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.
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