Expand knowledge base

4 files changed, 576 insertions, 4 deletions

diff --git a/content/know/concept/heisenberg-picture/index.pdc b/content/know/concept/heisenberg-picture/index.pdc
index c49169f..03a386d 100644
--- a/content/know/concept/heisenberg-picture/index.pdc
+++ b/content/know/concept/heisenberg-picture/index.pdc
@@ -19,11 +19,15 @@ mechanics, and is equivalent to the traditionally-taught Schrödinger equation.
 In the Schrödinger picture, the operators (observables) are fixed
 (as long as they do not depend on time), while the state
 $\ket{\psi_S(t)}$ changes according to the Schrödinger equation,
-which can be written using the generator of translations
-$\hat{U}(t) = \exp\!(- i t \hat{H} / \hbar)$ like so:
+which can be written using the generator of translations $\hat{U}(t)$ like so,
+for a time-independent $\hat{H}_S$:
 
 $$\begin{aligned}
     \ket{\psi_S(t)} = \hat{U}(t) \ket{\psi_S(0)}
+    \qquad \quad
+    \boxed{
+        \hat{U}(t) \equiv \exp\!\bigg(\!-\! i \frac{\hat{H}_S t}{\hbar} \bigg)
+    }
 \end{aligned}$$
 
 In contrast, the Heisenberg picture reverses the roles:
diff --git a/content/know/concept/imaginary-time/index.pdc b/content/know/concept/imaginary-time/index.pdc
new file mode 100644
index 0000000..68e4e02
--- /dev/null
+++ b/content/know/concept/imaginary-time/index.pdc
@@ -0,0 +1,170 @@
+---
+title: "Imaginary time"
+firstLetter: "I"
+publishDate: 2021-11-11
+categories:
+- Physics
+- Quantum mechanics
+
+date: 2021-11-05T15:19:29+01:00
+draft: false
+markup: pandoc
+---
+
+# Imaginary time
+
+
+Let $\hat{A}_S$ and $\hat{B}_S$ be time-independent in the Schrödinger picture.
+Then, in the [Heisenberg picture](/know/concept/heisenberg-picture/),
+consider the following expectation value
+with respect to thermodynamic equilibium
+(as found in [Green's functions](/know/concept/greens-functions/) for example):
+
+$$\begin{aligned}
+    \expval*{\hat{A}_H(t) \hat{B}_H(t')}
+    &= \frac{1}{Z} \Tr\!\Big( \exp\!(-\beta \hat{H}_{0,S}(t)) \: \hat{A}_H(t) \: \hat{B}_H(t') \Big)
+\end{aligned}$$
+
+Where the Hamiltonian $\hat{H}_{0,S}$ is time-independent.
+Suppose a time-dependent $\hat{H}_{1,S}$ is added,
+so that the total Hamiltonian is $\hat{H}_S = \hat{H}_{0,S} + \hat{H}_{1,S}$.
+Then it is easier to consider the expectation value
+in the [interaction picture](/know/concept/interaction-picture/):
+
+$$\begin{aligned}
+    \expval*{\hat{A}_H(t) \hat{B}_H(t')}
+    &= \frac{1}{Z} \Tr\!\Big( \exp\!(-\beta \hat{H}_S(t)) \: \hat{K}_I(0, t) \hat{A}_I(t) \hat{K}_I(t, t') \hat{B}_I(t') \hat{K}_I(t', 0) \Big)
+\end{aligned}$$
+
+Where $\hat{K}_I(t, t_0)$ is the time evolution operator of $\hat{H}_{1,S}$.
+In front, we have $\exp\!(-\beta \hat{H}_S(t))$,
+while $\hat{K}_I$ is an exponential of an integral of $\hat{H}_{1,I}$, so we are stuck.
+Keep in mind that exponentials of operators
+cannot just be factorized, i.e. in general
+$\exp\!(\hat{A} \!+\! \hat{B}) \neq \exp\!(\hat{A}) \exp\!(\hat{B})$
+
+To get around this, a useful mathematical trick is
+to use an **imaginary time** variable $\tau$ instead of the real time $t$.
+Fixing a $t$, we "redefine" the interaction picture along the imaginary axis:
+
+$$\begin{aligned}
+    \boxed{
+        \hat{A}_I(\tau)
+        \equiv \exp\!\bigg(\frac{\tau \hat{H}_{0,S}}{\hbar}\bigg) \: \hat{A}_S \: \exp\!\bigg( \!-\! \frac{\tau \hat{H}_{0,S}}{\hbar}\bigg)
+    }
+\end{aligned}$$
+
+Ironically, $\tau$ is real; the point is that this formula
+comes from the real-time definition by replacing $t \to -i \tau$.
+The Heisenberg and Schrödinger pictures can be redefined in the same way.
