Time-independent perturbation theory: why $i$'th order perturbations are orthogonal to base state? I have been learning about time independent perturbation theory (non-degenerate for the moment), and am not satisfied about a particular point: the justification for setting $\langle n^i|n^0\rangle = 0$ for $i>0$ where $i$ denotes the order of the approximation.
I should think that we can set this to zero iff it has no observable consequences. And since the only quantities that we observe/measure are the eigenvalues, this is iff the eigenvalues remain unchanged. Now it is far from obvious to me as to how we can make a claim that the eigenvalues of ANY operator we apply to the subsequent kets will be unchanged, and therefore that we really can never know by ANY measurement whether the $i>0$ order corrections to the $n$th eigenvalue, $|n\rangle$, really are entirely orthogonal to $|n^0\rangle$. I would appreciate a comment on this, but even just considering the Hamiltonian operator it seems not to be true.
If we have $$(H^0+\lambda H^1)(|n^0\rangle + \lambda |n^1\rangle + \lambda ^2|n^2\rangle + ...) = (E_n^0 +\lambda E_n^1 +\lambda ^2 E_n^2+...)(|n^0\rangle + \lambda |n^1\rangle + \lambda ^2|n^2\rangle + ...)$$
and from the first order approximation
$$E_n^1= \langle n^0 | H^1 | n^0\rangle$$
which is independent on $|n^1\rangle$.
One can easily find the coefficients $\langle m^0|n^1 \rangle = \frac{\langle m^0|H^1|n^0\rangle}{E_n^0-E_m^0}$ for $m \neq n$ such that
$$|n^1\rangle = \alpha |n^0\rangle + \Sigma _{m\neq n} |m^0\rangle \frac{\langle m^0|H^1|n^0\rangle}{E_n^0-E_m^0}.$$
And that is about as much as one can say from considering the first order terms. I appreciate that the independence of $E_n^1$ on $|n^1\rangle$ means that, working to first order, and if the Hamiltonian's spectrum are the only quantities we care about, we can set $\alpha = 0$. And in fact, it is convenient that way because then the $|n\rangle = |n^0\rangle+\lambda |n^1\rangle$ is normalised to $O(\lambda ^2)$.
However, the value of $\alpha$ does have second order consequences (as demonstrated below), which is why I do not understand why authors use $\langle n^1|n^0\rangle = 0$ in finding second-order uentities, i.e. $E_n^2$.
To second order:
$$H^1|n^1\rangle + H^0|n^2\rangle = E_n^0|n^2\rangle + E_n^1|n^1\rangle + E_n^2|n^0\rangle$$
and then taking the inner product with $\langle n^0|$, one gets
$$E_n^2 = \langle n^0|H^1|n^1 \rangle - E_n^1\langle n_0|n^1\rangle$$
and substituting in for $E_n^1$ and $|n^1\rangle$, and after some rearranging, one finds
$$E_n^2 = \langle n^0|H^1 [|n^1\rangle - \alpha |n^0\rangle]$$
which does depend on $\alpha$!
 A: $\newcommand{\bra}[1]{\left< #1 \right|}
\newcommand{\ket}[1]{\left| #1 \right>}
\newcommand{\bk}[2]{\left< #1 \middle| #2 \right>}
\newcommand{\bke}[3]{\left< #1 \middle| #2 \middle| #3 \right>}$The justification for setting $\bk{n^i}{n^0} = 0$ is as follows:
$$\ket{n} = \ket{n^0} + \lambda\ket{n^1} + \lambda^2\ket{n^2}+\cdots.\tag{1}\label{1}$$
We can always assume that the eigenstates of the unperturbed Hamiltonian are normalized. Thus, $$\bk{n^0}{n^0} = 1 \tag{2}\label{2}.$$ From \ref{1}, we have $$\begin{aligned}\bk{n^0}{n} &= \bk{n^0}{n^0} + \lambda\bk{n^0}{n^1} + \lambda^2\bk{n^0}{n^2}+\cdots\\
\bk{n^0}{n} &= 1 + \lambda\bk{n^0}{n^1} + \lambda^2\bk{n^0}{n^2}+\cdots\end{aligned} $$
Now, since $\ket{n}$ is not normalized, we are at liberty to normalize it as we want. So we choose $\bk{n^0}{n} = 1$. Note that this does not mean that $\ket{n^0}$ and $\ket{n}$ are the same vectors or lie in the same direction. Our normalization just ensures that the projection of $\ket{n}$ on $\ket{n^0}$ is unity. Thus, $$0 = \lambda\bk{n^0}{n^1} + \lambda^2\bk{n^0}{n^2}+\cdots$$ for all $\lambda$. This means that $\bk{n^i}{n^0} = 0$.

