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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 $\delta<<E$, 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.

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  • $\begingroup$ 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 $H_1$. $\endgroup$
    – secavara
    Oct 31, 2019 at 23:15
  • $\begingroup$ 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? $\endgroup$ Nov 6, 2019 at 7:59

1 Answer 1

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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.

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