2
$\begingroup$

Context:

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?

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ 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? $\endgroup$ – Lucas Baldo Oct 22 '20 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? $\endgroup$ – Lucas Baldo Oct 22 '20 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 '20 at 20:22
  • $\begingroup$ I see. Does this mean that the system that originated an $U$ with negative determinant is "unphysical" in some sense? $\endgroup$ – Lucas Baldo Oct 22 '20 at 21:21
  • $\begingroup$ no if one has an odd number od particles they cannot all be paired. $\endgroup$ – mike stone Oct 23 '20 at 0:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.