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Probability Current: from plane wave to wave packet?

  1. Oct 25, 2011 #1
    Hello, (for those in a hurry: the last paragraph contains the essence)

    I'm in my last year of bachelor of physics and following a QM class and as is standard we calculated the reflection R and transmission T coefficients for a plane wave in some potential well/barrier situation.

    Our teacher said "of course, the plane wave is not a good genuine wave function, not being normalizable, but due to superposition we can make a wave packet and the R and T are also applicable to this wave packet."

    I had some objections to this quote and upon inquiring more, he said that for a wave packet described as [itex]\psi = \int g(k) e^{i(kx-\omega(k) t)} \mathrm d k [/itex] the reflection coefficient (for example) is [itex]R = \int g(k) R(k) \mathrm d k [/itex] where R(k) is the earlier calculated reflection coefficient for a plane wave with parameter k.

    I get the intuitive plausibility that the above formula has, but when I try to think about it more exactly and try to argue it mathematically it seems to be untrue.

    My reasoning is as follows: how does "R" acquire its meaning as some number characterizing how much goes back? Due to the notion of probability current. So for the above formula to work, the probability current of all the plane waves must also adhere to the superposition principle, i.e. [itex]J = \int g(k) J(k) \mathrm d k[/itex], and this is what my professor (literally) said when I asked him. However, when I look at the definition of J, namely something proportional to [itex]J \propto \psi^* \nabla \psi - \psi \nabla \psi^* [/itex], that looks very unlinear (w.r.t. to psi), so how can we possibly rely on the superposition principle?

    Thank you!
     
  2. jcsd
  3. Oct 26, 2011 #2

    tom.stoer

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    You must not calculate currents for different wave functions and sum over the currents, but of course you can sum over wave functions in order to form wave packets and then calculate the current for the wave packet. This is the same as for the probability density.
     
    Last edited: Oct 26, 2011
  4. Oct 26, 2011 #3
    But then what is the meaning of the probability current for plane waves if you can't use them for a (more physical) wave packet? Any QM book studies the plane wave case for reflection and transmission, but if the superposition principle doens't hold for these coefficients/probability current, then there's no indication that this is a good approximation for very unlocalized particles, is there?
     
  5. Oct 26, 2011 #4

    tom.stoer

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    Suppose you have plane waves (or some other basis in a function space)

    [tex]u_n(x)[/tex]

    Then you can calculate the probability current and of course the probability density

    [tex]\rho_n(x) = |u_n(x)|^2[/tex]

    Now suppose you have a wave packet (which is not necessarily a solution of the Schrödinger equation!)

    [tex]\psi(x) = \sum_n \psi_n u_n(x)[/tex]

    The corresponding probability density is

    [tex]\rho(x) = |\psi(x)|^2 = |\sum_n \psi_n u_n(x)|^2[/tex]

    and not (!!!!!!)

    [tex]\rho(x) = \sum_n \rho_n(x) = \sum_n f(\psi_n)\,|u_n(x)|^2[/tex]
     
    Last edited: Oct 26, 2011
  6. Oct 26, 2011 #5
    Hm, I don't seem to understand how your post is an answer to my question?
     
  7. Oct 26, 2011 #6

    tom.stoer

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    I only wanted to explain what the superposition principle is - and what it is not; it shows how to calculate the probability density for plane waves and for wave packets; you can do exactly the same for the currents
     
  8. Oct 26, 2011 #7
    So if psi is a sum of psi_a and psi_b, then the probability current is the sum of current_a and current_b?

    That was my main question anyway; if it seemed otherwise, then I must've confused you by using the wrong terminology in some places, my apologies.
     
  9. Oct 26, 2011 #8

    tom.stoer

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    No, you can see that from my calculation: psi is the sum of un's, but the psi-density is not the sum of the u-densites!

    [tex]\psi = \psi_1 u_1 + \psi_2 u_2[/tex]

    [tex]\rho_1 = |u_1|^2[/tex]
    [tex]\rho_2 = |u_2|^2[/tex]

    [tex]\rho_\psi = |\psi|^2 = |\psi_1 u_1 + \psi_2 u_2|^2 \neq |\psi_1 u_1|^2 + |\psi_2 u_2|^2[/tex]
     
  10. Oct 26, 2011 #9
    Oh yes I see, of course, you were implicity using the continuity equation, gotcha.

    But then why do we study the unphysical case of plane waves when it comes to reflection coefficients? There's no way to get physical information out of it, correct? (in spite of what my professor said, who claimed that we can just use superposition to get the coefficients/currents for wave packets [and so justifying studying the plane wave case], but I didn't believe that and you indeed show that it's incorrect)
     
  11. Oct 26, 2011 #10

    tom.stoer

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    No, algebra; (a+b)² = a² + b² + 2ab
     
  12. Oct 26, 2011 #11

    Bill_K

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    mr. vodka, The reason you're off track on this is that you're focused on the reflection coefficient defined in terms of the probabilities and probability currents, which are quadratic. Normally we talk about reflection/transmission coefficients R, T defined by the wave amplitudes. These are complex, linear in ψ and add linearly when we superpose plane waves to make wave packets. The unitarity condition is |R|2 + |T|2 = 1.
     
  13. Oct 26, 2011 #12
    I think you misunderstood me. I meant so say you used the continuity equation in the sense that showing the superposition doesn't hold for the density is the same as showing it doesn't hold for the current because the continuity equation relates the two.

    @ Bill K, and also a bit at Tom.Stoer:

    I thank you both for your replies and I get what you are saying, but I think you're missing my point/question... I do know what the superposition is and note that I never claimed that superposition gives that if you sum some wave functions, you may also sum the density current or transmission coefficients, rather I was objecting to it, because my profesor claimed it and I didn't believe it (for the reasons you guys state). (Talking of which: should I bring it up with my professor to point out that he was mistaken? Or is that a no-no?)

    But this still leaves my other question, which hasn't gotten any response, and this might be because I hadn't formulated it clearly enough. Let me try to make it clear, cause it's an important question for me:
    in a typical upper undergraduate course on quantum mechanics, like I am following, one calculates the reflection and transmission coefficients |R|² and |T|² for plane waves. These are basically the coefficients (in square modulus) in front of the plane waves and acquire their meaning of "the fraction which reflects/transmits" due to the notion of probability current. The case of a plane wave is of course unphysical, and when I aksed my professor "why do we study this unphysical case?" he said "because to get to the physical case of a wave packet, you can just sum up the probability currents". We now know, however, that he was wrong, so my question again is: "why do we study this unphysical case?".
    More concretely: we use the expressions for |T|² in exercises without any justification for why the number should approximately hold for physical cases.

    I hope my point is clear now :)
     
  14. Oct 26, 2011 #13

    kith

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    |T|² may not have physical content, but T has because of the superposition principle (as Bill K has pointed out). Given this, the only remaining question I see, is why one should calculate |T|² from T for a plane wave.

    This is probably just done for illustrative reasons. Although plane waves are not physical, you can illustrate what happens for plane-wave-like wavepackets.
     
  15. Oct 26, 2011 #14
    Not even then, I think, cause even that reasoning requires some sort of superpositioning for |T|², but I get your general point.
     
  16. Oct 26, 2011 #15

    kith

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    Sure but if the weight function of your wave packet is almost a delta function, the error in taking just |T0|² instead of |ƩTi|² will be very small.
     
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