Categories: 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) \\ %&= \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 $$\Theta$$ 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}

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.

© Marcus R.A. Newman, a.k.a. "Prefetch". Available under CC BY-SA 4.0.