# Non-Degenerate Perturbation theory in Sakurai

As stated in the title, I'm studying Non-degenerate Perturbation Theory with the book 'Modern Quantum Mechanics' by J.J. Sakurai. The problem to solve is $$(H_0+\lambda V)|n\rangle = E_n |n\rangle$$ where the exact eigenkets and energy eigenvalues are known: $$H_0 |n^{(0)}\rangle= E_n^{(0)}|n^{(0)}\rangle$$. He defines the energy shift $$\Delta_n\equiv E_n-E_n^{(0)}$$, and transforms the original equation to this: $$(E_n^{(0)}-H_0)|n\rangle = (\lambda V-\Delta_n)|n\rangle$$ Also, he defines $$\phi_n\equiv 1-|n^{(0)}\rangle \langle n^{(0)}|$$ and says the inverse operator $$\frac{1}{E_n^{(0)}-H_0}$$ is well defined when it multiplies $$\phi_n$$ on the right. But then, he says the equation can't be rewrited as $$|n\rangle = \frac{1}{E_n^{(0)}-H_0}\phi_n (\lambda V-\Delta_n)|n\rangle$$ and it needs to be like this: $$|n\rangle = c_n(\lambda)|n^{(0)}\rangle +\frac{1}{E_n^{(0)}-H_0}\phi_n (\lambda V-\Delta_n)|n\rangle$$ where $$\lim_{\lambda \to 0} \, c_n(\lambda)=1$$.

My question is, why can't you write the equation in the first form, and where does the term $$c_n(\lambda)$$ come from?

It's pretty obvious that you can't write it in the first form because $$\phi_n \neq 1$$ but $$1 = \phi_n + | n^{(0)}\rangle \langle n^{(0)}|$$. If you insert this definition of $$1$$ to the right side of the second equation you will get that additional term in the last equation with that ket $$| n^{(0)}\rangle$$. If you also carefully derive the equation you can even figure out what $$c_n(\lambda)$$ is and why $$lim_{\lambda \to 0} \, c_n(\lambda)=1$$
Recall your objective is to solve for $$|n\rangle$$ and $$\Delta_n$$. For an n, the projection operator $$\phi_n\equiv 1-|n^{(0)}\rangle \langle n^{(0)}|= \phi_n^2$$ splits $$|n\rangle= |n^{(0)}\rangle \langle n^{(0)} | n\rangle + \phi_n |n\rangle \equiv c_n(\lambda) |n^{(0)}\rangle + \phi_n |n\rangle$$ with, of course, $$\lim_{\lambda \to 0} \, c_n(\lambda)=1$$.
The equation $$(E_n^{(0)}-H_0)|n\rangle = (\lambda V-\Delta_n)|n\rangle$$ gives you 0 when dotted on the left by $$\langle n^{(0)}|$$, so you really have $$\phi_n (E_n^{(0)}-H_0)\phi_n|n\rangle = \phi_n(\lambda V-\Delta_n)|n\rangle ~.$$ It constrains $$\phi_n|n\rangle$$ but leaves $$c_n$$ arbitrary/undetermined. If your original space were N-dimensional, this final vector equation is an N-1-dimensional one. So you have $$\phi_n|n\rangle = \phi_n \frac{1}{E_n^{(0)}-H_0}\phi_n (\lambda V-\Delta_n)|n\rangle ,$$ which allows you to merge the projectors into one on the numerator unambiguously, $$= \frac{\phi_n}{E_n^{(0)}-H_0} (\lambda V-\Delta_n)|n\rangle .$$ The leading term with $$c_n$$ is so far unconstrained, save for the boundary condition at the unperturbed system. It is easiest to illustrate with a 2×2 matrix system, taking $$\phi_1=$$diag(0,1), $$H_0=\operatorname{diag} (E^{(0)}_1, E^{(0)}_2\equiv E^{(0)}_1-a)$$, so that $$\phi_1 \frac{1}{E_1^{(0)}-H_0}\phi_1=1/a$$, for all practical purposes, i.e. you are effectively projected to a 2-1=1-dimensional space.