Time Independent Perturbation Theory - Degenerate

Similar to the non-degenerate case, we take our Hamiltonian $$H = H_0 + H'$$ except now, the initial unperturbed Hamiltonian has degenerate states. Assuming it is q- fold degenerate, if the symmetry giving rise to the degeneracy is disturbed, it will produce distinct 'q' non degenerate states. Our motive is to calculate the energy levels of these newly produced distinct energy levels. Let us consider two different initial eigenstates $E_1^0$ and $E_{q+1}^0$ [it is equal to the zeroth order term in the expansion] where $E_1^0$ is q-fold degenerate with states $$E_1^0 = E_2^0 = .... = E_q^0$$. As we have seen in non-degenerate perturbation theory, the coefficients $$c_{jn} = \frac{H'_{jn}}{E_n^0 - E_j^0}\,\,\,j\neq{n}\,\,and\,\, j,n\leq{q}$$ will become infinite for two different energy levels which has the same energy value. This situation can be avoided only if the numerator term is zero for all $j,n \leq {q}$ (Not the diagonal terms because $j\neq{n}$). It is equivalent to diagonalizing the sub-matrix $H'_{jn}\,\,\,\,for j,n\leq{q}$. This can be achieved by transforming to the new set of eigenfunctions from the initial set of eigenfunctions.
Thus, our problem for solving the energy values of the degenerate case reduced to diagonalizing the sub-matrix of H'. The diagonal elements are the change in energy levels of the initial degenerate states due to the perturbation. 
Representing the 'q' new eigenfunctions that diagonalize the $H'_{jn}$ as $\tilde{\psi_n}$ we write, $$ \tilde{\psi_n} = \sum_{m=1}^q a_{nm} \psi_m^0$$ Since, it will diagonalize, $$ \langle\tilde{\psi_n}\vert\tilde{\psi_j}\rangle = H'_{nj}\delta_{nj} $$ So, our new set of basis functions are, $$ Basis \,= \left\{\tilde{\psi_1},\tilde{\psi_2},\cdots\tilde{\psi_q},\psi_{q+1}^0,\cdots\right\}$$The diagonal element of the sub-matrix H' is the first order correction to the energy i.e. $$E'_n = \langle\tilde{\psi_n}\vert\tilde{\psi_n}\rangle = H'_{nn} \,\,n\leq{q}$$
Corresponding eigenfunctions are determined from $$H'\tilde{\psi_n} = E'_n\tilde{\psi_n}$$ substituting for new eigenfunctions in terms of initial wavefunctions, $$H' \sum_{m=1}^q a_{nm}\psi_m^0 = E'_n \sum_{m=1}^q a_{nm} \psi_m^0$$ Taking inner product with $\psi_j^0$ we have, $$\sum_{m=1}^q \left(H'_{jm} - E'_n\delta_{jm}\right)a_nm = 0\,\,\,n,j\leq{q} $$ It could be written simply in matrix form. For non-trivial solution for {a_nm} the determinant of coefficient matrix should vanish gives the secular equation in this case as, $$determinant\,\,\vert{H'_{jm}}-E'_n\mathbb{I}\vert = 0 $$ Using these the new eigenfunctions and eigenvalues are calculated.

Time Independent Perturbation Theory - Non-degenerate

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 $d_{nn} = 0$