Heterojunction

Using the Schrödinger equation for a finite well with width of ${\displaystyle l_{w}}$and center at 0, the equation for the achieved quantum well can be written as:

${\displaystyle -{\frac {\hbar ^{2}}{2m_{b}^{*}}}{\frac {\mathrm {d} ^{2}\psi (z)}{\mathrm {d} z^{2}}}+V\psi (z)=E\psi (z)\quad \quad {\text{ for }}z<-{\frac {l_{w}}{2}}\quad \quad (1)}$

${\displaystyle \quad \quad -{\frac {\hbar ^{2}}{2m_{w}^{*}}}{\frac {\mathrm {d} ^{2}\psi (z)}{\mathrm {d} z^{2}}}=E\psi (z)\quad \quad {\text{ for }}-{\frac {l_{w}}{2}}<z<+{\frac {l_{w}}{2}}\quad \quad (2)}$

${\displaystyle -{\frac {\hbar ^{2}}{2m_{b}^{*}}}{\frac {\mathrm {d} ^{2}\psi (z)}{\mathrm {d} z^{2}}}+V\psi (z)=E\psi (z)\quad {\text{ for }}z>+{\frac {l_{w}}{2}}\quad \quad (3)}$

Solution for above equations are well-known, only with different(modified) k and ${\displaystyle \kappa }$ ^[15]

${\displaystyle k={\frac {\sqrt {2m_{w}E}}{\hbar }}\quad \quad \kappa ={\frac {\sqrt {2m_{b}(V-E)}}{\hbar }}\quad \quad (4)}$.

At the z = ${\displaystyle +{\frac {l_{w}}{2}}}$ even-parity solution can be gained from

${\displaystyle A\cos({\frac {kl_{w}}{2}})=B\exp(-{\frac {\kappa l_{w}}{2}})\quad \quad (5)}$.

By taking derivative of (5) and multiplying both sides by ${\displaystyle {\frac {1}{m^{*}}}}$

${\displaystyle -{\frac {kA}{m_{w}^{*}}}\sin({\frac {kl_{w}}{2}})=-{\frac {\kappa B}{m_{b}^{*}}}\exp(-{\frac {\kappa l_{w}}{2}})\quad \quad (6)}$.

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

${\displaystyle f(E)=-{\frac {k}{m_{w}^{*}}}\tan({\frac {kl_{w}}{2}})-{\frac {\kappa }{m_{b}^{*}}}=0\quad \quad (7)}$.

Similarly, for odd-parity solution,

${\displaystyle f(E)=-{\frac {k}{m_{w}^{*}}}\cot({\frac {kl_{w}}{2}})+{\frac {\kappa }{m_{b}^{*}}}=0\quad \quad (8)}$.

For

even parity:

${\displaystyle {\frac {df}{dE}}={\frac {1}{m_{w}^{*}}}{\frac {dk}{dE}}\tan({\frac {kl_{w}}{2}})+{\frac {k}{m_{w}^{*}}}\sec ^{2}({\frac {kl_{w}}{2}})\times {\frac {l_{w}}{2}}{\frac {dk}{dE}}-{\frac {1}{m_{b}^{*}}}{\frac {d\kappa }{dE}}\quad \quad (9-1)}$

odd parity:

${\displaystyle {\frac {df}{dE}}={\frac {1}{m_{w}^{*}}}{\frac {dk}{dE}}\cot({\frac {kl_{w}}{2}})-{\frac {k}{m_{w}^{*}}}\csc ^{2}({\frac {kl_{w}}{2}})\times {\frac {l_{w}}{2}}{\frac {dk}{dE}}+{\frac {1}{m_{b}^{*}}}{\frac {d\kappa }{dE}}\quad \quad (9-2)}$

where ${\displaystyle {\frac {dk}{dE}}={\frac {\sqrt {2m_{w}^{*}}}{2{\sqrt {E}}\hbar }}\quad \quad \quad {\frac {d\kappa }{dE}}=-{\frac {\sqrt {2m_{b}^{*}}}{2{\sqrt {V-E}}\hbar }}}$.

The difference in effective mass between materials results in a larger difference in ground state energies.