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Homework Help: Particle in potential well

  1. Jun 6, 2008 #1
    If you think this will get more replies in the intro phys please let me know and i will move it:

    A particle of total energy 9Vo is incident from the -x axis on a potential given by:
    V \left\{8Vo, x<0\right\}
    \left\{0, 0<x<a\right\}
    \left\{5Vo, x>a\right\}
    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]


    [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:
    this gives me:
    Solving the above equations with only constants, i find the constants in terms of A:
    B = \frac{A(k_{1}-k_{2}}{k_{1}+k_{2}}
    C = \frac{2k_{1}A}{k_{1}+k_{2}}
    The transmission coefficient is:
    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}}
    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:
    T_{2} =\frac{4k_{2}k_{3}}{(k_{2}+k_{3})^{2}}
    If so, will my final transmission probability be the product of the two?
    Any help is appreciated!
  2. jcsd
  3. Jun 6, 2008 #2
    I'm no expert, so don't take what I say as absolute truth. But I'll try to help anyway.

    I don't think you can do this. Set up all of your boundary conditions, then solve the resulting equations for the coefficients simultaneously.

    Otherwise, you're approach looks good.

    And yes, In the end, the total transmission is the product of the two.
  4. Jun 7, 2008 #3
    If your method is correct, then the width of the middle region doesn't matter: you get the same transmission no matter how wide. I don't think this is correct physically. Think of tunneling situations.

    I think the easier way to do these kinds of problems is to start at the second interface, assuming an arbitrary value for the transmitted wave, and work backwards.
  5. Jun 7, 2008 #4

    Dr Transport

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    Gold Member

    Continuity of the wave function and its derivative at each potential discontinuity is the only way to solve this problem correctly. If you set [itex] F = 0 [/itex], you will have 4 equations and 5 variables to solve for, so if you set you incident wave to be unity, you get 4 equations and 4 unknowns.
  6. Jun 7, 2008 #5
    Thank you for all the help! I dont think the width of the well should matter at all though, because we have a complex exponential everywhere so the wave function wont be decaying at all. I will solve it again and put it up, solving all of the equations simultaneously. once again, thank you guys for the help.
  7. Jun 7, 2008 #6
    I just did a quick run through of the problem on maple with these conditions, but i actually have a concern about this. When I solve the first set of conditions to get A + B = C + D, I am assuming that wave D, the wave reflected at x = a, is already there at the first potential interface at x = 0 which seems wrong. If i do this, i Ill be matching up wave D with A and B which will change the amplitudes of the wavefunctions while it seems to me that D is irrelevant to the first boundary condition at first. The first wave that will emerge to the right of x = 0 will just be exp(ik2x), and then after it hits the interface at x = a, i will get wave D, exp(-ik2x). Am i just thinking about this wrong? Thanks for the help.
  8. Jun 7, 2008 #7
    EDIT: i guess more specifically, should I just pretend that all the wave functions are present at the start, or should I do it so that first, the wavefunction originates at x<0, then meets the interface at x = 0, then it reflects and transmits, then the transmitted ray hits the interface at x = a, then it reflects and transmits. thanks for any help
  9. Jun 7, 2008 #8
    I understand why it seems strange to include the D wave in the boundary condition at x=0, but that is the correct way to do it. I wish I could explain why, but I don't have a firm understanding myself. Just remember that you are calculating the coefficients AFTER the scattering has occurred, and imagine that you have taken a time-lapsed picture of what has happened. In that picture, you will see the D part at x=0. Anyway, I think it's something like that. I may be totally of in my interpretation.

    Do it like Dr. Transport says. Set up all the boundary conditions at once and solve for everthing simultaneously in terms of the incident wave amplitude A.
  10. Jun 7, 2008 #9
    Yes. Althought it may not satisfy your desire to understand exaclty what is going on, it will give you the correct answer.
  11. Jun 7, 2008 #10
    Don't forget that the wave associated with D is phase-shifted because it had to propagate back to get to the first interface.

    I think it's possible to solve for all the waves at once but it's a lot easier to do what I suggested and arbitrarily give the final outgoing wave a value of 1, and then work backwards. But don't forget the phase shifts.
  12. Jun 7, 2008 #11
    Thanks for the help guys. How would I deal with the phase shift? Is it not enough to just have a negative on the argument of the exponential, or would I have to add pi to the argument (not sure if this is right)?
  13. Jun 7, 2008 #12
    Well, you've shown the right way to calculate the C,D, and E coefficients except you've done it for the A, B and C terms instead. But your method is OK, so starting from the right hand side you're able to calculate what happens at the second boundary. Now work backwards to the left...at x=a, you have the C and D waves in phase with each other. But in physical terms, they diverge as you move to the left: if your wave number in the middle zone is k2 (I'm too lazy to code in k-subscript-2!!), then k2*a gives you a positive phase shift for one of them and a negative phase shift for the other.

