The Bogoliubov-de-Gennes (BdG) formalism of a Hamiltonian reduces the dimension of the Hilbert space we work on. For example, in 1D superconducting Hamiltonians with $N$ lattice sites, the actual number of basis states is $2^N$. When BdG is applied, it reduces to $2N$. What is the justification for the equivalence of the two Hamiltonians?

  • $\begingroup$ The number of basis states isn't reduced. It's still an N-body problem. It's just that the single particle orbitals happen to be the eigenvectors of a certain 2N×2N matrix. That doesn't mean your actual Hilbert space is 2N dimensional. $\endgroup$ Aug 4, 2017 at 14:32
  • $\begingroup$ thanks for the comment. could you explain a bit on how the term c†c† correspond to in single particle Hamiltonian? does that mean the basis states in the N-body problems will be the anti-symmetrized wave function produced from singleparticle eigenstates? $\endgroup$
    – Prince M S
    Aug 5, 2017 at 14:57
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    $\begingroup$ When you use the BDG transformation, your Hamiltonian becomes $\sum_i E_i\gamma_i^\dagger \gamma_i$, where the $\gamma_i$ are the single particle (really single quasiparticle) creation operators. Each $\gamma_i$ is a linear combination of the $c,c^\dagger$. There are $N$ of them. Your Hilbert space is spanned by $|0\rangle, \gamma_1^\dagger|0\rangle, ...,\gamma_N^\dagger|0\rangle, \gamma_1^\dagger\gamma_2^\dagger|0\rangle, ...,\gamma_1^\dagger\gamma_N^\dagger|0\rangle,...$. If you count the number of basis states, there are then $2^N$ basis states. $\endgroup$ Aug 5, 2017 at 17:09
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    $\begingroup$ The terms $c^\dagger c^\dagger$ don't correspond to single particle Hamiltonians. But the whole point of a BDG transformation is to come up with quasiparticle operators $\gamma$ which are linear combinations of $c$ and $c^\dagger$ so that in this basis you do have a single (quasi)particle Hamiltonian. $\endgroup$ Aug 5, 2017 at 17:12

1 Answer 1


The Bogoliubov-de-Gennes Hamiltonian is a mean-field Hamiltonian, that is, a one-body (quadratic) Hamiltonian: it is by no means equivalent to the original many-body Hamiltonian. The two-body forces present in the many-body Hamiltonian, which lead to correlations in the big Hilbert space of size $2^N$, are replaced by effective one-body potentials (e.g. the superconducting order parameter) in the Bogoliubov-de-Gennes Hamiltonian: correlations are lost and the Hilbert space size is reduced to the size of a one-particle Hilbert space.

Let me try to be more accurate. First, one should distinguish the Hilbert space (in which all states have the same number of particles) from the Fock space (which contains states with all possible particle numbers); second, one should distinguish the many-body mean-field Hamiltonian (which contains the $c^+c^+$ terms) from the Bogoliubov-de Gennes Hamiltonian (which does not). I assume that the question is about superconductivity of electrons. A 1-D chain of $N$ sites hosting spin-1/2 fermions (electrons) has a Fock space of size $4^N$, because the Fock space of a single site has dimension 4: it can be either empty, occupied by one electron with up or down spin, or occupied by two electrons. The size of the Hilbert space depends on how many electrons live on the chain, and is generally not $2^N$. It is $2^N$ if one imposes the constraint that there is exactly one electron per site, with up or down spin (like in insulating spin chains). If one simply imposes that there are $N$ electrons on the chain (half-filling), the size of the Hilbert space is $(2N)!/(N!)^2$. The mean-field theory of superconductivity (Bardeen-Cooper-Schrieffer, Bogoliubov-de Gennes, Gor'kov) is formulated in the Fock space. One typically starts from an interacting Hamiltonian with a local attraction $$ H=H_0-g\sum_{i=1}^N c^{\dagger}_{i\uparrow}c^{\dagger}_{i\downarrow}c^{\phantom\dagger}_{i\downarrow}c^{\phantom\dagger}_{i\uparrow}, $$ where $H_0$ is the quadratic part. This Hamiltonian conserves the particle number and will not mix different Hilbert spaces. Then one writes $c^{\dagger}_{i\uparrow}c^{\dagger}_{i\downarrow}=\langle c^{\dagger}_{i\uparrow}c^{\dagger}_{i\downarrow}\rangle+[c^{\dagger}_{i\uparrow}c^{\dagger}_{i\downarrow}-\langle c^{\dagger}_{i\uparrow}c^{\dagger}_{i\downarrow}\rangle]$, where the term in brackets describes fluctuations about the average, similarly for $c_{i\downarrow}c_{i\uparrow}$, and one moves on to the mean-field approximation by discarding terms of second order in the fluctuations: $$ H_{\mathrm{MF}}=H_0+\sum_{i=1}^N \left(\Delta^{\phantom*}_ic^{\dagger}_{i\uparrow}c^{\dagger}_{i\downarrow}+\Delta^*_ic^{\phantom\dagger}_{i\downarrow}c^{\phantom\dagger}_{i\uparrow}+\frac{|\Delta_i|^2}{g}\right), $$ with $\Delta_i=-g\langle c_{i\downarrow}c_{i\uparrow}\rangle$. This is a many-body Hamiltonian, as Jahan correctly emphasizes. Unlike $H$, it does not conserve particle number: applying $H_{\mathrm{MF}}$ to a state with exactly $N$ particles, one gets a mixture of states with $N$, $N+2$, and $N-2$ particles. I would not call $H_{\mathrm{MF}}$ the ''Bogoliubov-de Gennes Hamiltonian'', because we have not made the Bogoliubov transformation at this stage. It is perhaps more common to call it the ''BCS mean-field Hamiltonian''. Now, one performs the Bogoliubov transformation by introducing the Bogoliubov quasiparticles $\gamma^{\dagger}_{n\sigma}=\sum_{i=1}^N\left(u_{in}c^{\dagger}_{i\sigma}+\sigma v_{in}c^{\phantom\dagger}_{i,-\sigma}\right)$. They are not quasiparticles in the sense of Landau quasiparticles, but rather emergent particles carrying unit spectral weight. If the amplitudes $u_{in}$ and $v_{in}$ satisfy $$ H_{\mathrm{BdG}}\begin{pmatrix}u_n\\v_n\end{pmatrix}=E_n\begin{pmatrix}u_n\\v_n\end{pmatrix}, $$ then the Bogoliubov quasiparticles diagonalize the BCS Hamiltonian, which becomes $H_{\mathrm{MF}}=E_0+\sum_{n\sigma}E_n\gamma^{\dagger}_{n\sigma}\gamma^{\phantom\dagger}_{n\sigma}$. This shows that the $\gamma$'s are excitations of $H_{\mathrm{MF}}$ above its ground state of energy $E_0$. $H_{\mathrm{BdG}}$ is what I guess is generally called the ''Bogoliubov-de Gennes Hamiltonian''. It is a one-particle Hamiltonian of size $2N\times2N$ with a symmetric spectrum. It is important to notice that the index $n$ of the Bogoliubov quasiparticles only runs over the $N$ positive eigenvalues $E_n$ of $H_{\mathrm{BdG}}$, such that there are $2N$ operators $\gamma^{\dagger}_{n\sigma}$ like there are $2N$ operators $c^{\dagger}_{i\sigma}$.