+
+In fact, by substituting $t \to -i \tau$,
+all the key results of the interaction picture can be updated,
+for example the Schrödinger equation for $\ket{\psi_S(\tau)}$ becomes:
+
+$$\begin{aligned}
+    \hbar \dv{t} \ket{\psi_S(\tau)}
+    = - \hat{H}_S \ket{\psi_S(\tau)}
+    \quad \implies \quad
+    \ket{\psi_S(\tau)}
+    = \exp\!\bigg( \!-\! \frac{\tau \hat{H}_S}{\hbar} \bigg) \ket{\psi_H}
+\end{aligned}$$
+
+And the interaction picture's time evolution operator $\hat{K}_I$
+turns out to be given by:
+
+$$\begin{aligned}
+    \boxed{
+        \hat{K}_I(\tau, \tau_0)
+        = \mathcal{T} \bigg\{ \exp\!\bigg( \!-\! \frac{1}{\hbar} \int_{\tau_0}^\tau \hat{H}_{1,I}(\tau') \dd{\tau'} \bigg) \bigg\}
+    }
+\end{aligned}$$
+
+Where $\mathcal{T}$ is the
+[time-ordered product](/know/concept/time-ordered-product/)
+with respect to $\tau$.
+This operator works as expected:
+
+$$\begin{aligned}
+    \ket{\psi_I(\tau)}
+    = \hat{K}_I(\tau, \tau_0) \ket{\psi_I(\tau_0)}
+\end{aligned}$$
+
+Where $\ket{\psi_I(\tau)}$ is related to
+the Schrödinger and Heisenberg pictures as follows:
+
+$$\begin{aligned}
+    \ket{\psi_I(\tau)}
+    \equiv \exp\!\bigg(\frac{\tau \hat{H}_{0,S}}{\hbar}\bigg) \ket{\psi_S(\tau)}
+    = \exp\!\bigg(\frac{\tau \hat{H}_{0,S}}{\hbar}\bigg) \exp\!\bigg( \!-\! \frac{\tau \hat{H}_S}{\hbar}\bigg) \ket{\psi_H}
+\end{aligned}$$
+
+It is interesting to combine this definition
+with the action of time evolution $\hat{K}_I(\tau, \tau_0)$:
+
+$$\begin{aligned}
+    \ket{\psi_I(\tau)}
+    &= \hat{K}_I(\tau, \tau_0) \ket{\psi_I(\tau_0)}
+    \\
+    \exp\!\bigg(\frac{\tau \hat{H}_{0,S}}{\hbar}\bigg) \exp\!\bigg( \!-\! \frac{\tau \hat{H}_S}{\hbar}\bigg) \ket{\psi_H}
+    &= \hat{K}_I(\tau, \tau_0) \exp\!\bigg(\frac{\tau_0 \hat{H}_{0,S}}{\hbar}\bigg) \exp\!\bigg( \!-\! \frac{\tau_0 \hat{H}_S}{\hbar}\bigg) \ket{\psi_H}
+\end{aligned}$$
+
+Rearranging this leads to the following useful
+alternative expression for $\hat{K}_I(\tau, \tau_0)$:
+
+$$\begin{aligned}
+    \boxed{
+        \hat{K}_I(\tau, \tau_0)
+        = \exp\!\bigg(\frac{\tau \hat{H}_{0,S}}{\hbar}\bigg)
+        \exp\!\bigg(\!-\! \frac{(\tau \!-\! \tau_0) \hat{H}_{S}}{\hbar}\bigg)
+        \exp\!\bigg(\!-\! \frac{\tau_0 \hat{H}_{0,S}}{\hbar}\bigg)
+    }
+\end{aligned}$$
+
+Returning to our initial example,
+we can set $\tau = \hbar \beta$ and $\tau_0 = 0$,
+so $\hat{K}_I(\tau, \tau_0)$ becomes:
+
+$$\begin{aligned}
+    \hat{K}_I(\hbar \beta, 0)
+    &= \exp\!\big(\beta \hat{H}_{0,S}\big) \exp\!\big(\!-\! \beta \hat{H}_{S}\big)
+    \\
+    \implies \quad
+    \exp\!\big(\!-\! \beta \hat{H}_{S}\big)
+    &= \exp\!\big(\!-\! \beta \hat{H}_{0,S}\big) \hat{K}_I(\hbar \beta, 0)
+\end{aligned}$$
+
+Using the easily-shown fact that
+$\hat{K}_I(\hbar \beta, 0) \hat{K}_I(0, \tau) = \hat{K}_I(\hbar \beta, \tau)$,
+we can therefore rewrite the thermodynamic expectation value like so:
+
+$$\begin{aligned}
+    \expval*{\hat{A}_H(\tau) \hat{B}_H(\tau')}
