We normally solve the Bogoliubov-de Gennes (BdG) equations in order to compute the energy spectrum of a superconductor. The Nambu spinor is a common object that is used in formulating these equations. In other words, the problem is formulated with the a priori assumption that the superconductors obeys particle-hole symmetry. But how do we know that it will obey particle-hole symmetry? I'm sure that by assuming particle-hole symmetry, computing the spectrum by solving the BdG equations, and comparing with experiments we get a good agreement. So we know it's true. I just don't understand why that is the case. Is there any physical reason behind it?


3 Answers 3


The particle-hole symmetry is not exactly a property of the superconductors. It's rather a property of the free electron gas, with the Fermi level -- separating all the empty states above from all the occupied below it -- cutting a parabola. The property to have the spectrum symmetry $\varepsilon_{\mathbf{k}}=\varepsilon_{-\mathbf{k}}$ (see the quotation below for the notations meaning) is mandatory to have a particle-hole symmetry also.

I quote below the book by C. Kittel, Quantum theory of solids (1963), chapter 5.

In treating the electron system it is particularly convenient to redefine the vacuum state as the filled Fermi sea, rather than the state with no particles present. With the filled Fermi sea as the vacuum we must provide separate fermion operators for processes which occur above or below the Fermi level. The removal of an electron below the Fermi level is described in the new scheme as the creation of a hole. We consider first a system of $N$ free non-interacting fermions having the Hamiltonian $$H_{0}=\sum_{\mathbf{k}}\varepsilon_{\mathbf{k}}c_{\mathbf{k}}^{+}c_{\mathbf{k}}$$ where $\varepsilon_{\mathbf{k}}$ is the energy of a single particle having $\varepsilon_{\mathbf{k}}=\varepsilon_{-\mathbf{k}}$ . We agree to measure $\varepsilon_{\mathbf{k}}$ from the Fermi level $\varepsilon_{F}$ .

In the ground state [... $\Phi_{0}$ ...] all one-particle states are filled to the energy $\varepsilon_{F}$ and above $\varepsilon_{F}$ all states are empty. We regard the state $\Phi_{0}$ as the vacuum of the problem: it is then convenient to repreent the annihilation of an electron in the Fermi sea as the creation of a hole. Thus we deal only with electrons (for states $k > k_{F}$) and holes (for states $k < k_{F}$). The act of taking an electron from $\mathbf{k}'$ within the sea to $\mathbf{k}''$ outside the sea involves the creation of an electron-hole pair. The language of the theory has a similarity to positron theory, and there is a complete formal similarity between particles and holes.

We introduce the electron operators $\alpha^{+}$ , $\alpha$ by $$\alpha_{\mathbf{k}}^{+}=c_{\mathbf{k}}^{+}\;;\;\alpha_{\mathbf{k}}=c_{\mathbf{k}}\;\;\text{for}\;\;\varepsilon_{\mathbf{k}}>\varepsilon_{F}$$ and the hole operators $\beta^{+}$ , $\beta$ by $$\beta_{\mathbf{k}}^{+}=c_{-\mathbf{k}}\;;\;\beta_{\mathbf{k}}=c_{-\mathbf{k}}^{+}\;\;\text{for}\;\;\varepsilon_{\mathbf{k}}<\varepsilon_{F}$$ The $-\mathbf{k}$ introduced for the holes is a convention which gives correctly the net change of wave-vector or momentum: the annihilation $c_{-\mathbf{k}}$ of an electron at $-\mathbf{k}$ leaves the Fermi sea with a momentum $\mathbf{k}$ . Thus $\beta_{\mathbf{k}}^{+}\equiv c_{-\mathbf{k}}$ creates a hole of momentum $\mathbf{k}$ .

So it is essentially a matter of convention in defining the operators. Obviously, it is no more strange than calling all the excitations of a solid a particle I believe.

For superconductor, though, the particle-hole symmetry plays a more prominent role than just a redefinition of what is called quasi-electron and quasi-hole. Since the phonons (as in the BCS model) couple electrons at $\mathbf{k}$ and $-\mathbf{k}$, they couple electrons and holes in the prescription cited above. Hence the crucial role played by the Nambu spinor.

This is nevertheless not an exact symmetry as Lubos Motl mentioned, but it can hardly be contradict. Indeed, the ratio $\Delta / \varepsilon_{F}$ is so small ($10^{-3} -10^{-5}$ for BCS superconductors) that the linearisation of the spectrum close to the Fermi energy is a really good approximation, and so the relation $\varepsilon_{\mathbf{k}}=\varepsilon_{-\mathbf{k}}$ is well verified.

