# the degenerate case of the non-Hermitian perturbation theory

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Let us take the free Hamiltonian $H_0 = \frac{\partial^2 }{\partial x^2}$, on $x \in S^1$. Then the eigenfunctions are just free waves, $\psi_n = e^{inx}$. Note that the energy level has two-fold degeneracy since $E_n = E_{-n}$.

Now if we have a non-Hermitian perturbation $H_1$, can anybody teach me how to do the perturbation theory for the spectrum and the wavefunction?

I know when there is no degeneracy I can split $H_1 = A+B$ into the Hermitian piece and anti-Hermitian piece, and then do the usual perturbative expansions for both. But it seems not to work for the degenerate case. asked Apr 20, 2017

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It doesn't matter whether or not the perturbation is Hermitian. One poses the eigenvalue problem in block form with one block corresponding to each unperturbed eigenvalue and obtains a coupled system for the block coefficients. For simplicity assume that in a basis where the unperturbed Hamiltonian is diagonal, the perturbed Hamiltonian is $H=\pmatrix{H_{11} & H_{12} \cr H_{21} & H_{22}}$, where $H_{11}$ is the block corresponding to the degenerate eigenvalue $E_0$ of interest. Then the eigenvalue problem takes the form $$(E-H_{11})\psi_1- H_{12} \psi_2=0,$$    $$-H_{21}\psi_1+(E- H_{22})\psi_2=0.$$  Here the dimension of $\psi_1$ is the algebraic multiplicity of the unperturbed eigenvalue $E_0$. By construction, $E-H_{22}$ is nonsingular for $E$ close to the unperturbed eigenvalue $E_0$. Thus we can formally solve the second equation for $\psi_2$ and insert the result into the first equation. This results in a nonlinear eigenvalue problem $H_{red}(E)\psi_1=0$, which is still exact. First order perturbation theory amounts to linearizing $H_{red}(E)$ around $E=E_0$ and then solving the resulting linear eigenvalue problem.

A detailed exposition of perturbation theory in the presence of degeneracy is given in

Klein, D. J. "Degenerate perturbation theory." The Journal of Chemical Physics 61.3 (1974): 786-798.

answered Apr 20, 2017 by (15,747 points)
edited Apr 25, 2017

But how do you diagonalize order by order? For example let us say $\langle \alpha \vert H_1 \vert \beta \rangle = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$, for $\alpha, \beta \in \mathcal{H}_n$. Then I cannot solve the first order equation $(H_0 - E^{(0)} ) \psi_\alpha ^{(1)} + (H_1 - E^{(1)} ) \psi_\alpha ^{(0)} =0$ since the $H_1$ is not diagonalizable.

@genideal: It is enough to have everything block-diagonalized. The relevant diagonal blocks are invertible.

But if I apply $\langle 1 \vert$ from the left and choose $\psi_\alpha ^{(0)} = \vert 2 \rangle$, on the equation above, I get $1=0$. Could you explain in more detail? Thanks!

@genideal: I don't understand your notation. In the correct setting, you get a linear system in block form. Each partial vector must have length 2 (the eigenvalue multiplicity in your case) and each coefficient must be a 2x2 matrix. Please look at the reference given!

@Arnold: I am sorry.. I have read through the paper but I still don't understand why the non-hermiticity does not make any difference. Can I ask you about a very simple specific example?

If $H_0 = \frac{\partial^2}{\partial x^2}$ on $x \in S^1$, the eigenfunctions are $\vert n \rangle = e^{inx}$. Let us say we have the non-hermitian perturbation $H_1 = e^{2ix}$. We know that $D=\{ \vert -1 \rangle, \vert 1 \rangle \}$ form a degenerate subspace of Hilbert space with the energy $E^{(0)}=-n^2 =-1$, at the 0th order. Now the full Schrodinger equation is $(H_0 + \lambda H_1 ) (\vert \psi_\alpha ^{(0)} \rangle + \lambda \vert \psi_\alpha ^{(1)} \rangle + \cdots ) = (E^{(0)} + \lambda E^{(1)} + \cdots)(\vert \psi_\alpha ^{(0)} \rangle + \lambda \vert \psi_\alpha ^{(1)} \rangle + \cdots )$. At the 0th order, $\vert \psi_\alpha ^{(0)} \rangle \in D$ and the first order equation is $(H_0 - E^{(0)}) \vert \psi_\alpha ^{(1)} \rangle + (H_1 - E^{(1)} )\vert \psi_\alpha ^{(0)} \rangle =0$. If I apply $\langle \psi_\beta ^{(0)} \vert$ from the left, I get $\langle \psi_\beta ^{(0)} \vert H_1 - E^{(1)} \vert \psi_\alpha ^{(0)} \rangle =0$. If $H_1$ is Hermitian, this is just a problem of diagonalization. But in this example $H_1 = e^{2ix}$ and is represented on $D$ as $H_1 \sim \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$. Namely the first order equation does not make sense.

I should be missing something here for the whole perturbation theory to work.. Could you let me know what it is?

@genideal: I updated my answer to show in more detail why nothing problematic happens when one proceeds correctly.

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