# 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. Jan 5, 2015 at 13:22
• Dear Djamillah, for a better understanding see Landau&Lifshitz quantum mechanics section 38, here's the link (on archive). Jan 5, 2015 at 13:41

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.

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