Time-Independent Perturbation Theory
A casual intro to time-independent perturbation theory — what 'perturbed' even means, why the real world is messy, and how approximation theory saves us.
Time-independent perturbation theory. Here we go.
“Perturbation” — what a word, right? Sounds way fancier than what it actually is, which is basically: shake the system a little. So we’ll be talking about “perturbed” systems vs “unperturbed” systems, and that’s the whole vibe.
Now — why time-independent first? Couple of reasons. One, we kinda have to crawl before we walk; the time-dependent case comes later. And two — get this — apparently most physical setups in nature are actually time-independent. So Chapter 6 is no joke. They literally say it’s the most important chapter in the book.
Let’s see why.
Remember the time-independent Schrödinger equation? One of the very first things we ever solved with it — like, way back in Chapter 2 — was the infinite square well.
Eigenfunctions and eigenvalues, served on a silver platter. No drama. No tears. Everything came out clean.
That right there — the no-drama, clean situation — is exactly what we mean by an unperturbed system. (This will land harder once we see the perturbed version. Just trust me for a sec.)
To keep our notation straight going forward, let’s stick a little superscript 0 on everything to mark “unperturbed”:
Got it?
OK so. The world isn’t actually that nice. The world is messy. That’s just how it is… sigh.
Like — instead of a nice clean square well, picture an infinite square well that’s juuust ever so slightly tilted. A little crooked. Skewed.
And honestly? That kind of slightly-messed-up well is way more common in nature than the perfect one. Welcome to reality.
THIS is why Chapter 6 matters. A system that’s just a tiny bit shaken, a tiny bit skewed — that’s a perturbed system.
So our ultimate goal: find the eigenfunction $\psi_n$ and eigenvalue $E_n$ in that slightly-messed-up system. That’s it. That’s perturbation theory.
Honestly though — can we ever find an exactly, perfectly precise state function and energy for the messy system? Like spot-on?
Of course not. Nope nope nope nope.
So if we can’t get it exactly, what is this whole “perturbation theory” thing actually doing for us?
It’s an approximation theory. It hands us an approximate formula. That’s the deal.
And by the way — Chapters 6, 7, 8, 9? All approximation theory, lol. From here on out, brace yourself: we approximate. Magnificently.
Alright. Let’s get into it.
(In we gooooooo!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!)
Look at the situation. Time to make some guesses. Time to predict.
“The perturbed system probably isn’t that far off from the unperturbed one… right?”
Yeah, let’s go with that. Let’s say the Hamiltonians of the two systems differ only by a tiny bit. And same goes for the eigenfunctions and eigenvalues — only slightly different.
※Assumption time※
$$\text{Perturbed Hamiltonian } H \\ = \text{unperturbed Hamiltonian } H^{0} \text{ plus only } \lambda H', \quad (\lambda \ll 1)$$Same vibe for the eigenfunctions and eigenvalues:
※Assumption time※
$$\psi_{n} = \psi_{n}^{0} + \lambda\psi_{n}^{1} + \lambda^{2}\psi_{n}^{2} + \cdots \\ E_{n} = E_{n}^{0} + \lambda E_{n}^{1} + \lambda^{2}E_{n}^{2} + \cdots$$Here, the red term is the 1st-order perturbation.
The blue term is the 2nd-order perturbation.
cf.) Heads up: those superscript numbers up there are NOT exponents. Not squared. Not cubed. Just labels. Just labels.
The exponents on $\lambda$ — those are real exponents. So as the order goes up, the term gets smaller and smaller.
And from here on, we’ll call those terms (starting from 1st order) the correction terms. Because that’s what they do — they correct.
Now we plug all this into the time-independent Schrödinger equation.
$$H\psi_{n} = E\psi_{n}\\ \left(H^{0} + \lambda H'\right)\left(\psi_{n}^{0} + \lambda\psi_{n}^{1} + \lambda^{2}\psi_{n}^{2} + \cdots\right) = \left(E_{n}^{0} + \lambda E_{n}^{1} + \lambda^{2}E_{n}^{2} + \cdots\right)\left(\psi_{n}^{0} + \lambda\psi_{n}^{1} + \lambda^{2}\psi_{n}^{2} + \cdots\right)$$We’re going to handle this like a coefficient-matching problem.
