1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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).
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Free Particle Understanding
  1. Free particle (Replies: 25)

  2. Free Particle (Replies: 19)

  3. Free Particle (Replies: 1)

  4. The Free particle (Replies: 18)

  5. The free particle (Replies: 6)