perturbation theory - Perturbative Quantum Mechanics

I am, in full generality, confused about perturbation theory in quantum mechanics.

My textbook and Wikipedia have the same general approach to explaining it: given some Hamiltonian $H=H^{(0)} + H^\prime$, we can break down each eigenfunction $\left\vert n \right\rangle$ into a power series in an invented constant $\lambda$ and the eigenenergies likewise:

$\left\vert n \right\rangle = \sum\lambda^i\left\vert n^{(i)}\right\rangle$

$E_n = \sum \lambda^i E_n^{(i)}$

$\left(H^{(0)} + H^\prime\right) \left(\left\vert n^{(0)}\right\rangle + \lambda \left\vert n^{(0)}\right\rangle + \cdots \right) = \left(E^{(0)}+ \lambda E^{(1)} + \cdots\right) \left(\left\vert n^{(0)}\right\rangle + \lambda \left\vert n^{(1)}\right\rangle + \cdots \right)$

... and then they take $\lambda\to1$.

My question is - what's the logic here? Where did this come from? What purpose does $\lambda$ serve, given that the actual size of each contribution will be determined by the $E^{(i)}$'s and $\left\vert n^{(i)}\right\rangle$'s?

Answer

Firstly, I refer you to Prof. Binney's textbook (see below) which covers perturbation theory in quantum mechanics in explicit detail. When doing perturbation theory, we perturb the Hamiltonian $H^{(0)}$ of a system which has been solved analytically, i.e. the eigenstates and eigenvalues are known. Specifically,

$$H^{(0)}\to H^{(0)} + \lambda H'$$

where $H'$ is the perturbation, and $\lambda$ is a coupling constant. Why include such a constant? As Binney says, it provides us a 'slider' which when gradually increased to unity increases the strength of the perturbation. When $\lambda = 0$, the system is unperturbed, and when $\lambda=1$ we 'fully perturb the system.'

Introducing a coupling constant $\lambda$ also provides us with a manner to refer to a particular order of perturbation theory; $\mathcal{O}(\lambda)$ is first order, $\mathcal{O}(\lambda^2)$ is second order, etc. As we increase in powers of the coupling constant, we hope the corrections decrease. (The series may not even converge.)

A caveat: the demand that a coupling $\lambda \ll1$ may not be sufficient or correct to ensure that the coupling is small; this is only the case when the coupling is dimensionless. For example, if the coupling, in units where $c=\hbar=1$, had a mass (or equivalently energy) dimension of $+1$, then to ensure a weak coupling we would need to demand, $\lambda/E \ll 1$, where $E$ had dimensions of energy. Such couplings are known as relevant as at low energies they are high, and at high energies the coupling is low.