# Basics of Non-degenerate time-independent quantum perturbation theory (Rayleigh-Schrödinger’s)

Some prerequisites: series expansion, quantum mechanics, bra-ket notation.

1. Introduction

I just recently discovered a satisfied explanation of the manipulations in non-degenerate time-independent quantum perturbation theory. I believe my discovery is not original. Even if it is I believe it is not of the level of publication in physics journals. Maybe, after some further polishes and additions, it might be publishable in some physics teaching journals. Well, I have no experience in publishing in this type of journals and do not wish to try it any time soon. More importantly, I believe it would be beneficial to students who start learning this topic. So it is more urgent to put it in my blog (although not many people would read mine anyway). I am sure that if I am satisfied with the analysis, students would be also satisfied.

2. Analysis often found in textbooks

The starting point of non-degenerate time-independent quantum perturbation theory is an eigenvalue equation $\displaystyle H|\psi_n\rangle = E_n|\psi_n\rangle, \ \ \ \ \ (1)$

where the Hamiltonian is expressed as $\displaystyle H = H^0 + {\lambda} H^1, \ \ \ \ \ (2)$

where ${H^0}$ is the Hamiltonian that we know the spectrum, ${H^1}$ is the perturbation Hamiltonian, and ${{\lambda}}$ is a book-keeping parameter. Knowing the spectrum of ${H^0}$ means knowing all the eigenvalues ${E^0_n}$ and eigenstates ${|\psi^0_n\rangle}$ from $\displaystyle H^0|\psi_n^0\rangle = E_n^0|\psi_n^0\rangle. \ \ \ \ \ (3)$

The eigenstates are orthonormal, i.e. ${\langle\psi_n^0|\psi_m^0\rangle = {\delta}_{nm}.}$ Note that we have limited the analysis to the non-degenerate case only.

Then in order to solve eq.(1), one first expands $\displaystyle E_n =E_n^0 + {\lambda} E_n^1 + {\lambda}^2 E_n^2 + \cdots, \ \ \ \ \ (4)$ $\displaystyle |\psi_n\rangle =|\psi_n^0\rangle + {\lambda} |\psi_n^1\rangle + {\lambda}^2 |\psi_n^2\rangle + \cdots. \ \ \ \ \ (5)$

Substituting eq.(2), (4), (5) into eq.(1) gives $\displaystyle (H^0 + {\lambda} H^1)(|\psi_n^0\rangle + {\lambda} |\psi_n^1\rangle + {\lambda}^2 |\psi_n^2\rangle + \cdots) = (E_n^0 + {\lambda} E_n^1 + {\lambda}^2 E_n^2 + \cdots)(|\psi_n^0\rangle + {\lambda} |\psi_n^1\rangle + {\lambda}^2 |\psi_n^2\rangle + \cdots). \ \ \ \ \ (6)$

Expanding both sides of the equation and comparing the orders in ${{\lambda}}$ gives $\displaystyle {\lambda}^0: H^0 |\psi_n^0\rangle = E_n^0|\psi_n^0\rangle, \ \ \ \ \ (7)$ $\displaystyle {\lambda}^1: H^1 |\psi_n^0\rangle+H^0 |\psi_n^1\rangle = E_n^1|\psi_n^0\rangle+E_n^0|\psi_n^1\rangle, \ \ \ \ \ (8)$ $\displaystyle {\lambda}^2: H^2 |\psi_n^0\rangle+H^1 |\psi_n^1\rangle +H^0 |\psi_n^2\rangle = E_n^2|\psi_n^0\rangle+E_n^1|\psi_n^1\rangle+E_n^0|\psi_n^2\rangle, \ \ \ \ \ (9)$ $\displaystyle etc. \ \ \ \ \ (10)$

One sees that the order ${{\lambda}^0}$ simply coincides with eq.(3). So let us consider the order ${{\lambda}^1.}$ Acting on the equation (8) by ${\langle\psi_n^0|}$ and rearranging gives $\displaystyle E_n^1 =\langle\psi_n^0|H^1|\psi_n^0\rangle. \ \ \ \ \ (11)$

