## Q. 3.3

Electrons in the conduction band of different semiconductors have differing effective electron mass. In addition, the effective electron mass, $m_{j}$ , is almost never the same as the free electron mass, $m_{0}$ . These facts are usually accommodated in simple models of semiconductor heterojunction barrier transmission and reflection by requiring $\psi _{1} \mid _{0-\delta } = \psi _{2} \mid _{0+\delta }$ and $(1/m_{1} )\nabla \psi _{1}\mid _{0-\delta } = (1/m_{2} )\nabla \psi _{2}\mid _{0+\delta }$ at the heterojunction interface. Why are these boundary conditions used? Calculate the transmission and reflection coefficient for an electron of energy $E$, moving from left to right, impinging normally to the plane of a semiconductor heterojunction potential barrier of energy $V_{0}$. The effective electron mass on the left-hand side is $m_{1}$, and the effective electron mass on the right-hand side is $m_{2}$. Under what conditions will there be no reflection? Express this condition in terms of electron velocities on either side of the interface.

## Step-by-Step

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To accommodate the fact that electrons in the conduction band of different semiconductors labeled 1 and 2 have differing effective electron mass, simple models of semiconductor heterojunction barrier transmission and reflection require that the electron wave function satisfy $\psi _{1} \mid _{0-\delta } = \psi _{2} \mid _{0+\delta }$ and $(1/m_{1} )\nabla \psi _{1}\mid _{0-\delta } = (1/m_{2} )\nabla \psi _{2}\mid _{0+\delta }$ at the heterojunction interface. This ensures conservation of current across the heterointerface. In this limited model, current conservation is taken to be more important than ensuring continuity in the first spatial derivative of the wave function.

We now calculate the transmission and reflection coefficient for an electron of energy $E$, moving from left to right, impinging normally to the plane of a semiconductor heterojunction potential barrier of energy $V_{0}$, where the effective electron mass on the left-hand side is $m_{1}$ and the effective electron mass on the right-hand side is $m_{2}$.

At the boundary between regions 1 and 2 at position $x_{0}$, we require continuity in $\psi$ and $\nabla \psi /m_{j}$, so that

$\psi _{1} \mid_{x_{0}} =\psi _{1} \mid_{x_{0}}$

and

$\frac{1}{m_{1} }\frac{d}{dx} \psi _{1} \mid _{x_{0}} =\frac{1}{m_{2}}\frac{d}{dx} \psi _{2} \mid _{x_{0}}$

$A+B = C +D$

and

$A-B=\frac{m_{1}k_{2} }{m_{2}k_{1} } C-\frac{m_{1}k_{2} }{m_{2}k_{1} }D$

If we know that the particle is incident from the left, then $A = 1$ and $D = 0$, giving

$1+B = C$

$1-B=\frac{m_{1} k_{2} }{m_{2} k_{1} } C$

Recognizing velocity $v_{1} =\hbar k_{1} /m_{1}$ and $v_{2} =\hbar k_{2} /m_{2}$ and solving for the transmission probability $\left|C\right| ^{2}$ and the reflection probability $\left|B\right| ^{2}$ gives

$\left|C\right| ^{2} =\frac{4}{\left(1+\frac{m_{1}k_{2} }{m_{2} k_{1} } \right)^{2} } =\frac{4}{\left(1+\frac{v_{2} }{v_{1} } \right)^{2} }$

$\left|B\right| ^{2} =\frac{\left(1-\frac{m_{1}k_{2} }{m_{2} k_{1} } \right)^{2} }{\left(1+\frac{m_{1}k_{2} }{m_{2} k_{1} } \right)^{2}} =\frac{\left(1-\frac{v_{2} }{v_{1} } \right)^{2}}{\left(1+\frac{v_{2} }{v_{1} } \right)^{2}}$

From this it is clear that there is no reflection when particle velocity is the same in each of the two regions. This is an example of an impedance-matching condition that, in this case, will occur when particle energy $E=V_{0} /(1-m_{1}/m_{2} )$.
It is worth making a few comments about our approach to solving this exercise. We selected two boundary conditions to guarantee current conservation and no discontinuity in the electron wave function. However, with this choice, we were unable to avoid the possibility of kinks in the wave function. Such an approach was first discussed by Bastard.$^{10}$ Unfortunately, the Schrödinger equation cannot be solved correctly using different effective electron masses for electrons on either side of an interface. A more complex, but correct, way to proceed is to use the atomic potentials of atoms forming the heterointerface and solve using the bare electron mass and the usual boundary conditions

$\psi \mid _{x_{0}-\delta }=\psi \mid _{x_{0}+\delta }$

and

$\frac{d\psi }{dx} \mid _{x_{0}-\delta } =\frac{d\psi }{dx} \mid _{x_{0}+\delta }$

A simplified version of this situation might use the Kronig–Penney potential, which will be discussed in Chapter 4.

There are a number of additional reasons why the model we have used should only be considered elementary. For example, the electron can be subject to other types of scattering. Such scattering may destroy either momentum or both momentum and energy conservation near the interface. This may take the form of diffuse electron scattering due to interface roughness or high phonon emission probability. The presence of nonradiative electron recombination due to impurities and traps, or, typically in direct band gap semiconductors, radiative processes may be important. These effects may result in a nonunity sum of transmission and reflection coefficients $(R+T ≠ 1)$.
Situations often arise when an effective electron mass cannot be used to describe electron motion. In this case the electron dispersion relation is nonparabolic over the energy range of interest. In addition, when the character (or $s, p, d,$ etc., symmetry) of electron wave function is very different on both sides of a potential step, the simple approach used above is inappropriate and can lead to incorrect results.

10. G. Bastard, Phys. Rev. B24, 5693 (1981).

Question: 3.7

Question: 3.2

Question: 3.5

Question: 3.6

Question: 3.4

Question: 3.1