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Quantum Mechanics

  1. Oct 27, 2010 #1
    b]1. The problem statement, all variables and given/known data[/b]

    Consider a particle of mass m incident on a rectangular barrier with a potential

    V(x) = V0 for 0 [tex]\leq[/tex] x < w
    V(x) = 0 otherwise,

    where V0 is a positive constant. The particle has an energy E > V0. Find expressions in terms of E and V0 for the magnitudes of the associated wavenumbers inside and outside the barrier, ki and ko, respectively.

    2. Relevant equations

    3. The attempt at a solution

    E = p/2m = (h-cross*k)2/2m

    Therefore, E = [tex]\frac{(h-cross*k_{o})^{2}}{2m}[/tex] gives k[tex]_{o}[/tex] and
    E - V[tex]_{o}[/tex] = [tex]\frac{(h-cross*k_{i})^{2}}{2m}[/tex] gives k[tex]_{i}[/tex].

    Am I right?
     
  2. jcsd
  3. Oct 27, 2010 #2

    fzero

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    Looks good to me.
     
  4. Oct 27, 2010 #3
    Thanks!

    Next I have been asked the following:

    For a particle incident from the left, write down appropriate general forms of wavefunctions
    which satisfy the time-independent Schrodinger equation for each of the
    three regions, namely x < 0, 0 [tex]\leq[/tex] x < w, and w [tex]\leq[/tex] x.

    This is my answer:

    x < 0 : u(x) = Aeik0x + B-ik0x

    0 [tex]\leq[/tex] x < w : No idea

    w [tex]\leq[/tex] x : No idea

    Thanks in advance for any help.
     
  5. Oct 27, 2010 #4

    fzero

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    Try using sums of [tex] e^{\pm ikx}[/tex] in both locations. Use the Schrodinger equation to relate k in each region to ko or ki and boundary conditions to relate the coefficients to A and B.
     
  6. Oct 27, 2010 #5
    Thanks!

    So, to begin, I'll write the solution as follows:

    0 [tex]\leq[/tex] x < w : u2 = Ceikix + De-ikix

    w [tex]\leq[/tex] x : u3 = Eeikix + Fe-ikix.

    I don't think any of the coefficients are zero, so I'm guessing the psi-squared is sinusoidal in all three regions.

    I don't understand what you have meant when you said "Use the Schrodinger equation to relate k in each region to ko or ki ".

    I'll express C, D, E and F in terms of A and B later.

    Anyway, to continue, the next part of the question goes as follows:

    For the particular case where the particle energy is [tex] E = V_{0} + \frac{(hcross*\pi)^{2}}{2mw^{2}} [/tex] ,
    calculate the wavenumber k[tex]_{i}[/tex] , and hence wavelength, of the wavefunction in terms of the barrier width w.

    The wavefunction and its derivative must be continuous at x = 0 and x = w. For
    a particle with the energy given in part 3 above, use these continuity requirements
    to find an expression for the ratio of the transmitted wavefunction amplitude to the
    incoming wavefunction amplitude.

    Find the value of the barrier transmission coefficient for this energy.

    This is how I've answered it.

    Firstly, substituting the given expression for E into the formula for ki gives a value for ki of pi/w.

    Next, I have used the boundary conditions as follows.

    u1 = u2 at x = 0 : A + B = C + D
    Same for derivatives : A - B = [tex]\frac{k0}{ki}[/tex] (C - D)

    u2 = u 3 at x = w: C - D = Eeikix + Fe-ikix
    Same for derivatives : C + D = [tex]\frac{k0}{ki}[/tex] Eeikix + Fe-ikix


    I'm not sure what to do next.

    PS: I'm sorry if this is along question. There are only three more parts to it.
     
  7. Oct 27, 2010 #6

    fzero

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    u2 and u3 must be solutions to the Schrodinger equation in the appropriate regions. You need to check this and, if necessary, modify them so that they are. Hint: u3 is not a solution.

    As far as whether or not all coefficients are nonzero, you need to consider the physical situation of this being a problem about an incident particle on the barrier. After you decide whether the incident particle is right or left moving, you will need to decide which boundaries allow reflection and whether you will have right and left moving components in every region.

    The 2nd equation is wrong, you should recheck your algebra and get the ratio of wavenumbers correctly. When you've got the correct equations you can solve for C and D in terms of A and B.

    Since u3 is not a correct solution, you'll need to redo these when you get that sorted out. In your first equation you haven't evaluated the RHS at x = w. When you sort the solution and algebra out, you will be able to solve for E and F in terms of C and D, however you still need to sort out whether all of the A, B,...F are nonzero on physical grounds.
     
  8. Oct 27, 2010 #7
     
  9. Oct 27, 2010 #8

    fzero

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  10. Oct 27, 2010 #9
    Thank you very much!

    So, I understand that everything so far has been correct.

    I'm pretty sure C and D are both non-zero, aren't they? :confused:

    So, to move on, I have calculated E/A = [tex]\frac{2k^{2}_{0}}{k^{2}_{0}+k^{2}_{i}} e^{-ik_{0}w}[/tex].

