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+---
+title: "Kubo formula"
+date: 2021-09-23
+categories:
+- Physics
+- Quantum mechanics
+- Perturbation
+layout: "concept"
+---
+
+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:
+
+$$\begin{aligned}
+    \hat{K}_I^\dagger \hat{A}_I \hat{K}_I
+    &\approx \bigg( 1 + \frac{i}{\hbar} \int_{t_0}^t \hat{H}_{1,I}(t') \dd{t'} \bigg) \hat{A}_I(t)
+    \bigg( 1 - \frac{i}{\hbar} \int_{t_0}^t \hat{H}_{1,I}(t') \dd{t'} \bigg)
+    \\
+    &\approx \hat{A}_I(t)
+    - \frac{i}{\hbar} \int_{t_0}^t \hat{A}_I(t) \hat{H}_{1,I}(t') \dd{t'}
+    + \frac{i}{\hbar} \int_{t_0}^t \hat{H}_{1,I}(t') \hat{A}_I(t) \dd{t'}
+\end{aligned}$$
+
+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 $\hat{H}_1$:
+
+$$\begin{aligned}
+    \boxed{
+        \delta\!\expval{\hat{A}}\!(t)
+        = \int_{t_0}^\infty C^R_{A H_1}(t, t') \dd{t'}
+    }
+\end{aligned}$$
+
+Where we have defined the **retarded correlation function** $C^R_{A H_1}(t, t')$ as follows:
+
+$$\begin{aligned}
+    \boxed{
+        C^R_{A H_1}(t, t')
+        \equiv - \frac{i}{\hbar} \Theta(t \!-\! t') \Expval{\Comm{\hat{A}_I(t)}{\hat{H}_{1,I}(t')}}_0
+    }
+\end{aligned}$$
+
+Note that observables are bosonic,
+because in the [second quantization](/know/concept/second-quantization/)
+they consist of products of even numbers
+of particle creation/annihiliation operators.
+Therefore, this correlation function
+is a two-particle [Green's function](/know/concept/greens-functions/).
+
+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:
+
+$$\begin{aligned}
+    \hat{H}_{1,S}(t)
+    = \hat{B}_S \: f(t)
+    \quad \implies \quad
+    C^R_{A H_1}(t, t')
+    = C^R_{A B}(t, t') f(t')
+\end{aligned}$$
+
+Since $C_{AB}^R$ is a Green's function,
+we know that it only depends on the difference $t - t'$,
+as long as the system was initially in thermodynamic equilibrium,
+and $\hat{H}_{0,S}$ is time-independent:
+
+$$\begin{aligned}
+    C^R_{A B}(t, t')
+    = C^R_{A B}(t - t')
+\end{aligned}$$
+
+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,
+    *Advanced solid state physics: linear response theory*,
+    2013, unpublished.