Categories: Perturbation, Physics, Quantum mechanics.

# 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:

\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 $\Theta$ 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 they consist of products of even numbers of particle creation/annihiliation operators. Therefore, this correlation function is a two-particle Green’s function.

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 states that the 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.