Consider the step potential defined by(adsbygoogle = window.adsbygoogle || []).push({});

V(x) = 0 if x < 0

Vb > 0 if x=> 0

a) Compute te relfection and tranmission coefficients for a particle of energy E > Vb approaching from th left

For x < 0

Schrodinger equaion since V = 0

[tex] -\frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2} = E \psi(x) [/tex]

[tex] \frac{\partial^2 \psi}{\partial x^2} = -\frac{2mE}{\hbar^2} \psi(x) [/tex]

define [tex] k_{1}^2 = \frac{2mE}{\hbar^2} [/tex]

[tex] \frac{\partial^2 \psi}{\partial x^2} = -k^2 \psi(x) [/tex]

[tex] \psi(x) = A \cos(k_{1}x) + B \sin(k_{1} x) [/tex]

for x=> 0

[tex] -\frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2} = (E-V_{B}) \psi(x) [/tex]

[tex] k_{2}^2 = \frac{2m}{\hbar^2} (E - V_{B}) [/tex]

[tex] \frac{\partial^2 \psi}{\partial x^2} = k_{2}^2 \psi(x) [/tex]

[tex] \psi_(x) = C \exp(ik_{2} x) + D\exp(-ik_{2}x) [/tex]

but we only care about the one with the negative exponential

hence C = 0

[tex] \psi_(x) = D\exp(-ik_{2}x) [/tex]

are these solutions correct? Im just afraid i may have made a stupid mistake by switching the negative sign somewhere :(

ok to proceed i apply the boundary conditions s.t.

[tex] \psi_{I}(0) = \psi_{T}(0) [/tex]

[tex] \psi_{I}'(0) = \psi_{T}'(0) [/tex]

we calculate probability current density j

[tex] j_{trans} = \frac{\hbar k_{2}}{m} |D|^2 [/tex]

[tex] j _{inc} = \frac{\hbar k_{1}}{m} |A|^2 [/tex]

[tex] j_{refl} = \frac{\hbar k_{1}}{m} |B|^2 [/tex]

and transmission coefficient is calculated like this

[tex] T = \left| \frac{j_{trans}}{j_{inc}} \right| [/tex]

[tex] R = \left| \frac{j_{refl}}{j_{inc}} \right| [/tex]

thank you for your input

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# Transmission and Reflection Coefficients

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