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.

  • $\begingroup$ Possible duplicate: physics.stackexchange.com/q/17893/2451 $\endgroup$ – Qmechanic Jun 18 '13 at 11:43
  • $\begingroup$ Could you give a more precise example where quasi-particles are assumed to be fermionic, "without giving proper justifications". $\endgroup$ – Trimok Jun 18 '13 at 14:29
  • $\begingroup$ @Trimok Take the case of BCS theory, when we use the Bogoliubov-Valatin transformation, we assume that the new operators are also fermionic operators. My question is :what is the validity of this assumption ? $\endgroup$ – cleanplay Jun 18 '13 at 17:20
  • $\begingroup$ In BCS theory, you have an electron-phonon basic interaction (Fröhlich Hamiltonian), and with some transformation , you have an effective electron-electron interaction. The electron is not a quasi-particle, it is a particle, and it is a fermion, so it obeys anti-commutation rules. See for instance this reference Chapter 1 $\endgroup$ – Trimok Jun 18 '13 at 17:39
  • $\begingroup$ @Trimok I know that the quasi-particles are the ones we obtain after diagonalizing the Hamiltonian and those quasi-particles are assumed also as fermions. My question was not for the electrons we started with, but the quasi-particles we obtained in the process of diagonalizing the Hamiltonian. When we diagonalize the Hamiltonian using the Bogoliubov transformation, we say that 'taking the nature of the new operators to be fermionic, lets impose the anti-commutation relations'. My question is for this step that how it is justified to assume the new operators to obey anti-commutation relations. $\endgroup$ – cleanplay Jun 19 '13 at 4:39

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.


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."

  • $\begingroup$ Please note that $u_{k}^{2}+v_{k}^{2}=1$ is the condition for the transformation to be canonical such that the operators $ b_{k\sigma}$ and $ b_{k\sigma}^{+}$ are fermionic. Here you are supposing that in order to prove the statement which I questioned. So basically you want the new operators to be fermionic for the so-called canonical transformation. My question is why you want the new operators to be fermionic. Why should the transformation be canonical ? $\endgroup$ – cleanplay Jun 20 '13 at 5:46
  • $\begingroup$ I made an edit. The idea is that it is possible to find $u_k,v_k$ such as it is a canonical transformation, and that the hamiltonian, at least in the mean field approximation, is diagonal. $\endgroup$ – Trimok Jun 20 '13 at 6:59
  • $\begingroup$ Let me put it this way. I can also try to find the diagonalized form of the Hamiltonian by assuming the new operators to be bosonic and thus getting a different condition on the form of $u_{k}$ and $v_{k}$ by taking the commutation conditions this time instead of the anti-commutations (fermions). What prevents me from doing that ? (Preferably, a physical argument.) $\endgroup$ – cleanplay Jun 20 '13 at 7:12
  • 1
    $\begingroup$ If you follow the chapter 8.1, a physical explanation is done : "These are not the original electrons making up the metal together with the atomic nuclei. Rather they are electrons or missing electrons surrounded by polarization clouds of other electrons and nuclei in which nearly all the Coulomb interaction is absorbed. We do not precisely know these excitations nor do we know the ground state (although a quite elaborate theory exists for them which we ignore here).We just assume that they may be represented by fermionic operators with properties like those in the (Fermi)gas" $\endgroup$ – Trimok Jun 20 '13 at 7:20
  • $\begingroup$ Moreover, with the Bogoliubov transformation, the nature of the final operators is the same as the nature of the initial operators. So beginning with fermionic operators $c_k$, the only possible choice is fermionic final operators $b_k$ . $\endgroup$ – Trimok Jun 20 '13 at 9:39

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