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4 files changed, 593 insertions, 2 deletions

diff --git a/content/know/concept/drude-model/index.pdc b/content/know/concept/drude-model/index.pdc
new file mode 100644
index 0000000..a738dff
--- /dev/null
+++ b/content/know/concept/drude-model/index.pdc
@@ -0,0 +1,236 @@
+---
+title: "Drude model"
+firstLetter: "D"
+publishDate: 2021-09-23
+categories:
+- Physics
+- Electromagnetism
+- Optics
+
+date: 2021-09-23T16:22:51+02:00
+draft: false
+markup: pandoc
+---
+
+# Drude model
+
+The **Drude model** classically predicts
+the dielectric function and electric conductivity of a gas of free charge carriers,
+as found in metals and doped semiconductors.
+
+
+## Metals
+
+An [electromagnetic wave](/know/concept/electromagnetic-wave-equation/)
+has an oscillating [electric field](/know/concept/electric-field/)
+$E(t) = E_0 \exp\!(- i \omega t)$
+that exerts a force on the charge carriers,
+which have mass $m$ and charge $q$.
+They thus obey the following equation of motion,
+where $\gamma$ is a frictional damping coefficient:
+
+$$\begin{aligned}
+    m \dv[2]{x}{t} + m \gamma \dv{x}{t}
+    = q E_0 \exp\!(- i \omega t)
+\end{aligned}$$
+
+Inserting the ansatz $x(t) = x_0 \exp\!(- i \omega t)$
+and isolating for the displacement $x_0$ yields:
+
+$$\begin{aligned}
+    - x_0 m \omega^2 - i x_0 m \gamma \omega
+    = q E_0
+    \quad \implies \quad
+    x_0
+    = - \frac{q E_0}{m (\omega^2 + i \gamma \omega)}
+\end{aligned}$$
+
+The polarization density $P(t)$ is therefore as shown below.
+Note that the dipole moment $p$ goes from negative to positive,
+and the electric field $E$ from positive to negative.
+Let $N$ be the density of carriers in the gas, then:
+
+$$\begin{aligned}
+    P(t)
+    = N p(t)
+    = N q x(t)
+    = - \frac{N q^2}{m (\omega^2 + i \gamma \omega)} E(t)
+\end{aligned}$$
+
+The electric displacement field $D$ is thus as follows,
+where $\varepsilon_r$ is the unknown relative permittivity of the gas,
+which we will find shortly:
+
+$$\begin{aligned}
+    D
+    = \varepsilon_0 \varepsilon_r E
+    = \varepsilon_0 E + P
+    = \varepsilon_0 \bigg( 1 - \frac{N q^2}{\varepsilon_0 m} \frac{1}{\omega^2 + i \gamma \omega} \bigg) E
+\end{aligned}$$
+
+The parenthesized expression is the desired dielectric function $\varepsilon_r$,
+which depends on $\omega$:
+
+$$\begin{aligned}
+    \boxed{
+        \varepsilon_r(\omega)
+        %= 1 - \frac{N q^2}{\varepsilon_0 m} \frac{1}{\omega^2 + i \gamma \omega}
+        = 1 - \frac{\omega_p^2}{\omega^2 + i \gamma \omega}
+    }
+\end{aligned}$$
+
+Where we have defined the important so-called **plasma frequency** like so:
+
+$$\begin{aligned}
+    \boxed{
+        \omega_p
+        \equiv \sqrt{\frac{N q^2}{\varepsilon_0 m}}
+    }
+\end{aligned}$$
+
+If $\gamma = 0$, then $\varepsilon_r$ is
+negative $\omega < \omega_p$,
+positive for $\omega > \omega_p$,
+and zero for $\omega = \omega_p$.
+Respectively, this leads to
+an imaginary index $\sqrt{\varepsilon_r}$ (high absorption),
+a real index tending to $1$ (transparency),
+and the possibility of self-sustained plasma oscillations.