Rip open all the parentheses on both sides, and group everything by powers of $\lambda$. Logic: “coefficients of $\lambda$ at the same order have to match!” Group, group, group:
$$H^{0}\psi_{n}^{0} = E_{n}^{0}\psi_{n}^{0} \quad \cdots\cdots\cdots \quad \text{0th-order}$$$$H^{0}\psi_{n}^{1} + H'\psi_{n}^{0} = E_{n}^{0}\psi_{n}^{1} + E_{n}^{1}\psi_{n}^{0} \quad \cdots\cdots \quad \text{1st-order}$$$$H^{0}\psi_{n}^{2} + H'\psi_{n}^{1} = E_{n}^{0}\psi_{n}^{2} + E_{n}^{1}\psi_{n}^{1} + E_{n}^{2}\psi_{n}^{0} \quad \cdots\cdots\cdots \quad \text{2nd-order}$$3rd, 4th, 5th order… I mean, you could keep writing these foreverrrr. But word is, even just going up to 2nd order gets you something close enough to the truth. Good enough.
Alright. So finding the nth-order wave function and nth-order energy in general would be massive overkill. Our actual goal is to find the wave functions and energy correction terms up to 2nd order — exactly these guys:
$$\psi_{n}^{1}, \quad \psi_{n}^{2}, \quad E_{n}^{1}, \quad E_{n}^{2}$$Annnnnd we’ll cut one more from the list, lol. This one:
$$\psi_{n}^{2}$$is apparently a nightmare to compute by hand. So we’ll skip it.
Which leaves us with:
$$\psi_{n}^{1}, \quad E_{n}^{1}, \quad E_{n}^{2}$$— deriving the general formulas for these three is the goal of this post.
Let’s gogogogogogogogogogo.
OK first up: the 1st-order energy correction.
$$\text{1st-order:} \\ H^{0}\psi_{n}^{1} + H'\psi_{n}^{0} = E_{n}^{0}\psi_{n}^{1} + E_{n}^{1}\psi_{n}^{0}$$Trick: take the inner product of both sides with $\langle\psi_{n}^{0}|$.
$$\langle\psi_{n}^{0}|H^{0}\psi_{n}^{1}\rangle + \langle\psi_{n}^{0}|H'\psi_{n}^{0}\rangle = E_{n}^{0}\langle\psi_{n}^{0}|\psi_{n}^{1}\rangle + E_{n}^{1}\langle\psi_{n}^{0}|\psi_{n}^{0}\rangle \\ H^{0} \text{ is Hermitian — for sure.} \\ \text{And the blue piece is } 1 \text{ (it's normalized).} \\ \langle[H^{0}\psi_{n}^{0}]|\psi_{n}^{1}\rangle + \langle\psi_{n}^{0}|H'\psi_{n}^{0}\rangle = E_{n}^{0}\langle\psi_{n}^{0}|\psi_{n}^{1}\rangle + E_{n}^{1} \\ \text{And in } H^{0}\psi_{n}^{0}, \text{ that wave function is an eigenfunction.} \\ \text{So } H^{0}\psi_{n}^{0} = E_{n}^{0}\psi_{n}^{0}. \\ E_{n}^{0}\langle\psi_{n}^{0}|\psi_{n}^{1}\rangle + \langle\psi_{n}^{0}|H'\psi_{n}^{0}\rangle = E_{n}^{0}\langle\psi_{n}^{0}|\psi_{n}^{1}\rangle + E_{n}^{1} \\ \text{The red ones cancel.} \\ \therefore E_{n}^{1} = \langle\psi_{n}^{0}|H'\psi_{n}^{0}\rangle$$Boom. One out of three goals — done.
$E_{n}^{1}$ — got it.
OK now for $\psi_{n}^{1}$. How do we get this guy?
Pull the 1st-order equation back down:
$$\text{1st-order:} \\ H^{0}\psi_{n}^{1} + H'\psi_{n}^{0} = E_{n}^{0}\psi_{n}^{1} + E_{n}^{1}\psi_{n}^{0}$$Let’s mess around with it a bit:
$$H^{0}\psi_{n}^{1} + H'\psi_{n}^{0} = E_{n}^{0}\psi_{n}^{1} + E_{n}^{1}\psi_{n}^{0}\\ H^{0}\psi_{n}^{1} - E_{n}^{0}\psi_{n}^{1} = E_{n}^{1}\psi_{n}^{0} - H'\psi_{n}^{0}\\ \left(H^{0} - E_{n}^{0}\right)\psi_{n}^{1} = -\left(H' - E_{n}^{1}\right)\psi_{n}^{0}$$Now — we trust that the unperturbed wave functions form a complete basis, right? Whatever wave function you hand me, I can write it as a linear combo of
$$\psi_{1}^{0}, \quad \psi_{2}^{0}, \quad \psi_{3}^{0}, \quad \psi_{4}^{0}, \quad \ldots\psi_{n}^{0}$$these. As a linear sum. Why? Completeness, baby.
