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diff --git a/source/know/concept/time-evolution-operator/index.md b/source/know/concept/time-evolution-operator/index.md new file mode 100644 index 0000000..f489ac6 --- /dev/null +++ b/source/know/concept/time-evolution-operator/index.md @@ -0,0 +1,184 @@ +--- +title: "Time evolution operator" +sort_title: "Time evolution operator" +date: 2024-10-15 +categories: +- Quantum mechanics +- Physics +layout: "concept" +--- + +In general, given a system whose governing equation is known, +the **time evolution operator** $$\hat{U}(t, t_0)$$ +transforms the state at time $$t_0$$ to the one at time $$t$$. +Although not specific to it, +this is most often used in quantum mechanics, +as governed by the Schrödinger equation: + +$$\begin{aligned} + i \hbar \dv{}{t} \ket{\psi(t)} + = \hat{H}(t) \ket{\psi(t)} +\end{aligned}$$ + +Such that the definition of $$\hat{U}(t)$$ is as follows, +where we have set $$t_0 = 0$$: + +$$\begin{aligned} + \ket{\psi(t)} + = \hat{U}(t) \ket{\psi(0)} +\end{aligned}$$ + +Clearly, $$\hat{U}(t)$$ must be unitary. +The goal is to find an expression that satisfies this relation. + + + +## Time-independent Hamiltonian + +We start by inserting the definition of $$\hat{U}(t)$$ +into the Schrödinger equation: + +$$\begin{aligned} + \dv{}{t} \hat{U}(t) \ket{\psi(0)} + = - \frac{i}{\hbar} \hat{H} \: \hat{U}(t) \ket{\psi(0)} +\end{aligned}$$ + +If we hide the state $$\ket{\psi(0)}$$, +then $$\hat{U}(t)$$ can be said to satisfy the equation in its own right: + +$$\begin{aligned} + \dv{}{t} \hat{U}(t) + = - \frac{i}{\hbar} \hat{H} \: \hat{U}(t) +\end{aligned}$$ + +If the Hamiltonian $$\hat{H}$$ is time-independent, +this is straightforward to integrate, yielding: + +$$\begin{aligned} + \boxed{ + \hat{U}(t) + = \exp\!\bigg( \!-\! \frac{i}{\hbar} t \hat{H} \bigg) + } +\end{aligned}$$ + +And the generalization to $$t_0 \neq 0$$ is trivial, +since we can just shift the time axis: + +$$\begin{aligned} + \hat{U}(t, t_0) + = \exp\!\bigg( \!-\! \frac{i}{\hbar} (t - t_0) \hat{H} \bigg) +\end{aligned}$$ + + + +## Time-dependent Hamiltonian + +Even when $$\hat{H}$$ is time-dependent, +$$\hat{U}(t)$$ can be said to satisfy the Schrödinger equation: + +$$\begin{aligned} + \dv{}{t} \hat{U}(t) + = - \frac{i}{\hbar} \hat{H}(t) \: \hat{U}(t) +\end{aligned}$$ + +Integrating from $$0$$ to $$t$$, +and using $$\hat{U}(0) = 1$$ (which should be clear from its definition): + +$$\begin{aligned} + \hat{U}(t) + = 1 + \frac{1}{i \hbar} \int_0^t \hat{H}(\tau_1) \: \hat{U}(\tau_1) \dd{\tau_1} +\end{aligned}$$ + +This is a self-consistent equation for $$\hat{U}(t)$$. +We can recursively insert it into itself, yielding: + +$$\begin{aligned} + \hat{U}(t) + &= 1 + \frac{1}{i \hbar} \int_0^t \hat{H}(\tau_1) + \bigg( 1 + \frac{1}{i \hbar} \int_0^{\tau_1} \hat{H}(\tau_2) \: \hat{U}(\tau_2) \dd{\tau_2} \bigg) \dd{\tau_1} + \\ + &= 1 + \frac{1}{i \hbar} \int_0^t \hat{H}(\tau_1) \dd{\tau_1} + + \frac{1}{(i \hbar)^2} \int_0^t \hat{H}(\tau_1) \int_0^{\tau_1} \hat{H}(\tau_2) \: \hat{U}(\tau_2) \dd{\tau_2} \dd{\tau_1} + \\ + &= 1 + \frac{1}{i \hbar} \int_0^t \hat{H}(\tau_1) \dd{\tau_1} + + \frac{1}{(i \hbar)^2} \int_0^t \hat{H}(\tau_1) \int_0^{\tau_1} \hat{H}(\tau_2) \dd{\tau_2} \dd{\tau_1} + + \frac{1}{(i \hbar)^3} \int_0^t \cdots \: \dd{\tau_1} +\end{aligned}$$ + +And so on. +Let us take a closer look at the third (i.e. second-order) term in this series, +noting that the integrals are ordered such that $$\tau_2 < \tau_1$$ always. +We can exploit this fact to introduce several +[Heaviside step functions](/know/concept/heaviside-step-function/) $$\Theta(t)$$: + +$$\begin{aligned} + &\quad \int_0^t \hat{H}(\tau_1) \int_0^{\tau_1} \hat{H}(\tau_2) \dd{\tau_2} \dd{\tau_1} + \\ + &= \frac{1}{2} \int_0^t \hat{H}(\tau_1) \int_0^{\tau_1} \hat{H}(\tau_2) \dd{\tau_2} \dd{\tau_1} + + \frac{1}{2} \int_0^t \hat{H}(\tau_2) \int_0^{\tau_2} \hat{H}(\tau_1) \dd{\tau_1} \dd{\tau_2} + \\ + &= \frac{1}{2} \int_0^t \! \hat{H}(\tau_1) + \int_0^{\tau_1} \! \Theta(\tau_1 \!-\! \tau_2) \hat{H}(\tau_2) \dd{\tau_2} \dd{\tau_1} + + \frac{1}{2} \int_0^t \! \hat{H}(\tau_2) + \int_0^{\tau_2} \! \Theta(\tau_2 \!-\! \tau_1) \hat{H}(\tau_1) \dd{\tau_1} \dd{\tau_2} + \\ + &= \frac{1}{2} \int_0^t \int_0^t + \bigg( \Theta(\tau_1 \!-\! \tau_2) \: \hat{H}(\tau_1) \: \hat{H}(\tau_2) + + \Theta(\tau_2 \!-\! \tau_1) \: \hat{H}(\tau_1) \: \hat{H}(\tau_2) \bigg) \dd{\tau_1} \dd{\tau_2} + \\ + &= \frac{1}{2} \int_0^t \int_0^t \mathcal{T} \Big\{ \hat{H}(\tau_1) \: \hat{H}(\tau_2) \Big\} \dd{\tau_1} \dd{\tau_2} +\end{aligned}$$ + +Where we have recognized the +[time-ordering meta-operator](/know/concept/time-ordered-product/) $$\mathcal{T}$$. +The above procedure is easy to generalize to the higher-order terms, +so we arrive at the following expression for $$\hat{U}(t)$$: + +$$\begin{aligned} + \hat{U}(t) + &= 1 + \frac{1}{i \hbar} \int_0^t \hat{H}(\tau_1) \dd{\tau_1} + + \frac{1}{2} \frac{1}{(i \hbar)^2} \iint_0^t \mathcal{T} \Big\{ \hat{H}(\tau_1) \: \hat{H}(\tau_2) \Big\} \dd{\tau_2} \dd{\tau_1} + \\ + &\qquad+ \frac{1}{6} \frac{1}{(i \hbar)^3} \iiint_0^t + \mathcal{T} \Big\{ \hat{H}(\tau_1) \: \hat{H}(\tau_2) \: \hat{H}(\tau_3) \Big\} \dd{\tau_3} \dd{\tau_2} \dd{\tau_1} + + \: ... + \\ + &= 1 + \sum_{n = 1}^\infty \frac{1}{n!} \frac{1}{(i \hbar)^n} \int_0^t \!\cdots\! \int_0^t + \mathcal{T} \Big\{ \hat{H}(\tau_1) \cdots \hat{H}(\tau_n) \Big\} \dd{\tau_n} \cdots \dd{\tau_1} +\end{aligned}$$ + +This result is sometimes called a **Dyson series**. +Convention allows us to write it as follows, +despite such a use of $$\mathcal{T}$$ looking a bit strange: + +$$\begin{aligned} + \hat{U}(t) + &= 1 + \sum_{n = 1}^\infty \frac{1}{n!} \frac{1}{(i \hbar)^n} + \mathcal{T} \bigg\{ \bigg( \int_0^t \hat{H}(\tau) \dd{\tau} \bigg)^n \bigg\} +\end{aligned}$$ + +Here, we recognize the Taylor expansion of $$\exp(x)$$, +leading us to the desired result: + +$$\begin{aligned} + \boxed{ + \hat{U}(t) + = \mathcal{T} \bigg\{ \exp\!\bigg( \!-\! \frac{i}{\hbar} \int_0^t \hat{H}(\tau) \dd{\tau} \bigg) \bigg\} + } +\end{aligned}$$ + +Where once again $$\mathcal{T}$$ is being used according to convention. +Finally, the time axis can be shifted arbitrarily, +so many authors write the evolution operator from $$t_0$$ to $$t$$ as $$\hat{U}(t, t_0)$$: + +$$\begin{aligned} + \hat{U}(t, t_0) + = \mathcal{T} \bigg\{ \exp\!\bigg( \!-\! \frac{i}{\hbar} \int_{t_0}^t \hat{H}(\tau) \dd{\tau} \bigg) \bigg\} +\end{aligned}$$ + + + +## References +1. H. Bruus, K. Flensberg, + *Many-body quantum theory in condensed matter physics*, + 2016, Oxford. |