+    &= \frac{1}{Z} \Tr\!\Big(\! \exp\!(-\beta \hat{H}_{0,S}) \hat{K}_I(\hbar \beta, \tau)
+    \hat{A}_I(\tau) \hat{K}_I(\tau, \tau') \hat{B}_I(\tau') \hat{K}_I(\tau', 0) \!\Big)
+\end{aligned}$$
+
+Assuming $\tau > \tau'$,
+we introduce a time-ordering $\mathcal{T}$,
+allowing us to reorder the operators inside,
+and thereby reduce the expression considerably:
+
+$$\begin{aligned}
+    \expval*{\hat{A}_H \hat{B}_H}
+    &= \frac{1}{Z} \Tr\!\Big( \mathcal{T} \Big\{ \hat{K}_I(\hbar \beta, \tau) \hat{K}_I(\tau, \tau') \hat{K}_I(\tau', 0)
+    \hat{A}_I(\tau) \hat{B}_I(\tau') \Big\} \exp\!(-\beta \hat{H}_{0,S}) \Big)
+    \\
+    &= \frac{1}{Z} \Tr\!\Big( \mathcal{T} \Big\{ \hat{K}_I(\hbar \beta, 0) \hat{A}_I(\tau) \hat{B}_I(\tau') \Big\} \exp\!(-\beta \hat{H}_{0,S}) \Big)
+\end{aligned}$$
+
+Where $Z = \Tr\!\big(\exp\!(-\beta \hat{H}_S)\big) = \Tr\!\big(\hat{K}_I(\hbar \beta, 0) \exp\!(-\beta \hat{H}_{0,S})\big)$.
+For another application of imaginary time,
+see e.g. the [Matsubara Green's function](/know/concept/matsubara-greens-function/).
+
+
+
+## References
+1.  H. Bruus, K. Flensberg,
+    *Many-body quantum theory in condensed matter physics*,
+    2016, Oxford.
diff --git a/content/know/concept/interaction-picture/index.pdc b/content/know/concept/interaction-picture/index.pdc
index 1ce330d..45950ff 100644
--- a/content/know/concept/interaction-picture/index.pdc
+++ b/content/know/concept/interaction-picture/index.pdc
@@ -40,8 +40,10 @@ With $\hat{H}_S(t)$ the full Schrödinger Hamiltonian.
 We define the unitary conversion operator:
 
 $$\begin{aligned}
-    \hat{U}(t)
-    \equiv \exp\!\Big( i \frac{\hat{H}_{0,S} t}{\hbar} \Big)
+    \boxed{
+        \hat{U}(t)
+        \equiv \exp\!\bigg( i \frac{\hat{H}_{0,S} t}{\hbar} \bigg)
+    }
 \end{aligned}$$
 
 The interaction-picture states $\ket{\psi_I(t)}$ and operators $\hat{L}_I(t)$
diff --git a/content/know/concept/matsubara-greens-function/index.pdc b/content/know/concept/matsubara-greens-function/index.pdc
new file mode 100644
index 0000000..a89ec43
--- /dev/null
+++ b/content/know/concept/matsubara-greens-function/index.pdc
@@ -0,0 +1,396 @@
+---
+title: "Matsubara Green's function"
+firstLetter: "M"
+publishDate: 2021-11-12
+categories:
+- Physics
+- Quantum mechanics
+
+date: 2021-11-09T14:20:16+01:00
+draft: false
+markup: pandoc
+---
+
+# Matsubara Green's function
+
+The **Matsubara Green's function** is an
+[imaginary-time](/know/concept/imaginary-time/) version
+of the real-time [Green's functions](/know/concept/greens-functions/).
+We define as follows in the imaginary-time
+[Heisenberg picture](/know/concept/heisenberg-picture/):
+
+$$\begin{aligned}
+    \boxed{
+        C_{AB}(\tau, \tau')
+        \equiv -\frac{1}{\hbar} \expval{\mathcal{T} \big\{ \hat{A}(\tau) \hat{B}(\tau') \big\}}
+    }
+\end{aligned}$$
+
+Where the expectation value $\expval{}$ is with respect to thermodynamic equilibrium,
+and $\mathcal{T}$ is the [time-ordered product](/know/concept/time-ordered-product/) pseudo-operator.