About quasiparticles

In a system described by a quadratic Hamiltonian like $H_0$, the electrons are independent: they may be perturbed by a local potential, e.g. a trap or the periodic potential of a crystal, but they do not interact with one another. As a result, when you extract one electron from such a system (for instance using photoemission – the photoelectric effect), you can only gain information about that particular electron: its momentum $\hbar k$ and energy $\varepsilon_k$. Fixing the momentum and scanning the energy $E$, one sees a peak at $E=\varepsilon_k$: all the weight of the single-electron excitations at momentum $\hbar k$ is concentrated at energy $\varepsilon_k$; or, using the jargon, the spectral function $A(k,E)$ is a delta function, $A(k,E)=\delta(E-\varepsilon_k)$, and the spectral weight is unity, $\int_{-\infty}^{\infty}dE\,A(k,E)=1$. Things change when the electrons interact: the spectral function for removal of an electron is spread in energy. This is because the extracted electron is a bare electron, but this bare electron does not exist as such in the system (the elementary excitations of a system of interacting electrons are not bare electrons). Therefore, when the system is photoexcited, a bunch of bare electrons are emitted at momentum $\hbar k$ with a distribution of energies rather than a single energy $\varepsilon_k$. This distribution is precisely the spectral function $A(k,E)$. If the interactions are weak enough, $A(k,E)$ may still have a peak at a certain energy $E_k\approx\varepsilon_k$, on top of a broad background, however the spectral weight of this peak is smaller than unity, because the total weight, including the background, must remain unity. This peak is a Landau quasiparticle: it has a reasonably well-defined momentum and energy, but a spectral weight smaller than one. The Bogoliubov quasiparticles are not Landau quasiparticles, because they emerge as the elementary excitations of a quadratic Hamiltonian. They are a coherent superposition of an electron and a hole, hence their spectral function is made of two delta peaks, $A(k,E)=u_k^2\delta(E-E_k)+v_k^2\delta(E+E_k)$. Clearly these two peaks carry unit spectral weight because $u_k^2+v_k^2=1$.

For completeness, in the last expressions the symbols mean: $E_k=\sqrt{(\varepsilon_k-\mu)^2+\Delta^2}$, $u_k^2=[1+(\varepsilon_k-\mu)/E_k]/2$, and $v_k^2=[1-(\varepsilon_k-\mu)/E_k]/2$.

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    $\begingroup$ This is misleading. It's true that correlations are lost, but that doesn't reduce the size of the Hilbert space. Even a system of N noninteracting particles has a Hilbert space of dimension $2^N$. $\endgroup$ Aug 4, 2017 at 14:34
  • $\begingroup$ thanks @Berthod. I started off from $H_{MF}$ itself so it was quite confusing. I could not quite understand what you mean by "unit spectral weight" could you please explain? Thanks in advance... $\endgroup$
    – Prince M S
    Aug 7, 2017 at 18:53
  • $\begingroup$ 600 characters is too short for explaining this. I have added a paragraph about quasiparticles in the answer. $\endgroup$
    – C. Berthod
    Aug 8, 2017 at 22:58

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