One can also determine ${|\psi_n^1\rangle}$ from eq.(8) by first expressing it in the basis of ${|\psi_m^0\rangle.}$ That is $\displaystyle |\psi_n^1\rangle =c_{nn}|\psi_n^0\rangle +\sum_{m\neq n}c_{mn}|\psi_m^0\rangle. \ \ \ \ \ (12)$

Substituting this expression into eq.(8) gives $\displaystyle H^1 |\psi_n^0\rangle+\sum_{m\neq n}(E_m^0 - E_n^0) c_{mn}|\psi_m^0\rangle = E_n^1|\psi_n^0\rangle. \ \ \ \ \ (13)$

Acting on this equation by ${\langle\psi_n^0|}$ with ${k\neq n}$ gives $\displaystyle \langle\psi_k^0| H^1 |\psi_n^0\rangle+(E_m^0 - E_n^0) c_{kn} = 0. \ \ \ \ \ (14)$

So $\displaystyle c_{kn} = -\frac{\langle\psi_k^0| H^1 |\psi_n^0\rangle}{E_k^0 - E_n^0};\qquad k\neq n. \ \ \ \ \ (15)$

Note however that the expression of ${c_{nn}}$ is not fixed by eq.(8). So up to this point, one has $\displaystyle |\psi_n^1\rangle =c_{nn}|\psi_n^0\rangle -\sum_{m\neq n}\frac{\langle\psi_m^0| H^1 |\psi_n^0\rangle}{E_m^0 - E_n^0}|\psi_m^0\rangle. \ \ \ \ \ (16)$

In principle, ${c_{nn}}$ can be fixed by using an additional condition which does not contradict to eq.(7)(10). The usual choice is ${c_{nn} = 0.}$ I have found from textbooks two alternative explanations for this:

1. The equation (1) we are allowed to choose any normalisation and phase of ${|\psi_n\rangle.}$ So let us demand the normalisation ${\langle\psi_n|\psi_n\rangle = 1.}$ By expanding in ${{\lambda},}$ one obtains ${\langle\psi_n^0|\psi_n^1\rangle = -\langle\psi_n^1|\psi_n^0\rangle = -\langle\psi_n^0|\psi_n^1\rangle^*.}$ Furthermore, we can choose the phase such that ${\langle\psi_n^0|\psi_n^1\rangle}$ is a real number. But this gives ${c_{nn} = \langle\psi_n^0|\psi_n^1\rangle = 0.}$
2. Alternatively, we do not yet wish to normalise ${|\psi_n\rangle.}$ But let us choose its scale such that ${\langle\psi_n^0|\psi_n\rangle=1.}$ By expanding in ${{\lambda}}$ this gives ${c_{nn} = \langle\psi_n^0|\psi_n^1\rangle = 0.}$

I have always been worried with these explanations as it is not clear to me (except after I learn of the analysis to be given in the next section) why we are still allowed to choose normalisation and phase of ${|\psi_n\rangle}$ after we have fixed the normalisation of its zeroth order ${|\psi_n^0\rangle.}$ How can we see this explicitly? Or maybe how can we count the number of independent conditions against the number of coefficients to really see if we are allowed to impose further conditions; after all these numbers seem to be infinite anyway?

So although the analysis is still need to be continue in order to obtain ${E_n^2, E_n^3,\cdots, |\psi_n^2\rangle, |\psi_n^3\rangle, \cdots,}$ let us end this section here and present an alternative analysis in which the above questions do not arise.