+For metals, $\omega_p$ lies in the UV.
+
+We can refine this result for $\varepsilon_r$,
+by recognizing the (mean) velocity $v = \dv*{x}{t}$,
+and rewriting the equation of motion accordingly:
+
+$$\begin{aligned}
+    m \dv{v}{t} + m \gamma v = q E(t)
+\end{aligned}$$
+
+Note that $m v$ is simply the momentum $p$.
+We define the **momentum scattering time** $\tau \equiv 1 / \gamma$,
+which represents the average time between collisions,
+where each collision resets the involved particles' momentums to zero.
+Or, more formally:
+
+$$\begin{aligned}
+    \dv{p}{t}
+    = - \frac{p}{\tau} + q E
+\end{aligned}$$
+
+Returning to the equation for the mean velocity $v$,
+we insert the ansatz $v(t) = v_0 \exp\!(- i \omega t)$,
+for the same electric field $E(t) = E_0 \exp\!(-i \omega t)$ as before:
+
+$$\begin{aligned}
+    - i m \omega v_0 + \frac{m}{\tau} v_0 = q E_0
+    \quad \implies \quad
+    v_0 = \frac{q \tau}{m (1 - i \omega \tau)} E_0
+\end{aligned}$$
+
+From $v(t)$, we find the resulting average current density $J(t)$ to be as follows:
+
+$$\begin{aligned}
+    J(t)
+    = - N q v(t)
+    = \sigma E(t)
+\end{aligned}$$
+
+Where $\sigma(\omega)$ is the **AC conductivity**,
+which depends on the **DC conductivity** $\sigma_0$:
+
+$$\begin{aligned}
+    \boxed{
+        \sigma
+        = \frac{\sigma_0}{1 - i \omega \tau}
+    }
+    \qquad \quad
+    \boxed{
+        \sigma_0
+        = \frac{N q^2 \tau}{m}
+    }
+\end{aligned}$$
+
+We can use these quantities to rewrite
+the dielectric function $\varepsilon_r$ from earlier:
+
+$$\begin{aligned}
+    \boxed{
+        \varepsilon_r(\omega)
+        = 1 + \frac{i \sigma(\omega)}{\varepsilon_0 \omega}
+    }
+\end{aligned}$$
+
+
+## Doped semiconductors
+
+Doping a semiconductor introduces
+free electrons (n-type)
+or free holes (p-type),
+which can be treated as free particles
+moving in the bands of the material.
+
+The Drude model can also be used in this case,
+by replacing the actual carrier mass $m$
+by the effective mass $m^*$.
+Furthermore, semiconductors already have
+a high intrinsic permittivity $\varepsilon_{\mathrm{int}}$
+before the dopant is added,
+so the diplacement field $D$ is:
+
+$$\begin{aligned}
+    D
+    = \varepsilon_0 E + P_{\mathrm{int}} + P_{\mathrm{free}}
+    = \varepsilon_{\mathrm{int}} \varepsilon_0 E - \frac{N q^2}{m^* (\omega^2 + i \gamma \omega)} E
+\end{aligned}$$
+
+Where $P_{\mathrm{int}}$ is the intrinsic undoped polarization,
+and $P_{\mathrm{free}}$ is the contribution of the free carriers.
+The dielectric function $\varepsilon_r(\omega)$ is therefore given by:
+
+$$\begin{aligned}
+    \boxed{
+        \varepsilon_r(\omega)
+        %= \varepsilon_{\mathrm{int}}(\omega) - \frac{N q^2}{\varepsilon_0 m^*} \frac{1}{\omega^2 + i \gamma \omega}
+        = \varepsilon_{\mathrm{int}} \Big( 1 - \frac{\omega_p^2}{\omega^2 + i \gamma \omega} \Big)
+    }
+\end{aligned}$$
+
+Where the plasma frequency $\omega_p$ has been redefined as follows
+to include $\varepsilon_\mathrm{int}$:
+
+$$\begin{aligned}
+    \boxed{
+        \omega_p
+        = \sqrt{\frac{N q^2}{\varepsilon_{\mathrm{int}} \varepsilon_0 m^*}}
+    }
+\end{aligned}$$
+
+The meaning of $\omega_p$ is the same as for metals,
+with high absorption for $\omega < \omega_p$.