So that means $\psi_{n}^{1}$ — yep, that one — can also be written as a linear sum of those same guys.
Following so far?
$$\psi_{n}^{1} = \sum_{m\neq n} C_{m}^{(1)}\psi_{m}^{0}$$The (1) up there in $C_{m}^{(1)}$ means “1st-order correction.” (A (2) version will show up later — no panic.)
And the $m \neq n$? It’s because $\psi_{n}^{1}$ should have the original $\psi_{n}^{0}$ component subtracted out. Look back at the equation up there — we already have a $\psi_{n}^{0}$ floating around, so we drop that one from the expansion.
(Honestly, you could leave the $m=n$ term in. It would just mean $C_n = 0$ anyway, lol. Same difference.)
Alright. New goal: find all those $C_m^{(1)}$ constants. And it’s actually not that bad.
We take this re-expressed $\psi_{n}^{1}$ and plug it into our equation:
$$\left(H^{0} - E_{n}^{0}\right)\psi_{n}^{1} = -\left(H' - E_{n}^{1}\right)\psi_{n}^{0}$$In we go. Substitute:
$$\left(H^{0} - E_{n}^{0}\right)\sum_{m\neq n} C_{m}^{(1)}\psi_{m}^{0} = -\left(H' - E_{n}^{1}\right)\psi_{n}^{0}\\ \sum_{m\neq n}\left(H^{0} - E_{n}^{0}\right)C_{m}^{(1)}\psi_{m}^{0} = -\left(H' - E_{n}^{1}\right)\psi_{n}^{0}\\ \text{Rewrite the red part as an eigenvalue:}\\ \sum_{m\neq n}\left(E_{m}^{0} - E_{n}^{0}\right)C_{m}^{(1)}\psi_{m}^{0} = -\left(H' - E_{n}^{1}\right)\psi_{n}^{0}\\ \text{Now the trick: take the inner product of both sides with } \langle\psi_{\ell}^{0}|.\\ \left\langle\psi_{\ell}^{0}\,\bigg|\,\sum_{m\neq n}\left(E_{m}^{0} - E_{n}^{0}\right)C_{m}^{(1)}\psi_{m}^{0}\right\rangle = \left\langle\psi_{\ell}^{0}\,\bigg|\,\left(-\left(H' - E_{n}^{1}\right)\psi_{n}^{0}\right)\right\rangle$$
OK let’s stop calling it “ell” — just rename it back to $m$. Same thing.
$$\therefore C_{m}^{(1)} = \frac{\langle\psi_{m}^{0}|H'\psi_{n}^{0}\rangle}{\left(E_{n}^{0} - E_{m}^{0}\right)}$$So our 1st-order wave function correction is:
$$\psi_{n}^{1} = \sum_{m\neq n} C_{m}^{(1)}\psi_{m}^{0} = \sum_{m\neq n}\frac{\langle\psi_{m}^{0}|H'\psi_{n}^{0}\rangle}{\left(E_{n}^{0} - E_{m}^{0}\right)}\psi_{m}^{0}$$Yoooooo we got it. Both the 1st-order wave function correction AND the 1st-order energy correction. Done.
Aaaaand one more thing. That little $m \neq n$ in the sigma? That automatically saves us from dividing by zero in the denominator. Which is also basically saying: we’ve been quietly assuming we’re in the non-degenerate case this whole time. (The degenerate case needs its own treatment — we’ll get there.)
OK now, on to 2nd order. Let’s just dive in.
