# Why is perturbation method possible?

Following is a generic statement of time independent perturbation theory.

Given: $H^o\left|\phi_n\right> = E^o_n\left|\phi_n\right>$

Find : $H\left|\psi_n\right> = E_n\left|\psi_n\right>\qquad$ (1)

where $H=H^o + V$

Surely as soon as someone reads this, one thinks that can solve this probably by using expansions in unperturbed basis.

$$\left|\psi_n\right> = \sum_l c_{nl} \left|\phi_l\right>$$

By plugging it in the equation of interest we get.

$$(H^o+V)(\sum_l c_{nl} \left|\phi_l\right>) = E_n (\sum_l c_{nl} \left|\phi_l\right>)$$

Let's take inner product of this equation with $\left<\phi_k\right|$

$$c_{nk} E^o_k + \sum_l c_{nl} V_{kl} = c_{nk} E_n\tag{2}$$

Here $k\in (1, N) \cup \mathbb{N}$ with $N$ being the dimensions of the Hilbert space.

So as I can see it to solve the problem we simply solve coupled $N$ equations in eq (2).

But this is where I am confused. $E_n$ and $c_{nl}$ are the $2N$ unknowns but we only have $N$ equations. I would like to point out that equation (1) can be solved if we know the $V_{nk}$ matrix. Then equation (1) is simply an eigenvalue problem.

Where is fault in my argument?

• The way people do perturbation theory isn't like what you have mentioned. See textbooks for more information. In fact you haven't used any perturbation methods, nothing is an approximate in your statements. – Hosein Sep 25 '16 at 11:45
• if your states are orthogonal (true for states of unique eigenvalues) then <\phi_m | \phi_n > = \delta_(m,n). This is the first place you can simplify your problem. – anon01 Sep 25 '16 at 12:08
• It helps to know that sequential approximation methods have a long history and were done in Hamiltonian form in classical mechanics long before they were done in quantum mechanics. In that case you end up writing explicit differential equations so the notation isn't as neat, but the process is very similar. – dmckee --- ex-moderator kitten Sep 25 '16 at 16:15
• @Hosein I deliberately neglected to show that the perturbative methods should not work. Since this simple method does not. – The Imp Sep 26 '16 at 2:06

When you are solving for $E_n$ (at a specific value of $n$), you have only $N+1$ unknowns, namely $c_{nl}$ and $E_n$. But you have $N+1$ equations given by the $N$ equation you pointed out in addition to the one equation which comes from normalising $\psi_{n}$ and thus putting a condition on $\sum_{l}c_{nl}c_{nl}^{*}$ = $c$, where $c$ is some constant (equal to $1$ if the basis vectors are orthonormal). I have assumed that $l$ takes on $N$ possible values, since you have used both $l$ and $k$ to label the basis vectors, and have said that $k$ takes on integer values between $1$ and $N$.