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Free Particle Understanding

  1. Oct 27, 2007 #1
    I am reading One-dimensional examples from Bransden and Joachain.For the free particle solutions:Ψ=A exp [i(kx-ωt)] +B exp [–i(kx+ωt)]
    they say that for |A|=|B|,the probability current density=0.This is OK.Then they say we can associate the standing wave with a free particle along the x axis with a momentum whose magnitude is p=ћk but the direction is unknown...

    My problem is I cannot understand what they say regarding momentum.If j=0,how can momentum be non-zero?

    In fact, if A=0 or, B=0 I can see there is a momentum of precise value p=ћk.There j is non-zero and j=vP where P is the probability...

    Also, in the very next example of a potential step they show j=0 everywhere and concludes that no net momentum in the state...

    I tried to solve the problem by thinking that in the latter case, j=0 for the entire state.So, by conservation of momentum, p=0 everywhere...

    But in the former case, j is not zero everywhere.So, p must conserve its non-zero value...!!!Or, may be that They meant momentum direction is unknown as there is no net momentum in free particle if |A|=|B|?
  2. jcsd
  3. Oct 27, 2007 #2
    Let me approach mathematically:

    Ψ=A exp [i(kx-ωt)] +B exp [–i(kx+ωt)]

    For A=B, Ψ=C cos kx exp(-iωt)

    Operating this with momentum operator,we are getting
    -(ћk/i) C sin(kx) exp(-iωt)

    Sin and cos differ by a phase factor only...

    So,momentum value matches...and the direction is "unknown" for the "i"...

    what about the potential step?
  4. Oct 28, 2007 #3


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    If the two terms are present, then your particle is in a superposition of having momentum +hbar k and momentum - hbar k.

    This is the momentum equivalent of being in a superposition of positions (like the wavefunction indicates you).
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