1
$\begingroup$

The second-order correction to the energy of an eigenstate due to a perturbation $H'$ is given by:

$$E_n^2=\sum_{m\not=n}^{} \frac{\left \lvert \left \langle \psi_m^0|H'|\psi_n^0 \right \rangle \right \rvert^2}{E_n^0-E_m^0}$$

If $E_n^0 \approx E_m^0$, the expression above blows up. In some undergraduate quantum mechanics textbooks like Griffiths' or Binney's, only the case where $E_n^0=E_m^0$ is analyzed using degenerate perturbation theory. What does one do in the case where $E_n^0 \approx E_m^0$ and not $E_n^0=E_m^0$? I am having trouble finding resources that explain how to systematically approach the case where $E_n^0 \approx E_m^0$. If you could please provide an explanation below or guide me toward the right resources, I would be grateful.

$\endgroup$
3
  • 2
    $\begingroup$ Simply follow similar steps in the degenerate case, but with the gap? $\endgroup$ Apr 19 at 17:08
  • 1
    $\begingroup$ Check out a toy case, e.g. H=diag ( a+ε, a, b) and a suitable simple off-diagonal H'... $\endgroup$ Apr 19 at 17:54
  • $\begingroup$ You treat the difference in energies as a perturbation. This is actually the (implicit) context of Griffiths Problem 7.12 (in the 3rd edition), and I'll think that you'll find that many quantum texts do treat this situation explicitly. $\endgroup$
    – march
    Apr 19 at 18:21

1 Answer 1

2
$\begingroup$

Here (see the file) is the discussion of the case you are interested in. It is referred as "nearly degenerate perturbation theory", and it is reduced to a usual degenerate perturbation theory by a simple rearrangement of terms in Hamiltonian. The resulting secular equation differs from the degenerate perturbation theory by a simple modification: \begin{equation} \begin{pmatrix} E_m^{(0)} + V_{mm} - \epsilon_\alpha & V_{mn}\\ V_{nm} & E_n^{(0)} + V_{nn} - \epsilon_\alpha \end{pmatrix} \begin{pmatrix} c_m^\alpha\\ c_n^\alpha \end{pmatrix} = 0. \end{equation} where $c^{\alpha}_k$ are the coefficients of the zero-order eigenstates $|\alpha\rangle = c^\alpha_m|\psi^{(0)}_m\rangle + c^\alpha_n|\psi_n^{(0)}\rangle$ with the first-order corrected energies $\epsilon_\alpha$.

The rearrangement is performed as follows. Consider a Hamiltonian $H = H_0 + V$, where \begin{equation} H_0 = \sum_i E_i^{(0)}|\psi_i^{(0)}\rangle\langle \psi_i^{(0)}|, \end{equation} \begin{equation} V = \sum_{ij} V_{ij}|\psi_i^{(0)}\rangle\langle \psi_j^{(0)}|, \end{equation} where $E_n^{(0)} \approx E_m^{(0)}$ for some $m$,$n$. Now, let me write \begin{equation} H = H_0' + V', \end{equation} where \begin{equation} H_0' = H_0 - \frac{E_m^{(0)} - E_n^{(0)}}{2}(|\psi_m^{(0)}\rangle\langle \psi_m^{(0)}| - |\psi_n^{(0)}\rangle\langle \psi_n^{(0)}|), \end{equation} \begin{equation} V' = V + \frac{E_m^{(0)} - E_n^{(0)}}{2}(|\psi_m^{(0)}\rangle\langle \psi_m^{(0)}| - |\psi_n^{(0)}\rangle\langle \psi_n^{(0)}|). \end{equation} The Hamiltonian $H'$ has two strictly degenerate eigenstates $|\psi_m^{(0)}\rangle$ and $|\psi_n^{(0)}\rangle$ with the energies $(E_m^{(0)} + E_n^{(0)})/2$, which allows to use usual degenerate perturbation theory by $V'$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.