# 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 $n$?
• I don't think the equation is correct; shouldn't all subscripts by the same, e.g., $n$? Otherwise, the right hand side is a matrix element, not an expectation value. – Alfred Centauri Jan 5 '15 at 13:22
• Dear Djamillah, for a better understanding see Landau&Lifshitz quantum mechanics section 38, here's the link (on archive). – Phonon Jan 5 '15 at 13:41

$E_n^1\quad=\quad\langle\psi_n^0|H^{'}|\psi_n^0\rangle$, 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.

What you do in perturbation theory is you assume the correct eigenvalue equation for a (hitherto) unknown correct wavefuction $|\psi_n\rangle$ and its associated eigenvalue $E_n$: $$H|\psi_n\rangle = E_n |\psi_n\rangle,$$ where $H$ is the full Hamiltionian.

What you have is a main contribution $H_0$ (e.g. the Coulomb potential) and a perturbation $H'$ which is small compared to $H_0$ and that will therefore just slightly change the main solution.

You then assume that you can expand the correct solution for the energy and the wavefunction as a perturbative series, i.e. in terms that are smaller and smaller: $$|\psi_n\rangle = |\psi_n\rangle^0 + |\psi_n\rangle^1 + |\psi_n\rangle^2 ...$$ $$E_n = E_n^0 + E_n^1 + E_n^2 ...$$ where the exponents signify the order of the term. Higher order terms are smaller, and therefore only needed in you want higher precision.

Now, the full TISE becomes: $$(H_0 + H')\,(|\psi_n\rangle^0 + |\psi_n\rangle^1 + |\psi_n\rangle^2 ... ) = (E_n^0 + E_n^1 + E_n^2 + ...)\,(|\psi_n\rangle^0 + |\psi_n\rangle^1 + |\psi_n\rangle^2+...)$$

Now you take the $0^{th}$ order equation -- where each term is of $O^{th}$ order:

$$H_0 |\psi_n\rangle^0\rangle = E_n^0 |\psi_n\rangle^0,$$ which is just the unperturbed equation, and therefore the starting point for any perturbative calculation.

Now look at the $1^{st}$ order equation -- remember that two $1^{st}$ order terms multiplied together give you a $2^{nd}$ order terms, whereas you only want to keep the $1^{st}$ order ones. $H'$ is first order:

$$H_0 |\psi_n\rangle^1 + H'|\psi_n\rangle^0 = E_n^0|\psi_n\rangle^1 + E_n^1 |\psi_n\rangle^0.$$

What you are after is $E_n^1$, i.e. the $1^{st}$ order contribution to the energy.

Multiply by $^0\langle \psi_n|$ from the left:

$$^0\langle \psi_n|H_0 |\psi_n\rangle^1 + ^0\langle \psi_n|H'|\psi_n\rangle^0 = ^0\langle \psi_n|E_n^0|\psi_n\rangle^1 + ^0\langle \psi_n|E_n^1 |\psi_n\rangle^0,$$

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

$$\implies (E_n^0 - E_n^0) \,^0\langle \psi_n|\psi_n\rangle^1 + ^0\langle \psi_n|H'|\psi_n\rangle^0 = E_n^1 \,^0\langle \psi_n|\psi_n\rangle^0.$$

Assuming normalised states, $^0\langle \psi_n|\psi_n\rangle^0 = 1$, so: $$E_n^1 = ^0\langle \psi_n|H'|\psi_n\rangle^0$$.

As other have noted, there's a mistake in you formula.

The same procedure applies for higher order corrections.