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diff --git a/content/know/concept/kubo-formula/index.pdc b/content/know/concept/kubo-formula/index.pdc new file mode 100644 index 0000000..9e52835 --- /dev/null +++ b/content/know/concept/kubo-formula/index.pdc @@ -0,0 +1,227 @@ +--- +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. |