+However, due to the lower carrier density $N$ in a semiconductor,
+$\omega_p$ lies in the IR rather than UV.
+
+However, instead of asymptotically going to $1$ for $\omega > \omega_p$ like a metal,
+$\varepsilon_r$ tends to $\varepsilon_\mathrm{int}$ instead,
+and crosses $1$ along the way,
+at which point the reflectivity is zero.
+This occurs at:
+
+$$\begin{aligned}
+    \omega^2
+    = \frac{\varepsilon_{\mathrm{int}}}{\varepsilon_{\mathrm{int}} - 1} \omega_p^2
+\end{aligned}$$
+
+This is used to experimentally determine the effective mass $m^*$
+of the doped semiconductor,
+by finding which value of $m^*$ gives the measured $\omega$.
+
+
+
+## References
+1.  M. Fox,
+    *Optical properties of solids*, 2nd edition,
+    Oxford.
+2.  S.H. Simon,
+    *The Oxford solid state basics*,
+    Oxford.
diff --git a/content/know/concept/fabry-perot-cavity/index.pdc b/content/know/concept/fabry-perot-cavity/index.pdc
new file mode 100644
index 0000000..2f1f84f
--- /dev/null
+++ b/content/know/concept/fabry-perot-cavity/index.pdc
@@ -0,0 +1,128 @@
+---
+title: "Fabry-Pérot cavity"
+firstLetter: "F"
+publishDate: 2021-09-18
+categories:
+- Physics
+- Optics
+
+date: 2021-09-18T00:42:59+02:00
+draft: false
+markup: pandoc
+---
+
+# Fabry-Pérot cavity
+
+In its simplest form, a **Fabry-Pérot cavity**
+is a region of light-transmitting medium
+surrounded by two mirrors,
+which may transmit some of the incoming light.
+Such a setup can be used as e.g. an interferometer or a laser cavity.
+
+
+## Modes of macroscopic cavity
+
+Consider a Fabry-Pérot cavity large enough
+that we can neglect the mirrors' thicknesses,
+which have reflection coefficients $r_L$ and $r_R$.
+Let $\tilde{n}_C$ be the complex refractive index inside,
+and $\tilde{n}_L$ and $\tilde{n}_R$ be the indices outside.
+The cavity has length $L$, centered on $x = 0$.
+
+To find the quasinormal modes,
+we make the following ansatz, with mode number $m$:
+
+$$\begin{aligned}
+    E_m(x)
+    =
+    \begin{cases}
+        A_m \exp\!(-i \tilde{n}_L \tilde{k}_m x) & \mathrm{if}\; x < -L/2 \\
+        B_m \exp\!(i \tilde{n}_C \tilde{k}_m x) + C_m \exp\!(-i \tilde{n}_C \tilde{k}_m x) & \mathrm{if}\; -\!L/2 < x < L/2 \\
+        D_m \exp\!(i \tilde{n}_R \tilde{k}_m x) & \mathrm{if}\; L/2 < x
+    \end{cases}
+\end{aligned}$$
+
+On the left, $B_m$ is the reflection of $C_m$,
+and on the right, $C_m$ is the reflection of $B_m$,
+where the reflected amplitude is determined
+by the coefficients $r_L$ and $r_L$, respectively:
+
+$$\begin{aligned}
+    B_m \exp\!(-i \tilde{n}_C \tilde{k}_m L/2)
+    &= r_L C_m \exp\!(i \tilde{n}_C \tilde{k}_m L/2)
+    \\
+    C_m \exp\!(-i \tilde{n}_C \tilde{k}_m L/2)
+    &= r_R B_m \exp\!(i \tilde{n}_C \tilde{k}_m L/2)
+\end{aligned}$$
+
+These equations might seem to contradict each other.
