# Why are unperturbed states valid basis for perturbed system?

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1. Feb 20, 2017

### bananabandana

1. The problem statement, all variables and given/known data
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:
$$\psi(\vec{r},t) = c_{1}(t) exp(-i\omega_{1}t) |\phi_{1}\rangle +c_{2}(t) exp(-i\omega_{2}t) |\phi_{2}\rangle$$

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?
2. Relevant equations

3. The attempt at a solution

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!

2. Feb 20, 2017

### Simon Bridge

If it makes physical sense, then that is where it is coming from. Physics is an empirical science.
Mathematically, the approach comes from the nature of vector spaces: so check - what is required for the old basis set to also be a basis for the new system?

3. Feb 20, 2017

### bananabandana

Physically, it doesn't necessarily make sense that there isn't an extra state generated by this process?

Mathematically
- The basis $\{ |\phi_{1}\rangle,|\phi_{2}\rangle \}$ is a valid basis for any state in a Hilbert space $H$ of dimensionality 2. Therefore, so long as I know that only two eigenstates exist before and after the perturbation then the expression is fine.

So, if I write the total Hamiltonian, $\mathbf{\hat{H'}}$ as $\mathbf{\hat{H'}} = \mathbf{\hat{H_{0}}} + \mathbf{\hat{H_{I}}}$, then for Foot's expression to be valid it's a requirement that for the perturbed system we have eigenstates $|\chi_{i}\rangle$ such that:
$$\mathbf{\hat{H'}} = (\mathbf{\hat{H_{0}}}+\mathbf{\hat{H_{I}}}) |\chi_{i}\rangle = E'_{i} | \chi_{i} \rangle$$
and $|\chi_{i} \rangle \ \in \ \{ |\chi_{1}\rangle, |\chi_{2}\rangle\}$ - $\mathbf{\hat{H'}}$ must have two and only two (independent) eigenstates.

But how do I prove that the operator $\mathbf{\hat{H'}}$ necessarily has this property?

4. Feb 23, 2017

### Simon Bridge

Sure, you could have a process that increases the depth of the potential well by a significant amount, for example.
The method you are using here is called "perturbation theory" for a reason - what is special about modifications that count as perturbations?

5. Feb 23, 2017

### Staff: Mentor

If the dimensionality of the Hilbert space should change, it would mean that the original Hilbert space was not big enough to start with, that the description of the quantum system was incomplete, and there was a hidden degeneracy.

Note that in the example you are citing, it is an approximation that there are only two levels involved. In reality, there are many more levels, but it is assumed that two can be singled out and the physics will not be affected by the presence of the other levels. By construction, there can only be two levels in this system.