quantum mechanics - What does the first order energy correction formula in non-degenerate perturbation theory means?

I'm studying for a test in quantum mechanics and I'm currently trying to learn about perturbation theory and I've realized that I don't quite understand what I'm doing when I'm doing my calculations.

Considering the case of non-degenerate perturbation theory, the formula for the first order energy correction is

$$E_n^1=\langle \psi_n^0|H'|\psi_n^0 \rangle.$$

• What exactly does this mean?


• I understand that it's some kind of expectation value of the perturbation but what more? And what is the meaning of $n$?

Answer

$E_n^1\quad=\quad\langle\psi_n^0|H^{'}|\psi_n^0\rangle$, tells you that the first order correction to energy is nothing but, the average value of your unperturbed wavefunction $(|\psi_m^0\rangle)$, with the perturbing Hamiltonian $(H^{'})$. The subscript on $\psi$ tells you the excitation of the wavefunction, viz. $n=0$, means its the ground state, if $\psi$ represents the 1-D Harmonic Oscillator; $n=1$, represents the first excited state, so on and so forth. The superscript $0$, tells me, that its the wavefunction of the unperturbed Hamiltonian $H^{0}$. Remember my total Hamiltonian $(H)$ consists of 2 parts, the unperturbed part $(H^0)$(the one that you have been doing till now, which solve the time-independent Sch. Eq.), and the second part, is the perturbed Hamiltonian $(H^{'})$. So $H = H^0 + H^{'}$.

And as to how this comes, I will tell some simple and very precise mathematical steps. Look up Griffiths for more information. Its Chapter 6, 2nd Ed.