagger}_{\sigma}\} & = & 2 + 2 b^{\dagger}_2 b_2 + (b^{\dagger}_1 b_1 - b^{\dagger}_2 b_2) \sum_{\sigma} f^{\dagger}_{\sigma} f_{\sigma} \\ & = & 2 + b^{\dagger}_1b_1 + b^{\dagger}_2 b_2 + (b^{\dagger}_2b_2 - b^{\dagger}_1b_1)\left(\sum_{\sigma}f^{\dagger}_{\sigma}f_{\sigma}+1\right) \end{array} $$ which has some similar features to the condition for slave-particles, but it's not quite it! Good luck with that, I'll keep an eye on it.

  • $\begingroup$ Thank you for your effort. your answer has some problems: 1. we are going to show that the anticommutation relation of electrons is satisfied under the slave-boson decomposition and thereby the equality of second equation is under question and can not be employed. I also did not figure out how you have obtained the right-hand side of the first line. 2. As $b(b^{\dagger})$ is an operator, you are not allowed to treat them as a number and consequently, the right-hand side of the last equation is not correct. $\endgroup$
    – Shasa
    Commented Jun 30, 2017 at 9:12
  • $\begingroup$ 1. So you want to prove the anticommutation relation $\{c_{\sigma},c^{\dagger}_{\sigma}\} = 1$ using only the condition for slave-particles to preserve $SU(2)$ symmetry? To get this condition you will need the summation on spin. 2. I wrote it like this because the nominator and denominator commute but I meant the inverse. $\endgroup$
    – gingras.ol
    Commented Jun 30, 2017 at 16:43

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