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Scattering Within Rectangular Waveguide With A Step Obstacle

  1. Aug 8, 2014 #1
    Dear PF:

    I'm currently working in a problem that has had me stranded for several weeks now. The problem reads as follows:

    (See attachment)

    Consider a beam of quantum particles (that is, the particles are small enough to exhibit non-negligible quantum effects) that propagates through a two-dimensional waveguide of width [itex] H [/itex] from [itex]x=-\infty [/itex] to [itex]x=+\infty [/itex]. At [itex] x=0 [/itex] the particles encounter a step of height [itex]0<H_0<H[/itex]. All walls are impenetrable. Calculate the reflection and transmission coefficients.


    The potential within the waveguide can be described as:

    \infty & \text{at} \ AO, OH_0, H_0D, CE, OF\\
    0 & \text{elsewhere}

    A particular solution I worked out was:

    A_1^+\sin k_1x\sin\frac{\pi y}{H}, & x<0, 0<y<H_0 \\
    A_1^+e^{ik_1x}\sin \frac{\pi y}{H}+ \sum\limits_mA_m^-e^{-ik_mx}\sin\frac{m\pi y}{H},& x<0, H_0<y<H\\
    \sum\limits_mB_m^+e^{ik'_mx}\sin\frac{m\pi(y-H_0)}{H-H_0}, & x>0, H_0<y<H

    where [itex] k_1=\sqrt{K^2-(\pi/H)^2}, k_m = \sqrt{K^2-(m\pi/H)^2}, k'_m=\sqrt{K^2-(m\pi/(H-H_0))^2}[/itex], [itex]E=\hbar^2K^2/2m[/itex].

    The wavefunction above must satisfy the following boundary conditions:

    [tex] \psi(AO, OH_0, H_0D, CE, OF)=0 ,

    From the above, I can calculate the constants [itex]A[/itex] and [itex]B[/itex], but all I get is nonsense. In particular, the above solution is not continuous along the boundary [itex]BH_0[/itex], but as hard as I try, I cannot make a satisfactory modification such that this discontinuity is healed.

    What am I doing wrong? Could someone direct me to a similar problem? I'm sure there has to be a treatise for presence of steps in rectangular waveguides, but I can't seem to find any.

    Thanks in advance.

    Attached Files:

    • step.jpg
      File size:
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    Last edited: Aug 8, 2014
  2. jcsd
  3. Aug 8, 2014 #2
    Hi skujesco2014!
    It looks like for the solutions in regions I and II to be continuous along BHo you need the same solution for both regions with the reflected wave perhaps modified by a exp -(y-Ho) type term. It's just a hunch as I haven't looked at a waveguide problem before.

    I'm not sure why you need to introduce the summations where you have though. I thought it was implicit anyway In a general solution.
  4. Aug 8, 2014 #3
    Hi, Jilang. Thanks for your input.

    If you notice, the incident waves are the form [itex]A_1^+e^{ik_1x}\sin(\pi y/H)[/itex] but waves in region I are fully reflected whereas in region II they're partially reflected. The sums are introduced because I'm expanding in the basis of the complete sets [itex]\{\sin m\pi y/H\}[/itex] and [itex]\{\sin m\pi (y-H_0)/(H-H_0)\}[/itex] for [itex]x<0[/itex] and [itex]x>0[/itex] respectively. I could do what you say, but the functions would still not match essentially because of the presence of the [itex]e^{ik_1x}[/itex] term.
  5. Aug 8, 2014 #4
    Yes I see what you are saying, I think both solutions need to include the reflected term. The waves will bounce off the the bottom wall and head upwards, no?
  6. Aug 8, 2014 #5
    I just reread your last post. Are your first and second expressions the wrong way round then for the segments?
  7. Aug 8, 2014 #6
    The reflected wave from the [itex]0<y<H_0[/itex] region is included: the incoming wave is [itex]A_1^+e^{ik_1x}\sin(\pi y/H)[/itex] and when hitting the step it is 100% reflected; thus, the reflected wave consists of [itex]\sum_m A_m^-e^{-ik_mx}\sin(m\pi y/H)[/itex]. Since the wavefunction has to be zero on [itex]OH_0[/itex], we conclude [itex]A_m^-=-A_1^+\delta_{m1}[/itex]. Combining exponentials we get [itex]\sin k_1x[/itex].

    The wavefunction as it was given satisfies being zero along all the boundaries (i.e., all the walls). If the BCs (boundary conditions) are implemented on the interface [itex] H_0H[/itex] (see attachment) one can determine expressions for the coefficients [itex]A_m^-[/itex] and [itex]B_m^+[/itex] in terms of [itex]A_1^+[/itex] (I can show you this math if you want).

    Given a number of particles coming in (so far, there is only one particle incoming in quantum state 1, but this is arbitrary), I'm interested in computing an expression that gives me the number of particles transmitted through the interface [itex] H_0H[/itex] and reflected. I have, in fact, computed these expressions, but I obtain a very strange answer. I discovered that [itex]\psi[/itex] is not continuous along [itex]BH_0[/itex], thus my expressions are wrong. I can't, however, come up with a satisfactory wavefunction for the problem. That's where I'm stuck.
    Last edited: Aug 8, 2014
  8. Aug 9, 2014 #7
    Have you tried including a third term in region II for a backwards reflection at the second frequency?
  9. Aug 9, 2014 #8
    That term is [itex]\sum_m A^−_m e^{−ik_mx}\sin(m\pi y/H)[/itex].
  10. Aug 10, 2014 #9
    I meant a sin mπ(y-Ho)/(H-Ho) type term.
  11. Aug 10, 2014 #10
    If we do so, then the wavefunction wouldn't be zero on the [itex]AO[/itex] wall. :frown:
  12. Aug 10, 2014 #11
    It would be problematic in region I, but would it work just in region II?
  13. Aug 10, 2014 #12
    Looking at it holistically, I think there is going to be total reflection unless n= MH/(H-Ho). Only waves of certain frequencies are going to make it through. Is this a filter?
  14. Aug 10, 2014 #13
    I don't understand what you meant there.

    How do you arrive to this conclusion?
  15. Aug 11, 2014 #14
    What I am thinking is for continuity at the line H Ho you need to fit an integer number of half wavelengths between H and Ho and to also be able to fit an integer number of half wavelengths between H and O. This will only be possible for certain special wavelengths.
  16. Aug 11, 2014 #15
    I see what you mean now.

    So, because we have to have an integer number of half-waves in the right-half of the waveguide, only those will that have a node along [itex] BH_0[/itex] will make it through. That would imply then that the wavefunction is forced to have a node along the [itex]BH_0[/itex] line, would you agree?
  17. Aug 11, 2014 #16
    Yes, that's the way I see it, anything else would be reflected. I can't see how else you avoid the discontinuities.
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