Imposing anti-commutation relations on fermionic quasi-particles

Question

In many theories of CMT, we assume the nature of quasi-particles (without giving proper justifications). For example, we assume nature of quasi-particles to be fermionic in case of a interacting fermion system we began with and impose anti-commutation relations accordingly. Like in BCS theory, while using the Bogoliubov-Valatin transformation to diagonalize the Hamiltonian, we assume that the new operators are also fermionic in nature. Please explain more on this step and how is it justified.

Answer 1

Everything traced back to the Landau theory of the Fermi liquid, when Landau supposed that the excited states of a Fermi liquid (a Fermi liquid is a Fermi gas with an additional two-body interaction, or electron-phonon interaction, ...) obeys Fermi-Dirac statistic. Landau coined the term quasi-particles for the dressed electrons: a conventional electron surrounded by an interacting cloud of screening charges, or electron-phonon composite particle (called plasmons). Any book about metal would talk about that. The most famous ones are

• A.A. Abrikosov Fundamentals of the Theory of Metals North-Holland (1988)
• A. A. Abrikosov, L. P. Gor’kov, & I. E. Dzyaloshinsky Methods of quantum fiel theory in statistical physics Prentice Hall (1963).
• Philippe Nozières & David Pines Theory Of Quantum Liquids Westview Press (1999).

for the first generation books talking about that topics. I would avoid as much as possible modern books regarding your question, since they are usually very sloppy about that. [NB: For a good reason: modern developments of condensed-matter exhibit sometimes quasi-particles which are neither bosons nor fermions, but that's an other story.]

A really pedagogical introduction to the quasi-particle (what he notes particle) topic is in

• R.D. Mattuck A Guide to Feynman Diagrams in the Many-Body Problem Dover (1992)

especially chapters 2, 4 and 8.

Good literature for superconductivity, especially regarding the Bogoliubov transformation, are, in addition to the original literature (quite difficult to follow so I do not give you the references)

• P.G. de Gennes Superconductivity of Metals and Alloys, Westview (1966).
• A.I. Fetter and J.D. Walecka, Quantum theory of many-particle systems Dover Publications (2003, first edition 1971)

That was the details Trimok forget in her/his excellent answer.

Answer 2

Here I am following this reference

We consider here pairs made of 2 fermionic partners. We associate a different value of a parameter $\sigma$ for each of the partner.

The creation/anihilation fermionic operators verify :

$[c_{k,\sigma},c_{k',\sigma'}]_+ = 0$ and $[c_{k,\sigma},c^+_{k',\sigma'}]_+ = \delta(k - k')\delta(\sigma - \sigma')$

The Bogoliubov-Valatin transformation is :

$b_{k,\sigma} = (u_k ~c_{k,\sigma} - \sigma ~v_k~ c^+_{-k,-\sigma})$, $b^+_{k,\sigma} = (u_k ~c^+_{k,\sigma} - \sigma ~v_k~ c_{-k,-\sigma})$

For simplicity, here $u_k$ and $v_k$ are supposed real.

So, we have :

$[b_{k,\sigma},b_{k',\sigma'}]_+ = - u_kv_{k'}\sigma'[c_{k,\sigma},c^+_{-k',-\sigma'}]_+ - v_{k}u_{k'}\sigma[c^+_{-k,-\sigma},c_{k',\sigma'}]_+$

$[b_{k,\sigma},b_{k',\sigma'}]_+ = - (u_kv_{k'}\sigma'+ v_{k}u_{k'}\sigma)\delta(k + k')\delta(\sigma + \sigma')$

$[b_{k,\sigma},b_{k',\sigma'}]_+ = \sigma (u_kv_{k'} - v_{k}u_{k'})\delta(k + k')\delta(\sigma + \sigma')$

$[b_{k,\sigma},b_{k',\sigma'}]_+ = \sigma (u_kv_{-k} - u_{-k}v_{k})\delta(k + k')\delta(\sigma + \sigma')~~~~~~~~~~~~~~~$ $(1)$

The same relation holds for $[b^+_{k,\sigma},b^+_{k',\sigma'}]_+$

We have also:

$[b_{k,\sigma},b^+_{k',\sigma'}]_+ = u_ku_{k'}[c_{k,\sigma},c^+_{k',\sigma'}]_+ +\sigma \sigma' v_{k}v_{k'}[c^+_{-k,-\sigma},c_{-k',-\sigma'}]_+$

$[b_{k,\sigma},b^+_{k',\sigma'}]_+ = (u_ku_{k'} + \sigma \sigma' v_{k}v_{k'}) \delta(k - k')\delta(\sigma - \sigma')$

$[b_{k,\sigma},b^+_{k',\sigma'}]_+ = (u_k^2 + v_k^2) \delta(k - k')\delta(\sigma - \sigma')~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$ $(2)$

Now, supposing :$$u_k = u_{-k}, v_k = v_{-k}, (u_k^2 + v_k^2) = 1~~~~~~~~~~~~~~(3)$$: This is a canonical transformation.

From equation $(1)$, We get :

$$[b_{k,\sigma},b_{k',\sigma'}]_+ = [b^+_{k,\sigma},b^+_{k',\sigma'}]_+=0$$

From equation $(2)$, we get :

$$[b_{k,\sigma},b^+_{k',\sigma'}]_+ = \delta(k - k')\delta(\sigma - \sigma')$$

This shows that the operators $b_{k,\sigma}, b^+_{k,\sigma}$ are fermionic operators verifying anti-commutation relations.

See reference - Chapitre 8-4, page 46

[EDIT] Now we may show that it is possible to find $u_k and v_k$, such that they obey the equation (3), that is it corresponds to a canonical transformation.

We are only here give the logic followed by the reference, and citing the precise equation and page.

1) Write a hamiltonian with the new operators $b_k, b^+_k$ :

$$Formula~ (156)~ page~ 47$$

2) Introduction of the operator number $n_k$, expression of the hamiltonian with these operators, and search for a eigenvalue $E$:

$$Formula~ (157 - 158)~ page~ 48$$

3) Minimizing E relatively to $u_k$

$$Formula~ (159)~ page~ 48$$

4) Expression of $u_k,v_k$ function of energies $\epsilon_k$, chemical potential $\mu$, and a quantity $\Delta$ (this last quantity depends on $u_k,v_k,n_k$)

$$Formula~ (160, 161)~ page~ 48$$

5) At this point, the exigence of $u_k,v_k$ representing a canonical transformation, give an equation for the quantity $\Delta$

$$Formula~ (162)~ page~ 48$$

6) Visualisation of the parameters $u_k,v_k$.

$$Figure ~ (36)~ page~ 49$$

7) Mean-Field Approximation : The last term of the hamiltonian is modifyed and the mean-field hamiltonian appears diagonal:

$$Formula~ (164)~ page~ 49$$

8) Conclusion of the reference (begining of page 50)

"The fact that the Bogoliubov-Valatin transformation diagonalizes the BCS-Hamiltonian at least in mean-field approximation justifies a posteriori our assumption that the ground state may be found as an eigenstate of the ˆb- occupation number operators. In the literature, the key relations (160) are often derived as diagonalizing the mean-field BCS-Hamiltonian instead of minimizing the energy expression (158). In fact both connections are equally important and provide only together the solution of that Hamiltonian. Clearly, the BCS-theory based on that solution is a mean-field theory."