+Because the Hamiltonian $\hat{H}$ cannot depend on the imaginary time,
+$C_{AB}$ is a function of the difference $\tau \!-\! \tau'$ only:
+
+$$\begin{aligned}
+    C_{AB}(\tau, \tau')
+    &= - \frac{1}{\hbar Z} \Tr\!\Big( e^{-\beta \hat{H}} \hat{A}(\tau) \hat{B}(\tau') \Big)
+    \\
+    &= - \frac{1}{\hbar Z} \Tr\!\Big( e^{-\beta \hat{H}} e^{\tau \hat{H} / \hbar} \hat{A} e^{-\tau \hat{H} / \hbar}
+    e^{\tau' \hat{H} / \hbar} \hat{B} e^{-\tau' \hat{H} / \hbar} \Big)
+    \\
+    &= - \frac{1}{\hbar Z} \Tr\!\Big( e^{-\beta \hat{H}} e^{(\tau - \tau') \hat{H} / \hbar} \hat{A} e^{-(\tau - \tau') \hat{H} / \hbar} \hat{B} \Big)
+\end{aligned}$$
+
+For $\tau > \tau'$, we see by expanding in the many-particle eigenstates $\ket{n}$
+that we need to demand $\hbar \beta > \tau \!-\! \tau'$ to prevent
+$C_{AB}$ from diverging for increasing temperatures:
+
+$$\begin{aligned}
+    C_{AB}(\tau \!-\! \tau')
+    &= - \frac{1}{\hbar Z} \sum_{n} \matrixel**{n}{e^{-\beta \hat{H}} e^{(\tau - \tau') \hat{H} / \hbar}
+    \hat{A} e^{-(\tau - \tau') \hat{H} / \hbar} \hat{B}}{n}
+    \\
+    &= - \frac{1}{\hbar Z} \sum_{n} \matrixel**{n}{\hat{A} e^{-(\tau - \tau') \hat{H} / \hbar} \hat{B}}{n} e^{-\beta E_n} e^{(\tau - \tau') E_n / \hbar}
+\end{aligned}$$
+
+And likewise, for $\tau < \tau'$,
+we must demand that $\tau \!-\! \tau' > -\hbar \beta$
+for the same reason:
+
+$$\begin{aligned}
+    C_{AB}(\tau \!-\! \tau')
+    &= \mp \frac{1}{\hbar Z} \Tr\!\Big( e^{-\beta \hat{H}} \hat{B}(\tau') \hat{A}(\tau) \Big)
+    \\
+    &= \mp \frac{1}{\hbar Z} \Tr\!\Big( e^{-\beta \hat{H}} e^{-(\tau - \tau') \hat{H} / \hbar} \hat{B} e^{(\tau - \tau') \hat{H} / \hbar} \hat{A} \Big)
+    \\
+    &= \mp \frac{1}{\hbar Z} \sum_{n} \matrixel**{n}{\hat{B} e^{(\tau - \tau') \hat{H} / \hbar} \hat{A}}{n} e^{-\beta E_n} e^{- (\tau - \tau') E_n / \hbar}
+\end{aligned}$$
+
+With $-$ for bosons, and $+$ for fermions,
+due to the time-ordered product for $\tau > \tau'$.
+
+On this domain $[-\hbar \beta, \hbar \beta]$,
+the Matsubara Green's function $C_{AB}$
+obeys a useful shift relation:
+it is $\hbar \beta$-periodic for bosons,
+and $\hbar \beta$-antiperiodic for fermions:
+
+$$\begin{aligned}
+    \boxed{
+        C_{AB}(\tau \!-\! \tau') =
+        \begin{cases}
+            \pm C_{AB}(\tau \!-\! \tau' \!+\! \hbar \beta)
+            & \mathrm{if\;} \tau \!-\! \tau' < 0
+            \\
+            \pm C_{AB}(\tau \!-\! \tau' \!-\! \hbar \beta)
+            & \mathrm{if\;} \tau \!-\! \tau' > 0
+        \end{cases}
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-period"/>
+<label for="proof-period">Proof</label>
+<div class="hidden">
+<label for="proof-period">Proof.</label>
+First $\tau \!-\! \tau' < 0$.