The 2nd-order equation:
$$\text{2nd-order:}\\ H^{0}\psi_{n}^{2} + H'\psi_{n}^{1} = E_{n}^{0}\psi_{n}^{2} + E_{n}^{1}\psi_{n}^{1} + E_{n}^{2}\psi_{n}^{0}$$Same trick — take the inner product with $\langle\psi_{n}^{0}|$ on both sides:
$$\langle\psi_{n}^{0}|H^{0}\psi_{n}^{2}\rangle + \langle\psi_{n}^{0}|H'\psi_{n}^{1}\rangle = E_{n}^{0}\langle\psi_{n}^{0}|\psi_{n}^{2}\rangle + E_{n}^{1}\langle\psi_{n}^{0}|\psi_{n}^{1}\rangle + E_{n}^{2}\langle\psi_{n}^{0}|\psi_{n}^{0}\rangle\\ \text{Blue is } 1.\\ H^{0} \text{ is Hermitian.}\\ \langle H^{0}\psi_{n}^{0}|\psi_{n}^{2}\rangle + \langle\psi_{n}^{0}|H'\psi_{n}^{1}\rangle = E_{n}^{0}\langle\psi_{n}^{0}|\psi_{n}^{2}\rangle + E_{n}^{1}\langle\psi_{n}^{0}|\psi_{n}^{1}\rangle + E_{n}^{2}\\ \text{Rewrite red as eigenvalue:}\\ \langle E_{n}^{0}\psi_{n}^{0}|\psi_{n}^{2}\rangle + \langle\psi_{n}^{0}|H'\psi_{n}^{1}\rangle = E_{n}^{0}\langle\psi_{n}^{0}|\psi_{n}^{2}\rangle + E_{n}^{1}\langle\psi_{n}^{0}|\psi_{n}^{1}\rangle + E_{n}^{2}\\ \text{Green ones cancel.}\\ \text{Pink? Recall } \psi_{n}^{1} = \sum_{m\neq n} C_{m}^{(1)}\psi_{m}^{0} \text{ — meaning } \psi_{n}^{0} \text{ is orthogonal to every piece of } \psi_{n}^{1}. \text{ Pink is zero.}\\ \therefore E_{n}^{2} = \langle\psi_{n}^{0}|H'\psi_{n}^{1}\rangle$$But we don’t stop here — we want this in terms of unperturbed quantities. So plug in our 1st-order wave function correction:
$$\psi_{n}^{1} = \sum_{m\neq n}\frac{\langle\psi_{m}^{0}|H'\psi_{n}^{0}\rangle}{\left(E_{n}^{0} - E_{m}^{0}\right)}\psi_{m}^{0}$$To make this easier to follow, I’ll just shuffle the order — constants in the back, wave function up front:
$$\psi_{n}^{1} = \sum_{m\neq n}\frac{\psi_{m}^{0}\langle\psi_{m}^{0}|H'\psi_{n}^{0}\rangle}{\left(E_{n}^{0} - E_{m}^{0}\right)}$$Then plug it in:
$$E_{n}^{2} = \langle\psi_{n}^{0}|H'\psi_{n}^{1}\rangle = \left\langle\psi_{n}^{0}\,\bigg|\,H'\left(\sum_{m\neq n}\frac{\psi_{m}^{0}\langle\psi_{m}^{0}|H'\psi_{n}^{0}\rangle}{\left(E_{n}^{0} - E_{m}^{0}\right)}\right)\right\rangle$$I personally got stuck going from here to the next step. So I’m gonna write out every single piece. Painstakingly. T_T T_T T_T T_T — hopefully this helps:
$$E_{n}^{2} = \langle\psi_{n}^{0}|H'\psi_{n}^{1}\rangle = \left\langle\psi_{n}^{0}\,\bigg|\,H'\left(\sum_{m\neq n}\frac{\psi_{m}^{0}\langle\psi_{m}^{0}|H'\psi_{n}^{0}\rangle}{\left(E_{n}^{0} - E_{m}^{0}\right)}\right)\right\rangle\\ = \left\langle\psi_{n}^{0}\,\bigg|\,H'\left(\frac{\sum_{m\neq n}\psi_{m}^{0}\langle\psi_{m}^{0}|H'\psi_{n}^{0}\rangle}{\sum_{m\neq n}\left(E_{n}^{0} - E_{m}^{0}\right)}\right)\right\rangle\\ = \frac{\left\langle\psi_{n}^{0}\,\bigg|\,H'\sum_{m\neq n}\psi_{m}^{0}\langle\psi_{m}^{0}|H'\psi_{n}^{0}\rangle\right\rangle}{\sum_{m\neq n}\left(E_{n}^{0} - E_{m}^{0}\right)}\\ = \frac{\left\langle\psi_{n}^{0}\,\bigg|\,H'\left[\psi_{1}^{0}\langle\psi_{1}^{0}|H'\psi_{n}^{0}\rangle + \psi_{2}^{0}\langle\psi_{2}^{0}|H'\psi_{n}^{0}\rangle + \cdots + \psi_{k}^{0}\langle\psi_{k}^{0}|H'\psi_{n}^{0}\rangle + \cdots\right]\right\rangle}{\sum_{m\neq n}\left(E_{n}^{0} - E_{m}^{0}\right)}\\ = \frac{\langle\psi_{n}^{0}|H'\psi_{1}^{0}\langle\psi_{1}^{0}|H'\psi_{n}^{0}\rangle\rangle + \langle\psi_{n}^{0}|H'\psi_{2}^{0}\langle\psi_{2}^{0}|H'\psi_{n}^{0}\rangle\rangle + \cdots + \langle\psi_{n}^{0}|H'\psi_{k}^{0}\langle\psi_{k}^{0}|H'\psi_{n}^{0}\rangle\rangle + \cdots}{\sum_{m\neq n}\left(E_{n}^{0} - E_{m}^{0}\right)}\\ = \frac{\langle\psi_{n}^{0}|H'\psi_{1}^{0}\rangle\langle\psi_{1}^{0}|H'\psi_{n}^{0}\rangle + \langle\psi_{n}^{0}|H'\psi_{2}^{0}\rangle\langle\psi_{2}^{0}|H'\psi_{n}^{0}\rangle + \cdots + \langle\psi_{n}^{0}|H'\psi_{k}^{0}\rangle\langle\psi_{k}^{0}|H'\psi_{n}^{0}\rangle + \cdots}{\sum_{m\neq n}\left(E_{n}^{0} - E_{m}^{0}\right)}\\ = \frac{\sum_{m\neq n}\langle\psi_{n}^{0}|H'\psi_{m}^{0}\rangle\langle\psi_{m}^{0}|H'\psi_{n}^{0}\rangle}{\sum_{m\neq n}\left(E_{n}^{0} - E_{m}^{0}\right)} = \frac{\sum_{m\neq n}\langle\psi_{m}^{0}|H'\psi_{n}^{0}\rangle^{\dagger}\langle\psi_{m}^{0}|H'\psi_{n}^{0}\rangle}{\sum_{m\neq n}\left(E_{n}^{0} - E_{m}^{0}\right)}\\ = \frac{\sum_{m\neq n}\left|\langle\psi_{m}^{0}|H'\psi_{n}^{0}\rangle\right|^{2}}{\sum_{m\neq n}\left(E_{n}^{0} - E_{m}^{0}\right)}$$$$\therefore E_{n}^{2} = \sum_{m\neq n}\frac{\left|\langle\psi_{m}^{0}|H'\psi_{n}^{0}\rangle\right|^{2}}{\left(E_{n}^{0} - E_{m}^{0}\right)}$$Quick admin — let’s all agree on a piece of shorthand right here:
(Ugh, I should’ve made this agreement BEFORE writing out all those equations above. T_T T_T T_T T_T)
$$\langle\psi_{i}^{0}|H'|\psi_{j}^{0}\rangle \equiv H'_{ij}$$What’s this even mean? You’ll see — once you start grinding through actual problems, it’ll click. Usually you’re handed $H'$, you do the integral, and you stuff the result into the sigma above. Out pops your 2nd-order energy correction.
Wait wait wait!!! What about the degenerate case?!
Yep, like I said — everything we just did was secretly assuming non-degeneracy. So now we need to redo this for when degeneracy shows up.
Let’s just follow the book here.
A degenerate situation = different states that share the same energy. (For now we’ll keep it simple — two-fold degeneracy.)
In equation form: unperturbed state $a$ and unperturbed state $b$ have equal energies.
$$\begin{cases} H^{0}\psi_{a}^{0} = E_{0}\psi_{a}^{0} \\ H^{0}\psi_{b}^{0} = E_{0}\psi_{b}^{0} \end{cases}\\ \text{Also, } a \text{ and } b \text{ are orthogonal.} \quad \langle\psi_{a}^{0}|\psi_{b}^{0}\rangle = 0$$Look at this for a sec — any linear combination of $a$ and $b$ is also an eigenstate of $H^{0}$ with that same energy:
$$H^{0}\left(\alpha\psi_{a}^{0} + \beta\psi_{b}^{0}\right) = E_{0}\left(\alpha\psi_{a}^{0} + \beta\psi_{b}^{0}\right)$$So our unperturbed degenerate state, call it $\psi^{0}$, is actually:
$$\psi^{0} = \alpha\psi_{a}^{0} + \beta\psi_{b}^{0}$$That’s how we’ll think of it.