  • $\begingroup$ With spin, we should define $\beta_{k\uparrow}^\dagger=c_{-k\downarrow}$, right? For removing a particle of spin up will leave the fermi sea with spin down. If this is true, what is the relation of this transformation with the particle hole transformation in real space, see this article eqn(3). I am confused. What is particle hole symmetry exactly? $\endgroup$ Commented Apr 28, 2015 at 3:35
  • $\begingroup$ @Buzhidao Perhaps it's better to ask an other question if yours is a bit different from this one. Particle-hole symmetry is defined as the anti-unitary operation which anti-commutes with your Hamiltonian. When the spin is added, it's simpler I feel to discuss the pure mathematical root of this symmetry. In the real space, the Fermi surface is not defined, so please do not use the schematic picture hole is a withdrawal of an electron from the Fermi sphere one in the real space. The particle-hole symmetry is nicely reviewed in the first sections of arxiv.org/abs/1407.2131 $\endgroup$
    – FraSchelle
    Commented Apr 28, 2015 at 10:30
  • $\begingroup$ Why do you claim p-h symmetry is an exactly symmetry for free electron gas? It seems that it is also an approximate symmetry just as superconductors. $\endgroup$ Commented Jul 4, 2015 at 5:22
  • $\begingroup$ Are you measuring $\mathbf{k}$ from $k_F$ or something. Since in your answer, it seems that you equate p-h symmetry with $\varepsilon_k=\varepsilon_{-k}$, see your last paragraph. $\endgroup$ Commented Jul 4, 2015 at 5:24
  • $\begingroup$ @buzhidao I do not see any mention of PH being an exact symmetry in my answer, so I can not comment on your first comment. I think you confound symmetry and property : having $\beta^\dagger=c$ is a property of the operator. You should try to write a theory with particles and anti-particles well separated (à la Dirac, i.e. with spinors) and you will realise that this property is verified for quadratic dispersion relation. Once you make all the replacements $\beta \sim c$ the PH symmetry is the transformation of the Hamiltonian onto itself (modulo a phase, which can be e.g. -1) (...) $\endgroup$
    – FraSchelle
    Commented Jul 5, 2015 at 10:11

I think OP's question is about the p-h symmetry in BdG equation of superconductivity. This is really an exact p-h symmetry (in mathematics), but it is however a redundancy of description. Since it doubles the degree of freedom.


In general, solids – conductors and superconductors, but not semiconductors – tend to have an approximate electron-hole symmetry because they have conduction or valence bands and the density of states $\rho(E)$ is symmetric around the Fermi surface, $$\rho (E_f+\Delta E) = \rho(E_f-\Delta E)$$ So for every occupied state that is $\Delta E$ below the Fermi energy, one finds an empty state that is $\Delta E$ above the Fermi energy, and there's a symmetry between the creation operators for the holes and electrons $$ c^\dagger_h(E_f-\Delta E) \leftrightarrow c^\dagger_e(E_f+\Delta E)$$ where $h,e$ stand for hole, electron. With this pairing, adding a hole beneath the Fermi surface changes the energy by the same amount as an added electron above the Fermi surface. Also, the interactions with the phonons, for example, respect this symmetry, as one may directly verify.

This symmetry arises in the approximation in which the quantities such as the density of states are linearized functions of energy and the Fermi energy is a "generic" value of a possible electron energy. For non-conductors, these assumptions are strongly broken. For semiconductors, they're "somewhat" broken because the structure of the energy bands isn't $E\to 2E_0 - E$ (upside-down) symmetric. There are also papers claiming that the symmetry breaking is important even for conductors or superconductors.

At any rate, this symmetry isn't exact because materials obviously do break the symmetry between electrons and holes, electrons and positrons. The only place where the electron-hole symmetry is exact is the vacuum where the holes (in the Dirac sea) are called positrons and the electron-hole symmetry is the C-symmetry (which may still be broken, but if we talk about the CPT-symmetry, it holds in every quantum field theory). In materials, the symmetry is an artifact of the linearization near the Fermi surface.

  • $\begingroup$ Hello Luboš, could you elaborate why a semiconductor or a conductor can't have particle-hole symmetry but a conductor does? Why do you say that $E$ is not -> $2E_0 - E$? Also, what about excitons? They should come close to a Bogoliubov description. $\endgroup$
    – Mike
    Commented Apr 20, 2013 at 7:18
  • $\begingroup$ @Matthias As I explain in my answer physics.stackexchange.com/a/67584/16689 below, particle-hole symmetry in condensed matter physics is a property of the free electron gas. Semi-conductor are not explained by a free electron gas theory (you need the underneath lattice to understand the band structure), and therefore do not verify electron-hole symmetry. $\endgroup$
    – FraSchelle
    Commented Jun 10, 2013 at 12:38
  • $\begingroup$ [continued] Say differently, the Fermi level of a metal is somewhere cutting a parabola, and then define a Fermi sea below and an empty space above, whereas the Fermi level of a semi-conductor locates in a gap: there is no reason for the electrons and the holes to have the same (inverted) dispersion relation. $\endgroup$
    – FraSchelle
    Commented Jun 10, 2013 at 12:39

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