Evolution Operators and Dyson Series

March 14, 2015

In the Schrödinger picture, states evolve according to the Schrödinger equation

(1) $\begin{equation*}i\frac{d}{dt}\ket{\psi(t)}=\hat{H}(x,p,t)\ket{\psi(t)}\end{equation*}$

where $\hat{H}(x,p,t)$ is the Hamiltonian of the system (and $\hbar=1$ ). Because the Hamiltonian is a hermitian operator, one can define a unitary evolution operator $\hat{U}(t,t')$ which takes $\ket{\psi(t')}$ to $\ket{\psi(t)}$ for all states. The evolution operator obeys the differential equation

$i\frac{d}{dt}U(t,t')=H(t)U(t,t')$

Simple case
If $\hat{H}$ is constant, then the differential equation is easy, and the evolution operator can be written

$\hat{U}(t,t')=e^{-i\hat{H}(t-t')}$

If $\hat{H}(t)$ does depend on time, but at least commutes with itself across different times, then $\hat{U}$ can be generalized to

$\hat{U}(t,t')=e^{-i\int_{t'}^t\hat{H}(t_0)dt_0}$

General case
If we have no such guarantees on $\hat{H}(t)$ , then things get more interesting. We can integrate the Schrödinger equation

$\ket{\psi(t)}=\ket{\psi(t')}-i\int_{t'}^t dt_0\hat{H}(x,p,t_0)\ket{\psi(t_0)}$

And then iterate by expanding the $\ket{\psi(t_0)}$ part:

(2) $\begin{equation*}\hat{U}(t,t_0)=1-i\int_{t_0}^t dt_1\hat{H}(t_1) -\int_{t_0}^t dt_1\int_{t_0}^{t_1}dt_2\hat{H}(t_1)\hat{H}(t_2)+\cdots \end{equation*}$

This gives us an infinite sum of integrals to evaluate to find $\hat{U}(t,t')$ . But we can play some games to make it prettier. We could for instance take the second-order term in that expansion and extend the upper limit of the inner integral to $t$ provided we insert a $\theta(t_1-t_2)$ to kill off the $t_2>t_1$ region.

$\text{second-order term}=-\int_{t_0}^t dt_1\int_{t_0}^{t}dt_2\theta(t_1-t_2)\hat{H}(t_1)\hat{H}(t_2)$

Or further, we could recognize that the integral in the $t_2>t_1$ region is the same as the integral in the $t_1>t_2$ region but with the operators flipped. So let’s allow both regions but divide by two.

$\begin{align*} \text{second-order term}&=\frac{-1}{2}\left\{ \int_{t_0}^t dt_1\int_{t_0}^{t}dt_2\theta(t_1-t_2)\hat{H}(t_1)\hat{H}(t_2)+ \int_{t_0}^t dt_1\int_{t_0}^{t}dt_2\theta(t_2-t_1)\hat{H}(t_2)\hat{H}(t_1) \right\}\\ &=\frac{-1}{2} \int_{t_0}^t dt_1\int_{t_0}^{t}dt_2\mathcal{T}\left[\hat{H}(t_1)\hat{H}(t_2)\right]\\ &=\mathcal{T}\left[\frac{-1}{2} \left(\int_{t_0}^t dt_1\hat{H}(t_1)\right)^2\right] \end{align*}$

where $\mathcal{T}$ is the time-ordering “operation”: it just rearranges the operators inside its brackets to be ordered latest-on-the-left without any regard for whether those operators commute.

We can do this funny business for every term of the expansion 2: for the $n$ -th term, extend all the $n-1$ limits of integration, put in the time-ordering operator, and divide by the $n!$ permutations of the operators. We find

(3) $\begin{align*} \hat{U}(t, t')=\sum_n \mathcal{T} \frac{\left(-i\int_{t'}^{t}dt_0\hat{H}(t_0)\right)^n}{n!} =\mathcal{T}\exp\left\{-i\int_{t'}^{t}dt_0\hat{H}(t_0)\right\} \end{align*}$

where the last equality is just a definition. This is known as the Dyson series.

Double-check
We can “check ourselves” that this is valid. Since $\mathcal{T}$ rearranges operators freely, anything inside a $\mathcal{T}[]$ effectively commutes. So we can treat everything like numbers when we differentiate by $t$ .

$\frac{d}{dt}\hat{U}(t,t')=\mathcal{T}\left[-i\hat{H}(t)\exp\left\{-i\int_{t'}^{t}dt_0\hat{H}(t_0)\right\}\right]=-i\hat{H}(t)\mathcal{T}\left[\exp\left\{-i\int_{t'}^{t}dt_0\hat{H}(t_0)\right\}\right]=-i\hat{H}(t)\hat{U}(t,t')$

where, in the middle equality, we can pull $\hat{H}(t)$ out of the $\mathcal{T}$ on the left because $t$ is the latest time.