+We insert the argument $\tau \!-\! \tau' \!+\! \hbar \beta$,
+and use the cyclic property:
+
+$$\begin{aligned}
+    C_{AB}(\tau \!-\! \tau' \!+\! \hbar \beta)
+    &= - \frac{1}{\hbar Z} \Tr\!\Big( e^{-\beta \hat{H}} e^{(\tau - \tau' + \hbar \beta) \hat{H} / \hbar}
+    \hat{A} e^{-(\tau - \tau' + \hbar \beta) \hat{H} / \hbar} \hat{B} \Big)
+    \\
+    &= - \frac{1}{\hbar Z} \Tr\!\Big( e^{(\tau - \tau') \hat{H} / \hbar} \hat{A} e^{-(\tau - \tau') \hat{H} / \hbar} e^{-\beta \hat{H}} \hat{B} \Big)
+    \\
+    &= - \frac{1}{\hbar Z} \Tr\!\Big( e^{-\beta \hat{H}} e^{\tau' \hat{H} / \hbar} \hat{B} e^{-\tau' \hat{H} / \hbar}
+    e^{\tau \hat{H} / \hbar} \hat{A} e^{-\tau \hat{H} / \hbar} \Big)
+    \\
+    &= - \frac{1}{\hbar Z} \Tr\!\Big( e^{-\beta \hat{H}} \hat{B}(\tau') \hat{A}(\tau) \Big)
+\end{aligned}$$
+
+Since $\tau < \tau'$ by assumption,
+we can bring back the time-ordered product $\mathcal{T}$:
+
+$$\begin{aligned}
+    C_{AB}(\tau \!-\! \tau' \!+\! \hbar \beta)
+    &= \mp \frac{1}{\hbar Z} \Tr\!\Big( e^{-\beta \hat{H}} \mathcal{T}\big\{ \hat{A}(\tau) \hat{B}(\tau') \big\} \Big)
+    \\
+    &= \pm C_{AB}(\tau \!-\! \tau')
+\end{aligned}$$
+
+Moving on to $\tau \!-\! \tau' > 0$, the proof is perfectly analogous:
+
+$$\begin{aligned}
+    C_{AB}(\tau \!-\! \tau' \!-\! \hbar \beta)
+    &= \mp \frac{1}{\hbar Z} \Tr\!\Big( e^{-\beta \hat{H}} e^{-(\tau - \tau' - \hbar \beta) \hat{H} / \hbar}
+    \hat{B} e^{(\tau - \tau' - \hbar \beta) \hat{H} / \hbar} \hat{A} \Big)
+    \\
+    &= \mp \frac{1}{\hbar Z} \Tr\!\Big( e^{-(\tau - \tau') \hat{H} / \hbar} \hat{B} e^{(\tau - \tau') \hat{H} / \hbar} e^{-\beta \hat{H}} \hat{A} \Big)
+    \\
+    &= \mp \frac{1}{\hbar Z} \Tr\!\Big( e^{-\beta \hat{H}} e^{\tau \hat{H} / \hbar} \hat{A} e^{-\tau \hat{H} / \hbar}
+    e^{\tau' \hat{H} / \hbar} \hat{B} e^{-\tau' \hat{H} / \hbar} \Big)
+    \\
+    &= \mp \frac{1}{\hbar Z} \Tr\!\Big( e^{-\beta \hat{H}} \hat{A}(\tau) \hat{B}(\tau') \Big)
+    \\
+    &= \mp \frac{1}{\hbar Z} \Tr\!\Big( e^{-\beta \hat{H}} \mathcal{T}\big\{ \hat{A}(\tau) \hat{B}(\tau') \big\} \Big)
+    \\
+    &= \pm C_{AB}(\tau \!-\! \tau')
+\end{aligned}$$
+</div>
+</div>
+
+Due to this limited domain $\tau \in [-\hbar \beta, \hbar \beta]$,
+the [Fourier transform](/know/concept/fourier-transform/)
+of $C_{AB}(\tau)$ consists of discrete frequencies
+$k_n \equiv n \pi / (\hbar \beta)$.
+The forward and inverse Fourier transforms
+are therefore defined as given below (with $\tau' = 0$).
+It is convention to write $C_{AB}(i k_n)$ instead of $C_{AB}(k_n)$:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            C_{AB}(i k_n)
+            &\equiv \frac{1}{2} \int_{-\hbar \beta}^{\hbar \beta} C_{AB}(\tau) \: e^{i k_n \tau} \dd{\tau}
+            \\
+            C_{AB}(\tau)
+            &= \frac{1}{\hbar \beta} \sum_{n = -\infty}^\infty C_{AB}(i k_n) e^{-i k_n \tau}
+        \end{aligned}
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-FT-def"/>
+<label for="proof-FT-def">Proof</label>
+<div class="hidden">
+<label for="proof-FT-def">Proof.</label>
+We will prove that one is indeed the inverse of the other.