+We recast them into matrix form:
+
+$$\begin{aligned}
+    \begin{bmatrix}
+        1 & - r_L \exp\!(i \tilde{n}_C \tilde{k}_m L) \\
+        - r_R \exp\!(i \tilde{n}_C \tilde{k}_m L) & 1
+    \end{bmatrix}
+    \cdot
+    \begin{bmatrix}
+        B_m \\ C_m
+    \end{bmatrix}
+    =
+    \begin{bmatrix}
+        0 \\ 0
+    \end{bmatrix}
+\end{aligned}$$
+
+Now, we do not want to be able to find values for $B_m$ and $C_m$
+that satisfy this for a given $\tilde{k}_m$.
+Instead, we only want specific values of $\tilde{k}_m$ to be allowed,
+corresponding to the cavity's modes.
+We thus demand that the determinant to zero:
+
+$$\begin{aligned}
+    0
+    &= 1 - r_L r_R \exp\!(i 2 \tilde{n}_C \tilde{k}_m L)
+\end{aligned}$$
+
+Isolating this for $\tilde{k}_m$ yields the following modes,
+where $m$ is an arbitrary integer:
+
+$$\begin{aligned}
+    \boxed{
+        \tilde{k}_m
+        = - \frac{\ln\!(r_L r_R) + i 2 \pi m}{i 2 \tilde{n}_C L}
+    }
+\end{aligned}$$
+
+These $\tilde{k}_m$ satisfy the matrix equation above.
+Thanks to linearity, we can choose one of $B_m$ or $C_m$,
+and then the other is determined by the corresponding equation.
+
+Finally, we look at the light transmitted through the mirrors,
+according to $1 \!-\! r_L$ and $1 \!-\! r_R$:
+
+$$\begin{aligned}
+    A_m \exp\!(i \tilde{n}_L \tilde{k}_m L/2)
+    &= (1 - r_L) C_m \exp\!(i \tilde{n}_C \tilde{k}_m L/2)
+    \\
+    D_m \exp\!(i \tilde{n}_R \tilde{k}_m L/2)
+    &= (1 - r_R) B_m \exp\!(i \tilde{n}_C \tilde{k}_m L/2)
+\end{aligned}$$
+
+We simply isolate for $A_m$ and $D_m$ respectively,
+yielding the following amplitudes:
+
+$$\begin{aligned}
+    A_m
+    &= (1 - r_L) C_m \exp\!\big( i (\tilde{n}_C \!-\! \tilde{n}_L) \tilde{k}_m L/2 \big)
+    \\
+    D_m
+    &= (1 - r_R) B_m \exp\!\big( i (\tilde{n}_C \!-\! \tilde{n}_R) \tilde{k}_m L/2 \big)
+\end{aligned}$$
+
+
+
+## References
+1.  P.T. Kristensen, K. Herrmann, F. Intravaia, K. Busch,
+    [Modeling electromagnetic resonators using quasinormal modes](https://doi.org/10.1364/AOP.377940),
+    2020, Optical Society of America.
