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Quantum mechanics probability

  1. Feb 1, 2013 #1
    In the one moment you accept something in the other you're not sure why? Why probability density in quantum mechanics is defined as ##|\psi|^2## and no just ##|\psi|## if we know that ##|\psi|## is also positive quantity.
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  3. Feb 1, 2013 #2

    Simon Bridge

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    Because - in general, ##\psi## is a complex number ... go look at how complex numbers work. i.e., ##|\psi(\vec{r},t)|## is not, in general, positive everywhere.
    Consider a particle in a 1D unit box, prepared in the n=2 state.

    Mainly the model is the way it is because that is what nature does.
    Just taking the modulus does not give the correct statistics.
    Last edited: Feb 1, 2013
  4. Feb 2, 2013 #3


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    Of course [itex]|\psi(t,\vec{x})|[/itex] is a positive semidefinite real expression everywhere, since the modulus of a complex number is a positive real number (or 0 if the number itself is 0).

    Born's rule is, as far as we know, one of the fundamental postulates of quantum theory. It has not yet been deduced from other more fundamental postulates yet. So we cannot explain, why this interpretation of the wave function works to describe nature.

    Recently, however it has been tested experimentally that it holds with very high accuracy:

  5. Feb 2, 2013 #4
    I believed that theory is consistent. But why Bohm took ##|\psi(x,t)|^2##? Is there some bond with
  6. Feb 2, 2013 #5


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    A Google search for "Born rule" led me to the Wikipedia page


    which refers to Born's 1954 Nobel Prize lecture


    in which he talks about the origin of his famous rule.

    Classically, the square of the electric field (or of the magnetic field) is proportional to the energy density carried by the field, and therefore also to the number-density of photons (photons per m3)in the field. For a weak field (small number of photons), we have to talk about probability rather than number-density.
  7. Feb 3, 2013 #6


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    First indication to use [itex] |\psi(x,t)|^2[/itex] is that this quantity is normalised such that the integral is = 1 which means that the probablity to find the particle anywhere is 100%.

    The second indication is that the density

    [tex]\rho = |\psi|^2[/tex]

    plus the current


    fulfill a continuity equation

    [tex]\frac{\partial}{\partial t}\rho + \nabla \vec{j} = 0[/tex]

    which means that the probablity is conserved. This follows in a rather general way using the Lagrangian corresponding to the Schrödinger equation plus Noether's theorem. It is not special for Bohm's interpretation.

    All this works for [itex] |\psi(x,t)|^2[/itex] not for [itex] |\psi(x,t)|[/itex]
    Last edited: Feb 3, 2013
  8. Feb 3, 2013 #7
    Well I agree with second reason. But you can normalized state in such way that
    And the second one. Why current is defined like
  9. Feb 3, 2013 #8


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    No, because we are using the space L^2(R^3) for the space of wavefunctions. Not every L^2 function is also a L^1 function. We can't use L^1, because it's not a Hilbert space.

    Because this definition turns out to give you a conserved current.
  10. Feb 3, 2013 #9


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    Correct; L^2 is the only Lp-Banach-Space which is also a Hilbert space.

    Using Noether's theorem this current can be derived.


    You can normalize the density for one time t, but time evolution (Schrödinger equation) will not respect this normalization, i.e. it will be violated for t' > t.

    As I said
    Last edited: Feb 3, 2013
  11. Feb 3, 2013 #10


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    Yeah, given Schrodinger's equation alone, with no assumption about the meaning of [itex]\vert \Psi \vert^2[/itex], you can prove that the latter is a conserved quantity (when integrated over all space). So it's pretty natural to interpret it as something conserved: probability, or charge, or energy...
  12. Feb 3, 2013 #11


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    Yes, you can prove that, but you cannot derive it w/o Noether's theorem; you have to guess it.
  13. Apr 17, 2013 #12
    Hello, I have a question concerning normalization. I start from integral |psi|^2 dV = 1, which means somehow that the particle should be in the universe with probability 1. Now using the the Gauss theorem in such a way that if we have a vector function

    Fi: div Fi = |psi|^2,
    integral |psi|^2 dV = integral div Fi dv =(Gauss) integral Fi dA = 1 so the normalization is transformed from volumetric to surface integral.

    The unit of psi is [psi]=1/m^(3/2) so [psi^2]=1/m^3 and vector [Fi] = 1/m^2.

    So we prepared a "surface probability density vector" from the scalar volumetric probability density |psi|^2.

    I would like to know whether this trick with mathematics could have some physical meaning at all? It looks like that the internal probability appeared on the surface of the universe somehow, also I am aware that probability vector sounds quite unphysical.

    Thank you for your answer.
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