--- 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.