diff --git a/content/know/concept/heisenberg-picture/index.pdc b/content/know/concept/heisenberg-picture/index.pdc
index 9e4887d..c49169f 100644
--- a/content/know/concept/heisenberg-picture/index.pdc
+++ b/content/know/concept/heisenberg-picture/index.pdc
@@ -20,7 +20,7 @@ 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:
+$\hat{U}(t) = \exp\!(- i t \hat{H} / \hbar)$ like so:
 
 $$\begin{aligned}
     \ket{\psi_S(t)} = \hat{U}(t) \ket{\psi_S(0)}
@@ -100,7 +100,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 This equation is closer to classical mechanics than the Schrödinger picture:
-inserting the position $\hat{X}$ and momentum $\hat{P} = - i \hbar \: d/d\hat{X}$
+inserting the position $\hat{X}$ and momentum $\hat{P} = - i \hbar \: \dv*{\hat{X}}$
 gives the following Newton-style equations:
 
 $$\begin{aligned}
diff --git a/content/know/concept/kubo-formula/index.pdc b/content/know/concept/kubo-formula/index.pdc
new file mode 100644
index 0000000..9e52835
--- /dev/null
+++ b/content/know/concept/kubo-formula/index.pdc
@@ -0,0 +1,227 @@
+---
+title: "Kubo formula"
+firstLetter: "K"
+publishDate: 2021-09-23
+categories:
+- Physics
+- Quantum mechanics
+
+date: 2021-09-23T16:21:51+02:00
+draft: false
+markup: pandoc
+---
+
+# Kubo formula
+
+Consider the following quantum Hamiltonian,
+split into a main time-independent term $\hat{H}_{0,S}$
+and a small time-dependent perturbation $\hat{H}_{1,S}$,
+which is turned on at $t = t_0$:
+
+$$\begin{aligned}
+    \hat{H}_S(t)
+    = \hat{H}_{0,S} + \hat{H}_{1,S}(t)
+\end{aligned}$$
+
+And let $\ket{\psi_S(t)}$ be the corresponding solutions to the Schrödinger equation.
+Then, given a time-independent observable $\hat{A}$,
+its expectation value $\expval*{\hat{A}}$ evolves like so,
+where the subscripts $S$ and $I$
+respectively refer to the Schrödinger
+and [interaction pictures](/know/concept/interaction-picture/):
+
+$$\begin{aligned}
+    \expval*{\hat{A}}(t)
+    = \matrixel{\psi_S(t)}{\hat{A}_S}{\psi_S(t)}
+    &= \matrixel{\psi_I(t)}{\hat{A}_I(t)}{\psi_I(t)}
+    \\
+    &= \matrixel{\psi_I(t_0)}{\hat{K}_I^\dagger(t, t_0) \hat{A}_I(t) \hat{K}_I(t, t_0)}{\psi_I(t_0)}
+\end{aligned}$$
+
+Where the time evolution operator $\hat{K}_I(t, t_0)$ is as follows,
+which we Taylor-expand:
+
+$$\begin{aligned}
+    \hat{K}_I(t, t_0)
+    = \mathcal{T} \bigg\{ \exp\!\bigg( \frac{1}{i \hbar} \int_{t_0}^t \hat{H}_{1,I}(t') \dd{t'} \bigg) \bigg\}
+    \approx 1 - \frac{i}{\hbar} \int_{t_0}^t \hat{H}_{1,I}(t') \dd{t'}
+\end{aligned}$$
+
+With this, the following product of operators (as encountered earlier) can be written as:
+
+$$\begin{aligned}
+    \hat{K}_I^\dagger \hat{A}_I \hat{K}_I
+    &\approx \bigg( 1 + \frac{i}{\hbar} \int_{t_0}^t \hat{H}_{1,I}(t') \dd{t'} \bigg) \hat{A}_I(t)
+    \bigg( 1 - \frac{i}{\hbar} \int_{t_0}^t \hat{H}_{1,I}(t') \dd{t'} \bigg)
+    \\
+    %&= \bigg( \hat{A}_I + \frac{i}{\hbar} \int_{t_0}^t \hat{H}_{1,I}(t') \hat{A}_I \dd{t'} \bigg)
+    %\bigg( 1 - \frac{i}{\hbar} \int_{t_0}^t \hat{H}_{1,I}(t') \dd{t'} \bigg)
+    %\\
+    &\approx \hat{A}_I(t)
+    - \frac{i}{\hbar} \int_{t_0}^t \hat{A}_I(t) \hat{H}_{1,I}(t') \dd{t'}
+    + \frac{i}{\hbar} \int_{t_0}^t \hat{H}_{1,I}(t') \hat{A}_I(t) \dd{t'}
+\end{aligned}$$
+
+Where we have dropped the last term,
+because $\hat{H}_{1}$ is assumed to be so small
+that it only matters to first order.
