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Units of j

  1. Aug 2, 2007 #1
    1. The problem statement, all variables and given/known data
    What are the units of the probability flux j?

    [tex]\vec j = \frac{\hbar}{2mi}\left(\Psi^* \vec \nabla \Psi - \Psi \vec \nabla \Psi^*\right)[/tex]


    2. Relevant equations



    3. The attempt at a solution

    The units of psi are probability amplitude and the units of del(psi) are probability amplitude per unit area, so I would say the units are psi* del psi are probability density per unit area?
     
  2. jcsd
  3. Aug 2, 2007 #2
    Think about normalization equation for units of Psi.
    Also, the del operator is a fancy sort of derivative (d/dx). It reduces dimension by one power of length, since x has the units of lenght.
     
  4. Aug 2, 2007 #3

    nrqed

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    First off, psi is a probability amplitude but we can't say that the units are "probability amplitude". This does not quite make sense. Its' like saying that the units of "m" are "mass". Mass is the name of the quantity, it does not give its dimensions. Dimensions are expressible in terms of length, time and mass units (meter, second and kilogram in MKSA). So the first question is: what are the units of psi? The answer is provided by noticing that a normalized psi obeys [itex] \int dV | \psi|^2 = 1 [/itex]. That should give you the units of psi.

    Not per unita area, per unit length. [itex] \frac{\partial }{\partial x} [/itex] for example introduced a factor fo 1/meter in the dimensions of the quantity differentiated.
     
  5. Aug 2, 2007 #4
    Are they sqrt(1/dV) then? So, then del * psi would have units 1/dV^2? That does not seem right.
     
  6. Aug 3, 2007 #5

    Dick

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    del(psi) will have units of sqrt(1/V)/L. del is a derivative wrt length L.
     
  7. Aug 3, 2007 #6
    So psi* del psi has units 1/V/L? How does this translate into probability density per unit length which I thought nrqed implied were the correct units of j?
     
  8. Aug 3, 2007 #7

    olgranpappy

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    he didn't imply that. anyways, a probability is a pure number... so the dimension of "probability density" is just 1/L^3. thus a "probability density per unit length" would have dimensions of 1/L^4 which is the wrong units for j.

    Perhaps what he was implying was that the units of psi*del psi are 1/L^4 which is correct. Now, what are the units of the rest of the things in your equation...
     
  9. Aug 3, 2007 #8
    Planck's constant is J * s, mass is kg, so we have

    kg * m^2 / s^2 * 1/kg * 1/ m^4 = 1/m^2/s^2

    where m is meters. So how do you put that in words: length * probability density per second per second?
     
  10. Aug 3, 2007 #9

    olgranpappy

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    you left off a factor of seconds... you said "Joule seconds" but you only put in the "joule" part.
     
  11. Aug 3, 2007 #10
    Sorry.

    kg * m^2 / s * 1/kg * 1/ m^4 = 1/m^2/s
     
  12. Aug 3, 2007 #11

    olgranpappy

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    thats right
    [tex]
    \frac{1}{TL^2}
    [/tex]
     
  13. Aug 3, 2007 #12

    nrqed

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    Yes, psi has units of q/sqrt(volume), so the units of [itex] \psi [/itex] are [itex] \frac{1}{L^{3/2}} [/itex] where L stands for length.

    Therefore, the units of del psi are [itex] \frac{1}{L^{5/2}}[/itex], so the units of psi del psi are [itex] \frac{1}{L^4} [/itex] as someone else pointed out.
     
    Last edited: Aug 3, 2007
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