+We demand that the inverse FT of the forward FT of $C_{AB}(\tau)$
+is simply $C_{AB}(\tau)$ again:
+
+$$\begin{aligned}
+    C_{AB}(\tau)
+    &= \frac{1}{\hbar \beta} \sum_{n = -\infty}^\infty
+    \bigg( \frac{1}{2} \int_{-\hbar \beta}^{\hbar \beta} C_{AB}(\tau') \: e^{i k_n \tau'} \dd{\tau'} \bigg) e^{-i k_n \tau}
+    \\
+    &= \frac{1}{\hbar \beta} \int_{-\hbar \beta}^{\hbar \beta} C_{AB}(\tau')
+    \bigg( \frac{1}{2} \sum_{n = -\infty}^\infty e^{i k_n (\tau' - \tau)} \bigg) \dd{\tau'}
+    \\
+    &= \frac{\pi}{\hbar \beta} \int_{-\hbar \beta}^{\hbar \beta} C_{AB}(\tau')
+    \bigg( \frac{1}{2 \pi} \sum_{n = -\infty}^\infty e^{i \pi n (\tau' - \tau) / \hbar \beta} \bigg) \dd{\tau'}
+\end{aligned}$$
+
+Here, the inner expression turns out to be
+a [Dirac delta function](/know/concept/dirac-delta-function/):
+
+$$\begin{aligned}
+    \frac{1}{2 \pi} \sum_{n = -\infty}^\infty e^{i n x}
+    = \delta(x)
+\end{aligned}$$
+
+From which the rest of the proof follows straightforwardly:
+
+$$\begin{aligned}
+    C_{AB}(\tau)
+    &= \frac{\pi}{\hbar \beta} \int_{-\hbar \beta}^{\hbar \beta} C_{AB}(\tau') \: \delta\big( (\tau' \!-\! \tau) \pi / \hbar \beta \big) \dd{\tau'}
+    \\
+    &= \frac{\pi \hbar \beta}{\pi \hbar \beta} \int_{-\hbar \beta}^{\hbar \beta} C_{AB}(\tau') \: \delta(\tau' \!-\! \tau) \dd{\tau'}
+    \\
+    &= \int_{-\hbar \beta}^{\hbar \beta} C_{AB}(\tau') \: \delta(\tau' \!-\! \tau) \dd{\tau'}
+    \\
+    &= C_{AB}(\tau)
+\end{aligned}$$
+</div>
+</div>
+
+Let us now define the **Matsubara frequencies** $\omega_n$
+as a species-dependent subset of $k_n$:
+
+$$\begin{aligned}
+    \boxed{
+        \omega_n \equiv
+        \begin{cases}
+            \displaystyle\frac{2 n \pi}{\hbar \beta}
+            & \mathrm{bosons}
+            \\
+            \displaystyle\frac{(2 n + 1) \pi}{\hbar \beta}
+            & \mathrm{fermions}
+        \end{cases}
+    }
+\end{aligned}$$
+
+With this, we can rewrite the definition of the forward Fourier transform as follows:
+
+$$\begin{aligned}
+    \boxed{
+        C_{AB}(i \omega_n)
+        = \int_0^{\hbar \beta} C_{AB}(\tau) \: e^{i \omega_n \tau} \dd{\tau}
+        = \int_{-\hbar \beta}^0 C_{AB}(\tau) \: e^{i \omega_n \tau} \dd{\tau}
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-FT-alt"/>
+<label for="proof-FT-alt">Proof</label>
+<div class="hidden">
+<label for="proof-FT-alt">Proof.</label>
+We split the integral, shift its limits,
+and use the (anti)periodicity of $C_{AB}$:
+
+$$\begin{aligned}
+    C_{AB}(i k_n)
+    &= \frac{1}{2} \int_0^{\hbar \beta} C_{AB}(\tau) \: e^{i k_n \tau} \dd{\tau}
+    + \frac{1}{2} \int_{-\hbar \beta}^0 C_{AB}(\tau) \: e^{i k_n \tau} \dd{\tau}
+    \\
+    &= \frac{1}{2} \int_0^{\hbar \beta} C_{AB}(\tau) \: e^{i k_n \tau} \dd{\tau}
+    + \frac{1}{2} \int_0^{\hbar \beta} C_{AB}(\tau \!-\! \hbar \beta) \: e^{i k_n (\tau - \hbar \beta)} \dd{\tau}
+    \\
+    &= \frac{1}{2} \int_0^{\hbar \beta} \Big( C_{AB}(\tau) \pm C_{AB}(\tau) \: e^{-i k_n \hbar \beta} \Big) \: e^{i k_n \tau} \dd{\tau}
+    \\
+    &= \frac{1}{2} \big( 1 \pm e^{-i k_n \hbar \beta} \big) \int_0^{\hbar \beta} C_{AB}(\tau) \: e^{i k_n \tau} \dd{\tau}
+\end{aligned}$$
+
+With $+$ for bosons, and $-$ for fermions. Since $k_n \equiv n \pi / (\hbar \beta)$,
+we know $e^{-i k_n \hbar \beta} \in \{-1, 1\}$,
+so for bosons all odd $n$ vanish, and for fermions all even $n$,
+yielding the desired result.
