SC determinant

.gitignore

@@ -11,3 +11,4 @@ determinant.pdf
 fig?.jpg
 perturbative.pdf
 determinant-v1.1.pdf
+determinant-v1.2.pdf

determinant.tex

@@ -17,7 +17,7 @@
 
 
 %\def\fileversion{\InputIfFileExists{.git/current_version}{\it version~}{\it unversioned}%
-\def\fileversion{\it version~1.1\,%
+\def\fileversion{\it version~1.2\,%
 %(\pdffilesize{\jobname.tex} bytes)}
 \if\pdfstrcmp{\pdffilesize{\jobname.tex}}{21446}0{(original)}\else{(changed)}\fi}
 
@@ -374,6 +374,147 @@ $\langle c_{m}c_{n}^{\dagger}e^{i\lambda\mathcal{Q}}\rangle = \langle e^{i\lambd
 \end{align}
 
 
+
+\section{Superconducting case}
+
+Let us consider the hermitian operator of type
+\begin{gather}
+\widetilde{\mathcal{C}} =
+\sum_{ij}\left[
+a^{\dagger}_{i} C^{(0)}_{ij} a_{j}
+- 
+\frac12 a_{i}C_{ij}^{(1)*}a_{j} + \frac12 a_{i}^{\dagger}C_{ij}^{(1)}a_{j}^{\dagger}
+\right]
+=
+\frac12
+\Tr C^{(0)}
++
+\frac12
+{\mathcal{C}}
+\\
+{\mathcal{C}} =
+\begin{pmatrix}
+a^{\dagger}
+&
+a
+\end{pmatrix}
+C
+\begin{pmatrix}
+a
+\\
+a^{\dagger}
+\end{pmatrix},
+\quad
+C=
+\begin{pmatrix}
+C^{(0)} & C^{(1)}
+\\
+-{C^{(1)}}^{*} & -{C^{(0)}}^\mathsf{T}
+\end{pmatrix}
+\end{gather}
+Bogolyubov transformation will bring us to
+\begin{equation}
+\begin{pmatrix}
+a
+\\
+a^{\dagger}
+\end{pmatrix}
+=
+U
+\begin{pmatrix}
+c
+\\
+c^{\dagger}
+\end{pmatrix}
+\quad
+\mathcal{C} =
+\sum_{i}\lambda_{i}\left[
+c^{\dagger}_{i} c_{i}
+- 
+c_{i} c_{i}^{\dagger}
+\right]
+\end{equation}
+Thus,
+\begin{multline}
+\left[\Tr e^{\mathcal{C}/2}\right]^{2} = \left[\prod_{i=1}^{n} (e^{\lambda_{i}/2}+ e^{-\lambda_{i}/2})\right]^{2}
+= \prod_{i=1}^{n} (e^{\lambda_{i}/2}+ e^{-\lambda_{i}/2})(e^{-\lambda_{i}/2}+ e^{\lambda_{i}/2})
+=\prod_{i=1}^{n} (1+ e^{-\lambda_{i}})(1+ e^{\lambda_{i}})
+=\\=  \det\left(1+\exp\left\{
+\begin{pmatrix}
+\Lambda & 0 \\ 0 & -\Lambda 
+\end{pmatrix}
+\right\}\right) = \det\left(1+\exp\left\{
+\begin{pmatrix}
+C^{(0)} & C^{(1)} \\ -{C^{(1)}}^{*} & -{C^{(0)}}^\mathsf{T}
+\end{pmatrix}
+\right\}\right)
+\end{multline}
+So,
+\begin{equation}
+\Tr e^{\widetilde{\mathcal{C}}} = e^{\frac12 \Tr C^{(0)}}\left[\det(1+e^{C})\right]^{1/2}=
+\left[\det(1+e^{C})\det e^{C^{(0)}}\right]^{1/2}
+\end{equation}
+
+
+\subsection{Commutation rules}
+
+\begin{gather}
+\mathcal{A} = \sum_{mn} \left[ A_{mn}^{(0)}c_{m}^{\dagger} c_{n} + A_{mn}^{(1)}c_{m}^{\dagger} c_{n}^{\dagger} + A_{mn}^{(2)}c_{m} c_{n}\right],
+\qquad
