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Probability Current of a Wave Function

  1. Nov 19, 2012 #1
    Find the probability current of Ae^i(kx - ωt) + Be^-i(kx+ωt)


    Ok, to my understanding the probability current is the probability that you will find a certain particle as it moves with time, thus the probability of finding it changes with time. Quantum physics is a tricky one to grasp, I've got to say.

    I have an equation for probability current:

    J = (i*hbar / 2*m)( ψ∇ψ* - ψ*∇ψ)

    So, I have figured that ψ = Ae^i(kx - ωt) + Be^-i(kx+ωt)
    Thus ψ* = Ae^-i(kx - ωt) + Be^i(kx+ωt)

    Not too sure if I have done the complex conjugate correctly here, I'm used to simple uses of it.. z = a + bi, z* = a - bi ... for example

    Next, I'm assuming ∇ is del. I've used this function before, it works out as d/dx + d/dy + d/dz .. otherwise known as gradψ for this example.

    I've tried differentiating it but ψ∇ψ* is some horrible maths, should i expect that? Wolfram alpha shows that d/dx of ψ is:

    Ake^i(kx - ωt) * i'(kx - ωt) - Bke^-i(kx - ωt) * i'(kx - ωt)

    I dont really know what i' is? and multiplying that out looks horrific and makes me think I've gone down the wrong path...
     
  2. jcsd
  3. Nov 19, 2012 #2
    Could someone confirm for me that I have the complex conjugate correct?

    if ψ = Ae^i(kx - ωt) + Be^-i(kx+ωt)
    ψ* = Ae^-i(kx - ωt) + Be^i(kx+ωt)

    When I work it through with the equation above it looks very close to having lots of values cancel out and leave me with just [i*hbar / 2*m][ikA^2 + ikB^2]

    But with the complex conjugate I have at the moment I'm a couple of negative signs away from that
     
  4. Nov 19, 2012 #3

    TSny

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    You complex conjugate looks correct. What do you get for the derivative with respect to x of ψ*?
     
  5. Nov 19, 2012 #4
    Alright, to my delight it does cancel down somewhat but not quite enough.

    The derivatives I found were:

    ∇ψ = ikAe^i(kx-wt) - ikBe^-i(kx+wt)
    ∇ψ* = -ikA*e^-i(kx-wt) + ikB*e^i(kx+wt)

    where the complex conjugate of A is A* and likewise for B*
    Not sure if this is a necessary thing to do, but I found it somewhere and it seemed correct.

    Anyway, I wont type it all out because it will take forever but basically I found that:

    ψ∇ψ* - ψ*∇ψ = 2[ψ∇ψ*] = 2[-ikAA* + ikBB* +ikAB*e^2kxi - A*Be^-2kxi]

    Now I've put it into my equation for J = (i*hbar / 2*m)( ψ∇ψ* - ψ*∇ψ)
    The i and 2 cancel out and I'm left with:

    J = hbar/m[ψ∇ψ*]

    Don't really know where to go from here, some of the people on my course have told me that it should eventually result in zero.
     
  6. Nov 19, 2012 #5

    TSny

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    That looks correct.
    I don't think that's right. Certain terms in ψ∇ψ* will cancel certain terms in ψ*∇ψ when you combine ψ∇ψ* - ψ*∇ψ.
    The term Ae^i(kx-wt) represents a wave traveling to the right with amplitude A. Be^-i(kx+wt) travels to the left with amplitude B. If |A| ≠ |B| would you expect the current to be zero?
     
  7. Nov 20, 2012 #6
    When i calculate ψ∇ψ* - ψ*∇ψ i wind up with:

    ψ∇ψ* = [Ae^i(a-b) + Be^-i(a+b)][-ikA*e^-i(a-b) + ikB*e^i(a+b)]
    where a = kx , b = wt (for simplicity)

    ψ∇ψ* = -ikAA* + ikBB* + ikAB*e^[i(a-b)+i(a+b)] - ikA*Be^[-i(a-b)-i(a+b)]

    i(a-b)+i(a+b) = 2ai
    -i(a-b)-i(a+b) = -2ai

    ψ∇ψ* = -ikAA* + ikBB* + ikAB*e^2ai - ikA*Be^-2ai

    Next,

    ψ*∇ψ = [A*e^-i(a-b) + B*e^i(a+b)][ikAe^i(a-b) - ikBe^-i(a+b)]
    = ikAA* - ikBB* - ikA*Be^[-i(a-b)-i(a+b)] + ikAB*e^[i(a+b)+i(a-b)]
    = ikAA* - ikBB* - ikA*Be^-2ai + ikAB*e^2ai

    So, when these values are subtracted:

    [-ikAA* + ikBB* + ikAB*e^2ai - ikA*B^-2ai] - [ikAA* - ikBB* - ikA*Be^-2ai + ikAB*e^2ai]

    This looks to me like it results in 2[-ikAA* + ikBB* + ikAB*e^2ai - ikA*Be^-2ai] = 2[ψ∇ψ*]

    I follow your logic, however this is where the maths leads me. Perhaps I've made a mistake somewhere in this?
     
  8. Nov 20, 2012 #7
    There is an easy way out for calculating probability current by using the eqn.
    ∇.j=-∂ρ/∂t,since every thing is dependent on x you can write it,
    ∂j/∂x=-∂ρ/∂t,where ρ=|ψ|2,and then integrate with respect to x.see if it helps.
     
  9. Nov 20, 2012 #8

    TSny

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    Do the corresponding colored terms cancel? Maybe I'm overlooking a sign.
     
    Last edited: Nov 20, 2012
  10. Nov 20, 2012 #9


    Feel like a fool for missing that, yeah it does cancel down nicely!
    Thanks a lot for your help, got a few more tricky quantum questions so I will probably post if/when I get stuck on those haha
     
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