The hermicity of a k.p matrix?

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
2 replies · 3K views
lichen1983312
Messages
85
Reaction score
2
I am trying to use the k.p method to study quantum well band structure. One example Hamiltonian look like this [J. Appl. Phys., 116, 033709(2014)]

Hamiltonian.png
where
##{{\hat k}_ \pm } = {{\hat k}_x} \pm i{{\hat k}_y}##
and the matrix elements are function of ##{{\hat k}_i}##
and if quantum well is grown along z direction
the envelop functions have the form
\[F(x,y,z) = {e^{i{{\bf{k}}_\parallel } \cdot {\bf{r}}}}f(z)\]

and therefore

##\left\{ \begin{array}{l}
{{\hat k}_x} = - i\frac{\partial }{{\partial x}} \to {k_x}\\
{{\hat k}_y} = - i\frac{\partial }{{\partial y}} \to {k_y}\\
{{\hat k}_z} = - i\frac{\partial }{{\partial z}}
\end{array} \right.##

this Hamiltonian is an hermitian operator, but I am confused about this. For example, if I look at the (1,4) and (4,1) terms and use ${{\hat k}_z}=-i{\partial _z}$, the (1,4) term becomes
##-i{\partial _z}{P_1}##
and the (4,1) term becomes
##- i{P_1}{\partial _z}##

If I use finite difference method to turn these two terms into a matrix block, the factor -i is not going to flip sign and the matrix cannot he a hermitian matrix?

there must be something wrong with my understanding, please help.
 

Attachments

  • Hamiltonian.png
    Hamiltonian.png
    17 KB · Views: 1,408
Physics news on Phys.org
I have to look at the paper first, the matrix is hermitian, but I will say that [itex]P_1[/itex] and [itex]P_2[/itex] are constants, not operators if I remember my [itex]\vec{k} \cdot \vec{p}[/itex] theory correctly.
 
Dr Transport said:
I have to look at the paper first, the matrix is hermitian, but I will say that [itex]P_1[/itex] and [itex]P_2[/itex] are constants, not operators if I remember my [itex]\vec{k} \cdot \vec{p}[/itex] theory correctly.
Thanks very much, the reference is this one :
https://aip.scitation.org/doi/10.1063/1.4890585