The HFB equations

In terms of the density matrix $\rho$ and pairing tensor $\kappa$, defined as

$$\rho =V^{\ast }V^{T},\quad \kappa =V^{\ast }U^{T}=-UV^{\dagger },$$ (7)

the HFB energy is expressed as an energy functional:

$$E[\rho ,\kappa ] = \frac{\langle \Phi \vert H\vert\Phi \rangle }{\langle \Phi \vert\Phi \rangle }$$

$$= {\rm Tr}\left[ \left(e+\tfrac{1}{2}\Gamma \right)\rho \right] -\tfrac{1}{2}{\rm Tr} \left[ \Delta \kappa^{\ast }\right] ,$$ (8)

where

$$\Gamma_{nm} = \sum_{n'm'}V_{nn'mm'} \rho_{m'n'},$$ (9)

$$\Delta_{nn'} = \tfrac{1}{2}\sum_{mm'}V _{nn'mm'}\kappa_{mm'}.$$ (10)

The variation of the HFB energy (8) with respect to $\rho$ and $\kappa$ yields the HFB equations:

$${\cal H}\left( \begin{array}{c} U_k \\ V_k \end{array} \right) =E_k\left( \begin{array}{c} U_k \\ V_k \end{array} \right) ,$$ (11)

where

$${\cal H}=\left( \begin{array}{cc} e+\Gamma -\lambda & \Delta \\ -\Delta^{\ast } & -(e+\Gamma )^{\ast }+\lambda \end{array} \right) ,$$ (12)

$U_k$ and $V_k$ are the $k$th columns of matrices $U$ and $V$, respectively, and $E_k$ is a positive quasiparticle energy eigenvalue. Since the HFB state $\vert\Phi\rangle$ violates the particle-number symmetry, the Fermi energy $\lambda$ is introduced to fix the average particle number.