0
$\begingroup$

Suppose I want to solve the 1D, time-independent Schrödinger-equation for a metal-semiconductor junction.

In the metal region $0 \leq x \leq x_{0} $ the Schrödinger equation reads:

$(-\frac{\hbar^2}{2m} \frac{d^2}{dx^2}+V(x)) = E\psi $

In the semiconductor region $x_{0}<x\leq L$ the Schrödinger equation reads:

$(-\frac{\hbar^2}{2m^{*}} \frac{d^2}{dx^2}+\Delta E_{SM} + V(x)) \psi = E\psi $

Here $\Delta E_{SM}$ is the offset between the conduction band edges in the metal- and semiconductor region. My question is: Is there a way to "integrate out" the metal region or replace it with an effective boundary condition at the semiconductor-metal interface?

$\endgroup$
0
$\begingroup$

In the semiconductor, you are using the so called effective mass or (envelope wave function) Schrödinger equation which uses a parabolic approximation for the energy vs wave vector dispersion relation near the conduction band minimum. In the metal, your conduction band minimum (band edge) is probably many electron volts below the conduction band edge of the semiconductor. Thus, you probably cannot use the effective mass (envelope wave function) Schrödinger equation approximation there. You will most likely have to start from the full Schrödinger equation with periodic potential to find a meaningful approximation at the interface.

$\endgroup$
0
$\begingroup$

Using the condition that the boundary condition must be linear, self adjoint, and involve no more than $\psi$ and its first derivative (these requirements follow from the full Schroedinger equation) the most general boundary condition for an interface between effective mass $m_L$ and $m_R$ is $$ \left(\matrix{ \psi_{2L}\cr \partial_x \psi_{2L}}\right) = \left(\matrix{ a& b\cr c&d }\right) \left(\matrix{ \psi_{2R}\cr \partial_x \psi_{2R}}\right). $$ where $$ \left(\matrix{ a& b\cr c&d }\right) = e^{i\phi}\sqrt{\frac{m_L}{m_R}}\left(\matrix{ A& B\cr C&D }\right). $$ Here $A$, $B$, $C$, $D$, and the phase $\phi$ are real and $(AD-BC)=1$. These conditions are used in
heterojunctions etc. (see T.Ando, S.Mori, Surface Science
113 (1982) 124). The derivation is a bit too long for an answer here.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.