+Here, we notice a commutator, so we can rewrite:
+
+$$\begin{aligned}
+    \hat{K}_I^\dagger \hat{A}_I \hat{K}_I
+    &= \hat{A}_I(t) - \frac{i}{\hbar} \int_{t_0}^t \comm{\hat{A}_I(t)}{\hat{H}_{1,I}(t')} \dd{t'}
+\end{aligned}$$
+
+Returning to $\expval*{\hat{A}}$,
+we have the following formula,
+where $\expval{}$ is the expectation value for $\ket{\psi(t)}$,
+and $\expval{}_0$ is the expectation value for $\ket{\psi_I(t_0)}$:
+
+$$\begin{aligned}
+    \expval*{\hat{A}}(t)
+    = \expval*{\hat{K}_I^\dagger \hat{A}_I \hat{K}_I}_0
+    = \expval*{\hat{A}_I(t)}_0 - \frac{i}{\hbar} \int_{t_0}^t \expval{\comm{\hat{A}_I(t)}{\hat{H}_{1,I}(t')}}_0 \dd{t'}
+\end{aligned}$$
+
+Now we define $\delta\expval*{\hat{A}}(t)$
+as the change of $\expval*{\hat{A}}$ due to the perturbation $\hat{H}_1$,
+and insert $\expval*{\hat{A}}(t)$:
+
+$$\begin{aligned}
+    \delta\expval*{\hat{A}}(t)
+    \equiv \expval*{\hat{A}}(t) - \expval*{\hat{A}_I}_0
+    = - \frac{i}{\hbar} \int_{t_0}^t \expval{\comm{\hat{A}_I(t)}{\hat{H}_{1,I}(t')}}_0 \dd{t'}
+\end{aligned}$$
+
+Finally, we introduce
+a [Heaviside step function](/know/concept/heaviside-step-function) $\Theta$
+and change the integration limit accordingly,
+leading to the **Kubo formula**
+describing the response of $\expval*{\hat{A}}$ to first order in $\hat{H}_1$:
+
+$$\begin{aligned}
+    \boxed{
+        \delta\expval*{\hat{A}}(t)
+        %= - \frac{i}{\hbar} \int_{t_0}^\infty \Theta(t \!-\! t') \expval{\comm{\hat{A}_I(t)}{\hat{H}_{1,I}(t')}}_0 \dd{t'}
+        = \int_{t_0}^\infty C^R_{A H_1}(t, t') \dd{t'}
+    }
+\end{aligned}$$
+
+Where we have defined the **retarded correlation function** $C^R_{A H_1}(t, t')$ as follows:
+
+$$\begin{aligned}
+    \boxed{
+        C^R_{A H_1}(t, t')
+        \equiv - \frac{i}{\hbar} \Theta(t \!-\! t') \expval{\comm{\hat{A}_I(t)}{\hat{H}_{1,I}(t')}}_0
+    }
+\end{aligned}$$
+
+This result applies to bosonic operators,
+whereas for fermionic operators
+the commutator would be replaced by an anticommutator.
+
+A common situation is that $\hat{H}_1$ consists of
+a time-independent operator $\hat{B}$
+and a time-dependent function $f(t)$,
+allowing us to split $C^R_{A H_1}$ as follows:
+
+$$\begin{aligned}
+    \hat{H}_{1,S}(t)
+    = \hat{B}_S \: f(t)
+    \quad \implies \quad
+    C^R_{A H_1}(t, t')
+    = C^R_{A B}(t, t') f(t')
+\end{aligned}$$
+
+Conveniently, it can be shown that in this case
+$C^R_{AB}$ only depends on the difference $t - t'$,
+if we assume that the system was initially in thermodynamic equilibrium:
+
+$$\begin{aligned}
+    C^R_{A B}(t, t')
+    = C^R_{A B}(t - t')
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-time-difference"/>
+<label for="proof-time-difference">Proof</label>
+<div class="hidden">
+<label for="proof-time-difference">Proof.</label>
+This is trivial for $\Theta(t\!-\!t')$,
+so the challenge is to prove that
+$\expval*{\comm*{\hat{A}_I(t)}{B_I(t')}}$
+depends only on the time difference $t - t'$.