    Now when you rewrite your boundary conditions for x=0, you'll have C and D known and your unknowns will be A and B. So you should have enough information to solve for both of them.
    Your A value will be out of whack because you'd like it to be unity, so you just divide everything out to renormalize.
    Last edited: Jun 7, 2008
  14. Jun 7, 2008 #13
    Thanks for the help, i will try this bellow:
    Ok, I solve the equations satisfying continuity at x = a:

    Solving these two i get:
    [tex]C = \frac{De^{-ik_{2}a}(k_{3}-k_{2})}{e^{ik_{2}a}(k_{2}-k_{3})}[/tex]
    I will set E = 1 (for now)

    Now, i solve my equations at x = 0:
    [tex] A + B = C + D[/tex]
    [tex] k_{1}(A-B) = k_{2}(C-D)[/tex]
    [tex]B = A - \frac{k_{2}}{k_{1}}(C-D)[/tex]
    plugging this in i get:
    [tex] 2A = \frac{k_{2}}{k_{1}}(C-D)+C+D[/tex]
    I now plug in my value for D (which I get by rearranging the answer for C in terms of D)
    D = \frac{Ce^{ik_{2}a}(k_{2}-k_{3})}{e^{-ik_{2}a}(k_{3}-k_{2})}
    2A = \frac{k_{2}}{k_{1}}C(1-\frac{e^{ik_{2}a}(k_{2}-k_{3})}{e^{-ik_{2}a}(k_{3}-k_{2}})+C(1+\frac{e^{ik_{2}a}(k_{2}-k_{3})}{e^{-ik_{2}a}(k_{3}-k_{2})})[/tex]

    is this right to this point? I can solve for the transmission coefficient at x = 0 now, since I only need A in terms of C (or visa versa), is that right?
    Last edited: Jun 7, 2008
  15. Jun 8, 2008 #14
    I'm quoting here from your very first post in which you solved for the transmitted and reflected waves at the first transition, where x=0. If I compare this to what you've done in your most recent post, working now at the second transition (x=a) you've got an expression that looks like your first formula quoted above but you don't have anything that corresponds to your second formula.

    If you look carefully at your formula for C in terms of D and you should be able to see how it comes directly from the formula for B in terms of A which I've quoted above. So by analogy, you should be able to write out a formula for E in terms of C using your second formula.

    Then, since you've set E to unity, you can solve explicitly for C, and then express both C and D in terms of the k's and the exponential propagation factors.

    No, you don't want separate transmission coefficients at both barriers, you want one overall transmission coefficient which is E in terms of A. What you should have so far is an explicity formula for A which came from setting E=1. Just invert this value and you have the amplitude for the wave to get from A to E.
  16. Jun 8, 2008 #15
    Thanks again for all of the help Marty, and everyone else, I really appreciate it. Also, i believe the formulas I gave first for B and C in terms of A are incorrect because at that point I wasn't taking into consideration the reflected wave in the area 0<x<a. But, i unfortunately realized that I messed up the very first solution where I solve for C. The constants turn out to be:
    C = \frac{De^{-ik_{2}a}(k_{3}+k_{2})}{e^{ik_{2}a}(k_{2}-k_{3})}
    2A = \frac{k_{2}}{k_{1}}C(1-\frac{e^{ik_{2}a}(k_{2}-k_{3})}{e^{-ik_{2}a}(k_{3}+k_{2}})+C(1+\frac{e^{ik_{2}a}(k_{2}-k_{3})}{e^{-ik_{2}a}(k_{3}+k_{2})})
    Now, i solve for C in terms of E ( where E = 1):
    C = \frac{e^{ik_{3}a}e^{-ik_{2}a}(k_{3}+k_{2})}{e^{ik_{2}a}(k_{2}-k_{3})+(k_{3}+k_{2})}
    2A = \frac{k_{2}}{k_{1}}(\frac{e^{ik_{3}a}e^{-ik_{2}a}(k_{3}+k_{2})}{e^{ik_{2}a}(k_{2}-k_{3})+(k_{3}+k_{2})})(1-\frac{e^{ik_{2}a}(k_{2}-k_{3})}{e^{-ik_{2}a}(k_{3}+k_{2}})+C(\frac{e^{ik_{3}a}e^{-ik_{2}a}(k_{3}+k_{2})}{e^{ik_{2}a}(k_{2}-k_{3})+(k_{3}+k_{2})})(1+\frac{e^{ik_{2}a}(k_{2}-k_{3})}{e^{-ik_{2}a}(k_{3}+k_{2})})
    2\,A={\frac {k_{{2}} \left( {e^{-ik_{{2}}a}}{e^{ik_{{3}}a}} \left( k_{
    {3}}+k_{{2}} \right) -{e^{ik_{{2}}a}}{e^{ik_{{3}}a}} \left( k_{{2}}-k_
    {{3}} \right) \right) }{k_{{1}} \left( k_{{3}}+k_{{2}}+{e^{ik_{{2}}a}
    } \left( k_{{2}}-k_{{3}} \right) \right) }}+{\frac {{e^{-ik_{{2}}a}}{
    e^{ik_{{3}}a}} \left( k_{{3}}+k_{{2}} \right) +{e^{ik_{{2}}a}}{e^{ik_{
    {3}}a}} \left( k_{{2}}-k_{{3}} \right) }{k_{{3}}+k_{{2}}+{e^{ik_{{2}}a
    }} \left( k_{{2}}-k_{{3}} \right) }}
    And you said that I should just do the total transmission coefficient, so would I have to calculate:
    T_{f} = \frac{v_{3}}{v_{1}}\frac{E^{*}E}{A^{*}A}
    v_{3}\propto k_{3}
    v_{1}\propto k_{1}
    Looking at my equations, though, I don't think this will give me a real answer. Once again, thank you for the help
  17. Jun 8, 2008 #16
    Not exactly. They were good formulas for the case of no incoming wave (D) from the right. So they are a good template for what your formulas should look like at the second interface.

