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What're the relations and differences between slave-fermion and slave-boson formalism?

+ 7 like - 0 dislike

As we know, in condensed matter theory, especially in dealing with strongly correlated systems, physicists have constructed various "peculiar" slave-fermion and slave-boson theories. For example,

For Heisenberg model, we have Schwinger-fermion $\vec{S_i}=\frac{1}{2}f^\dagger_i\vec{\sigma }f_i $ and Shwinger-boson $\vec{S_i}=\frac{1}{2}b^\dagger_i\vec{\sigma }b_i $ approaches, with constraints $f^\dagger_{1i}f_{1i}+f^\dagger_{2i}f_{2i}=1$ and $b^\dagger_{1i}b_{1i}+b^\dagger_{2i}b_{2i}=2S$, repectively.

For $t-j$ model, there are slave-fermion $C^\dagger_{i\sigma}=b^\dagger_{i\sigma}f_i$ and slave-boson $C^\dagger_{i\sigma}=f^\dagger_{i\sigma}b_i$ methods, with constraints $f^\dagger_{i}f_{i}+\sum b^\dagger_{i\sigma}b_{i\sigma}=1$ and $b^\dagger_{i}b_{i}+\sum f^\dagger_{i\sigma}f_{i\sigma}=1$, repectively.

For Hubbard model, we have slave-fermion $C^\dagger_{i\sigma}=b^\dagger_{i\sigma}f_{1i}+\sigma f^\dagger_{2i}b_{i\sigma}$ and slave-boson $C^\dagger_{i\sigma}=f^\dagger_{i\sigma}b_{1i}+\sigma b^\dagger_{2i}f_{i\sigma}$ methods, with constraints $\sum f^\dagger_{\alpha i}f_{\alpha i}+\sum b^\dagger_{i\sigma}b_{i\sigma}=1$ and $\sum b^\dagger_{\alpha i}b_{\alpha i}+\sum f^\dagger_{i\sigma}f_{i\sigma}=1$, repectively.

And my questions are:

(1) Whatever it's spin or electron system, the slave-fermion and slave-boson constructions have very similar forms, simply interchanging bosonic and fermionic operators in one we would get the other one. So is there any deep connection between these two formalism? Does this similarity have something to do with supersymmetry?

(2) From the mathematical point of view, both the slave-fermion and slave-boson constructions are correct. But physically, when should we use slave-fermion methods and when should we use slave-boson methods? What are the differences between these two approaches when we deal with a particular physical model?

This post imported from StackExchange Physics at 2014-03-09 08:45 (UCT), posted by SE-user K-boy
asked May 2, 2013 in Theoretical Physics by Kai Li (975 points) [ no revision ]

1 Answer

+ 7 like - 0 dislike

The slave particle approach is based on the assumption of spin-charge separation in the strongly correlated electron systems (typically Mott insulators). It was proposed that the electrons can decay into spinons and chargons (holons/doublons). But to preserve the fermion statistics of the electrons, the spinon-chargon bound state must be fermionic, so the simplest way is to ascribe the fermion statistics to one of them: if the spinon is fermionic then the chargon should be bosonic (slave-boson), or if the chargon is fermionic then the spinon should be bosonic (slave-fermion). The differences are just a matter of which degrees of freedom (spin or charge) should the fermion statistics be ascribed to.

Within one kind of slave particle formalism, the supersymmetry is possible but not necessary. Whether or not the spinon and chargon are suppersymmetric to each other depends on their spectrums (which are the details). If the spinon and chargon have different spectrums (which is always the case), then the effective theory has no supersymmetry.

It seems that there is a certain kind of duality between the slave-boson and slave-fermion approaches, one may conjecture if the two approaches can be unified into a single theory. And it was indeed the case. Now we know that the two approaches are just two low-energy effective theories of the complete-fractionalization theory, which has two equivalent versions: the Chern-Simons version by Kou, Qi and Weng Phys. Rev. B 71, 235102 (2005), or the Majorana version by Xu and Sachdev Phys. Rev. Lett. 105, 057201 (2010).

In the complete-fractionalization theory, electron is fully fractionalized into bosonic spinon, bosonic chargon, and a mutual Chern-Simons gauge field (or Majorana fermion) to take care of the fermion statistics. Spinons and chargons are treated equally in this theory, both are bosonic degrees of freedom. In the mutual Chern-Simons theory, the spinons and chargons are coupled to a Chern-Simons theory of the K matrix $$K=\left(\begin{matrix}0&2\\2&0\end{matrix}\right),$$ meaning that the spinon and the chargon will mutually see each other as a $\pi$-vortex. This mutual $\pi$-vortex binding renders the spinon-chargon bound state a fermion, corresponding to the electron. If both the spinons and chargons are gapped, the remaining low-energy effective theory will be a $\mathbb{Z}_2$ gauge theory, which supports 3 topological excitations: electric charges, magnetic monopoles and fermions; corresponding to chargons, spinons and electrons respectively. Thus in this phase, the relation between the chargon and the spinon is just like that between the charge and the monopole, i.e. the spin and charge degrees of freedom are electromagnetically dual to each other, and the duality is protected by the underlying $\mathbb{Z}_2$ topological order. This exotic topological phase was proposed to be the basic physics behind the pseudo-gap phase of the cuprates superconductors Phys. Rev. Lett. 106, 147002 (2011) (although more complexity should be added to fully explain the phenomena, but I personally believe that the basic principles are captured by this theory).

