Assumptions in basic perturbation theory I am teaching myself the basics of perturbation theory, mainly from Sakurai's 'Modern Quantum Mechanics', but also looking up lecture notes online. I am puzzled by one thing from the start of the discussion of perturbation theory in Sakurai. It seems that it is assumed that the perturbation of each eigenstates as a linear combination of the existing eigenstates does not include a component  in the direction of this eigenstate. i.e. the perturbation of the nth eigenstates is assumed to be a linear combination of all eigenstates except the nth (specifically, for the two state case, this assumption can be seen in page 287 of the revised edition of Sakurai, equation 5.1.5)
I suppose strictly it doesn't matter whether an additional term is included or not. It just affects the normalisation of the state (which it wouldn't be normalised either way); but I still don't see a convincing reason as to why this factor is omitted. I have looked at other sources and some excellent lectures on Youtube, but they have all presented the form of the purturbation (with the parallel component of the eigenstate omitted) without motivating this. I see in some calculations it is useful to take the inner product with the original eigenket, in which case the perturbation vanishes which can simplify a calculation. But surely this can't be the only reason? It doesn't seem like it makes life much easier than using all terms in the perturbation...
 A: Ooofff, there's a lot going on here.


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*For one, the choice in the test case in Sakurai's (5.1.5), which is seen more clearly in the equivalent equation (5.1.6),
$$ H = \begin{pmatrix} E_1^{(0)} & \lambda V_{12} \\ \lambda V_{21} & E_2^{(0)}\end{pmatrix}, \tag{5.1.6}$$
in the choice to have $V$ be purely off-diagonal, does not reflect the body of your question. Here the choice to have an off-diagonal $V$ is only speaking about the perturbation itself, and not about the eigenvectors' response to that perturbation; the choice in this limited example has been done simply to keep the exposition as simple as possible and ─ because it is a simple test case ─ it does not have any bearing on the general theory.

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I suppose strictly it doesn't matter whether an additional term is included or not. It just affects the normalisation of the state (which it wouldn't be normalised either way).

It's a red herring to ask whether the full state is normalized or not. Take the state $$v = N \begin{pmatrix} 1 \\ \lambda \end{pmatrix},$$ normalized to unity by having $N=(1+\lambda^2)^{-1/2}$, but taken only to first order in $\lambda$: this means that $N \approx 1-\frac12 \lambda^2 \approx 1$ doesn't actually kick in, as it only depends quadratically (to leading order) on $\lambda$. The full state can indeed be normalized (which may or may not be a good idea ─ it can be easier to account for the normalization elsewhere, i.e. when taking explicit expectation values, and Sakurai devotes an entire section to this) but that only comes in at the $\lambda^2$ level or higher. If you're doing second-order perturbation theory then it will matter, but if you're sticking to first-order calculations then it's a moot point.

*As an important point to notice, having diagonal elements for $V$ in the eigenbasis of $H_0$ doesn't do very much, as those elements only change the eigenvalues without actually altering the eigenvectors. This simplicity then feeds into the first-order change in the energy, in (5.1.37),
$$ \lambda \Delta_n^{(1)} = \langle n^{(0)} | \lambda V | n^{(0)} \rangle.\tag{5.1.37(a)}$$
Setting that matrix element to zero is equivalent to saying that $\Delta_n$ has no first-order dependence, which is obviously not the case in general.
Now, all that said, your intuition that this is related to the state's normalization does have a lot of conceptual depth behind it. The main point where this comes in is at the first full recursive expression for the perturbed eigenvector, eq. (5.1.27),
$$
|n\rangle = c_n(\lambda) |n^{(0)}\rangle + \frac{1}{E_n^{(0)}-H_0}  \phi_n (\lambda V- \Delta_n) |n\rangle,
\tag{5.1.27}
$$
where $\phi_n = 1- |n^{(0)}\rangle\langle n^{(0)}|$ is a projector that excludes the $|n^{(0)}\rangle$  direction (and which commutes with $H_0$, so the product $\frac{1}{E_n^{(0)}-H_0}  \phi_n $ also has no $|n^{(0)}\rangle$ component), which was introduced to deal with the fact that the shifted original hamiltonian $H_0 - E_n^{(0)}$ is singular at the $|n^{(0)}\rangle$ subspace.
The fact that the second term in (5.1.27) has no component along $|n^{(0)}\rangle$ is extremely important, because it means that to all orders in $\lambda$, the contributions to $|n\rangle$ along $|n^{(0)}\rangle$ are completely encased within the quasi-normalization function $c_n(\lambda)$. Now, as I said, it is perfectly possible to build a perturbation-theory formalism that keeps the full perturbed state normalized to unity, $\langle n|n\rangle = 1$, but this complicates a number of things, as the various orders of dependence in $c_n(\lambda)$ need to be included into the response along the other directions. Instead, Sakurai chooses to normalize his full state to
$$
\langle n^{0}|n\rangle = c_n(\lambda) = 1,
$$
which does simplify life significantly. (If you want to see how it simplifies life, set up a full Taylor expansion for $c_n(\lambda)$ and impose the normalization $\langle n|n\rangle = 1$, and then see what that does to (5.1.44) and its higher-order analogues.)
And, since you've now completely specified the behaviour of the component of $|n\rangle$ along $|n^{(0)}\rangle$ for all orders in $\lambda$, then you explicitly need to exclude it from the rest of your expansion.
