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This question was motivated by a Mathoverflow posting.

I am comparing two traces of fermionic creation/annihilation operators $a_n,a_n^\dagger$ ($n=1,2,\ldots N$): $$T_{\rm empty}= \operatorname{tr}\bigl(a_1 a_1^\dagger e^{a^\dagger X_1 a}a_1 a_1^\dagger e^{a^\dagger X_2 a} a_1 a_1^\dagger e^{a^\dagger X_3 a}\cdots a_1 a_1^\dagger e^{a^\dagger X_d a}\bigr),$$ $$T_{\rm filled}= \operatorname{tr} \bigl( a_1^\dagger a_1 e^{a^\dagger X_1 a}a_1^\dagger a_1 e^{a^\dagger X_2 a} a_1^\dagger a_1 e^{a^\dagger X_3 a}\cdots a_1^\dagger a_1 e^{a^\dagger X_d a}\bigr).$$ Here $a^\dagger X_i a\equiv\sum_{n,m=1}^N a^\dagger_n (X_i)_{nm}a_m$ and the matrices $X_1,X_2,\ldots X_d$ are arbitrary $N\times N$ complex matrices. So Gaussian operators are alternated with projections onto mode number 1, either empty or filled.

For the projection onto the empty mode I have an expression of the operator trace as a single determinant of an $N\times N$ matrix,$^\ast$ $$T_{\rm empty}=\det\bigl(I+Qe^{X_1}Qe^{X_2}\cdots Qe^{X_d}\bigr),\;\;Q=\text{diagonal}(0,1,1,\ldots 1).\tag{1}$$ For the projection onto the filled mode, I can substitute $a_1^\dagger a_1=1-a_1 a_1^\dagger $ and write $T_{\rm filled}$ as a sum of $2^d$ determinants (with $Q^0\equiv I$): $$T_{\rm filled}=\sum_{p_1,p_2,\ldots p_d=0}^1 (-1)^{\sum_i p_i}\det\bigl(I+Q^{p_1}e^{X_1}Q^{p_2}e^{X_2}\cdots Q^{p_d}e^{X_d}\bigr).\tag{2}$$ I do not have a single-determinant expression for $T_{\rm filled}$.

The different complexity of the evaluation of $T_{\rm empty}$ and $T_{\rm filled}$ puzzles me. Is there a high-level reason for this? What is a more efficient way to evaluate $T_{\rm filled}$ for large $d$ and large $N$?

$^\ast$ To derive the expression (1) for $T_{\rm empty}$ I represent the empty-mode projector $a_1a_1^\dagger$ as the limit of a Gaussian operator, $$a_1a_1^\dagger=\lim_{f\rightarrow \infty}e^{-fa_1^\dagger a_1}=\lim_{f\rightarrow\infty}e^{-fa^\dagger Pa},\;\;P=\operatorname{diag}(1,0,0,\ldots 0)=I-Q,$$ and then use the determinantal expression for the trace of a product of Gaussian operators: \begin{align} T_{\rm empty}={}&\lim_{f\rightarrow \infty}\operatorname{tr}\bigl(e^{-fa^\dagger Pa} e^{a^\dagger X_1 a}e^{-fa^\dagger Pa} e^{a^\dagger X_2 a} \cdots e^{-fa^\dagger Pa} e^{a^\dagger X_d a}\bigr)\\ ={}& \lim_{f\rightarrow \infty}\operatorname{det}\bigl(I+e^{-fP}e^{X_1}e^{-fP}e^{X_2}\cdots e^{-fP}e^{X_d}\bigr)= \operatorname{det}\bigl(I+Qe^{X_1}Qe^{X_2}\cdots Qe^{X_d}\bigr). \end{align}
Carlo Beenakker
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1 Answers1

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Let us consider the trace formula for three fermionic quadratic forms $X_1$, $X_2$, and $X_3$. We want to evaluate \begin{align} T_{\mathrm{empty}}&=\mathrm{tr}(a_{1}a_{1}^{\dagger}e^{a^{\dagger}X_{1}a}a_{1}a_{1}^{\dagger}e^{a^{\dagger}X_{2}a}a_{1}a_{1}^{\dagger}e^{a^{\dagger}X_{3}a})\\T_{\mathrm{filled}}&=\mathrm{tr}(a_{1}^{\dagger}a_{1}e^{a^{\dagger}X_{1}a}a_{1}^{\dagger}a_{1}e^{a^{\dagger}X_{2}a}a_{1}^{\dagger}a_{1}e^{a^{\dagger}X_{3}a}) \end{align} For $T_\mathrm{filled}$, we rewrite the quadratic forms on the exponential in a reverse order as \begin{align} T_{\mathrm{filled}}&=\mathrm{tr}(a_{1}^{\dagger}a_{1}e^{\mathrm{tr}X_{1}}e^{-aX_{1}^{T}a^{\dagger}}a_{1}^{\dagger}a_{1}e^{\mathrm{tr}X_{2}}e^{-aX_{2}^{T}a^{\dagger}}a_{1}^{\dagger}a_{1}e^{\mathrm{tr}X_{3}}e^{-a^{\dagger}X_{3}^{T}a})\\&=e^{\mathrm{tr}X_{1}+\mathrm{tr}X_{2}+\mathrm{tr}X_{3}}\mathrm{tr}(a_{1}^{\dagger}a_{1}e^{-aX_{1}^{T}a^{\dagger}}a_{1}^{\dagger}a_{1}e^{-aX_{2}^{T}a^{\dagger}}a_{1}^{\dagger}a_{1}e^{-a^{\dagger}X_{3}^{T}a}), \end{align} where the minus sign comes from the anti-commutation relation, and $e^{\mathrm{tr}X_{1}+\mathrm{tr}X_{2}+\mathrm{trX_3}}$ comes from the diagonal part of the anti-commutation relation $a_j^\dagger a_j=1-a_j a_j^\dagger$.

For fermionic operators, one can use the particle-hole symmetry to transform $a\leftrightarrow a^\dagger$, and there is a similar trace formula between the two above expressions \begin{align} T_{\mathrm{empty}}&=\det(I+Qe^{X_{1}}Qe^{X_{2}}Qe^{X_{3}})\\T_{\mathrm{filled}}&=e^{\mathrm{tr}X_{1}+\mathrm{tr}X_{2}+\mathrm{tr}X_{3}}\det(I+Qe^{-X_{1}^{T}}Qe^{-X_{2}^{T}}Qe^{-X_{3}^{T}}), \end{align} where $Q=\mathrm{diag}(0,1,1,...,1)$.

Jinfu Chen
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