# Two-Band k.p Model is not Hermitian for imaginary wavevectors

In E. O. Kane's original work on Zener Tunneling, he uses a two-band $k\cdot p$ model for the semiconductor bandstructure:

$$H=\begin{pmatrix}E_g+\frac{\hbar^2k^2}{2m_0}&(\hbar/m)kp\\(\hbar/m)kp&\frac{\hbar^2k^2}{2m_0}\end{pmatrix}$$

Moving along real $k$, one sees a conduction and valence band repelling each other via the momentum matrix element. But if you step onto the imaginary $k$ axis, there is a curve stretching through the band-gap to connect the extrema of the conduction and valence bands. These solutions for imaginary $k$ can be used to represent surface states, and they provide a smooth path for a semiclassical wavepacket to "tunnel" between bands while always having a well-defined position and wavevector.

But the above Hamiltonian is non-Hermitian when one plugs in complex values for $k$. (And if you make it Hermitian by conjugating one of the off-diagonal terms, the connection between the bands disappears.) So these spatially decaying states (which all have real energies within a certain $k$ range) are not actually eigenvectors of a Hermitian Hamiltonian...what are they?

I suppose I'm missing something fundamental about this procedure, but I don't see why this famous model assumes and works with a non-Hermitian Hamiltonian and what that means for the computed eigenvalues*. Has anyone come across any physics papers that address the subtle aspects of this?

.* For instance, I noticed that, at least in the range where all eigenvalues are real, the eigenvectors must be non-orthogonal (otherwise this non-Hermitian matrix could be real-diagonalized, which is a contradiction.)

• Not 100% sure, but I think that the matrix will have left and right eigenvectors. And then a wild guess that one of these eigenvector sets will represent asymptotically diverging solutions, and the other decaying solutions. – Mikael Kuisma Nov 29 '15 at 1:43
• The Hamiltonian $H$ is Hermitian while $e^{\kappa z} H e^{-\kappa z}$ does not have to be hermitian. The eigenvalues are real as long as the wave function is normalizable. For a half infinite surface $z>0$ this means a solution $\propto e^{-\kappa z}$ with $\kappa>0$. – Praan Nov 29 '15 at 10:22
• The lower left entry of your Hamiltonian should have $p^\dagger$ instead, or in position represenation, a minus sign. – Praan Nov 29 '15 at 16:11
• @MikaelKuisma The eigenstates of this matrix represent Bloch waves of whatever $k$ you plug in ($k$ being simply a parameter of the Hamiltonian), so the right eigenvectors represent decaying or growing solutions purely based on your choice of $\pm i k$. – Sam Bader Nov 29 '15 at 16:36
• @Praan I'm not sure what you're talking about with the $e^{kz}He^{-kz}$ expression, could you say more about where that came from? Also, the above matrix comes directly from Eq 19 of the cited paper. And the author says $p$, which is simply a number (momentum matrix element between Bloch basis states), can be taken pure real. One can't put in a minus sign there, else the matrix would be nonhermitian for real $k$. – Sam Bader Nov 29 '15 at 16:37