- #1
spaghetti3451
- 1,344
- 34
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
$$\theta_{n}(t)\equiv -\frac{1}{\hbar}\int_{0}^{t}E_{n}(t')dt'?$$
$$i\hbar\frac{\partial}{\partial t}|\alpha;t\rangle=H(t)|\alpha;t\rangle,$$
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
$$\theta_{n}(t)\equiv -\frac{1}{\hbar}\int_{0}^{t}E_{n}(t')dt'.$$
---
##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
$$\theta_{n}(t)\equiv -\frac{1}{\hbar}\int_{0}^{t}E_{n}(t')dt'?$$