# I Phase factors of eigenstates in time-dependent Hamiltonians

1. Nov 17, 2016

### spaghetti3451

For a time-dependent Hamiltonian, the Schrodinger equation is given by

$$i\hbar\frac{\partial}{\partial t}|\alpha;t\rangle=H(t)|\alpha;t\rangle,$$

where the physical time-dependent state $|\alpha;t\rangle$ is given by

$$|\alpha;t\rangle = \sum\limits_{n}c_{n}(t)e^{i\theta_{n}(t)}|n;t\rangle$$

and

$$\theta_{n}(t)\equiv -\frac{1}{\hbar}\int_{0}^{t}E_{n}(t')dt'.$$

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$e^{i\theta_{n}(t)}$ is the phase factor that has been pulled out from the eigenstate-expansion coefficients of $|\alpha;t\rangle$.

Why is $\theta_{n}(t)$ given by

$$\theta_{n}(t)\equiv -\frac{1}{\hbar}\int_{0}^{t}E_{n}(t')dt'?$$

2. Nov 17, 2016

### Staff: Mentor

Because the time evolution operator $\hat{U}(t)$ is a solution of the TDSE
$$i \hbar \frac{d\hat{U}(t)}{dt} = \hat{H}(t) \hat{U}(t)$$
hence
$$\hat{U}(t) = \exp\left[-\frac{i}{\hbar} \int_0^t \hat{H}(t') dt' \right]$$