By tuning the relative density of spinons and chargons (presumably achievable by doping in real materials), the system can be driven into the ordered phase by condensing one of the fractionalized degrees of freedom (note that now both spinons and chargons are bosons and can condense). Condensing spinon results in a spin ordered state (i.e. the Neel antiferromagnet), while condensing chargon results in a charge superfluid state (i.e. d-wave superconductivity). But the mutual Chern-Simons theory forbidden both spinon and chargon to condense at the same time, which is consistent with the fact that we can never Bose-condense the electrons.

The Majorana theory is similar, but more comprehensive. In SU(2) operator matrix form, the decomposition reads $$C=B\Xi Z,$$ where $C$, $B$, $Z$ collect electron, chargon and spinon operators in matrices $$C=\left(\begin{matrix}c_\uparrow & c_\downarrow\\-c_\downarrow^\dagger & c_\uparrow^\dagger\end{matrix}\right),B=\left(\begin{matrix}b_d & b_h^*\\-b_h & b_d^*\end{matrix}\right),Z=\left(\begin{matrix}z_\uparrow & z_\downarrow\\-z_\downarrow^* & z_\uparrow^*\end{matrix}\right),$$ and $\Xi=\xi_0\sigma_0+i\xi_1\sigma_1+i\xi_2\sigma_2+i\xi_3\sigma_3$ contains the Majorana operators. Both $b$ and $z$ are bosonic, and the fermion statistics is carried by the Majorana fermions $\xi$. This is the most general scheme of spin-charge separation for electrons. The emergent gauge structure of this theory is $O(4)$ (representing the rotation among 4 Majorana fermions), which can be factorized into two $SU(2)$ gauge structures, as $O(4)\simeq SU(2)_B\otimes SU(2)_Z$, coupling to the chargons and spinons respectively. Because the $SU(2)$ fluctuation is confining in (2+1)D spacetime, without any topological order, the fractionalized particles will all be confined into electrons. But if we condense one of the bosonic degrees of freedom, say the chargons $B$, then $SU(2)_B$ can be Higgs out, and the remaining $SU(2)_Z$ gauge fluctuation would confine the Majorna fermion $\Xi$ with the spinon $Z$ into a composite particle $\Xi Z$, rendering the bosonic spinon to fermionic and reducing the Majorana theory to the slave-boson theory (which is exactly the same as Wen's $SU(2)$ theory Quantum Orders and Symmetric Spin Liquids). If we condense the spinon first, then the story will be reversed, ending up with the slave-fermion theory. Therefore the choice of which slave-particle approach depends entirely on the phase we wish to study (the degrees of freedom that we wish to condense). Different orders in the ground state would support different low-energy effective theories, which appear to us as different slave-particle approaches.

This post imported from StackExchange Physics at 2014-03-09 08:45 (UCT), posted by SE-user Everett You
answered May 5, 2013 by Everett You (660 points) [ no revision ]
Such a beautiful answer! For folks without PROLA access, the arXiv links for the mentioned articles are, respectively, arxiv.org/abs/cond-mat/0410391, arxiv.org/abs/1004.5431, arxiv.org/abs/1007.2507

This post imported from StackExchange Physics at 2014-03-09 08:45 (UCT), posted by SE-user wsc
@ Everett You, Wow, what a wonderful, excellent and detailed answer, +1 and accept. This answer contains so rich and deep physics that I think it will take me a little long time to understand it, anyway, thank you very much.

This post imported from StackExchange Physics at 2014-03-09 08:45 (UCT), posted by SE-user K-boy
@ Everett You, one more naive question: what does "condense" mean for "fermions" ? So far I only know the ordinary Bose-condensation, but I always encounter the so called "fermions condense" in literatures too.

This post imported from StackExchange Physics at 2014-03-09 08:45 (UCT), posted by SE-user K-boy
@K-boy Yes, I think it was Prof. Wen and his former student Maissam who invented this term "anyon condensation"(arxiv.org/abs/1007.2030), which includes fermion condensation as a special case. By condensation, we mean the many-body system is in a superposition state of all possible configurations, a state of the strongest quantum fluctuation, a melted liquid state. Boson condensation, anyon condensation, and string-net condensation all share this meaning. Specifically fermion condensation simply means superconductivity, in which fermions are fluctuating strongly.

This post imported from StackExchange Physics at 2014-03-09 08:45 (UCT), posted by SE-user Everett You
@K-boy In my answer above, I mentioned electron can not Bose-condense, which literally means electron can not simply condense as bosons (in the most narrow sense). The reason that I stress "Bose"-condense is because I know there is the possibility of fermion condensation, which, however, is not relavent to our discussion of the slave-particle theory.

This post imported from StackExchange Physics at 2014-03-09 08:45 (UCT), posted by SE-user Everett You

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