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Derivation of reflection/transmission coefficients

  1. Dec 28, 2013 #1
    Greetings!

    Going throught Ballentine Ch.4 and the derivation of transmission/reflection coefficients. The math seems fairly straightforward: assuming a particle in a piecewise-constant potential, the solution of the correspondent time-independent SE is a piecewise-exponential wavefunction in a form ##ψ_i = A_i e^{k_ix} + B_i e^{-k_i x}##, where ##k_i=\sqrt{-2m(E-V_i)}/\hbar## is either real or imaginary depending on the sign ##E-V_i##. After adding corresponding boundary conditions, one can solve the system for the allowed values of ##E## and corresponding ##A_i## and ##B_i##. So far so good.

    Now here is the bit I'm not comfortable with: we treat the two parts ##Ae^{ikx}## and ##Be^{-ikx}## separately, and claim that one describes the particle going left-to-right and another right-to-left. Why? As far as I can see, there is just 1 particle and it is not going anywhere because the solution is time-independent. And it is just a sheer luck that I get a solution as a sum of "left" and "right" parts. What if I get quadratic potential wells instead of boxy ones, then the solutions will be gaussian-ish (energy states of harmonic oscillator to be exact), with no obvious split into "left" and "right" parts.

    I mean it all seems sort of right intuitively, but I have a feeling there is something they are not telling me. For example, problem 4.3 asks: electron with momentum ##p=\hbar k## going from left to right, impinges on a potential step of height V, what is the probability of it passing through. Ok I know (I think) how to calculate the answer they expect, which is the transnission coefficient , I just dont see why it should be the answer. They ask for probability and the only official recipe given so far in the book for calculating a probability is ##Prob\{R<a\}=\left\langle \theta(R-a) \right\rangle = Tr\{\rho \theta(R-a)\}## where ##\rho## is a state operator, ##\theta## is a unit step function and ##R## is an observable. I just don't see how to apply it to the problem at hand.

    Thanks for your help,
    DK
     
  2. jcsd
  3. Dec 29, 2013 #2
    I sort of had/have the same problem with the whole reflection and transmission coefficients bit. Griffiths has something to say on on the matter, and to paraphrase him, I think that your discomfort with the entire problem is the fact that a quintessentially time-dependent problem ( electron impinging on potential and then scattering ) has been analyzed using stationary states. There's nothing wrong mathematically with breaking the wave-function into left moving and right moving parts, and if every particle impinging on the potential is identical, it seems to me that everything you need to know about the scattering problem is located in the wavefunction of one particle. I mean, isn't the interference pattern of the double slit experiment contained entirely in the wavefunction of just one photon, even though the interference experiment is essentially a time-dependent problem, requiring a large no of photons to impinge upon the screen ?

    Further, if you tack on the characteristic time dependence exp(-iEt/h) it become clear these solutions actually represent right moving and left moving plane waves, and seeing as the probability of finding a particle in a region should be proportional to the wavefunction squared, the relative probabilities of finding the wave in states represented by Aexp(ikx)exp(-iEt/h) or Bexp(-ikx)exp(-iEt/h) or Fexp(ikx)exp(-iEt/h)( if the wavefunction is normalizable, these represent physically realizable states, otherwise the transmission and reflection coefficients are approximations in the vicinity of E ) should be given by the ratio of the squares of the amplitudes. And the probability of the particle being in the state Aexp(ikx)(-iEt/h) represents, loosely speaking, how much of the wavefunction can be decomposed into a right moving wave.

    The individual right moving and left moving components of the wave function do not represent particles moving to the left or right. That is absurd. The only thing that represents every particle is the superposition of stationary states with the associated time dependence. The overall effect though, is as if a right moving particle were represent by a right moving plane wave, and a left moving one by a left moving plane wave.
     
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