- #1

bananabandana

- 113

- 5

## Homework Statement

So we have a two state system, with unperturbed eigenstates ## |\phi_{1}\rangle##, ## |\phi_{2}\rangle ##, and Hamiltonian ## \mathbf{\hat{H_{0}}}## - i.e ##\mathbf{\hat{H_{0}}}|\phi_{1}\rangle = E_{1}|\phi_{1}\rangle##

We shine some z-polarized light on the system. This gives us an interaction Hamiltonian:

$$\mathbf{\hat{H_{I}}} = -\mathbf{\hat{d_{z}}}\mathcal{E}cos(\omega t) $$

Where ## \mathbf{\hat{d_{z}}} ## is the dipole operator in the z direction - ##\mathbf{\hat{d_{z}}} = -e\mathbf{\hat{z}}##.

Foot (C.J Foot,Atomic Physics, OUP 2005) states without proof in Chapter 7 that:

*"The wavefunctions at any instant of time [for the perturbed system] can be expressed as:*

$$ \psi(\vec{r},t) = c_{1}(t) exp(-i\omega_{1}t) |\phi_{1}\rangle +c_{2}(t) exp(-i\omega_{2}t) |\phi_{2}\rangle $$

$$ \psi(\vec{r},t) = c_{1}(t) exp(-i\omega_{1}t) |\phi_{1}\rangle +c_{2}(t) exp(-i\omega_{2}t) |\phi_{2}\rangle $$

How do we know that the original basis - ##\{ |\phi_{1}\rangle , |\phi_{2}\rangle \}## is

**still**a valid basis for the system in the case of the perturbation? Surely the perturbation could (potentially) increase the number of possible states of the system, so that our previous basis is no longer sufficient?

## Homework Equations

## The Attempt at a Solution

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Can we justify this by referring back to non-degenerate perturbation theory - where the perturbed eigenstates are always linear superpositions of the unperturbed eigenstates (for first order). But... then, if this is the case, surely we should be careful to state that this formula is only a first order approximation, and not a general representation??

EDIT - and it makes physical sense that ## c_{1} = c_{1}(t) ## but where is that coming from?

Thanks!