    This means that the transmission coefficient = square of the modulus of E/A = [tex]\frac{2k^{2}_{0}}{k^{2}_{0}+k^{2}_{i}}^{2}[/tex].

    I hope this is all right.

    I'd like to move on to the last part of the question:

    " Find the ratio of the probability density at the point x = w/2 to the incoming wavefunction
    probability density and state whether the magnitude of the ratio is greater
    than or smaller than one.

    Comment on the physical interpretation of the answer to part 6 above. "

    This is my attempt at it.

    [tex] \frac{ |\psi_{\tex{x = w/2}}|^2}{|\psi_{\tex{x = 0}} |^2} = \frac{ | Ce^{ik_{i}w/2} + De^{-ik_{i}w/2} |^2}{| A + B |^2}[/tex]

    I'm not sure what I should do next ?:confused:
     
  11. Oct 27, 2010 #10

    fzero

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    Since the incoming wavefunction is only the term proportional to A, the "ratio of the probability density at the point x = w/2 to the incoming wavefunction
    probability density" should be |E|^2/|A|^2. From the boundary conditions, you should also find that

    [tex]
    |E|^2 = | Ce^{ik_{i}w/2} + De^{-ik_{i}w/2} |^2
    [/tex]

    once you simplify the expression on the right-hand side and use the relationships between the coefficients that you found earlier.
     
  12. Oct 27, 2010 #11
    Erm... I don't really understand why the numerator has to be the square of the modulus of E. I mean, we are evaluating the wavefunction at x = w/2, which means that we should use [tex]u = Ce^{ik_{i}x} + De^{-ik_{i}x}[/tex] for the numerator, shouldn't we?
     
  13. Oct 27, 2010 #12
    Or do you mean that we start with [tex]u = Ce^{ik_{i}x} + De^{-ik_{i}x}[/tex] in the numerator and find that it equal to E at x = w/2?
     
  14. Oct 27, 2010 #13
    Ok... I have thought about it and I think I'm pretty sure that's what you meant. Ok, so we find that the ratio is |E|^2/|A|^2. This means that the ratio is [tex]\frac{2k^{2}_{0}}{k^{2}_{0}+k^{2}_{i}}^{2}[/tex]. This is less than 1. (Am I right?)

    To finish off, I have asked to comment on the physical interpretation of the previous answer. Well, I'm thinking that the since the ratio half-way into the barrier is the same as the transmission coefficient, there is no chance of a particle returning to x < 0 if it crosses x =w/2 and moves to the right. Am I right?
     
  15. Oct 27, 2010 #14

    fzero

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    No it was a mistake on my part, I didn't read carefully enough and thought they were talking about x = w.

    My point about A vs A+B stands, so I think you want to simplify

    [tex]
    \frac{ | Ce^{ik_{i}w/2} + De^{-ik_{i}w/2} |^2}{| A |^2} = \frac{ ( Ce^{ik_{i}w/2} + De^{-ik_{i}w/2} )( Ce^{-ik_{i}w/2} + De^{ik_{i}w/2} )}{| A |^2} .
    [/tex]
    You'll probably want to express the result in terms of trig functions.
     
  16. Oct 27, 2010 #15
    Well, we know from before that [tex]k_{i}w = \pi[/tex]. Therefore

    [tex] \frac{ |\psi_{\tex{x = w/2}}|^2}{|\psi_{\tex{x = 0}} |^2}
    = \frac{ | Ce^{ik_{i}w/2} + De^{-ik_{i}w/2} |^2}{| A + B |^2}
    = \frac{ | Ce^{i\pi/2} + De^{-i\pi/2} |^2}{| A |^2}
    = \frac{ | i (C - D) |^2}{| A |^2}
    = \frac{ | Ee^{ik_{0}w}e^{i\pi/2} |^2}{| A |^2}
    = \frac{ | E |^2}{| A |^2}
    [/tex].

    Am I right?
     
  17. Oct 27, 2010 #16

    fzero

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    That looks right, but I was trying to check the physical interpretation and found that your

    [tex]
    \frac{E}{A} = \frac{2k^{2}_{0}}{k^{2}_{0}+k^{2}_{i}} e^{-ik_{0}w}
    [/tex]

    has a mistake. I find

    [tex]
    \frac{E}{A} = \frac{2k_{0} k_i}{k^{2}_{0}+k^{2}_{i}} e^{-ik_{0}w}.
    [/tex]

    This is pretty important if you want to compare the probability of transmission to the probability of reflection.
     
  18. Oct 28, 2010 #17
    Thanks!

    So how do I actually physically interpret the result?
     
  19. Oct 28, 2010 #18
    And as an aside, I got your result in the end, but I don't understand why we have to compare the probability of transmission to the probability of reflection. I mean, A does not signify reflection. B does.
     
  20. Oct 28, 2010 #19

    fzero

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    The particle is either reflected or transmitted and there's an associated mathematical identity between the probabilities of transmission and reflection that has to be satisfied. I was just using that as a consistency check. I'm surprised that it isn't part of the problem.
     
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