Editing Heterojunction (section)

== Effective mass mismatch  ==

When a heterojunction is formed by two different [[semiconductor]]s, a [[quantum well]] can be fabricated due to difference in [[band structure]]. In order to calculate the static [[energy level]]s within the achieved quantum well, understanding variation or mismatch of the [[Effective mass (solid-state physics)|effective mass]] across the heterojunction becomes substantial. The quantum well defined in the heterojunction can be treated as a finite well potential with width of <math> l_w</math>. In addition, in 1966, Conley et al.<ref>{{cite journal|doi=10.1103/PhysRev.150.466|title=Electron Tunneling in Metal–Semiconductor Barriers|year=1966|last1=Conley|first1=J.|last2=Duke|first2=C.|last3=Mahan|first3=G.|last4=Tiemann|first4=J.|journal=Physical Review|volume=150|issue=2|pages=466|bibcode = 1966PhRv..150..466C }}</ref> and BenDaniel and Duke<ref>{{cite journal|doi=10.1103/PhysRev.152.683|title=Space-Charge Effects on Electron Tunneling|year=1966|last1=Bendaniel|first1=D.|last2=Duke|first2=C.|journal=Physical Review|volume=152|issue=2|pages=683|bibcode = 1966PhRv..152..683B }}</ref> reported a [[boundary condition]] for the [[Envelope (mathematics)|envelope function]] in a quantum well, known as BenDaniel–Duke boundary condition. According to them, the envelope function in a fabricated quantum well must satisfy a boundary condition which states that <math> \psi (z) </math> and <math> {\frac {1} {m^*} }{\partial \over {\partial z}} \psi (z) \,</math> are both continuous in interface regions.

{{hidden begin|border=solid 1px #aaa|title={{center|Mathematical details worked out for [[quantum well]] example.}}}}
Using the [[Schrödinger equation]] for a finite well with width of <math>l_w </math>and center at 0, the equation for the achieved quantum well can be written as:

::<math>-\frac{\hbar^2}{2m_b^*} \frac{\mathrm{d}^2 \psi(z)}{\mathrm{d}z^2} + V \psi(z) = E \psi(z) \quad \quad \text{ for } z < - \frac {l_w}{2} \quad \quad (1)</math>
::<math> \quad \quad -\frac{\hbar^2}{2m_w^*} \frac{\mathrm{d}^2 \psi(z)}{\mathrm{d}z^2} = E \psi(z) \quad \quad \text{ for } - \frac {l_w}{2} < z < + \frac {l_w}{2} \quad \quad (2)</math>
::<math>-\frac{\hbar^2}{2m_b^*} \frac{\mathrm{d}^2 \psi(z)}{\mathrm{d}z^2} + V \psi(z) = E \psi(z) \quad \text{ for } z > + \frac {l_w}{2} \quad \quad (3)</math>

Solution for above equations are well-known, only with different(modified) k and <math>\kappa </math> <ref>Griffiths, David J. (2004). ''Introduction to Quantum Mechanics'' (2nd ed.). Prentice Hall. {{ISBN|0-13-111892-7}}</ref>

::<math> k = \frac {\sqrt{2 m_w E}} {\hbar} \quad \quad \kappa = \frac {\sqrt{2 m_b (V-E)}} {\hbar} \quad \quad (4)</math>.

At the z = <math> + \frac {l_w} {2} </math> even-parity solution can be gained from
::<Math> A\cos(\frac {k l_w} {2}) = B \exp(- \frac {\kappa l_w} {2}) \quad \quad (5)</math>.

By taking derivative of (5) and multiplying both sides by <math> \frac {1} {m^*}</math>

::<Math> -\frac {kA} {m_w^*} \sin(\frac {k l_w} {2}) = -\frac {\kappa B} {m_b^*} \exp(- \frac {\kappa l_w} {2}) \quad \quad (6)</math>.

Dividing (6) by (5), even-parity solution function can be obtained,

::<Math> f(E) = -\frac {k} {m_w^*} \tan(\frac {k l_w} {2}) -\frac {\kappa } {m_b^*} = 0 \quad \quad (7)</math>.

Similarly, for odd-parity solution,

::<Math> f(E) = -\frac {k} {m_w^*} \cot(\frac {k l_w} {2}) +\frac {\kappa } {m_b^*} = 0 \quad \quad (8)</math>.

For [[numerical solution]], taking derivatives of (7) and (8) gives

even parity:
::<math> \frac {df}{dE} = \frac {1}{m_w^*} \frac {dk}{dE} \tan(\frac {k l_w} {2}) + \frac {k} {m_w^*} \sec^2(\frac {k l_w} {2}) \times \frac {l_w} {2} \frac {dk} {dE} - \frac {1}{m_b^*} \frac {d \kappa} {dE} \quad \quad (9-1)</math>
odd parity:
::<math> \frac {df}{dE} = \frac {1}{m_w^*} \frac {dk}{dE} \cot(\frac {k l_w} {2}) - \frac {k} {m_w^*} \csc^2(\frac {k l_w} {2}) \times \frac {l_w} {2} \frac {dk} {dE} + \frac {1}{m_b^*} \frac {d \kappa} {dE} \quad \quad (9-2)</math>

where <math> \frac {dk}{dE} = \frac {\sqrt {2 m_w^*}}{2 \sqrt E \hbar} \quad \quad \quad \frac {d \kappa}{dE} = - \frac {\sqrt {2 m_b^*}}{2 \sqrt {V-E} \hbar}</math>.

The difference in effective mass between materials results in a larger difference in [[ground state]] energies.
{{hidden end}}