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Finding general expression for probability current.

  1. Mar 19, 2017 #1
    The conservation of probability says:

    $$\partial_t J^{0} + \partial{i}J^{i} = 0$$

    Use the Schrodinger equation to obtain$$ J^{i} (\vec r)$$.

    I have no idea where to start this kind of problem because the notation makes no sense to me. I would appreciate a hint or nudge in the correct direction.
     
  2. jcsd
  3. Mar 19, 2017 #2

    vela

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    What about the notation confuses you?
     
  4. Mar 19, 2017 #3
    What do those partials mean? Im not used to seeing them this way? Also would I be using the general SE eq or the radial one since we are in r?
     
  5. Mar 19, 2017 #4

    vela

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    Typically ##x^0 = t##, ##x^1 = x##, ##x^2 = y##, and ##x^3 = z##, so ##\partial_1## would mean ##\partial_x##. The repeated index ##i## implies a summation from ##i=1## to ##i=3##.

    You have ##J^\mu = (\rho,J_x, J_y, J_z)##, so in more traditional notation, conservation of probability is
    $$\frac{\partial \rho}{\partial t} + \nabla \cdot \vec{j} = 0$$ where ##\vec{j} = (J_x, J_y, J_z)##.
     
  6. Mar 21, 2017 #5
    Thanks, I will look at this later today and see if it helps!
     
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