---
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.