+
+For the other case, we simply shift the first integral's limits instead of the seconds':
+
+$$\begin{aligned}
+    C_{AB}(i k_n)
+    &= \frac{1}{2} \int_{-\hbar \beta}^0 C_{AB}(\tau \!+\! \hbar \beta) \: e^{i k_n (\tau + \hbar \beta)} \dd{\tau}
+    + \frac{1}{2} \int_0^{\hbar \beta} C_{AB}(\tau) \: e^{i k_n \tau} \dd{\tau}
+    \\
+    &= \frac{1}{2} \int_{-\hbar \beta}^0 \Big( C_{AB}(\tau) \pm C_{AB}(\tau) \: e^{i k_n \hbar \beta} \Big) \: e^{i k_n \tau} \dd{\tau}
+    \\
+    &= \frac{1}{2} \big( 1 \pm e^{-i k_n \hbar \beta} \big) \int_{-\hbar \beta}^0 C_{AB}(\tau) \: e^{i k_n \tau} \dd{\tau}
+\end{aligned}$$
+</div>
+</div>
+
+If we actually evaluate this,
+we obtain the following form of $C_{AB}$,
+which is almost identical to the
+[Lehmann representation](/know/concept/lehmann-representation/)
+of the "ordinary" retarded and advanced Green's functions:
+
+$$\begin{aligned}
+    \boxed{
+        C_{AB}(i \omega_m)
+        = \frac{1}{Z} \sum_{n n'} \frac{\matrixel*{n}{\hat{A}}{n'} \matrixel*{n'}{\hat{B}}{n}}{i \hbar \omega_m + E_n - E_{n'}}
+        \Big( e^{-\beta E_n} \mp e^{- \beta E_{n'}} \Big)
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-Lehmann"/>
+<label for="proof-Lehmann">Proof</label>
+<div class="hidden">
+<label for="proof-Lehmann">Proof.</label>
+For $\tau \!-\! \tau' > 0$, we start by expanding
+in the many-particle eigenstates $\ket{n}$:
+
+$$\begin{aligned}
+    C_{AB}(\tau \!-\! \tau')
+    &= - \frac{1}{\hbar Z} \sum_{n}
+    \matrixel**{n}{e^{-\beta \hat{H}} e^{(\tau - \tau') \hat{H} / \hbar} \hat{A} e^{-(\tau - \tau') \hat{H} / \hbar} \hat{B}}{n}
+    \\
+    &= - \frac{1}{\hbar Z} \sum_{n n'} \matrixel**{n}{e^{-\beta \hat{H}} e^{(\tau - \tau') \hat{H} / \hbar} \hat{A}}{n'}
+    \matrixel**{n'}{e^{-(\tau - \tau') \hat{H} / \hbar} \hat{B}}{n}
+    \\
+    &= - \frac{1}{\hbar Z} \sum_{n n'} e^{-\beta E_n} \matrixel*{n}{\hat{A}}{n'}
+    \matrixel*{n'}{\hat{B}}{n} e^{(E_n - E_{n'})(\tau - \tau') / \hbar}
+\end{aligned}$$
+
+We take the Fourier transform by integrating over $[0, \hbar \beta]$:
+
+$$\begin{aligned}
+    C_{AB}(i \omega_m)
+    &= - \frac{1}{\hbar Z} \sum_{n n'} e^{-\beta E_n} \matrixel*{n}{\hat{A}}{n'}
+    \matrixel*{n'}{\hat{B}}{n} \int_0^{\hbar \beta} e^{(E_n - E_{n'}) \tau / \hbar} e^{i \omega_m \tau} \dd{\tau}
+    \\
+    &= - \frac{1}{\hbar Z} \sum_{n n'} e^{-\beta E_n} \matrixel*{n}{\hat{A}}{n'} \matrixel*{n'}{\hat{B}}{n}
+    \bigg[ \frac{\hbar e^{(i \hbar \omega_m + E_n - E_{n'}) \tau / \hbar}}{i \hbar \omega_m + E_n - E_{n'}} \bigg]_0^{\hbar \beta}
+    \\
+    &= - \frac{1}{Z} \sum_{n n'} e^{-\beta E_n} \frac{\matrixel*{n}{\hat{A}}{n'} \matrixel*{n'}{\hat{B}}{n}}{i \hbar \omega_m + E_n - E_{n'}}
+    \Big( e^{(i \hbar \omega_m + E_n - E_{n'}) \beta} - 1 \Big)
+    \\
+    &= - \frac{1}{Z} \sum_{n n'} \frac{\matrixel*{n}{\hat{A}}{n'} \matrixel*{n'}{\hat{B}}{n}}{i \hbar \omega_m + E_n - E_{n'}}
+    \Big( e^{i \hbar \omega_m \beta} e^{-\beta E_{n'}} - e^{-\beta E_n}  \Big)
+    \\
+    &= \frac{1}{Z} \sum_{n n'} \frac{\matrixel*{n}{\hat{A}}{n'} \matrixel*{n'}{\hat{B}}{n}}{i \hbar \omega_m + E_n - E_{n'}}
+    \Big( e^{-\beta E_n} \mp e^{- \beta E_{n'}} \Big)
+\end{aligned}$$
+
+Moving on to $\tau \!-\! \tau' < 0$,
