find Approximation transmission probability in quantum mechanics E<U
Answer
Let for electron has wave function
ψ∝eikxℏ\psi\propto e^\frac{ikx}{\hbar}ψ∝eℏikx
The effective momentum of electron is given by
ℏ2k22m∗=−qϕ\frac{\hbar^2 k^2}{2m*}=-q\phi2m∗ℏ2k2=−qϕ
k=i2m∗qϕℏ2k=i\sqrt{\frac{2m*q\phi}{\hbar^2}}k=iℏ22m∗qϕ
So wavefunction become
ψ∝ei2m∗qϕℏ2x\psi\propto e^{i\sqrt{\frac{2m*q\phi}{\hbar^2}}x}ψ∝eiℏ22m∗qϕx
ψ(x=d)ψ(x=0)∝e−2m∗qϕℏ2.de−2m∗qϕℏ2.0∝e−2m∗qϕℏ2d\frac{\psi (x=d) }{\psi(x=0) }\propto \frac{ e^{-\sqrt{\frac{2m*q\phi}{\hbar^2}}.d}}{e^{-\sqrt{\frac{2m*q\phi}{\hbar^2}}.0}}\propto e^{-\sqrt{\frac{2m*q\phi}{\hbar^2}}d}ψ(x=0)ψ(x=d)∝e−ℏ22m∗qϕ.0e−ℏ22m∗qϕ.d∝e−ℏ22m∗qϕd
Let JI
be the incident current and JT
be the transmitted current. Then
Transmission probability
T∝JTJ0∝(ψ(x=d)ψ(x=0))2∝e−2−2m∗qϕℏ2dT\propto \frac{J_T}{J_0}\propto (\frac{\psi(x=d) }{\psi(x=0) }) ^2\\\propto e^{-2\sqrt{-\frac{2m*q\phi}{\hbar^2}}d}T∝J0JT∝(ψ(x=0)ψ(x=d))2∝e−2−ℏ22m∗qϕd
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