# Quantum perturbation corrections

1. Feb 23, 2008

### T-7

1. The problem statement, all variables and given/known data

Substituting

$$|n> = |n^{0}> + \sum_{m}c_{nm}|m^{0}>$$

into the normalisation condition

$$<n|n> = 1$$

show that the correction $$c_{nn}^{(1)}$$ is zero.

3. The attempt at a solution

$$<n|n> = 1 => \left( <n^{0}| + \sum_{m}c_{nm}^{*}<m^{0}| \right) \left( |n^{0}> + \sum_{m}c_{nm}|m^{0}> \right) = 1$$

$$<n^{0}|n^{0}> + \sum_{m}c_{nm}<n^{0}|m^{0}> + \sum_{m}c_{nm}^{*}<m^{0}|n^{0}> + \sum_{m}c_{nm}^{*}\sum_{m'}c_{nm'}<m^{0}|m^{0}> = 1$$

$$1 + c_{nn} +c_{nn}^{*} + \sum_{m}c_{nm}^{*}\sum_{m'}c_{nm'} = 1$$

$$c_{nn} +c_{nn}^{*} + \sum_{m}c_{nm}^{*}\sum_{m'}c_{nm'} = 0$$

Now I believe I can say:

$$c_{nn} = \lambda c_{nn}^{(1)} + \lambda^{2} c_{nn}^{(2)} + ...$$

and so

$$c_{nn}^{*} = \lambda c_{nn}^{*(1)} + \lambda^{2} c_{nn}^{*(2)} + ...$$

And thus:

$$(\lambda c_{nn}^{(1)} + \lambda^{2} c_{nn}^{(2)} + ...) + (\lambda c_{nn}^{*(1)} + \lambda^{2} c_{nn}^{*(2)} + ...) + \sum_{m}(\lambda c_{nm}^{*(1)} + \lambda^{2} c_{nm}^{*(2)} + ...) \sum_{m'}(\lambda c_{nm'}^{(1)} + \lambda^{2} c_{nm'}^{(2)} + ...)= 0$$

Comparing coefficients of $$\lambda$$

$$\lambda c_{nn}^{(1)} + \lambda c_{nn}^{*(1)} = 0$$

Hence

$$Re(c_{nn}^{(1)}) = 0$$

Set imaginary part to zero; it is arbitrary.

So

$$c_{nn}^{(1)} = 0$$

Could someone tell me if this is along the right lines. I can't seem to do any more than prove that the real part is zero (if indeed I have proved this correctly), and it seems to me that the imaginary part, being arbitrary, could be set to zero. But perhaps this is too much hand waving, and something more should be said? Or perhaps my proof is wrong fullstop?

Cheers.