We are under no obligation to choose $\bk{n^0}{n} = 1$. Suppose that $\ket{n^0}$ and $\ket{n^i}$ are not orthogonal to each other. So we can take $$\ket{n^i} = \ket{n^i}' + a_i\ket{n^0}$$ where $\ket{n^i}'$ is orthogonal to $\ket{n^0}$. From \ref{1} we get that $$\begin{aligned}\ket{n} &= \ket{n^0} + \lambda\left[\ket{n^1}' + a_2\ket{n^0}\right] + \lambda^2\left[\ket{n^2}' + a_2\ket{n^0}\right] + \cdots\\
\ket{n} &= \left(1 + a_1\lambda + a_2\lambda^2 +\cdots \right)\ket{n^0} + \ket{n^1}' + \ket{n^2}' + \cdots \end{aligned}.$$ Dividing the above equation by $\left(1 + a_1\lambda + a_2\lambda^2 +\cdots \right)$ will yield us another physically identical vector which will lie in the same ray space as $\ket{n}$.

We can very well choose $\bk{n}{n} = 1$ instead of $\bk{n^0}{n} = 1$. We can also choose the phase of $\ket{n}$ such that $\bk{n}{n^0}$ is real. Upto first order, $$\begin{aligned}\ket{n} &= \ket{n^0} + \lambda\ket{n^1} + O(\lambda^2)\\
\bk{n}{n} &= \left[\bra{n^0} + \lambda\bra{n^1}\right]\left[\ket{n^0} + \lambda\ket{n^1}\right] + O(\lambda^2)\\
1 &= 1 + \lambda\left[\bk{n^1}{n^0} + \bk{n^0}{n^1}\right] + O(\lambda^2).\end{aligned}$$
Thus, $$\bk{n^1}{n^0} = \bk{n^0}{n^1} = 0.$$ For second order, we get $$\bk{n^0}{n^2} = \bk{n^2}{n^0} = -\frac{1}{2}\bk{n^1}{n^1}.$$ We can repeat the same process for $i^{\text{th}}$ order to get $$\bk{n^i}{n^0} = \bk{n^0}{n^i} = -\frac{1}{2}\left[\bk{n^{i-1}}{n^1} + \bk{n^{i-2}}{n^2} + \cdots + \bk{n^2}{n^{i-2}} + \bk{n^1}{n^{i-1}}\right].$$

From your question writeup, $$E_n^2 = \big\langle n^0 \vert H^1 \left[\ket{n^1} - \alpha\ket{n^0}\right]$$ and $$\ket{n^1} = \sum_{m\neq n}\ket{m^0}\frac{\bke{m^0}{H^1}{n^0}}{E_n^0 - E_m^0} + \alpha \ket{n^0}.$$ Thus, $$E_n^2 = \big\langle n^0 \vert H^1 \left[\sum_{m\neq n}\ket{m^0}\frac{\bke{m^0}{H^1}{n^0}}{E_n^0 - E_m^0} + \alpha \ket{n^0} - \alpha\ket{n^0}\right] = \big\langle n^0 \vert H^1 \left[\sum_{m\neq n}\ket{m^0}\frac{\bke{m^0}{H^1}{n^0}}{E_n^0 - E_m^0}\right].$$ We see that $E_n^2$ is independent of $\alpha$. One can calculate the energy change for any order in both the ways and convince oneself that it remains unchanged.
References:

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*Prof. Barton Zweibach. 8.06 Quantum Physics III. Spring 2018. Massachusetts Institute of Technology: MIT OpenCouseWare, https://ocw.mit.edu/courses/physics/8-06-quantum-physics-iii-spring-2018/lecture-notes/MIT8_06S18ch1.pdf. License: Creative Commons BY-NC-SA.


*Claude Cohen-Tannoudji; Bernard Diu; Franck Laloë (4 December 2019). Quantum Mechanics, Volume 2: Angular Momentum, Spin, and Approximation Methods. Wiley. ISBN 978-3-527-34554-0, p. 1119.