+we again expand in the many-particle eigenstates $\ket{n}$:
+
+$$\begin{aligned}
+    C_{AB}(\tau \!-\! \tau')
+    &= \mp \frac{1}{\hbar Z} \sum_{n}
+    \matrixel**{n}{e^{-\beta \hat{H}} e^{- (\tau - \tau') \hat{H} / \hbar} \hat{B} e^{(\tau - \tau') \hat{H} / \hbar} \hat{A}}{n}
+    \\
+    &= \mp \frac{1}{\hbar Z} \sum_{n n'} \matrixel**{n}{e^{-\beta \hat{H}} e^{-(\tau - \tau') \hat{H} / \hbar} \hat{B}}{n'}
+    \matrixel**{n'}{e^{(\tau - \tau') \hat{H} / \hbar} \hat{A}}{n}
+    \\
+    &= \mp \frac{1}{\hbar Z} \sum_{n n'} e^{-\beta E_n} \matrixel*{n}{\hat{B}}{n'}
+    \matrixel*{n'}{\hat{A}}{n} e^{-(E_n - E_{n'})(\tau - \tau') / \hbar}
+\end{aligned}$$
+
+Since $\tau \!-\! \tau' < 0$ this time,
+we take the Fourier transform over $[-\hbar \beta, 0]$:
+
+$$\begin{aligned}
+    C_{AB}(i \omega_m)
+    &= \mp \frac{1}{\hbar Z} \sum_{n n'} e^{-\beta E_n} \matrixel*{n}{\hat{B}}{n'}
+    \matrixel*{n'}{\hat{A}}{n} \int_{-\hbar \beta}^0 e^{-(E_n - E_{n'}) \tau / \hbar} e^{i \omega_m \tau} \dd{\tau}
+    \\
+    &= \mp \frac{1}{\hbar Z} \sum_{n n'} e^{-\beta E_n} \matrixel*{n}{\hat{B}}{n'} \matrixel*{n'}{\hat{A}}{n}
+    \bigg[ \frac{\hbar e^{(i \hbar \omega_m - E_n + E_{n'}) \tau / \hbar}}{i \hbar \omega_m - E_n + E_{n'}} \bigg]_{-\hbar \beta}^0
+    \\
+    &= \mp \frac{1}{Z} \sum_{n n'} e^{-\beta E_n} \frac{\matrixel*{n}{\hat{B}}{n'} \matrixel*{n'}{\hat{A}}{n}}{i \hbar \omega_m - E_n + E_{n'}}
+    \Big( 1 - e^{(-i \hbar \omega_m + E_n - E_{n'}) \beta} \Big)
+    \\
+    &= \mp \frac{1}{Z} \sum_{n n'} \frac{\matrixel*{n}{\hat{B}}{n'} \matrixel*{n'}{\hat{A}}{n}}{i \hbar \omega_m - E_n + E_{n'}}
+    \Big( e^{-\beta E_n} - e^{-i \hbar \omega_m \beta} e^{-\beta E_{n'}} \Big)
+    \\
+    &= \mp \frac{1}{Z} \sum_{n n'} \frac{\matrixel*{n}{\hat{B}}{n'} \matrixel*{n'}{\hat{A}}{n}}{i \hbar \omega_m - E_n + E_{n'}}
+    \Big( e^{- \beta E_n} \pm e^{-\beta E_{n'}} \Big)
+    \\
+    &= \frac{1}{Z} \sum_{n n'} \frac{\matrixel*{n}{\hat{B}}{n'} \matrixel*{n'}{\hat{A}}{n}}{i \hbar \omega_m - E_n + E_{n'}}
+    \Big( e^{- \beta E_{n'}} \mp e^{-\beta E_n} \Big)
+\end{aligned}$$
+
+Where swapping $n$ and $n'$ gives the desired result.
+</div>
+</div>
+
+This gives us the primary use of the Matsubara Green's function $C_{AB}$:
+calculating the retarded $C_{AB}^R$ and advanced $C_{AB}^A$.
+Once we have an expression for Matsubara's $C_{AB}$,
+we can recover $C_{AB}^R$ and $C_{AB}^A$ by substituting
+$i \omega_m \to \omega \!+\! i \eta$ and $i \omega_m \to \omega \!-\! i \eta$ respectively.
+
+In general, we can define the **canonical Green's function** $C_{AB}(z)$
+on the complex plane:
+
+$$\begin{aligned}
+    C_{AB}(z)
+    = \frac{1}{Z} \sum_{n n'} \frac{\matrixel*{n}{\hat{A}}{n'} \matrixel*{n'}{\hat{B}}{n}}{z + E_n - E_{n'}}
+    \Big( e^{-\beta E_n} \mp e^{- \beta E_{n'}} \Big)
+\end{aligned}$$
+
+This is a [holomorphic function](/know/concept/holomorphic-function/),
+except for poles on the real axis.
+It turns out that $C_{AB}(z)$ must have these properties
+for the substitution $i \omega_n \to \omega \!\pm\! i \eta$ to be valid.
+
+
+
+## References
+1.  H. Bruus, K. Flensberg,
+    *Many-body quantum theory in condensed matter physics*,
+    2016, Oxford.