+\mathcal{B} = \sum_{kl} \left[ B_{kl}^{(0)}c_{k}^{\dagger} c_{l} + B_{kl}^{(1)}c_{k}^{\dagger} c_{l}^{\dagger} + B_{kl}^{(2)}c_{k} c_{l} \right], 
+%\label{eq:commut2out1}
+\end{gather}
+The commutator is
+\begin{align*}
+[ \mathcal{A}, \mathcal{B} ]_{-} =&
+\sum_{mnkl} A_{mn}^{(0)} B_{kl}^{(0)}[c_{m}^{\dagger}c_{n},c_{k}^{\dagger}c_{l}]_{-}
+&&= \sum_{mn} \left([A^{(0)},B^{(0)}]_{-}\right)_{mn}c_{m}^{\dagger}c_{n}
+\\+&
+\sum_{mnkl} \left(A_{mn}^{(0)}B_{kl}^{(2)} - B_{mn}^{(0)}A_{kl}^{(2)}  \right) [c_{m}^{\dagger}c_{n},c_{k}c_{l}]_{-}
+&&= \sum_{mn} (-2) \left(B^{(2)} A^{(0)} + A^{(2)} B^{(0)} \right)_{mn}c_{m}c_{n}
+\\+&
+\sum_{mnkl} \left(A_{mn}^{(0)} B_{kl}^{(1)} + B_{mn}^{(0)} A_{kl}^{(1)}\right)[c_{m}^{\dagger}c_{n},c_{k}^{\dagger}c_{l}^{\dagger}]_{-}
+&&= \sum_{mn} (+2) \left(A^{(0)}B^{(1)} + B^{(0)} A^{(1)}\right)_{mn}c_{m}^{\dagger}c_{n}^{\dagger}
+\\+&
+\sum_{mnkl} \left(A_{mn}^{(1)} B_{kl}^{(2)} + B_{mn}^{(1)} A_{kl}^{(2)}\right)[c_{m}^{\dagger}c_{n}^{\dagger},c_{k}c_{l}]_{-}
+&&= \sum_{mn} (+2)\left([A^{(1)},B^{(2)}]_{-} + [B^{(1)},A^{(2)}]_{-}\right)_{mn}c_{m}^{\dagger}c_{n}
+\end{align*}
+If we compare with the expression
+\begin{multline}
+%\frac14
+\left[
+\begin{pmatrix}
+A^{(0)} & A^{(1)} \\ A^{(2)} & A^{(0)\mathsf{T}}
+\end{pmatrix},
+\begin{pmatrix}
+B^{(0)} & B^{(1)} \\ B^{(2)} & B^{(0)\mathsf{T}}
+\end{pmatrix}
+\right]_{-} = \\=
+\begin{pmatrix}
+[A^{(0)},B^{(0)}]_{-} + A^{(1)}B^{(2)} - B^{(1)}A^{(2)}  &
+ A^{(0)}B^{(1)} + A^{(1)}B^{(0)\mathsf{T}} - B^{(0)}A^{(1)} - B^{(1)}A^{(0)\mathsf{T}}
+ \\
+ A^{(2)}B^{(0)} + A^{(0)\mathsf{T}}B^{(2)} - B^{(2)} A^{(0)} - B^{(0)\mathsf{T}} A^{(2)} &
+ [A^{(0)\mathsf{T}},B^{(0)\mathsf{T}}]_{-} + A^{(2)}B^{(1)} - B^{(2)}A^{(1)}
+\end{pmatrix}
+\end{multline}
+we'll see that the commutator of operators is equivalent to the commutator of their Nambu matrices.
+Note that if $A^{(2)}=-A^{(1)*}$ and $B^{(2)}=-B^{(1)*}$ then the same is true for their commutator.
+
+\subsection{Charge on the line}
+
+\begin{equation}
+\langle e^{i\lambda\mathcal{Q}} \rangle = \frac{\Tr \left[e^{i\lambda\mathcal{Q}} e^{-\beta \mathcal{H}}\right]}{\Tr \left[e^{-\beta \mathcal{H}}\right]}
+= e^{i\lambda L/2} \left[ \frac{\det \left[1+e^{i\lambda Q} e^{-\beta H}\right]}{\det \left[1+ e^{-\beta H}\right]} \right]^{1/2}
+\qquad
+Q = 
+\begin{pmatrix}
+\widetilde{Q} & 0 \\ 0 & -\widetilde{Q}
+\end{pmatrix}
+\end{equation}
+
+
+
 \newpage
 \appendix