Wavefunction matching two different H, not just V

[SturgySturges]

Can the basic techniques of wavefunction matching that one would use to calculate the transmission through a step barrier potential and the Dirac hamiltonian of graphene be used for a situation where instead the fermi velocity changes in a step like fashion. i.e. instead of a Hamiltonian like

\begin{pmatrix}V(x) & k_x - i k_y \\ k_x + i k_y & V(x)\end{pmatrix}

where V(x)=V_0 \Theta(x) and \Theta(x) is the unit step function, you have a Hamiltonain like

v(x) \begin{pmatrix}0 & k_x - i k_y \\ k_x + i k_y & 0\end{pmatrix}

where v(x)=v_0 \Theta(x).

If not, what would be a way to approach this problem? Many thanks.

 

[Paul Colby]

I assume the $k_x+ik_y$ terms originate form a $\gamma^\mu \partial_\mu$ applied to the wave function. This being the case I would be concerned that the commutator, $[v(x),\gamma^\mu \partial_\mu]\ne 0$. This would make your Hamiltonian not hermitian? Just a thought.

 

[SturgySturges]

Hmm I'm not sure, I've described a much simpler generalisation of the problem that I'm working on just to get down to the key concept I'm uncertain about. Basically it's this business of having two different Hamiltonians either side of x=0 (normally the difference is just the `V' in H=H_0+V) and using wave function matching techniques at the discontinuity. It works for just a V but I want to know if/why/how these ideas can be extended to a situation like the above.

 

[Paul Colby]

It works for just a $V$ but I want to know if/why/how these ideas can be extended to a situation like the above.

I used to work with an old guy back in the 80's whose favorite question was; "what is it?"

if you write out the $V(x)$ case $H=H_o + V(x)$. Clearly, $[V(x),H_o]\ne 0$ because of the derivatives that appear in $H_o$. This is all fine and good because you are adding two hermitian operators which yields an hermitian $H$. If you write out the generalization, what does it look like in terms of derivatives (which is where the k's come from) and the new $v(x)$? Unless you are careful the result will not be hermitian. I assume that one could symmetrize any product and get a hermitian operators but the details might be different than what you've written, Don't know because it's unclear what you intend from what you have written out.