What is the point of degenerate perturbation theory in quantum mechanics?

What is the point of degenerate perturbation theory in quantum mechanics? Let's disregard for a moment the issue of constructing the perturbed wave functions and assume that the 1st order correction to energy is sufficiently accurate for our purposes. Do we really need degenerate perturbation theory?

For example, let's say I have some toy system with 4 possible states, $$|\Psi_i\rangle$$ $$i = \{1, 2, 3, 4\}$$, with energies $$E_i$$. But $$E_2= E_3$$. My question is this : is there any reason why we need to construct "good" wave functions? I understand that for second order we need to make sure we resolve the issue of $$\sum_{m \ne n}\frac{|H'_{mn}|^2}{E_n - E_m}$$ yielding infinities; but let's say we only need the first order corrections. Do we need to diagonalize the degenerate subspace of the perturbation as Griffiths (and my professor) say we do? Will $$E^0_n = \langle\Psi_n|H'|\Psi_n\rangle$$ yield the proper correction?

Secondly, can anyone shed any light on what it means to diagonalize that subspace? Is this a purely computational trick or is there some deeper intuition (either physical, or geometrical, relating to the notion of degenerate eigenspaces) that I can use to understand this better?

• I don't know enough QM to fully answer the question, but I will point out from a purely mathematical perspective that the ability to model the system you describe where $E_2=E_3$ frees us from proving that every state we choose must have a unique energy level before we go any further with the calculations. That opens up an entire body of tools for dividing the state space conveniently, such as making "there exists a X such that..." arguments trivial. Dec 16 '16 at 21:47

You don't need to use a made up toy example. Just take ordinary hydrogen.

If we model hydrogen as a spineless electron in a spherically symmetric Coloumb potential (as one does in quantum mechanics class), then hydrogen has enormous degeneracy. For any $n$, all the states $\ell,m$ with $0\leq\ell \leq n$ and $-\ell \leq m \leq \ell$ have the same energy, that's something like $n^2$ states with the same energy.

So, that's a partial answer--degenerate perturbation theory is important because that's actually what you need to deal with an important real world problem.

Oftentimes, degeneracies exist because there's a symmetry in the free Hamiltonian (this is true of hydrogen, in fact there's even more symmetry in hydrogen than you might think). When you add a perturbation, one of the most important things that tends to happen is that the perturbation breaks the symmetry of the free Hamiltonian.

That's what happens in the fine structure of hydrogen for example.

Note that this isn't about computing better wave functions. The splitting of energy levels can already be seen in the first order correction to the energies. The fine structure breaks the degeneracy between different $\ell$ values for fixed $n$.

Lastly, often you want to continue to work with energy eigenstates even after you've introduced the perturbation. Let's stick with hydrogen. To pick energy eigenstates I specify $n$, but then I can pick any basis I want in the degenerate subspabe and still have energy eigenstates. Once I introduce a perturbation that generically won't be true. Only certain special eigenstates will continue to have definite energy. Other states in that subspace will end up as a superposition over the perturbed, split energy levels, and thus not have definite energy.

Let's say we only need the first order corrections. Do we need to diagonalize the degenerate subspace of the perturbation?

Yes, in order to apply the perturbation theory we need the perturbation matrix to be in a diagonal form, in this way we can properly calculate the energies of the multiplet. This statement does not mean that we must always diagonalize a matrix by hand: many times we can exploit the symmetries of our problem in order to find some "good" eigenstates of $$H_0$$ to which we can apply the ordinary perturbation theory.

Is there some deeper intuition (either physical, or geometrical, relating to the notion of degenerate eigenspaces)?

Yes, one can search for an operator $$A$$ such that $$[A, H_0]=0$$ and $$[A, V]=0$$ ($$V$$ is the perturbation) and find the eigenstates of $$A$$ in the subspace in which $$H_0$$ has a degenerate spectrum. Then one can apply perturbation theory to these states. Here I think that an explanatory example could be useful: consider a particle in a 2-dimensional infinite potential well $$U(x, y)$$ such that

$$U(x, y)=\cases{0 \hspace{0.6cm} x\in [0, a], y\in [0, a] \\ \infty \hspace{0.6 cm} \text{otherwise}}$$

and a perturbation $$\lambda V(x, y)$$ which can be written as $$\lambda V(x, y)=\cases{\lambda \hspace{0.6cm} x\in [0, b], y\in [0, b]\\ 0 \hspace{0.6 cm} \text{otherwise}}$$ where $$b and $$\lambda$$ is a sufficiently small parameter with respect to the energy separation of the initial Hamiltonian $$H_0=\frac{p^2}{2m}+U(x, y)$$ eigenstates. Let $$|n_x n_y\rangle$$ be the eigenstates of $$H_0$$ and let $$\Psi^{(0)}_{n_x n_y}$$ be the projections of these states into the position basis $$\langle x|n_x n_y\rangle$$, we get

$$\Psi^{(0)}_{n_x n_y}=\sqrt{\frac{2}{a}}\sin \left(\frac{n_x \pi x}{a}\right)\sin \left(\frac{n_y \pi y}{a}\right)$$ The first two degenerate states are $$|1 2\rangle$$ and $$|2 1\rangle$$, in fact they share the same energy, namely $$\frac{5}{2}\frac{\pi^2 \hbar^2}{ma^2}$$. Let us call $$P_{xy}$$ the operator that carries out a reflection with respect to the bisector $$x=y$$ inverting the coordinates $$x$$ and $$y$$. We can note that $$|1 2\rangle$$ and $$|2 1\rangle$$ are not eigenstates of $$P_{xy}$$, but linear combinations such as $$\cases{|+\rangle =\frac{1}{\sqrt{2}}(|1 2\rangle +|2 1\rangle ) \\ |-\rangle =\frac{1}{\sqrt{2}}(|1 2\rangle -|2 1\rangle )}$$ produce two eigenstates for $$P_{xy}$$ with eigenvalues $$+1$$ and $$-1$$, hence $$P_{xy}$$ is diagonal in the representation of the $$\{|+\rangle, |-\rangle\}$$ basis. Since $$V(x, y)$$ is invariant under exchange of $$x$$ and $$y$$, we can deduce that $$[P_{xy}, V]=0$$. This last statement implicitly says that also $$V$$ is in diagonal form on the $$\{|+\rangle, |-\rangle\}$$ basis, in fact $$\langle +|[P_{xy}, V]| -\rangle=\langle +|P_{xy}V| -\rangle-\langle +|VP_{xy}| -\rangle=2\langle +| V| -\rangle=0 \Longrightarrow \langle +| V| -\rangle=0$$ Therefore we can express the energies of the two degenerate eigenstates using a first order correction: $$\cases{E_+=\frac{5}{2}\frac{\pi^2 \hbar^2}{ma^2}+\lambda \langle +| V| +\rangle \\ E_-=\frac{5}{2}\frac{\pi^2 \hbar^2}{ma^2}+\lambda \langle -| V| -\rangle }$$

From here it can be clearly seen that the perturbation removes the degeneration of the levels generating a so called "splitting".