- #1

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A particle of total energy 9Vo is incident from the -x axis on a potential given by:

[tex]

V \left\{8Vo, x<0\right\}

\left\{0, 0<x<a\right\}

\left\{5Vo, x>a\right\}

[/tex]

Find the probability that the particle will be transmitted on through the positive side of the x axis, x>a.

Because the particle initially has greater energy than all points in the well, i know it will be sinusoidal everywhere given by the following wave functions:

[tex]\Psi_{1} = Ae^{ik_{1}x} + Be^{-ik_{1}x}[/tex]

[tex]\Psi_{2} = Ce^{ik_{2}x} + De^{-ik_{2}x}[/tex]

[tex]\Psi_{3} = Fe^{ik_{3}x} + Fe^{-ik_{3}x}[/tex]

I will set F = 0, because there will be no wave traveling in the negative x direction there because there will be no reflected wave, giving me:

[tex]\Psi_{1} = Ae^{ik_{1}x} + Be^{-ik_{1}x}[/tex]

[tex]\Psi_{2} = Ce^{ik_{2}x} + De^{-ik_{2}x}[/tex]

[tex]\Psi_{3} = Fe^{ik_{3}x}[/tex]

with

[tex] k_{n} = \frac{2\pi}{\lambda_{n}}[/tex]

First, i find the transmission coefficient for the transition at x = 0, and I will set D = 0, because I will pretend that there is no reflected wave yet (CAN I DO THIS?)

[tex]\Psi_{1} = Ae^{ik_{1}x} + Be^{-ik_{1}x}[/tex]

[tex]\Psi_{2} = Ce^{ik_{2}x}[/tex]

To find the coefficients, i set:

[tex]\Psi_{1}|_{x=0} = \Psi_{2}|_{x=0}[/tex]

leading to:

[tex]A + B = C[/tex]

I also know that the derivatives will be continuous there because my potential doesn't approach infinity:

[tex]\frac{d\Psi_{1}}{dx}=\frac{d\Psi_{2}}{dx}[/tex]

[tex]

ik_{1}Ae^{ik_{1}x}|_{x=0}-ik_{1}Be^{-ik_{1}x}|_{x=0}=ik_{2}Ce^{ik_{2}x}|_{x=0}

[/tex]

this gives me:

[tex]k_{1}(A-B)=k_{2}C[/tex]

Solving the above equations with only constants, i find the constants in terms of A:

[tex]

B = \frac{A(k_{1}-k_{2}}{k_{1}+k_{2}}

[/tex]

[tex]

C = \frac{2k_{1}A}{k_{1}+k_{2}}

[/tex]

The transmission coefficient is:

[tex]

T_{1} = \frac{v_{2}C*C}{v_{1}A*A}=\frac{v_{2}}{v_{1}}\frac{2k_{1}}{k_{1}+k_{2}}^2=\frac{4k_{1}k_{2}}{(k_{1}+k_{2})^{2}}

[/tex]

At this point, i notice that the transmission coefficient for x = 0 doesn't depend on anything but the potential differences, whose information is provided in k:

[tex] k = \frac{\sqrt{2m(E-V)}}{\hbar}[/tex]

So, can I just immediately say that the transmission coefficient for the step at x =a (pretending the first step isn't there) is:

[tex]

T_{2} =\frac{4k_{2}k_{3}}{(k_{2}+k_{3})^{2}}

[/tex]

If so, will my final transmission probability be the product of the two?

Any help is appreciated!