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Probability Current Density

  1. Nov 12, 2012 #1
    hello, everyone. I have a question about Probability Current Density. I read a book which says that Probability Current Density equals velocity multiply wavefunction^2 .How to prove it? thank you!j=v*ψ^2.
    best wishes!
     
  2. jcsd
  3. Nov 12, 2012 #2

    Demystifier

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    Can you say what book it is?
     
  4. Nov 12, 2012 #3
    Quantum Mechanics-Landau
    chapter77 The required probability w is proportional to the current density along the z-axis. In
    the classically accessible region, this is vzψ^2

    thankyou!!
     
  5. Nov 12, 2012 #4
    The definition is much different but you know that j is defined as ρv.now ρ in quantum mechanics should be probability density which is |ψ|2,and hence current density
    should be (along z axis) |ψ|2vz
     
  6. Nov 12, 2012 #5

    Jano L.

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    einstein1921,

    the relation

    [tex]
    \mathbf j = \rho \mathbf v
    [/tex]

    was introduced already in continuum mechanics (Eulerian description). It gives amount of mass that will flow through a small planar surface of area [itex]\Delta S[/itex] and perpendicular to unit vector [itex]\mathbf n[/itex] after time interval [itex]\Delta t[/itex]:

    [tex]
    amount~of~mass = \mathbf j \cdot (\Delta S \mathbf n) \Delta t
    [/tex]

    Have a look into beginning chapters of some textbook on hydrodynamics, they explain this in greater length.

    The mass density [itex]\rho[/itex] and current density [itex]\mathbf j[/itex] satisfy the equation of continuity

    [tex]
    \partial_t \rho + \nabla \cdot \mathbf j = 0.
    [/tex]


    It turns out that Schroedinger's equation for one particle allows similar current density [itex]\mathbf f[/itex] to be defined, with the difference that now it gives the "amount of probability that flows through small area in unit time" in space, instead of giving directly amount of mass.

    In theory based on Schroedinger's equation, the equation of continuity (of "probability flow") is

    [tex]
    \partial_t (\psi^*\psi) + \nabla \cdot \mathbf f = 0,
    [/tex]

    where [itex]\mathbf f[/itex] is a triple of numbers given by

    [tex]
    \mathbf f = Re \{\frac{1}{m} \psi^* \hat{\boldsymbol \pi} \psi \}.
    [/tex]

    Here [itex]\hat{\boldsymbol \pi} = \hat{\mathbf p} - \frac{q}{c}\mathbf A(\mathbf r)[/itex] is operator of kinetic momentum (mv) of the particle. All this can be derived from Schroedinger's equation.

    We can even retain the same formula for current density as in continuum mechanics

    [tex]
    \mathbf f = \rho \mathbf v,
    [/tex]

    provided we define
    [tex]
    \rho = \psi^*\psi,
    [/tex]

    [tex]
    \mathbf v = \frac{Re\{ \frac{1}{m} \psi^* \hat{\boldsymbol \pi} \psi \}}{\rho}.
    [/tex]
     
  7. Nov 12, 2012 #6

    Demystifier

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    I have another question, for everybody here. Can this quantity in the probability current be interpreted as velocity if one does NOT adopt the Bohmian interpretation of quantum mechanics?
     
  8. Nov 12, 2012 #7

    Jano L.

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    Gold Member

    In Bohmian theory the above expression for [itex]\mathbf v[/itex] is understood directly as the velocity of the particle. But the actual value of velocity of the particle cannot be so simple; the velocity, if it exists, surely has to fluctuate (due to background radiation)). But I think we can understand the above expression in this way:

    The expression

    [tex]
    \mathbf v(\mathbf r) = Re\{ \frac{1}{m} \psi^*(\mathbf r) \hat{\boldsymbol{\pi}} \psi(\mathbf r) \}
    [/tex]

    gives expected average velocity for particle that is at [itex]\mathbf r[/itex]. So it may be a kind of conditional probabilistic description of velocity.
     
  9. Nov 12, 2012 #8
    thank you all!!!
     
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