Now turn on the perturbation:
$$H = H^{0} + \lambda H'$$And the wave function and energy each shift slightly:
$$\psi = \psi^{0} + \lambda\psi^{1} + \lambda^{2}\psi^{2} + \lambda^{3}\psi^{3} + \cdots\\ E = E^{0} + \lambda E^{1} + \lambda^{2}E^{2} + \lambda^{3}E^{3} + \cdots$$Then plug into the Schrödinger equation:
$$H\psi = E\psi\\ \left(H^{0} + \lambda H'\right)\left(\psi^{0} + \lambda\psi^{1} + \lambda^{2}\psi^{2} + \cdots\right) = \left(E^{0} + \lambda E^{1} + \lambda^{2}E^{2} + \cdots\right)\left(\psi^{0} + \lambda\psi^{1} + \lambda^{2}\psi^{2} + \cdots\right)$$Same coefficient-matching game as before. The 1st-order chunk:
$$H^{0}\psi^{1} + H'\psi^{0} = E^{0}\psi^{1} + E^{1}\psi^{0}\\ \text{Now take the inner product with } \langle\psi_{a}^{0}| \text{ on both sides — and separately with } \langle\psi_{b}^{0}|.\\ \text{Why? Because } \psi^{0} = \alpha\psi_{a}^{0} + \beta\psi_{b}^{0}, \text{ remember!}$$Inner-producting with $\langle\psi_{a}^{0}|$:
$$H^{0}\psi^{1} + H'\psi^{0} = E^{0}\psi^{1} + E^{1}\psi^{0}\\ \langle\psi_{a}^{0}|H^{0}\psi^{1}\rangle + \langle\psi_{a}^{0}|H'\psi^{0}\rangle = \langle\psi_{a}^{0}|E^{0}\psi^{1}\rangle + \langle\psi_{a}^{0}|E^{1}\psi^{0}\rangle\\ \text{Sub in } \psi^{0} = \alpha\psi_{a}^{0} + \beta\psi_{b}^{0}:\\ \langle\psi_{a}^{0}|H^{0}\psi^{1}\rangle + \langle\psi_{a}^{0}|H'\left(\alpha\psi_{a}^{0} + \beta\psi_{b}^{0}\right)\rangle = \langle\psi_{a}^{0}|E^{0}\psi^{1}\rangle + \langle\psi_{a}^{0}|E^{1}\left(\alpha\psi_{a}^{0} + \beta\psi_{b}^{0}\right)\rangle\\ H^{0} \text{ Hermitian, so:}\\ \langle H^{0}\psi_{a}^{0}|\psi^{1}\rangle + \alpha\langle\psi_{a}^{0}|H'\psi_{a}^{0}\rangle + \beta\langle\psi_{a}^{0}|H'\psi_{b}^{0}\rangle = E^{0}\langle\psi_{a}^{0}|\psi^{1}\rangle + \alpha E^{1}\langle\psi_{a}^{0}|\psi_{a}^{0}\rangle + \beta E^{1}\langle\psi_{a}^{0}|\psi_{b}^{0}\rangle\\ \text{Red is } 1, \text{ blue is orthogonal (=0), green ones cancel!}$$What’s left:
$$\alpha\langle\psi_{a}^{0}|H'\psi_{a}^{0}\rangle + \beta\langle\psi_{a}^{0}|H'\psi_{b}^{0}\rangle = \alpha E^{1}$$By the exact same logic, inner-producting with $\langle\psi_{b}^{0}|$ gives:
$$\alpha\langle\psi_{b}^{0}|H'\psi_{a}^{0}\rangle + \beta\langle\psi_{b}^{0}|H'\psi_{b}^{0}\rangle = \beta E^{1}$$Stack ’em side by side:
$$\alpha\langle\psi_{a}^{0}|H'\psi_{a}^{0}\rangle + \beta\langle\psi_{a}^{0}|H'\psi_{b}^{0}\rangle = \alpha E^{1}\\ \alpha\langle\psi_{b}^{0}|H'\psi_{a}^{0}\rangle + \beta\langle\psi_{b}^{0}|H'\psi_{b}^{0}\rangle = \beta E^{1}\\ \text{Remember our shorthand from earlier? Time to use it:}\\ \alpha H'_{aa} + \beta H'_{ab} = \alpha E^{1}\\ \alpha H'_{ba} + \beta H'_{bb} = \beta E^{1}\\ \text{Written like that, this is begging to be a matrix equation:}\\ \begin{pmatrix} H'_{aa} & H'_{ab} \\ H'_{ba} & H'_{bb} \end{pmatrix}\begin{pmatrix} \alpha \\ \beta \end{pmatrix} = E^{1}\begin{pmatrix} \alpha \\ \beta \end{pmatrix}\\ \text{Heh — so } E^{1} \text{ is just the eigenvalue of that 2x2 matrix on the left.