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Date: Sat, 25 Sep 2021 10:26:44 +0200
Subject: Expand knowledge base

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+---
+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.
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