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When we calculate the fermionic partition function, we use $\newcommand{\ket}[1]{| #1 \rangle} \newcommand{\bra}[1]{\langle #1 |} \newcommand{\braket}[1]{\langle #1 \rangle}$

$Z=tr(e^{-\beta H})=\int \bra{\bar \psi}e^{-\beta H}\ket{\psi} d\psi$

and in the bosonic case we use

$Z=tr(e^{-\beta H})=\int \bra{ \psi}e^{-\beta H}\ket{\psi} d\psi$

why $\braket{\bar \psi|\psi}=1$? Or following question, why is the resolution of identity given by $\int \ket {\bar \psi}\bra{\psi}d\psi e^{-\bar \psi \psi}d\bar \psi=1$

I understand that there is something to do with the fermionic statistics, but I am not sure how we see it here.

When we calculate the fermionic partition function, we use $\newcommand{\ket}[1]{| #1 \rangle} \newcommand{\bra}[1]{\langle #1 |} \newcommand{\braket}[1]{\langle #1 \rangle}$

$Z=tr(e^{-\beta H})=\int \bra{\bar \psi}e^{-\beta H}\ket{\psi} d\psi$

and in the bosonic case we use

$Z=tr(e^{-\beta H})=\int \bra{ \psi}e^{-\beta H}\ket{\psi} d\psi$

why $\braket{\bar \psi|\psi}=1$? Or following question, why is the resolution of identity given by $\int \ket {\bar \psi}\bra{\psi}d\psi e^{-\bar \psi \psi}d\bar \psi=1$ instead of $\int \ket { \psi}\bra{\psi}d\psi e^{-\bar \psi \psi}d\bar \psi=1$?

I understand that there is something to do with the fermionic statistics, but I am not sure how we see it here.

When we calculate the fermionic partition function, we use $\newcommand{\ket}[1]{| #1 \rangle} \newcommand{\bra}[1]{\langle #1 |} \newcommand{\braket}[2]{\langle #1 | #2 \rangle}$

$Z=tr(e^{-\beta H})=\int \bra{\bar \psi}e^{-\beta H}\ket{\psi} d\psi$

and in the bosonic case we use

$Z=tr(e^{-\beta H})=\int \bra{ \psi}e^{-\beta H}\ket{\psi} d\psi$

why $\braket{\bar \psi|\psi}=1$? Or following question, why is the resolution of identity given by $\int \ket {\bar \psi}\bra{\psi}d\psi e^{-\bar \psi \psi}d\bar \psi=1$

I understand that there is something to do with the fermionic statistics, but I am not sure how we see it here.

When we calculate the fermionic partition function, we use $\newcommand{\ket}[1]{| #1 \rangle} \newcommand{\bra}[1]{\langle #1 |} \newcommand{\braket}[1]{\langle #1 \rangle}$

$Z=tr(e^{-\beta H})=\int \bra{\bar \psi}e^{-\beta H}\ket{\psi} d\psi$

and in the bosonic case we use

$Z=tr(e^{-\beta H})=\int \bra{ \psi}e^{-\beta H}\ket{\psi} d\psi$

why $\braket{\bar \psi|\psi}=1$? Or following question, why is the resolution of identity given by $\int \ket {\bar \psi}\bra{\psi}d\psi e^{-\bar \psi \psi}d\bar \psi=1$

I understand that there is something to do with the fermionic statistics, but I am not sure how we see it here.

When we calculate the fermionic partition function, we use

$Z=tr(e^{-\beta H})=\int \bra{\bar \psi}e^{-\beta H}\ket{\psi} d\psi$

and in the bosonic case we use

$Z=tr(e^{-\beta H})=\int \bra{ \psi}e^{-\beta H}\ket{\psi} d\psi$

why $\braket{\bar \psi|\psi}=1$? Or following question, why is the resolution of identity given by $\int \ket {\bar \psi}\bra{\psi}d\psi e^{-\bar \psi \psi}d\bar \psi=1$

I understand that there is something to do with the fermionic statistics, but I am not sure how we see it here.

When we calculate the fermionic partition function, we use $\newcommand{\ket}[1]{| #1 \rangle} \newcommand{\bra}[1]{\langle #1 |} \newcommand{\braket}[2]{\langle #1 | #2 \rangle}$

$Z=tr(e^{-\beta H})=\int \bra{\bar \psi}e^{-\beta H}\ket{\psi} d\psi$

and in the bosonic case we use

$Z=tr(e^{-\beta H})=\int \bra{ \psi}e^{-\beta H}\ket{\psi} d\psi$

why $\braket{\bar \psi|\psi}=1$? Or following question, why is the resolution of identity given by $\int \ket {\bar \psi}\bra{\psi}d\psi e^{-\bar \psi \psi}d\bar \psi=1$

I understand that there is something to do with the fermionic statistics, but I am not sure how we see it here.