Consider a Bogoliubov-de Gennes Hamiltonian,

\begin{align} \hat{H}_{BdG} = \sum_{j,k} \hat{\Psi}_j^{\dagger}H_{jk}\hat{\Psi}_k, \end{align}

where $\hat{\Psi}$ is a $2n$-dimensional vector of fermionic creation operators and its annihilation counterparts, and $H$ is a $2n\times2n$ Hermitian matrix that obeys particle-hole symmetry: $\hat{P}\hat{H}\hat{P} = -\hat{H}^*$. The vacuum of the operators contained in $\hat{\Psi}$ can be defined as the state $\vert 0 \rangle$ such that $\hat{\Psi}_j\vert 0 \rangle = 0$ for all $j=1,...,n$.

In general this Hamiltonian can be put in its diagonal form by diagonalizing $\hat{H}$. If $U$ is the unitary matrix that does so, and we define $\hat{d} = U \hat{\Psi}$, then

\begin{align} \hat{H}_{BdG} &= \sum_{j} \left( \epsilon_j \hat{d}_j^{\dagger}\hat{d}_j - \epsilon_j \hat{d}_j \hat{d}_j^{\dagger} \right) \\ &= \sum_{j} \epsilon_j \left( 2 \hat{d}_j^{\dagger}\hat{d}_j - 1 \right), \end{align}

where $\epsilon_j$ are the positive eigenvalues of $H$.

Here comes the question:

  1. When does the vacuum of the new operators $\hat{d}_j$ exist?

By vacuum I mean the state such that $\hat{d}_j \vert 0 \rangle_{BdG} = 0 \quad \forall ~ j=1,...,n$.

By existing I mean being an element of the Fock space generated by the orignal vacuum $\vert 0 \rangle$ and the original fermionic operators $\hat{\Psi}^{\dagger}_j$.

  1. When it does exist (if it ever does), how can I write it in the original Fock basis, that is, using the original vacuum and the original operators?
  2. Is this vacuum the groundstate of the system?

I believe that these questions might be equivalent to answering

How does the Bogoliubov transformation affect the vacuum of a system?


1 Answer 1


I'm not sure whether you want just the linear algebra, or a more detailed functional analysis that worries about operator convergence. If it's just the former, the following extract from my notes may help:

Let $$ \hat H_{\rm Bogoliubov}= a^\dagger_i H_{ij}a_j +\frac 12 \Delta_{ij} a^\dagger_i a^\dagger_j +\frac 12 \Delta^{\dagger}_{ij} a_i a_j\nonumber\\ = \frac12 \left(\matrix{ a^\dagger_i &a_i}\right)\left(\matrix{ H_{ij}& \phantom {-}\Delta_{ij}\cr \Delta^{\dagger}_{ij}& -H^T_{ij}}\right) \left(\matrix{ a_j\cr a^\dagger_j}\right) +\frac 12 {\rm tr}(H). \nonumber $$

If we arrange for the positive eigenvalues of the BdG operator to be those for $(u,v)^T$ and set
$$ a_i= u_{i\alpha}b_\alpha +v^*_{i\alpha}b^\dagger_\alpha\nonumber\\ a^\dagger_i= v_{i\alpha} b_\alpha +u^*_{i\alpha}b^\dagger_\alpha.\nonumber $$ the mutual orthonormality and completeness of the eigenvectors ensure that the $b_\alpha$, $b^{\dagger}_\alpha$ have the same anti-commutation relations as the $a_i$ $a^\dagger_i$. In terms of the $b_\alpha$ $b^\dagger_\alpha$, the second-quantized Hamiltonian becomes
$$ \hat H_{\rm Bogoliubov} =\sum_{\alpha=1}^N E_\alpha b^\dagger_\alpha b_\alpha -\frac 12 \sum_{\alpha=1}^N E_\alpha +\frac 12\sum_{i=1}^N E^{(0)}_i. $$ Here the $E^{(0)}_i$ are the eigenvalues of $H$. Unlike the $E_\alpha$, these can be of either sign.

