Simple time-independent non-degenerate quantum perturbation

[dydxforsn]

I'm reading through this pdf (http://www.pa.msu.edu/~mmoore/TIPT.pdf) on simple quantum perturbation theory and I'm quite confused with equations 32 through 34.

They have E_{n}^{(2)} = <n^{(0)}|V|n^{(1)}> = - \sum_{m \neq 0}{\frac{|V_{mn}|^{2}}{E_{mn}}} but I would have done E_{n}^{(2)} = <n^{(0)}|V|n^{(1)}> - <n^{(0)}|E_{n}^{(1)}|n^{(1)}> and then plugged in E_{n}^{(1)} = V_{nn} from their earlier solution for first order terms. I don't know where I would have gone form there and I certainly couldn't even take a gander at how they end up with a summation either in this equation or in equations 33 and 34. Are there steps being omitted and/or can this be explained conceptually?

I have similar complaints about equations 33 and 34, though in equation 34 I have the first right hand side they end up with, but then again I have no idea about the summation that suddenly appears in the final answer. What am I overlooking/not thinking about?

 

[antibrane]

Looking at Equation (24),
<br /> \left\langle n^{(0)} \right.\left|n^{(1)}\right\rangle = -\frac{1}{2} \sum_{k=1}^{1-1}\left\langle n^{(1-k)}\right.\left| n^{(k)}\right\rangle = 0<br />
so the \left\langle n^{(0)} \right.\left|n^{(1)}\right\rangle term in what you wrote is zero.

 

[TSny]

They have E_{n}^{(2)} = &lt;n^{(0)}|V|n^{(1)}&gt; = - \sum_{m \neq 0}{\frac{|V_{mn}|^{2}}{E_{mn}}}

The pdf document has a small typo in that the summation index should have m ≠ n rather than m ≠ 0

but I would have done E_{n}^{(2)} = &lt;n^{(0)}|V|n^{(1)}&gt; - &lt;n^{(0)}|E_{n}^{(1)}|n^{(1)}&gt; and then plugged in E_{n}^{(1)} = V_{nn} from their earlier solution for first order terms.

See if you can show the last term on the right is zero using equation (31).

I don't know where I would have gone form there and I certainly couldn't even take a gander at how they end up with a summation either in this equation or in equations 33 and 34. Are there steps being omitted and/or can this be explained conceptually?

Note that any vector can be expanded in the basis set $\{|m^{(0)}>\}$. So, in particular the vector $|n^{(1)}>$ can be expanded as $|n^{(1)}> =\displaystyle \sum\limits_{m \neq n}{c_m|m^{(0)}>}$. Equation (31) allows the sum to be restricted to m≠n.

Now use equation (30) to identify the constants $c_m$. See what you get if you substitute this expansion of $|n^{(1)}>$ into $E_{n}^{(2)} = <n^{(0)}|V|n^{(1)}>$

 

[vela]

He skipped a couple of steps. Equations 30 and 31 are
\begin{align*}
\langle m^{(0)} | n^{(1)} \rangle &= -\frac{V_{mn}}{E_{mn}} \\
\langle n^{(0)} | n^{(1)} \rangle &= 0
\end{align*} If you expand $\lvert n^{(1)} \rangle$ in terms of the eigenstates of the unperturbed Hamiltonian, you get
$$\lvert n^{(1)} \rangle = \sum_{m} \lvert m^{(0)} \rangle\langle m^{(0)} \lvert n^{(1)} \rangle.$$ Using equations 30 and 31, you end up with
$$\lvert n^{(1)} \rangle = \sum_{m \ne n} -\frac{V_{mn}}{E_{mn}}\lvert m^{(0)} \rangle.$$ When you plug this into the first line of equation 32, you get the second line.

 

[dydxforsn]

Ok, that definitely cleared everything up. Thank you for everything, I especially wouldn't have guessed that they were expanding corrections to the eigenstates in terms of unperturbed eigenstates. Wow, you'd think that would have been a part of the derivations they would have spent more than nothing on...

The nuances are beginning to make sense..