---
title: "Kubo formula"
firstLetter: "K"
publishDate: 2021-09-23
categories:
- Physics
- Quantum mechanics
- Perturbation

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:

$$\begin{aligned}
    \hat{K}_I^\dagger \hat{A}_I \hat{K}_I
    &\approx \bigg( 1 + \frac{i}{\hbar} \int_{t_0}^t \hat{H}_{1,I}(t') \dd{t'} \bigg) \hat{A}_I(t)
    \bigg( 1 - \frac{i}{\hbar} \int_{t_0}^t \hat{H}_{1,I}(t') \dd{t'} \bigg)
    \\
    &\approx \hat{A}_I(t)
    - \frac{i}{\hbar} \int_{t_0}^t \hat{A}_I(t) \hat{H}_{1,I}(t') \dd{t'}
    + \frac{i}{\hbar} \int_{t_0}^t \hat{H}_{1,I}(t') \hat{A}_I(t) \dd{t'}
\end{aligned}$$

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.