Is the Correction \( c_{nn}^{(1)} \) Zero in Quantum Perturbation Theory?

[T-7]

Homework Statement

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.

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.

 

[Jhero]

Your approach is definitely on the right track. You correctly showed that the real part of c_{nn}^{(1)} must be zero, and it is also true that the imaginary part can be set to zero since it is arbitrary. However, to make your proof more rigorous, you could explicitly state that the imaginary part is arbitrary and that it can be set to zero without loss of generality. Additionally, you could also mention that this result holds for any value of n since the summation is over all possible values of m. Overall, your proof is correct and well-structured. Good job!