# Potential barrier in QM problem with coefficient of transmission

Gold Member

## Homework Statement

I'll try to recreate from my memory the problem we've been assigned on a test more than one month ago. They gave the solution but I either misunderstood or miscopied it.
An electron with kinetic energy 5 eV goes from a region with potential $V_0=6 eV$ (let's call this region I) to a region with potential 0 (let's call this region II). Calculate the coefficient of transmission.

## Homework Equations

The professor said we didn't need to have the explicit formula for the transmission. Rather we should write the expression of a plane wave (I guess she meant standing wave) and use the formula of probability current with j that follows.
$$j= \frac{1}{2im} \left ( \Psi ^* \frac{\partial \Psi }{\partial x} - \Psi \frac{\partial \Psi ^* }{\partial x} \right )$$
With the $\Psi _I$ of region I, this gives $j _{\text {incident} }+ j_{\text {reflected} }$ and for region II this gives $j_ \text {transmitted}$.
Here is my problem. The solution she gave was like $\Psi _I (x)=Ae^{ik_1x}+Be^{ik_2x}$ and $\Psi _{II}(x)=Ce^{i k_2 x}$ and that we should get $0.14$ for the coefficient of transmission.

## The Attempt at a Solution

So I tried to get $\Psi _I (x)$ but I don't get the same function at all. I get $\Psi _I (x) =Ae^{k_1 x}+Be^{-k_1 x}$ where $k_1 =\sqrt { \frac{2m (v_0 -E)}{\hbar ^2 } }$.
And even more than that, I'm almost sure that B must be worh 0, otherwise psi diverges when x tends to - infinity.
Am I right on this?!

vela
Staff Emeritus
Homework Helper
Yes. In region I, the energy of the electron is less than the height of the potential, so you should get real exponential solutions. Are you sure you remembered the energy of the electron correctly? Or perhaps the height of the potential is supposed to be lower.

Gold Member
Yes. In region I, the energy of the electron is less than the height of the potential, so you should get real exponential solutions. Are you sure you remembered the energy of the electron correctly? Or perhaps the height of the potential is supposed to be lower.

I just found what I copied from the solution. The transmission coeffient should give me 0.96 (not 0.14 like I said... Brehm's book gave an example with 0.14, hence my confusion). I am 100% sure that the kinetic energy of the electron was lower than the potential $V_0$; these numbers are right.
The solution assumes that the electron comes from the right side (x= infinity) toward the left side ( x = - infinity), unlike what I've done.
She gave $\Psi _I (x)=Ae^{-i k_1x}+B e^{ik_1x}$ and $\Psi _{II}=Ce^{-i k_2 x}$.
Then she wrote $j_1= j _{inc}+j_{ref}=\frac{\hbar ^2}{m} k_1 (|B|^2-|A|^2)$ and $j_2 = j_{trans}=-\frac{\hbar}{m} k_2 |C|^2$.
$\Rightarrow \frac{|B|^2}{|A|^2}+\frac{k_2}{k_1} \frac{|C|^2}{|A|^2}=1$.
$\underbrace { \big | \frac{j_{trans}}{j_{inc} } \big | } _{T}+ \underbrace { \big | \frac{j_{ref}}{j_{inc} } \big | } _{R}=1$.
$R=0.04$, $T=0.96$.

Now that I look at that, I'm guessing there are typos. Also how can you get $\big | \frac{j_{trans}}{j_{inc} } \big |$ when you only have $j_{inc}$ and $j _{inc}+j_{ref}$?

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Ok I asked the professor a copy of the test and he kindly gave it to me.
The problem is: An electron with kinetic energy 4eV enters suddenly a region where its potential energy drops by 5eV. Find the probability that the electron gets reflected and transmitted due to the violent drop of potential. Verify the conservation of probability.
So now her answers for the psi's makes sense.
What I've found is $j_{inc}+j_{ref}=\frac{\hbar \alpha }{m} (|A|^2-|B|^2)$ and $j_{trans}=\frac{\hbar \beta|C|^2}{m}$ with $\alpha = \sqrt {\frac {2m(E-V_0)}{\hbar ^2} }$ and $\beta =\sqrt {\frac {2mE}{\hbar ^2} }$. So my answer differs by a factor of -1 for the $j_{trans}$ compared to the one of my professor.
Now it seems somehow obvious to me that $j_{inc}=\frac{\hbar \alpha |A|^2}{m}$ and $j_{ref}=-\frac{\hbar \alpha |B|^2}{m}$ but I don't know if I have to demonstrate it. What do you say?
This would make $T= \big | \frac{j_{trans}}{j_{inc}} \big | =\frac{\beta} {\alpha } \cdot \frac{|C|^2}{|A|^2}$. Though I don't know how to get a numerical value for this. I can simplify the $\frac{\beta }{\alpha}=\sqrt {\frac{E}{E-V_0}}$ but how to do so for $\frac{|C|^2}{|A|^2}$?

vela
Staff Emeritus
Homework Helper
You need to solve for that ratio by matching the functions for region I and region II at the boundary.

Gold Member
You need to solve for that ratio by matching the functions for region I and region II at the boundary.

Thanks.
This gives me $A+B=C$ and $i \beta C = i \alpha (A-B)$.
I wrote B in terms of A and C and then I got a condition, which eventually gave me $C= \frac{2A}{\left ( 1+ \frac{\beta}{\alpha} \right ) }$
So I solved for T, but I am not sure what to put as values of $E$ and $V_0$. I got $\alpha \approx 10.522$. If I take E=9 eV and V_0=5eV, I reach $T\approx 0.09$. This doesn't match the answer.

Gold Member
I've checked what the professor consider as answer and the only difference with my answer is that I have an extra $1/\alpha$ factor. I don't know where I went wrong though.

Edit: I've rechecked my algebra and spotted the mistake. I carried the $1/ \alpha$ coefficient out of nowhere, so it shouldn't be here. I reach the answer given by the professor, problem solved. Thank you vela for all your help.

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