If all the $E_\alpha$ are strictly positive, the new ground state is non degenerate and is the unique state $|{0}\rangle_b$ annihilated by all the $b_\alpha$. If we could find a unitary operator ${\mathcal U}$ that acts on the $2^N$-dimensional Fock space such that $$ b_\alpha = a_iu^*_{i\alpha}+ a^\dagger_i v^*_{i\alpha}= {\mathcal U}a_i{\mathcal U}^{-1},\nonumber\\ b^\dagger_\alpha = a^\dagger_iu_{i\alpha}+ a_i v_{i\alpha}= {\mathcal U}a^\dagger_i {\mathcal U}^{-1},\nonumber $$ then we would have $ |{0}\rangle_b={\mathcal U}|{0}\rangle_a $, where $|{0}\rangle_a$ is the no-particle vacuum state annihilated by all the $a_i$. Except in the simplest cases, it is not easy to find a closed-form expression for ${\mathcal U}$. An alternative strategy for obtaining $|{0}\rangle_b$ begins by noting that if that the matrix $u_{i\alpha}$ is invertible then
the condition $b_i |{0}\rangle_b=0$ is equivalent to
$$ (a_i+a^\dagger_k v^*_{k\alpha}(u^*)^{-1}_{\alpha i})|{0}\rangle_b=0, \quad i=1,\ldots N. $$ We therefore introduce the skew-symmetric matrix $$ S_{ij}= v^*_{i\alpha}(u^*)^{-1}_{\alpha j} $$ which satisfies $$ \exp\left\{\frac 12 a^\dagger_ia^\dagger_jS_{ij}\right\} a_k \exp\left\{-\frac 12 a^\dagger_ia^\dagger_jS_{ij}\right\} =a_k+a^\dagger_iS_{ik}. $$ From this we conclude that we can take $ |{0}\rangle_b$ to be $$ |{0}\rangle_b ={\mathcal N} \exp\left\{\frac 12 a^\dagger_ia^\dagger_jS_{ij}\right\}|{0}\rangle_a $$ where $|{0}\rangle_a$ is the original no-particle state. This expression explicitly displays the superconducting ground state as a coherent superposition of Cooper-pair states, and allows us to identify $S_{ij}$ with the (unnormalized) pair wavefunction.

By assuming that the $E_\alpha$ are positive we have swept a potential problem under the rug. When we arrange for the positive energy BdG eigenvectors to be the $(u,v)^T$ and the negative eigenvectors to be $(v^*,u^*)^T$ we may have to interchange columns in the $2N$-by-$2N$ matrix $$ U= \left[\matrix{u &v^*\cr v&u^*}\right]. $$ Each interchange has the effect of changing the sign of ${\rm det} [U]$ and one can show that a negative sign for ${\rm det} [U]$ precludes the invertibility of the $N$-by-$N$ matrix $u$, and hence denies us the skew matrix $S_{ij}$. To avoid this issue we can keep ${\rm det} [U]$ positive, but at the price that one of the $E_\alpha$ --- let us call it $E_{\alpha_0} $ --- may have to remain negative. If so, the lowest energy state has the quasiparticle level $E_{\alpha_0}$ occupied
$$ |{0}\rangle_{\rm ground} \propto b^\dagger_{\alpha_0} \exp\left\{\frac 12 a^\dagger_ia^\dagger_jS_{ij}\right\} |{0}\rangle_a. $$ The state $|{0}\rangle_{\rm ground}$ is therefore a superposition of states with an odd number of particles, one of which is always unpaired.

To see that a negative determinant for $U$ prevents $u$ from the being invertible we consider some properties of $2N$-by-$2N$ unitary matrices of the form $$ U=\left[\matrix{u &v^*\cr v&u^*}\right], \quad U^\dagger= \left[\matrix{u^\dagger &v^\dagger\cr v^T&u^T}\right]. $$ The equations $U U^\dagger=1=U^\dagger U$ give us $$ uu^\dagger+v^*v^T=1= u^\dagger u+v^\dagger v,\nonumber\\ uv^\dagger+v^*u^T=0= u^\dagger v^*+v^\dagger u^*,\nonumber\\ vu^\dagger+u^*v^T=0= v^Tu+u^Tv,\nonumber\\ vv^\dagger +u^* u^T=1= v^Tv^*+u^Tu^*.\nonumber $$ These equations are symmetric under the interchange $u\leftrightarrow v$.