}\\ \text{Solve the eigenvalue equation:}\\ \begin{vmatrix} H'_{aa} - E^{1} & H'_{ab} \\ H'_{ba} & H'_{bb} - E^{1} \end{vmatrix} = 0$$$$\left(H'_{aa} - E^{1}\right)\left(H'_{bb} - E^{1}\right) - H'_{ab}H'_{ba} = 0\\ \left(E^{1}\right)^{2} - \left(H'_{aa} + H'_{bb}\right)E^{1} + \left(H'_{aa}H'_{bb} - H'_{ab}H'_{ba}\right) = 0\\ \text{Quadratic formula:}\\ E^{1} = \frac{\left(H'_{aa} + H'_{bb}\right) \pm \sqrt{\left(H'_{aa} + H'_{bb}\right)^{2} - 4\left(H'_{aa}H'_{bb} - H'_{ab}H'_{ba}\right)}}{2}\\ = \frac{1}{2}\left\{\left(H'_{aa} + H'_{bb}\right) \pm \sqrt{\left(H'_{aa} + H'_{bb}\right)^{2} - 4\left(H'_{aa}H'_{bb} - H'_{ab}H'_{ba}\right)}\right\}\\ = \frac{1}{2}\left\{\left(H'_{aa} + H'_{bb}\right) \pm \sqrt{\left(H'_{aa} - H'_{bb}\right)^{2} + 4H'_{ab}H'_{ba}}\right\}\\ = \frac{1}{2}\left\{\left(H'_{aa} + H'_{bb}\right) \pm \sqrt{\left(H'_{aa} - H'_{bb}\right)^{2} + 4H'_{ab}H'_{ab}^{\dagger}}\right\}\\ = \frac{1}{2}\left\{\left(H'_{aa} + H'_{bb}\right) \pm \sqrt{\left(H'_{aa} - H'_{bb}\right)^{2} + 4\left|H'_{ab}\right|^{2}}\right\}$$Aha.
$$E^{1} \text{ pops out as TWO values:}\\ E^{1}_{+}, \quad E^{1}_{-}\\ \text{Going from unperturbed to perturbed, the doubly-degenerate level}\\ \text{gets its 1st-order correction split into } E^{1}_{+} \text{ and } E^{1}_{-}.\\ \text{To us, it looks like the level cracks open into two!}$$
Graph and all — that’s literally how the degeneracy breaks apart.
OK, last little bit. Hey — what if the degeneracy isn’t 2 but more??
For two-fold degeneracy we did:
$$\begin{cases} H^{0}\psi_{a}^{0} = E_{0}\psi_{a}^{0} \\ H^{0}\psi_{b}^{0} = E_{0}\psi_{b}^{0} \end{cases} \to \begin{pmatrix} H'_{aa} & H'_{ab} \\ H'_{ba} & H'_{bb} \end{pmatrix}\begin{pmatrix} \alpha \\ \beta \end{pmatrix} = E^{1}\begin{pmatrix} \alpha \\ \beta \end{pmatrix} \to \begin{vmatrix} H'_{aa} - E^{1} & H'_{ab} \\ H'_{ba} & H'_{bb} - E^{1} \end{vmatrix} = 0 \quad \text{right?}$$So for higher-order degeneracy, you just blow it up to a bigger matrix:
$$\left\{H' - E^{1}\right\}\begin{pmatrix} \alpha \\ \beta \\ \gamma \\ \cdot \\ \cdot \\ \cdot \end{pmatrix} = 0$$Solve the bigger eigenvalue equation. That’s it.
We’ll do worked examples of higher-order degeneracy later. Don’t be too thrown^^
Ah, this got long^^ Theory’s done now^^ son of a…
Originally written in Korean on my Naver blog (2015-12). Translated to English for gdpark.blog.