    No, this is where you're going astray. The formula for C in terms of E is just like the "correct" formula you had in your very first post for C in terms of A except with the appropriate phase shifts. You had:

    C= 2k1*A/(k1+k2)

    at the left interface. So at the right interface C becomes your incoming wave and E your outgoing wave. Applying the phase shift you get:


    If you compare this to what you wrote, you'll see it's similar but you've got some extra junk in your version. Once you straighten this out, you should be able to solve for C and D explicitly in terms of the k's.

    Then you can go back to the first interface. In your original posting, you assumed that D was zero and A was 1. And you solved for B and C....two equations in two unknowns. That was OK then but not now. This time you have C and D known and your unknowns are A and B....again, two equations in two unknowns. It should work.

    I wish you wouldn't give up yet. You've done good work on individual sections of this problem but just haven't put it all together yet.
  18. Jun 8, 2008 #17
    I definitely wont be giving up until I figure this out (im not doing it for any class). I'm not sure why my new equation for C is wrong though-- All i did to get it was match the wf and its derivative at the boundaries at x = a. I dont see how you got your equation for E in terms of C so simply. Is there a way to do this from intuition? I noticed when you said you get a phase shift, you multiplied by the complex exponential -- do you always do this in such a case? I dont see why you dont have a bunch of other complex exponential terms in it since x = a there so you have to carry around all that mess. Thanks for helping me through this one.
  19. Jun 8, 2008 #18
    Your formula was right...at least the first one you wrote with D on the RHS. Then you did something to try and eliminate D and you ended up with what looks like a bunch of extra stuff. I don't know quite what went wrong.

    All I did was use the formula for simple transmission which you derived in your very first post, and applied it to C at x=a. But instead of just plugging C into your formula, I had to use c*exp(ik2a) to account for the propagation.

    Someone previously suggested you just set up the equations and solve for four unknowns in four equations. All I'm really doing that's different is I'm breaking up the equations in ways that let me solve for intermediate variables that have physical significance (at least to me). I think that's what you're calling intuition.
  20. Jun 8, 2008 #19
    But to set my my first equations, I set x = 0 which eliminated all of the exponentials -- i wont be able to do this with x = a though which is why i had so much extra junk. but i will work through it again to see where i made mistakes
  21. Jun 8, 2008 #20
    Alrighty, I worked through it again and found a couple mistakes which you pointed out (and I checked my work along with MAPLE). this is what I get for all the constants ( i had maple translate them to Latex so hopefully its in good format):

    A = 1/4\,{\frac {E{e^{-ia{\it k3}}} \left( {\it k2}\,{e^{-ia{\it k2}}}{
    \it k1}+{e^{ia{\it k2}}}{\it k1}\,{\it k2}+{{\it k2}}^{2}{e^{ia{\it k2
    }}}-{\it k1}\,{e^{-ia{\it k2}}}{\it k3}+{\it k2}\,{e^{ia{\it k2}}}{
    \it k3}+{\it k2}\,{e^{-ia{\it k2}}}{\it k3}-{{\it k2}}^{2}{e^{-ia{\it
    k2}}}+{e^{ia{\it k2}}}{\it k1}\,{\it k3} \right) }{{\it k2}\,{e^{ia{
    \it k2}}}{e^{-ia{\it k2}}}{\it k1}}}

    B = 1/4\,{\frac {E{e^{ia{\it k3}}} \left( {e^{ia{\it k2}}}{\it k1}\,{\it
    k2}+{\it k2}\,{e^{-ia{\it k2}}}{\it k1}-{{\it k2}}^{2}{e^{-ia{\it k2}}
    }-{e^{ia{\it k2}}}{\it k1}\,{\it k3}-{\it k2}\,{e^{-ia{\it k2}}}{\it
    k3}-{\it k2}\,{e^{ia{\it k2}}}{\it k3}+{{\it k2}}^{2}{e^{ia{\it k2}}}+
    {\it k1}\,{e^{-ia{\it k2}}}{\it k3} \right) }{{\it k2}\,{e^{ia{\it k2}
    }}{e^{-ia{\it k2}}}{\it k1}}}

    C = 1/2\,{\frac {E{e^{ia{\it k3}}} \left( {\it k2}+{\it k3} \right) }{{
    \it k2}\,{e^{ia{\it k2}}}}}

    D = 1/2\,{\frac {E{e^{ia{\it k3}}} \left( -{\it k3}+{\it k2} \right) }{{
    \it k2}\,{e^{-ia{\it k2}}}}}
    E = 1
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