Square Hill Barrier, weird question wording

[Raynor49]

Homework Statement

Consider a square hill barrier produced by a 10V potential. Incident upon this barrier is a steady stream of 5eV electrons.
(a) If half the electrons are transmitted, how thick is the barrier? (Please derive the transmission probability rather than merely quoting it.)

Homework Equations

TISE ΨE = H*Ψ
sorry I don't know how to put the ^ over the H for the hamiltonian operator.

The Attempt at a Solution

Normally when trying well and barrier questions I would start by drawing out the well or barrier in question. However in this question the issue I'm having is that I don't even know how to draw it. I know it's a barrier and I think it has total E=10V but really am unsure what I'm supposed to do with the stream of 5eV electrons. I've never had a question like this with electron volts before, all my previous questions will just say something like

"Total energy E=9V₀
V=0 for x<0 & V=5V₀ for x>0 "
which I know how to draw and visualize.
Any help is appreciated, thanks.

 

[DelcrossA]

On an energy graph, the potential energy of the barrier will look like rectangle of height 10 eV (remember to get potential energy from potential you simply multiply by the charge q ) and the incident electrons have a kinetic energy of 5 eV, which people tend to represent as just a horizontal line. The important thing to note is that the energy of the electrons is lower than the potential barrier - does "classically forbidden regions" sound familiar? Also, to solve this you need to the equation for probability current.

 

[Raynor49]

Thanks for replying!
So I have a drawing now of a rectangle of 10eV on the y-axis and length "w" on the x-axis, with a horizontal line at height 5eV. Classically forbidden regions sounds very familiar, that's where tunneling comes into play right? When classically a particle can't pass a barrier?
And is this the equation:
J = (i*ħ/2m )(Ψ dΨ*/dx - Ψ*dΨ/dx) ?
I managed to find it in my textbook but haven't actually heard of it before. So to solve it I'd just use the TISE with E=10eV right? Any hints what to use it for? I'm not sure what to do with that equation, I was thinking I'd have to split the energy graph into the three sections of before the barrier, in the barrier and past the barrier and use the TISE to somehow get the transmission probability equation (which would equal 1/2) and then somehow use that to find the width of the barrier.

 

[DelcrossA]

Yes, that is the probability current - it will be used to calculate the Transmission coefficient. Separate the problem into three regions: before the potential, inside the potential, and after the potential. Write down the general wave-functions for each of these three regions by solving the TISE, where E = energy of electron beam, and V is the potential. What conditions can you apply to the wave-functions to determine the coefficients?

If you do not know, the definition of transmission amplitude is the probability current of the wavefunction in region 3 divided by the probability current of the incidient wavefunction (region 1)

edit: j_trans/j_inc

 

[Raynor49]

So I have
Ψ1 = Aexp(i*k₁*x) + Bexp(-i*k₁*x)
Ψ2 = Fexp(i*k₂*x) + Gexp(-i*k₂*x)
Ψ3 = Cexp(i*k₃*x) + 0
the condition I have is that there won't be any particles going from region 3 in the negative x direction right so D is zero.
Then I get F and G in terms of C and can use those relations to get A in terms of C also.
Then you can get
T^-1= (k1)(A*)(A)/[(k3)(C*)(C)]
by dividing the probability currents like you said to do (which no, I didn't know about before so thank you). So did I get T correctly?
After all that I subbed in my values for A and A* and T=1/2 and solved for "a" (I had defined the barrier to go from x=0 to x=a) and got
a=1/k2 * ArcSin(±i)

That seems like kind of a strange width to have. What are your thoughts? See any obvious errors I've made?
Thanks again for all your help!

 

[DelcrossA]

So I have
Ψ1 = Aexp(i*k₁*x) + Bexp(-i*k₁*x)
Ψ2 = Fexp(i*k₂*x) + Gexp(-i*k₂*x)
Ψ3 = Cexp(i*k₃*x) + 0

This is not quite right.
Your solution for the wave-function inside the potential is should not be a plane wave - that would imply that the electrons are freely propagating through a forbidden zone. From the TISE, you have V - E > 0, so you do not get oscillatory behavior. Also, why do you have a k₁ and a k₃? Check again where they come from in the TISE.

I'm not really sure what you did to find A and C, so i can't tell you what problems there are with them. I'll just say that to find them, did you apply continuity conditions to the wavefunctions in each region?

 

[Raynor49]

Ya ok, this is the other issue, I've never dealt with a situation where the total energy E is less than V in a region as it is in region 2 of this case, so how do I get Ψ2?
k1 = k3 = √(2mE) / ħ and k2 = √(2m(E-V)) / ħ
where E = 5 and V = 10. Lastly, I found A in terms of C through substitution and solving of the three equations you quoted, where I said ψ1=Ψ2 for x=0 and Ψ2=Ψ3 for x=a, as well as equating their first derivatives with respect to x. Then when I went to solve the T^-1 equation with A and C I mentioned in my previous post, I subbed in my value for A and the C's cancelled.

 

[DelcrossA]

Remember you have a -ħ²/2m. Note the minus sign, so that you get -(E-V)=V-E > 0. This makes your solution not complex for Ψ₂.

And yes, they way you described finding A and C is correct. You will get something with along the lines of T^-1 = 1 + α sinh²(β a), where α and β are constants.

 

[Raynor49]

OH so
Ψ2 = Fexp(k2 * x) + Gexp(-k2 * x) where k2 = -√(10*m) / ħ
and then I just do more or less the same process I did earlier when I had Ψ2 as a complex function?

 

[Raynor49]

Ok now the main problem I am having is simplification. Because the second wave function wasn't complex, I got
A = C * exp(i*k3*a)/4 * [ exp(a*k2)*(1/k2 + i/k1)*(k2-i*k3) + exp(-a*k2)*(1/k2-i/k1)*(k2+i*k3) ]

which I can't simplify using cos and sin since the exponents inside the square brackets aren't complex.. and as you can imagine, trying to solve
T^-1 = (k1/k3) ((A*) (A)) / ((C*)(C))
is kind of absurd with the A I currently have. Does it still look like I'm doing this right or am I missing something to simplify this?

 

[DelcrossA]

Yes, your expression for A is correct! But, as I've hinted at before, $k_1\ =\ k_3$
That will simplify things greatly, and don't forget that the normal exponential can be expressed as the hyperbolic trig functions. cosh(x)=cos(ix) for example

 

[Raynor49]

Awesome, ok. Sorry for the late reply, been having to deal with midterms. So k1=k3, can I also say that k2=i*k1 ? Since

k1 = k3 = √(2mE) / ħ and k2 = √(2m(E-V)) / ħ
where E = 5 and V = 10.

I thought k2=i*k1=i*k3 but when I sub that in it ends up cancelling out everything and doesn't get me a function for T with the width.. :(
is there a reason I can't say k2=i*k1?

 

[DelcrossA]

Awesome, ok. Sorry for the late reply, been having to deal with midterms. So k1=k3, can I also say that k2=i*k1 ? Since

I thought k2=i*k1=i*k3 but when I sub that in it ends up cancelling out everything and doesn't get me a function for T with the width.. :(
is there a reason I can't say k2=i*k1?

You cannot equate $k_2$ with $k_3 \ or\ k_1$ in general.
your radical is flipped for $k_2$ it should be (V - E) - you chose your wavefunction in region 2 knowing $k_2$ is real. Just plug in numbers after you've simplified everything that way your solution works for all V > E. This is not a requirement, just good practice.

 

[Raynor49]

Ok cool, I think I got it. Thanks so much for all your help! :)