States with even and odd number of fermions

The ambiguity of choosing the occupied quasiparticle orbits is particularly important for odd-particle systems. Equations (1)-(8) look entirely the same irrespective of whether the state $\vert\Psi\rangle$ represents even- or odd-particle systems. While the particle number $N$ is not conserved by the product state $\vert\Psi\rangle$, i.e., this state is a linear superposition of components having different particle numbers, the parity of the particle number, $\pi_N=(-1)^N$, is conserved. That is, the decomposition of $\vert\Psi\rangle$ may contain either even- or odd-particle number components, but never both.

Of course, information on whether the particular vacuum $\vert\Psi\rangle$ is $\pi_N$-even or odd is contained in matrices $U$ and $V$ of the Bogoliubov transformation (1), i.e., it must depend on choices made for the occupied quasiparticle states. Specific choices of occupied quasiparticle states must be made in order to obtain even ($\pi_N=1$) or odd ($\pi_N=-1$) states.

In which way is the information on $\pi_N$ encoded in matrices $U$ and $V$? In the simplest situation, the set of occupied quasiparticle states are such that $U$ is non-singular, $\det(U)\neq0$. Then we can determine the matrix $Z=VU^{-1}$, and express the vacuum $\vert\Psi\rangle$ through the Thouless theorem [17]:

$$\vert\Psi\rangle = \vert\Phi\rangle_{\mbox{\rm\scriptsize {... ...um_{\mu\nu} Z^+_{\mu\nu} a^+_\mu a^+_\nu\right)\vert\rangle ,$$ (9)

where the normalization factor is ${\cal N}=\det^{1/2}(U^+)$. From this expression we see that $\vert\Psi\rangle$ has even number parity, $\pi_N=1$.

Let us now make another choice of selecting the occupied states: we replace one ($\mu$th) column of the matrix of occupied states $\varphi$ by the same ($\mu$th) column of the matrix of empty states $\chi$ (see Eq. (8)), i.e., In this way, we replace the quasiparticle energy of the occupied state $E_\mu$ by $-E_\mu$. We do not imply here that $E_\mu$ must have been positive, so we have replaced a positive quasiparticle energy by a negative one - we could have just replaced the negative one by the positive one. In fact, as discussed earlier, there is no such rule that occupations leading to the original non-singular matrix $U$ must correspond to all quasiparticle energies being positive.

Now it is a matter of simple algebra to see that for such a choice of occupied quasiparticle states, the vacuum state reads

$$\vert\Phi\rangle^{(\mu)}_{\mbox{\rm\scriptsize {odd}}} = {\... ...um_{\mu\nu} Z^+_{\mu\nu} a^+_\mu a^+_\nu\right)\vert\rangle ,$$ (10)

and is a manifestly odd state, $\pi_N=-1$, which we call a one-quasiparticle excitation of $\vert\Phi\rangle_{\mbox{\rm\scriptsize {even}}}$.

Note that in the odd state (12), the matrix $Z$ and normalization constant ${\cal N}$ are defined through the original matrices $U$ and $V$ of the even state (10), and not through those after the column replacement as in Eq. (11). Indeed, it is easy to see that the Thouless theorem does not work for the one-quasiparticle states because matrices $U^{(\mu)}_{\nu'\nu}$ are singular. This is obvious from Eq. (2d), whereby each column of matrix $V^*$ is a linear combination of columns of matrix $U$, i.e.,

$$V^* =- U \left(V^+(U^T)^{-1}\right) .$$ (11)

Therefore, after the column replacement as in Eq. (11), matrices of one-quasiparticle states, $U^{(\mu)}_{\nu'\nu}$, become singular and have null spaces of dimensions $D$=1. Consequently, the corresponding matrices $1-\rho^{(\mu)}=U^{(\mu)}U^{(\mu)+}$, cf. Eqs. (4) and (5), have exactly one eigenvalue equal to zero. Hence, the occupation numbers (eigenvalues of the density matrices $\rho^{(\mu)}$) of one of the single-particle states are in each case equal to 1. This fact is at the origin of the name ``blocked states'' attributed to one-quasiparticle states (12). These states contain fully occupied single-particle states that do not contribute to pairing correlations.

We can continue by building two-quasiparticle states

$$\vert\Phi\rangle^{(\mu\mu')}_{\mbox{\rm\scriptsize {even}}}... ...um_{\mu\nu} Z^+_{\mu\nu} a^+_\mu a^+_\nu\right)\vert\rangle ,$$ (12)

which are manifestly particle-number parity even, $\pi_N=1$. They correspond to two columns in $\varphi$ replaced by two columns of $\chi$, and the corresponding matrices $U^{(\mu\mu')}$ have null spaces of dimensionality $D$=0 or $D$=2. We now have a very definite prescription for telling which Bogoliubov transformations correspond to even and which to odd states, i.e., $\pi_N=(-1)^D$, where $D$ is the dimensionality of the null space of matrix $U$.

The main conclusion of this section is that vacuum states of given $\pi_N$ are obtained by making appropriate choices of occupied quasiparticle states. In particular, one should begin by selecting one even state (10) represented by a non-singular matrix $U$, which one can call a reference state, and than proceed by building on it one-, two-, or many-quasiparticle excitations. Note that constructing odd states is best realized by first building an even reference state and then making one-quasiparticle excitations thereof. This is best done by blocking specific quasiparticles, i.e., replacing columns of matrices as in Eq. (11). After the self-consistent procedure is converged for each blocked state, one may select the lowest one as the ground state of an odd system, and consider the higher ones as good approximations of the excited odd states. It is obvious that self-consistent polarization effects exerted by blocked states, which will be taken into account by iterating Eq. (7), may render reference states of every blocked configuration to be different from one another.

Note also that in the above analysis we did not talk about the average particle numbers, which can be even, odd, or fractional, depending on the value of the Fermi energy $\lambda$ in Eq. (6). Thus one can, in principle, consider odd states with even average particle numbers, or even states with odd average particle numbers. The latter ones provide especially useful reference states for building one-particle excitations on top of them, because they require the smallest readjustment of the Fermi energy between the reference state and a one-quasiparticle excitation.