Residues

Let us now discuss the residues of the integrand (35). From Eqs. (25) and (33), we see that the integrand is an odd function of $z$,

$${\cal{}E}_N(-z)=-{\cal{}E}_N(z) .$$ (36)

This is obvious for even particle numbers $N$, for which Eq. (25) has been derived, while for odd $N$, an additional power of $z$ appears when shifting the blocked HFB state,

$$\vert\Phi^{\mbox{\rm\scriptsize {odd}}}\rangle=a_{n_0}^+\pro... ...0}>0} \left(u_n+v_n{a}_n^+{a}_{\bar{n}}^+\right)\vert\rangle ,$$ (37)

i.e.,

$$\vert\Phi^{\mbox{\rm\scriptsize {odd}}}_N(z)\rangle=za_{n_0... ...left(u_n+z^2\,v_n{a}_n^+{a}_{\bar{n}}^+\right) \vert\rangle ,$$ (38)

which gives

$$\langle\Phi^{\mbox{\rm\scriptsize {odd}}}\vert\Phi^{\mbox{\r... ...rangle = z \prod_{n\neq{n_0}>0}\left(u_n^2+ z^2 v_n^2\right) ,$$ (39)

and renders the integrand (35) an odd function of $z$ also for odd systems.

Near the pole (28), the term in the integrand that produces the residue has the structure:

$${\cal{}E}_N(z)\simeq \frac{{\cal{}R}_n(z)}{u_n^2+z^2v_n^2},$$ (40)

where ${\cal{}R}_n(z)$ is an odd function of $z$, regular at the pole. Similarly, for the pole at $z$=0 we have

$${\cal{}E}_N(z)\simeq \frac{{\cal{}R}_0(z)}{z}.$$ (41)

Therefore, for pairs of poles that are symmetric with respect to $z$=0, the residues,

$$\begin{array}{rcl} \raisebox{-1.5ex}{$\stackrel{\mbox{Res}}... ..._n({i}\vert u_n/v_n\vert)}{2{i}\vert u_nv_n\vert} , \end{array}$$ (42)

have identical values. Hence, poles below and above the real axis yield the same contribution to the contour integral. Based on this consideration, the projected DFT energy (34), expressed in terms of residues, reads:

$$E_{\mbox{\rm\scriptsize {DFT}}}^N= \frac{{\cal{}R}_0(0)+2\... ...\mbox{\scriptsize {$z=0$}}}$}\,\langle\Phi\vert\Phi(z)\rangle}$$ (43)

or

$$E_{\mbox{\rm\scriptsize {DFT}}}^N= \sum_{n=0}^{\bar{n}}E_{\mbox{\rm\scriptsize {DFT}}}^N(n),$$ (44)

where $E_{\mbox{\rm\scriptsize {DFT}}}^N(n)$ denotes the contribution from the $n$th pole, including the $n$=0 pole at the origin up to $n=\bar{n}$ (last pole encircled by $C$).

As an example, we explicitly calculate the residues for a term that depends on the squared particle density,

$$E_{\mbox{\rm\scriptsize {DFT}}}(\rho_z)= C^\rho \int {\rm d}^3{\bf r}\rho^2_z({\bf r}) ,$$ (45)

with

$$\rho_z({\bf r}) = \displaystyle \sum_n \frac{z^2v_n^2} ... ...^2+z^2v_n^2}~ \sum_\sigma\vert\varphi_n({\bf r}\sigma)\vert^2$$ (46)

and $C^\rho$ being a coupling constant. Assuming a two-fold Kramers degeneracy, the corresponding residue at $\pm{i}\vert u_n/v_n\vert$ is:

$$\raisebox{-1.5ex}{$\stackrel{\mbox{Res}}{\mbox{\scriptsize ... ...>0}v_m^2\left(\frac{u_m^2}{v_m^2}-\frac{u_n^2}{v_n^2}\right) .$$ (47)

One can see that residues can be very large for poles corresponding to canonical states that have occupation numbers close to 1. These very large contributions to the projected DFT energy must be compensated by a similarly large contribution from the single pole at $z$=0. Therefore, within the DFT formalism, one cannot use the HFB expression (15) that involves only one residue at $z$=0.

Recall from our discussion in Sec. 3.3 that the poles have the order of $p-q$. In the above example, the polynomial order is $p$=2; hence, the residue (47) must vanish if the degeneracy factor $q\ge 2$. This is indeed the case as for $q>1$ $u^2_m=u^2_n$ for at least one value of $m\ne n$.