+
+Suppose that the system started in thermodynamic equilibrium
+(see [canonical ensemble](/know/concept/canonical-ensemble/)),
+so that its (unnormalized) [density operator](/know/concept/density-operator/)
+$\hat{\rho}$ was as follows before $\hat{H}_{1,I}$ was applied:
+
+$$\begin{aligned}
+    \hat{\rho} = \exp\!(- \beta \hat{H}_{0,S})
+\end{aligned}$$
+
+Let us assume that the perturbation $\hat{H}_{1,I}$
+does not affect the distribution of states,
+but only their individual evolutions in time.
+Note that, in general, this is not equilibrium.
+
+In that case, the expectation value of the product
+of two time-independent observables $\hat{A}$ and $\hat{B}$
+can be calculated as follows,
+where $Z_0 \equiv \Tr\!(\hat{\rho})$ is the partition function:
+
+$$\begin{aligned}
+    \expval*{\hat{A} \hat{B}}
+    = \frac{1}{Z_0} \Tr\!\big( \hat{\rho} \hat{A}_I \hat{H}_{1,I} \big)
+    = \frac{1}{Z_0} \Tr\!\Big( e^{-\beta \hat{H}_0} e^{i t \hat{H}_0 / \hbar} \hat{A}_S e^{-i t \hat{H}_0 / \hbar}
+    e^{i t' \hat{H}_0 / \hbar} \hat{B}_S e^{-i t' \hat{H}_0 / \hbar} \Big)
+\end{aligned}$$
+
+Where $\hat{H}_0 \equiv \hat{H}_{0,S}$ for brevity.
+Using that the trace $\Tr$ is invariant
+under cyclic permutations of its argument,
+and that functions of $\hat{H}_{0,S}$ always commute, we find:
+
+$$\begin{aligned}
+    \expval*{\hat{A} \hat{B}}
+    = \frac{1}{Z_0} \Tr\!\Big( e^{-\beta \hat{H}_0} e^{i (t - t') \hat{H}_0 / \hbar} \hat{A}_S
+    e^{-i (t - t') \hat{H}_0 / \hbar} \hat{B}_S \Big)
+\end{aligned}$$
+
+As expected, this clearly only depends on the time difference $t - t'$,
+because $\hat{H}_{0,S}$ is time-independent by assumption.
+</div>
+</div>
+
+With this, the Kubo formula can be written as follows,
+where we have set $t_0 = - \infty$:
+
+$$\begin{aligned}
+    \delta\expval*{A}(t)
+    = \int_{-\infty}^\infty C^R_{A B}(t - t') f(t') \dd{t'}
+    = (C^R_{A B} * f)(t)
+\end{aligned}$$
+
+This is a convolution,
+so the [convolution theorem](/know/concept/convolution-theorem/)
+states that the [Fourier transform](/know/concept/fourier-transform/)
+of $\delta\expval*{\hat{A}}(t)$ is simply the product
+of the transforms of $C^R_{AB}$ and $f$:
+
+$$\begin{aligned}
+    \boxed{
+        \delta\expval*{\hat{A}}(\omega)
+        = \tilde{C}{}^R_{A B}(\omega) \: \tilde{f}(\omega)
+    }
+\end{aligned}$$
+
+
+
+## References
+1.  H. Bruus, K. Flensberg,
+    *Many-body quantum theory in condensed matter physics*,
+    2016, Oxford.
+2.  K.S. Thygesen,
+    *Linear response theory*,
+    2013, unpublished.