# Propagators for time-dependent Hamiltonians

1. Aug 8, 2013

### AxiomOfChoice

Suppose I know

$$H \psi(x) = \left( -\frac{1}{2m} \Delta_x + V(x) \right) \psi(x) = E\psi(x).$$

Then

$$\psi(x,t) = e^{-iEt}\psi(x)$$

solves the time-dependent Schrodinger equation

$$\left( i \frac{\partial}{\partial t} + \frac{1}{2m} \Delta_x - V(x) \right)\psi(x,t) = 0.$$

I've done some computations, and it looks like

$$\Psi(x,t) = e^{-imvx}e^{-imv^2t/2}\psi(x+vt)$$

is a solution to the time-dependent Schrodinger equation

$$\left( i \frac{\partial}{\partial t} + \frac{1}{2m} \Delta_x - V(x+vt) \right)\Psi = 0.$$

1. What is going on here physically? That is, what are those two phase factors telling me?
2. What does this mean the propagator is? When $H$ is time-independent, $U(t) = e^{-iEt}$...but what is it in the time-dependent case? Is there a neat little formulation of it?

Last edited: Aug 8, 2013
2. Aug 8, 2013

### vanhees71

With the ansatz
$$\psi(t,\vec{x})=\exp(-\mathrm{i} E t) \phi_{E,\alpha}(\vec{x}),$$
you only get the energy-eigensolutions (stationary states) for a time-independent Hamiltonian,
$$\hat{H} \phi_{E,\alpha}(\vec{x})=E \phi_{E,\alpha}(\vec{x}).$$
Here $\alpha$ stands for all necessary additional observables, compatible with energy, to label the possible degeneracy of the energy eigenstates.

The most general solution of the time-dependent Schrödinger equation is then given as a superposition of energy eigensolutions
$$\psi(t,\vec{x})=\sum_{E,\alpha} \exp(-\mathrm{i} E t) \phi_{E,\alpha}(\vec{x}).$$

This is all for time-independent Hamiltonians. For time-dependent Hamiltonians, the problem is a bit more complicated. Here you get a formal solution by using the time-evolution operator for the state in the Schrödinger picture,
$$\hat{C}(t,t_0)=\mathcal{T}_c \exp[-\mathrm{i} \int_{t_0}^t \mathrm{d} t' \hat{H}(t)],$$
where $\mathcal{T}_c$ is the time-ordering operator that orders products of time-dependent operators such that the time arguments are ordered from right to left.

The exponential must be formaly expanded in terms of a power series to make sense of this time-ordering symbol. The nth-order contribution is
$$\hat{C}_n(t,t_0)=\frac{(-\mathrm{i})^n}{n!} \int_{t_0}^t \mathrm{d} t_1 \cdots \int_{t_0}^t \mathrm{d} t_n \mathcal{T}_c \hat{H}(t_1) \cdots \hat{H}(t_n).$$
Then
$$\hat{C}(t,t_0)=\sum_{n=0}^{\infty} \hat{C}_n(t,t_0).$$

The propagator of the Schrödinger equation is then given by
$$U(t,\vec{x};t_0,\vec{x}')=\langle \vec{x}|\hat{C}(t,t_0)|\vec{x}' \rangle,$$
and the solution of the time-dependent Schrödinger equation reads
$$\psi(t,\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x'} U(t,\vec{x};t_0,\vec{x}') \psi(t_0,\vec{x}').$$