*When a given Hamiltonian of a problem is not so much different from the Hamiltonian of an absolutely solvable problem, we use the perturbation theory to find the new eigenfunction from the old eigenfunctions with the assumption that the perturbing Hamiltonian is small compared to the original one. Let us jump into the mathematics by writing the total Hamiltonian as, $$ H = H_0 +\lambda{H' } \,\,\,\,...(1)$$ where H' is the perturbing Hamiltonian and $\lambda$ is used to denote how small the perturbation is with respect to the original. The eigenfunctions are expressed as, $$ H_0 \psi_n^0 = E_n \psi_n^0 \,\,\,\,...(2)$$ and $$H\psi_n = E_n \psi_n\,\,\,\,...(3)$$ When $\lambda\rightarrow{0}$ the perturbed eigenfunctions should reduce to the original eigenfunctions. So, we expand it in series as, $$\psi_n = \psi_n^0 + \lambda\psi_n^1+\lambda^2\psi_n^0+\cdots \,\,\,...(4)\\ E_n = E_n^0 +\lambda{En^1} + \lambda^2E_n^2+\cdots \,\,\,...(5)$$ where the upper index is used to denote the corresponding order correction on the specific eigenfunction and eigenvalue. Substituting this in (3) and to make this true for arbitrary $\lambda$ values we equate the corresponding terms of lambda in the equation to get, $$ H_0\psi_n^0 = E_n\psi_n^0\,\,\,...(6)\\ H_0\psi_n^1 +H'\psi_n^0 = E_n^1\psi_n^0 + E_n^0\psi_n^1\,\,\,...(7)\\ H_0\psi_n^2 + H'\psi_n^1 = E_n^1\psi_n^1 + E_n^2\psi_n^0+E_n^0\psi_n^2\,\,\,...(8) $$*

Rearranging them gives, $$ H_0\psi_n^0 = E_n\psi_n^0\\ \left(H_0 - E_n^0\right)\psi_n^1 = \left(E_n^1-H'\right)\psi_n^0 \\ \left(H_0-E_n^0\right)\psi_n^2 = \left(E_n^1-H'\right)\psi_n^1 + E_n^2\psi_n^0 $$ To avoid $\psi_n^1+ \alpha\psi_n^0$ being a solution with first order energy correction, we choose the correction terms orthogonal to the initial eigenfunctions. $$\langle\psi_n^k\vert\psi_n^0\rangle = 0 \,\,...k>0$$ Assuming all correction terms are in the same Hilbert space of the initial eigenfunctions, the correction terms are expressed as a linear combination. Substituing the linear combination $ \psi_n^k = \sum_j c_{jn} \psi_j^0 $ in the first equation and taking the inner-production with $\psi_l^0$ we get, $$ (E_l^0 - E_n^0) + H'_{ln} = E_n^1\delta_{ln}$$ The first order correction in Energy eigenvalue and eigenfunction is obtained to be, $$ E_n^1 = \langle\psi_n^0\vert{H'}\psi_n^0\rangle \\ \psi_n^1 = \sum_{j\neq{n}} \frac{H'_{jn}}{E_n^0 - E_j^0}\psi_j^0 $$

Similarly solving for second order perturbation,by assuming $\psi_n^2 = \sum_j d_{nj}\psi_j^0$we get, $$ E_n^2 = \sum_{j\neq{n}}\frac{|H'_{nj}|^2}{E_n^0-E_j^0} $$ and $$\psi_n^2 = \sum_{j\neq{n}} \left[\sum_{l\neq{n}}\frac{H'_{jk}H'{kn}}{\left(E_n^0-E_j^0\right)\left(E_n^0-E_l^0\right)}-\frac{H'_{nn}H'_{jn}}{\left(E_n^0-E_j^0\right)^2}\right]\psi_j^0$$ where $d_{nn} = 0$

Rearranging them gives, $$ H_0\psi_n^0 = E_n\psi_n^0\\ \left(H_0 - E_n^0\right)\psi_n^1 = \left(E_n^1-H'\right)\psi_n^0 \\ \left(H_0-E_n^0\right)\psi_n^2 = \left(E_n^1-H'\right)\psi_n^1 + E_n^2\psi_n^0 $$ To avoid $\psi_n^1+ \alpha\psi_n^0$ being a solution with first order energy correction, we choose the correction terms orthogonal to the initial eigenfunctions. $$\langle\psi_n^k\vert\psi_n^0\rangle = 0 \,\,...k>0$$ Assuming all correction terms are in the same Hilbert space of the initial eigenfunctions, the correction terms are expressed as a linear combination. Substituing the linear combination $ \psi_n^k = \sum_j c_{jn} \psi_j^0 $ in the first equation and taking the inner-production with $\psi_l^0$ we get, $$ (E_l^0 - E_n^0) + H'_{ln} = E_n^1\delta_{ln}$$ The first order correction in Energy eigenvalue and eigenfunction is obtained to be, $$ E_n^1 = \langle\psi_n^0\vert{H'}\psi_n^0\rangle \\ \psi_n^1 = \sum_{j\neq{n}} \frac{H'_{jn}}{E_n^0 - E_j^0}\psi_j^0 $$

Similarly solving for second order perturbation,by assuming $\psi_n^2 = \sum_j d_{nj}\psi_j^0$we get, $$ E_n^2 = \sum_{j\neq{n}}\frac{|H'_{nj}|^2}{E_n^0-E_j^0} $$ and $$\psi_n^2 = \sum_{j\neq{n}} \left[\sum_{l\neq{n}}\frac{H'_{jk}H'{kn}}{\left(E_n^0-E_j^0\right)\left(E_n^0-E_l^0\right)}-\frac{H'_{nn}H'_{jn}}{\left(E_n^0-E_j^0\right)^2}\right]\psi_j^0$$ where $d_{nn} = 0$