To get $U^*$ from $U$ we need to exchange even number of rows and columns; consequently ${\rm det}[U]= {\rm det}[U^*]$ is a real number. Further $1={\rm det}[U]{\rm det}[U^*]$ tells us that ${\rm det}[U]=\pm 1$. Under the interchange of $u$ and $v$, however, we have $$ \left|\matrix{u &v^*\cr v&u^*}\right| = (-1)^N \left|\matrix{v &u^*\cr u&v^*}\right|. $$ If $u$ is invertible, Schur's determinant identity
$$ \left|\matrix{A &B\cr C&D}\right| ={\rm det}[A] {\rm det}[D- CA^{-1}B] $$ tells us that $$ {\rm det}[U]= {\rm det}[u] {\rm det}[u^*-v u^{-1} v^*]\nonumber\\ = {\rm det}[u] {\rm det}[u^*+v v^\dagger (u^T)^{-1}]\nonumber\\ = {\rm det}[u] {\rm det}[u^*+(1-u^*u^T)(u^T)^{-1}]\nonumber\\ ={\rm det}[u] {\rm det}[(u^T)^{-1}]\nonumber\\ =1. $$ Similarly, if $v$ is invertible the $u\leftrightarrow v$ symmetry converts the above algebra to give $$ (-1)^N {\rm det}[U]=\left|\matrix{v &u^*\cr u&v^*}\right| = {\rm det}[v] {\rm det}[(v^T)^{-1}]=1. $$ We see that when $N$ is even and ${ \rm det}[U]=-1$ neither $u$ nor $v$ can be inverted. When $N$ is odd ${ \rm det}[U]=-1$ precludes $u$ from being inverted, while ${ \rm det}[U]=+1$ precludes $v$ from being inverted.

When $N$ is odd and $v$ is invertible we can define a "full" state that obeys $a^\dagger_i|{\rm full}\rangle=0$ for all $i$ and construct the odd-particle-number ground state $|{0}\rangle_{\rm ground}$ as a paired state of holes.

  • $\begingroup$ Perfect! This exactly what I wanted, and well explained too :) Thank you. I had difficulty in finding a material that talked about this at the level of linear algebra. May I ask for which course these notes were made? $\endgroup$ Oct 22, 2020 at 19:52
  • $\begingroup$ I also have a question regarding the $N even, det[U]=-1$ case. You mention that in order to make $u$ invertible we need to swap the columns so that one of that one $E_{\alpha_0}$ is negative. The columns we exchange are the ones due to $E_{\alpha_0}$ and $-E_{\alpha_0}$, right? And if so, I still don't get how to "choose" which of the $E_{\alpha}$ to leave negative. It doesn't look like it should be a choice, for by choosing different ones it looks like we change the groundstate of the system, which is absurd. But I don't see anything forcing us to pick a specific eigenvalue. What do I miss? $\endgroup$ Oct 22, 2020 at 20:01
  • 1
    $\begingroup$ I would pick the least negative, so the occupied state has lowest energy. The notes were part of an unpublished project on Berry phase of vortices and the Magnus force. I may put them on the Arxiv at some point. The negative determinat cases lead yo "improper" Bogoliubov transformations. $\endgroup$
    – mike stone
    Oct 22, 2020 at 20:22
  • $\begingroup$ I see. Does this mean that the system that originated an $U$ with negative determinant is "unphysical" in some sense? $\endgroup$ Oct 22, 2020 at 21:21
  • 2
    $\begingroup$ @Lucas Baldo Oh they are on my website: people.physics.illinois.edu/stone In the "longer notes" towards the bottom the page. $\endgroup$
    – mike stone
    Jan 12 at 14:43

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