# Fermionic vacuum under a Bogoliubov transformation

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?

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.

• 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? – Lucas Baldo Oct 22 '20 at 19:52
• 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? – Lucas Baldo Oct 22 '20 at 20:01
• 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. – mike stone Oct 22 '20 at 20:22
• I see. Does this mean that the system that originated an $U$ with negative determinant is "unphysical" in some sense? – Lucas Baldo Oct 22 '20 at 21:21
• no if one has an odd number od particles they cannot all be paired. – mike stone Oct 23 '20 at 0:14