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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.

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    $\begingroup$ Simply follow similar steps in the degenerate case, but with the gap? $\endgroup$ Commented Apr 19, 2023 at 17:08
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    $\begingroup$ Check out a toy case, e.g. H=diag ( a+ε, a, b) and a suitable simple off-diagonal H'... $\endgroup$ Commented Apr 19, 2023 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
    Commented Apr 19, 2023 at 18:21

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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 $V'$.

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