Categories: Physics, Quantum mechanics.

Time evolution operator

In general, given a system whose governing equation is known, the time evolution operator U^(t,t0)\hat{U}(t, t_0) transforms the state at time t0t_0 to the one at time tt. Although not specific to it, this is most often used in quantum mechanics, as governed by the Schrödinger equation:

iℏddt∣ψ(t)⟩=H^(t)∣ψ(t)⟩\begin{aligned} i \hbar \dv{}{t} \ket{\psi(t)} = \hat{H}(t) \ket{\psi(t)} \end{aligned}

Such that the definition of U^(t)\hat{U}(t) is as follows, where we have set t0=0t_0 = 0:

∣ψ(t)⟩=U^(t)∣ψ(0)⟩\begin{aligned} \ket{\psi(t)} = \hat{U}(t) \ket{\psi(0)} \end{aligned}

Clearly, U^(t)\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 U^(t)\hat{U}(t) into the Schrödinger equation:

ddtU^(t)∣ψ(0)⟩=−iℏH^ U^(t)∣ψ(0)⟩\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 ∣ψ(0)⟩\ket{\psi(0)}, then U^(t)\hat{U}(t) can be said to satisfy the equation in its own right:

ddtU^(t)=−iℏH^ U^(t)\begin{aligned} \dv{}{t} \hat{U}(t) = - \frac{i}{\hbar} \hat{H} \: \hat{U}(t) \end{aligned}

If the Hamiltonian H^\hat{H} is time-independent, this is straightforward to integrate, yielding:

U^(t)=exp⁡ ⁣( ⁣− ⁣iℏtH^)\begin{aligned} \boxed{ \hat{U}(t) = \exp\!\bigg( \!-\! \frac{i}{\hbar} t \hat{H} \bigg) } \end{aligned}

And the generalization to t0≠0t_0 \neq 0 is trivial, since we can just shift the time axis:

U^(t,t0)=exp⁡ ⁣( ⁣− ⁣iℏ(t−t0)H^)\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 H^\hat{H} is time-dependent, U^(t)\hat{U}(t) can be said to satisfy the Schrödinger equation:

ddtU^(t)=−iℏH^(t) U^(t)\begin{aligned} \dv{}{t} \hat{U}(t) = - \frac{i}{\hbar} \hat{H}(t) \: \hat{U}(t) \end{aligned}

Integrating from 00 to tt, and using U^(0)=1\hat{U}(0) = 1 (which should be clear from its definition):

U^(t)=1+1iℏ∫0tH^(τ1) U^(τ1)dτ1\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 U^(t)\hat{U}(t). We can recursively insert it into itself, yielding:

U^(t)=1+1iℏ∫0tH^(τ1)(1+1iℏ∫0τ1H^(τ2) U^(τ2)dτ2)dτ1=1+1iℏ∫0tH^(τ1)dτ1+1(iℏ)2∫0tH^(τ1)∫0τ1H^(τ2) U^(τ2)dτ2dτ1=1+1iℏ∫0tH^(τ1)dτ1+1(iℏ)2∫0tH^(τ1)∫0τ1H^(τ2)dτ2dτ1+1(iℏ)3∫0t⋯ dτ1\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 τ2<τ1\tau_2 < \tau_1 always. We can exploit this fact to introduce several Heaviside step functions Θ(t)\Theta(t):

∫0tH^(τ1)∫0τ1H^(τ2)dτ2dτ1=12∫0tH^(τ1)∫0τ1H^(τ2)dτ2dτ1+12∫0tH^(τ2)∫0τ2H^(τ1)dτ1dτ2=12∫0t ⁣H^(τ1)∫0τ1 ⁣Θ(τ1 ⁣− ⁣τ2)H^(τ2)dτ2dτ1+12∫0t ⁣H^(τ2)∫0τ2 ⁣Θ(τ2 ⁣− ⁣τ1)H^(τ1)dτ1dτ2=12∫0t∫0t(Θ(τ1 ⁣− ⁣τ2) H^(τ1) H^(τ2)+Θ(τ2 ⁣− ⁣τ1) H^(τ1) H^(τ2))dτ1dτ2=12∫0t∫0tT{H^(τ1) H^(τ2)}dτ1dτ2\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 T\mathcal{T}. The above procedure is easy to generalize to the higher-order terms, so we arrive at the following expression for U^(t)\hat{U}(t):

U^(t)=1+1iℏ∫0tH^(τ1)dτ1+121(iℏ)2∬0tT{H^(τ1) H^(τ2)}dτ2dτ1+161(iℏ)3∭0tT{H^(τ1) H^(τ2) H^(τ3)}dτ3dτ2dτ1+ ...=1+∑n=1∞1n!1(iℏ)n∫0t ⁣⋯ ⁣∫0tT{H^(τ1)⋯H^(τn)}dτn⋯dτ1\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 T\mathcal{T} looking a bit strange:

U^(t)=1+∑n=1∞1n!1(iℏ)nT{(∫0tH^(τ)dτ)n}\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)\exp(x), leading us to the desired result:

U^(t)=T{exp⁡ ⁣( ⁣− ⁣iℏ∫0tH^(τ)dτ)}\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 T\mathcal{T} is being used according to convention. Finally, the time axis can be shifted arbitrarily, so many authors write the evolution operator from t0t_0 to tt as U^(t,t0)\hat{U}(t, t_0):

U^(t,t0)=T{exp⁡ ⁣( ⁣− ⁣iℏ∫t0tH^(τ)dτ)}\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.