3. The analysis that I find it more satisfying

The basic setup of this alternative analysis is the same as that given in the previous section. Let us start from $\displaystyle (H^0 + {\lambda} H^1)|\psi_n\rangle = E_n|\psi_n\rangle. \ \ \ \ \ (17)$

Acting on this equation by ${\langle\psi_n^0|}$ gives $\displaystyle E_n =E_n^0 + {\lambda}\dfrac{\langle\psi_n^0|H^1|\psi_n\rangle}{\langle\psi_n^0|\psi_n\rangle}. \ \ \ \ \ (18)$

Next, let us act on eq.(17) by ${\langle\psi_m^0|}$ for ${m\neq n.}$ This gives $\displaystyle (E_m^0 - E_n)\langle\psi_m^0|\psi_n\rangle +{\lambda}\langle\psi_m^0|H^1|\psi_n\rangle =0. \ \ \ \ \ (19)$

So $\displaystyle \langle\psi_m^0|\psi_n\rangle =-{\lambda}\dfrac{\langle\psi_m^0|H^1|\psi_n\rangle}{E_m^0 - E_n}. \ \ \ \ \ (20)$

Applying completeness relation on ${|\psi_n\rangle,}$ we obtain $\displaystyle \begin{array}{rl} |\psi_n\rangle &=\displaystyle |\psi_n^0\rangle\langle\psi_n^0|\psi_n\rangle +\sum_{m\neq n}|\psi_m^0\rangle\langle\psi_m^0|\psi_n\rangle\\ &\\ &=\displaystyle |\psi_n^0\rangle\langle\psi_n^0|\psi_n\rangle -{\lambda}\sum_{m\neq n}\dfrac{\langle\psi_m^0|H^1|\psi_n\rangle}{E_m^0 - E_n}|\psi_m^0\rangle. \end{array} \ \ \ \ \ (21)$

The idea to obtain the spectrum of ${H}$ up to a desired accuracy is to make a repeated use of of eq.(18) and (21). However, we see that ${\langle\psi_n^0|\psi_n\rangle}$ is not restricted by any condition. So we are allowed to fix it. But instead of fixing, I am more comfortable with defining a new ket $\displaystyle |\varphi_n\rangle =\dfrac{|\psi_n\rangle}{\langle\psi_n^0|\psi_n\rangle}. \ \ \ \ \ (22)$

This definition automatically implies ${|\varphi_n^0\rangle = |\psi_n^0\rangle.}$ So eq.(18) and (21) now become $\displaystyle E_n =E_n^0 + {\lambda}\langle\varphi_n^0|H^1|\varphi_n\rangle. \ \ \ \ \ (23)$ $\displaystyle |\varphi_n\rangle =\displaystyle |\varphi_n^0\rangle -{\lambda}\sum_{m\neq n}\dfrac{\langle\varphi_m^0|H^1|\varphi_n\rangle}{E_m^0 - E_n}|\varphi_m^0\rangle. \ \ \ \ \ (24)$

In fact, it is better to use $\displaystyle (E_m^0-E_n)\langle\varphi_m^0|\varphi_n\rangle + {\lambda}\langle\varphi_m^0|H^1|\varphi_n\rangle =0;\qquad m\neq n. \ \ \ \ \ (25)$

Furthermore, the definition of ${|\varphi_n\rangle}$ gives $\displaystyle \langle\varphi_n^0|\varphi_n\rangle = 1. \ \ \ \ \ (26)$

In order to obtain ${E_n}$ and ${\varphi_n,}$ we expand in power of ${{\lambda}}$ and repeatedly solve eq.(23)(26) order-by-order. From eq.(23), it is easy to obtain $\displaystyle E_n^k = \langle\varphi_n^0|H^1|\varphi_n^{k-1}\rangle;\qquad k=1,2,3,\cdots. \ \ \ \ \ (27)$

Next, after some analysis eq.(25)(26) give [I might come back and show the detailed calculation later] $\displaystyle \displaystyle|\varphi_n^k\rangle =\sum_{m\neq n}\frac{|\varphi_m^0\rangle}{E_m^0-E_n^0} \left(\sum_{i=1}^{k-1}E_n^i\langle\varphi_m^0|\varphi_n^{k-i}\rangle - \langle\varphi_m^0|H^1|\varphi_n^{k-1}\rangle\right); \qquad k=1,2,3,\cdots \ \ \ \ \ (28)$

By recursively applying eq.(27)(28) starting from smaller values of ${k,}$ one can obtain order-by-order the solutions for ${E_n}$ and ${|\varphi_n\rangle.}$