Reflection amplitude at potential step

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Martin89
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Homework Statement


Typeset problem statement for a particle beam of energy E and mass m meeting a one-dimensional potential well: V(x) = 0 for x<0 (region I), -V0 for 0≤x≤L (region II), 0 for x>L (region III), with parts (d), (e) and (f) listed.


The Attempt at a Solution



Handwritten working for a quantum well: a sketch of a potential well with regions 1, 2 and 3 and width L, then wavefunction forms for each region, and boundary-condition equations at x=0 and x=L for continuity of psi and psi'.

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Hi All,

I'm having trouble answering part (f) of the above question. I have managed parts (d) and (e) fine but am not sure how to proceed with part (f). I am pretty sure that the amplitude of the reflected wave in region 1 will be zero but I don't know how to show it.

Thanks in advance for any help!
 
Attachments
  • Typeset problem statement for a particle beam of energy E and mass m meeting a one-dimensional potential well: V(x) = 0 for x<0 (region I), -V0 for 0≤x≤L (region II), 0 for x>L (region III), with parts (d), (e) and (f) listed.
    QM.png
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  • Handwritten working for a quantum well: a sketch of a potential well with regions 1, 2 and 3 and width L, then wavefunction forms for each region, and boundary-condition equations at x=0 and x=L for continuity of psi and psi'.
    20180526_165635.jpg
    18.6 KB · Views: 966
Your solutions to (d) and (e) look correct, except for one small detail--I can't tell if the right-hands side of your last equation is ##k_1 Te^{ik_1 L}## or ##k_2 Te^{ik_2 L}##. Obviously, one is correct and one is not. :)

If ##Lk_2 = n \pi##, then ##e^{iLk_2} = e^{-iLk_2} = \pm 1##. This should allow you to rewrite the boundary conditions at ##x = L## in a form that's easier to compare to the boundary conditions at ##x = 0##---in fact, you should be able to eliminate ##C## and ##D## from the equations involving ##A## and ##B## altogether (although ##T## will still be present). If you can get this far, it should be pretty clear what to do next. (Hint: your intuition is correct. ;))
 
VKint said:
Your solutions to (d) and (e) look correct, except for one small detail--I can't tell if the right-hands side of your last equation is ##k_1 Te^{ik_1 L}## or ##k_2 Te^{ik_2 L}##. Obviously, one is correct and one is not. :)

If ##Lk_2 = n \pi##, then ##e^{iLk_2} = e^{-iLk_2} = \pm 1##. This should allow you to rewrite the boundary conditions at ##x = L## in a form that's easier to compare to the boundary conditions at ##x = 0##---in fact, you should be able to eliminate ##C## and ##D## from the equations involving ##A## and ##B## altogether (although ##T## will still be present). If you can get this far, it should be pretty clear what to do next. (Hint: your intuition is correct. ;))
Handwritten derivation applying boundary conditions at x = 0 and x = L: e^(iknL) = e^(-iknL) = ±1, then C + D = Te^(ik2L), k2(C - D) = k2Te^(ik2L), A + B = Te^(ik2L), k1(A - B) = k2Te^(ik2L), leading to B = ((k1 - k2)/2k1) Te^(ik2L).


Thanks for the reply! I think I'm still missing something. Should k1 = k3 at x=0 or something?
 
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  • Handwritten derivation applying boundary conditions at x = 0 and x = L: e^(iknL) = e^(-iknL) = ±1, then C + D = Te^(ik2L), k2(C - D) = k2Te^(ik2L), A + B = Te^(ik2L), k1(A - B) = k2Te^(ik2L), leading to B = ((k1 - k2)/2k1) Te^(ik2L).
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Martin89 said:
View attachment 226253

Thanks for the reply! I think I'm still missing something. Should k1 = k3 at x=0 or something?
You have not yet imposed that the wave functions must satisfy Schrödinger's equation. Impose that in regions I and III, what does that tell you about k1 and k3? (note that the energy of the particle is fixed and is assigned the value E in the question)
 
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nrqed said:
You have not yet imposed that the wave functions must satisfy Schrödinger's equation. Impose that in regions I and II, what does that tell you about k1 and k3? (note that the energy of the particle is fixed and is assigned the value E in the question)

I've done as you said but I'm not sure what it shows me. K1 and K2 are different due to the potential in region 2. Looking at the problem physically, the Schrödinger equation must be identical in regions 1 and 3, and therefore K1=K3. Is this a correct assumption?
 
Martin89 said:
I've done as you said but I'm not sure what it shows me. K1 and K2 are different due to the potential in region 2. Looking at the problem physically, the Schrödinger equation must be identical in regions 1 and 3, and therefore K1=K3. Is this a correct assumption?
Well, it would be better to show things explicitly. What condition do you get when you plug in the wave function of region I into Schrödinger's equation?
 
nrqed said:
Well, it would be better to show things explicitly. What condition do you get when you plug in the wave function of region I into Schrödinger's equation?
Handwritten derivation of the region 1 time-independent Schrödinger equation: -ħ²/2m d²ψ/dx² = Eψ, with ψ = Ae^(ik1x) + Be^(-ik1x) differentiated twice, leading to E = k1²ħ²/2m, so k1 = √(2mE)/ħ.


That is the condition for K1. The expression for K3 should be identical right?
 
Attachments
  • Handwritten derivation of the region 1 time-independent Schrödinger equation: -ħ²/2m d²ψ/dx² = Eψ, with ψ = Ae^(ik1x) + Be^(-ik1x) differentiated twice, leading to E = k1²ħ²/2m, so k1 = √(2mE)/ħ.
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Dr Transport said:
do the calculation...the answer will pop right out.

Ok so I've done the calculation and K1 does indeed equal K3. Thanks guys!