Non-Degenerate Perturbation theory in Sakurai

Question

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?

Answer 1

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 $lim_{\lambda \to 0} \, c_n(\lambda)=1$

Answer 2

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.