---
title: "Kubo formula"
sort_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.