Quantum tunneling: T(E) graph for a potential barrier diagram

AI Thread Summary
The discussion focuses on the behavior of transmission (T) in quantum tunneling through potential barriers, specifically analyzing V(x) diagrams. It suggests that for a potential barrier, energy must exceed a certain threshold (V2) to initiate transmission, with T approaching 1 asymptotically as energy increases, while also experiencing resonances. The second potential scenario is expected to show a continuous increase in transmission from E=0, also approaching T=1 with resonances. Participants emphasize the importance of applying equations related to energy and potential barriers to understand the graph's behavior better. Overall, the conversation highlights the need for both reasoning and calculations to grasp the complexities of quantum tunneling.
Theodore0101
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Homework Statement
Problem: An even flux of particles is incident from left towards potentials as shown in the diagram (is below). Discuss what effects that can occur, especially with focus on what energies something 'new' occurs and how a function for the transmission depending on the particle energy, T(E) will look like
Relevant Equations
Potential barriers and transmission
69b5afaf8869408a8ab0560b2d75daac_A.jpeg


This is the V(x) diagrams and what I am thinking (really not sure though) is that for the first one you the energy has to reach V2 before it can start transmitting and the graph can take off from T=0, since there is an increase in energy potential that is V2. And as the energy increases, the transmission will start to approach 1 as an asymptote, since there will always be some amount of reflection. But what kind of function the graph takes after I don't know (logarithmic, exponential, etc), other than that it will start approaching 1 for big values of E and if I've understod things correctly, the graph will have resonances. Also, if there is anything that happens anywhere else in the graph, such as in E=V1, I don't know.

The second one I think will constantly increase since there is no step potential above the x-axis, so my thinking is that it constantly increases from E=0 onwards and again has an asymptote at T=1 and resonances. But if there is any point which something happens or how the function more specifically behaves, I don't know.

Any help is appreciated. Thanks!
 
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Hello Theodore, ##\qquad## :welcome: ##\qquad## !

Doing homework exercises usually involves applying the homework equations to the exercise at hand. Your homework equations are not mentioned, but I agree that the approach (like here) is sensible. What, exactly, is your problem in converting the V > 0 barrier situation to V < 0 ?

(reply inspired by PF culture as found iin the guidelines)
 
Hello BvU.

Thank you for your reply. Sorry for not giving any equations, I was thinking that this problem is meant to be solved with reasoning at what happens at different potential barriers, instead of calculations. But the equations i have with T depending on E are T=exp(-2a*√(2m(V0-E)/h^2)) and T=4k1k2/(k1+k2)^2 (k1=√(2mE/h^2) and k2=√(2m(E-V0)/h^2)).

I am guessing that the negative potential step will make the graph grow faster in some way, but I am feeling unsure about how it does, assuming it does. I'm not finding any parallell to this kind of problem in my textbook, and it isn't clear to me how you are supposed to think in this type of exercise.
 
Theodore0101 said:
T=exp(-2a*√(2m(V0-E)/h^2)) and T=4k1k2/(k1+k2)^2 (k1=√(2mE/h^2) and k2=√(2m(E-V0)/h^2))
What are these ? Don't see no V0 in post #1.

Theodore0101 said:
meant to be solved with reasoning at what happens at different potential barriers
Fair enough, but if you have no experience with such things, perhaps a few calculations may be needed. For example: from 'flux of particles incident from the left' you can deduce ##E > 0##. So for ##x<0## you have ##\Psi_L(x) = A_r e^{ik_0x} +A_l e^{-ik_0x}## (##k_0^2 = 2mE/\hbar^2##). Since every discontinuity may cause reflection, ##A_l## is probably ##\ne 0##.

And if your
Theodore0101 said:
make the graph grow faster in some way
means that you think that there ##k^2 = 2m(E-V_1)/\hbar^2## (##V_1<0##), I tend to agree.

Remainder of the reasoning goes as in the link. Some work unavoidable, I'm afraid.
 
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