# Perturbation Theory - Exact Solutions and Good Quantum States

I'm having a problem with the following question:

Problem: Consider the unperturbed, degenerate Hamiltonian $$H_0=\bigg(\begin{matrix} E &0\\ 0& E\end{matrix}\bigg)$$. Add the perturbation $$H_1=\bigg(\begin{matrix} 0 &\delta\\ \delta& 0\end{matrix}\bigg)$$, where $$\delta<, to form the full hamiltonian $$H$$.

a) Find the exact eigenstates and eigenvectors for $$H$$.

b) Use these to perform non-degenerate perturbation theory to first order. Why is this okay?

c) Consider the basis vectors $$v_1=\frac{1}{\sqrt2}\bigg(\begin{matrix}1\\i\end{matrix}\bigg)$$ and $$v_2=\frac{1}{\sqrt2}\bigg(\begin{matrix}1\\-i\end{matrix}\bigg)$$. Verify that these are eigenvectors of $$H_0$$. Use these to perform degenerate perturbation theory with the perturbation $$H_1$$. What are the resulting energy splittings? What are the “good" quantum states for the perturbed Hamiltonian? Compare these to the exact solution from part (a)

What I did/tried:

Part A is easy, just simple linear algebra: $$u_1=\frac{1}{\sqrt{2}}\bigg(\begin{matrix}1\\1\end{matrix}\bigg)$$ and $$u_2=\frac{1}{\sqrt{2}}\bigg(\begin{matrix}1\\-1\end{matrix}\bigg)$$, with $$\lambda_1=E+\delta$$, $$\lambda_2=E-\delta$$.

Part B is fine as well. One gets that the energy shift for the first eigenvector in a) is $$\delta$$, and for the one in b), $$-\delta$$. But I'm unsure if I get the 'why is this okay'. Is it because the eigenvalues in a) were different?

For Part C, I'm able to check that the proposed vectors are indeed eigenvectors of $$H_0$$. But, performing degenerate perturbation theory, I get the following $$W$$ matrix, $$W_{ij}=v_i^TH_1v_j$$, $$\bigg(\begin{matrix} 2\delta i & 0\\ 0&-2\delta i\end{matrix}\bigg)$$, so that $$E_{\pm} = \frac{1}{2}\big[W_{11}+W_{22} \pm \sqrt{(W_{11}-W_{22})^2+4|W_{12}|^2} \big]=\pm \delta i$$, which is complex... I'm also a bit confused whether I get what the good states would be, from $$W$$ it seems that $$v_1, v_2$$ would be these good states? If it weren't for the $$i$$, the shift would be the same as in a.

So basically, I'm uncomfortable with the notion of good states here, the value I got for the shift in C and the question on why it was okay to do the non degenerate perturbation theory in b.

• In your expression for $W_{ij}$, you should have instead $W_{ij} = \langle v_i |H_1| v_j \rangle = (v^T_i)^* \, H_1 \, v_j$. For part $b$, notice that the states $|u_i\rangle$ not only diagonalize $H_0 + H_1$, as you found out, but they also diagonalize $H_1$. Oct 31, 2019 at 23:15
• I do not agree with the proposal of closing this question: it contains a question about a specific physics concept (the concept of good states in the pesent context of perturbation thepry) and show some effort to work through the problem. So, why to close it? Nov 6, 2019 at 7:59

Ques. 1: Why is it okay to apply nondegenerate perturbation theory in b)?

It's okay because $$u_1$$ and $$u_2$$ are "good states" here. This means they diagonalize* $$H_1$$. That is if you express $$H_1$$ in the $$\{u_1,u_2 \}$$ basis, the diagonal elements are zero

$$H_1=\bigg(\begin{matrix} \delta &0\\ 0& -\delta\end{matrix}\bigg)$$

Ques. 2: the value I got for the shift in C?

The correct expression is $$W_{ij}=(v_i^{*})^{T} H_1v_j$$. You forgot to complex conjugate $$v_i$$. When applying the correct expression you will get

$$W=\bigg(\begin{matrix} 0 &i\delta\\ -i\delta& 0\end{matrix}\bigg)$$

and you get $$E_{\pm}=\pm \delta$$ as expected. Here $$\{v_1,v_2 \}$$ are bad states since the diagonal elements $$W_{12}$$ and $$W_{21}$$ are not zero, so you're forced to use nondegenerate perturbation theory.

*Caveat: The more precise definition of good states is this. Suppose you have $$N$$ eigenstates $$\{\psi_1,\psi_2,...,\psi_N \}$$ of $$H_0$$ where $$N$$ could be infinite. Among them, there are $$n$$ degenerate eigenstates $$\{\psi_k,\psi_{k+1},...,\psi_{k+(n-1)} \}$$ with same eigenvalue $$E$$. These degenerate states are said to be good states if they diagonalize your perturbation $$H_1$$ (i.e., $$W_{ij}=0$$ whenever $$i \ne j$$ where $$i,j\in\{k,k+1,...,k+(n-1) \})$$. You don't care if the set of all eigenstates $$\{\psi_1,\psi_2,...,\psi_N \}$$ diagonalize $$H_1$$, you only worry about the particular subset of degenrate eigenstates $$\{\psi_k,\psi_{k+1},...,\psi_{k+(n-1)} \}$$. Good states allow you to directly apply nondegenrate perturbation theory to find energy corrections.

Your example is very special since here $$N=n=2$$, so the set of all eigenstates is exactly the set of degenerate eigenstates of $$H_0$$.

To illustrate, suppose you have 3-level state (i.e., $$N=3$$) with two-fold degeneracy $$n=2$$. So it has three eigenstates$$\{\psi_1,\psi_{2},\psi_{3} \}$$ and suppose $$\psi_{1}$$ and $$\psi_{2}$$ are degenerate, then $$H_0$$ expressed in $$\{\psi_1,\psi_{2},\psi_{3} \}$$ basis will look like this

$$H_0=\Bigg(\begin{matrix} E &0 &0\\ 0& E &0\\ 0& 0 &E_3 \end{matrix}\Bigg)$$

$$\{\psi_1,\psi_{2} \}$$ are good states if they diagonalize $$H_1$$

$$W=\bigg(\begin{matrix} W_{11} &0\\ 0& W_{22}\end{matrix}\bigg)$$

You don't care if the whole set of eigenstates diagonalize $$H_1$$ which could look like this

$$H_1=\Bigg(\begin{matrix} W_{11} &0 &W_{13}\\ 0& W_{22} &W_{23}\\ W_{31}& W_{32} & W_{33} \end{matrix}\Bigg)$$

You only care that $$H_1$$ is diagonal in the degenerate subspace to be able to apply nondegenerate perturbation theory.