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Physical meaning for wavefunction

  1. Jul 22, 2005 #1
    The state of a photon can be represented by complex number or 2d vector, where the x-axis represents the electric field, and the y-axis represents the magnetic field. Interference can been seen as the addition of several of these vectors.

    Similarly, in quantum physics the wavefunction at a given point in space and time is a complex number/vector, and interference is the addition of such vectors. However, here it's often said that the wavefunction itself has no physical meaning (except that the square of its magnitude represents a probability), and there is only indirect evidence for its existence (such as interference experiments).

    Is there no physical significance to the x-axis of the wavefunction of an electron, like the electric field for photons? If not, why is there one for photons? Why are photons different?

    It seems that the situation should be the same for both particles, and we can clearly detect electric fields. Can someone explain this to me?
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  3. Jul 22, 2005 #2
    The kind of "vector" you are thinking of is too restrictive; the kind of vectors of a wavefunction are not component vectors in a Euclidean space, but rather vectors in a function space of square-integrable functions where the inner product is defined as
    [tex]\langle \phi | \psi \rangle = \int d\mathbf{x} \langle \phi| \mathbf{x} \rangle \langle \mathbf{x}| \psi \rangle = \int d\mathbf{x}~ \phi^*(\mathbf{x}) \psi(\mathbf{x}) [/tex]
    which is a different kind of vector space, what is referred to by mathematicians as a Hilbert space (even though the vector space in quantum mechanics isn't exactly a vector space, the vector concept is very useful).

    From a physical sense and with reference to wave mechanics, all the information about a physical system is contained in the wave function. If a particle is known to be in a state [tex]\psi(\mathbf{x}, t_0)[/tex] at time [tex]t_0[/tex], then the time evolution of the system is given, at least formally, by
    [tex]\psi(\mathbf{x},t) = e^{-\frac{\imath}{\hbar} \int_{t_0}^t \mathcal{H}(t') ~dt'} \psi(\mathbf{x},t_0) [/tex]
    where [tex]\mathcal{H}[/tex] is the Hamiltonian of the system, which may or may not depend on time (hence leaving the time integral in the exponential). What should be noted is that the exponential should indicate nested integrals, meaning that
    [tex]\left (\int_{t_0}^t \mathcal{H}(t') ~dt'\right)^2 = \int_{t'}^t\int_{t_0}^{t'} \mathcal{H}(t') \mathcal{H}(t'') ~dt''~dt'[/tex] and
    [tex] \left (\int_{t_0}^t \mathcal{H}(t') ~dt'\right )^3 = \int_{t'}^t\int_{t''}^{t'} \int_{t_0}^{t''} \mathcal{H}(t') \mathcal{H}(t'') \mathcal{H}(t''') ~dt''' ~dt''~dt' [/tex]
    and on. The physical interpretation of this is a discussion for another time, but you can simply take it as a mathematical result from partial differential equations.

    The wave function also accounts for interference nicely. Let's suppose we have a particle in a superposition of states given by
    [tex]\psi(\mathbf{x},t) = \frac{1}{\sqrt{2}}\left (\phi_1(\mathbf{x},t) + \phi_2(\mathbf{x},t)\right) [/tex]
    Then the probability distribution of the particles is given by
    [tex]|\psi(\mathbf{x},t)|^2 = |\phi_1(\mathbf{x},t)|^2 + |\phi_2(\mathbf{x},t)|^2 + \frac{1}{2} \left (\phi_1^*(\mathbf{x},t)\phi_2 (\mathbf{x},t)+ \phi_2(\mathbf{x},t) \phi_2^*(\mathbf{x},t) \right )[/tex]
    where the first two terms are analogous to classical probability amplitudes, and the term in parentheses represent quantum interference effects.

    The wave function clearly has a lot of physical meaning, it's just that the wave function itself is not directly measureable. It is consequences of the statistical interpretation of the modulus squared of the wave function that are measureable effects, and we can reproduce those results nicely using the wave function. The wave function also has a particular interpretation in quantum mechanics if you use a more abstract vector formulation of quantum mechanics (Dirac notation and all that). If you want to read more about the matrix mechanics formulation (as opposed to the wave mechanics formulation) I strongly recommend either Dirac's own book or the J.J. Sakurai Modern Quantum Mechanics .

    As for the difference, the example you are using is that you are specifying the state of a photon in a two-level system, namely polarization. If I had a two-level quantum system, I would have a similar interpretation (an example of such a system is the spin-1/2 particle). By chance, are you reading Baym's Lectures on Quantum Mechanics ?
    Last edited by a moderator: Jul 22, 2005
  4. Jul 23, 2005 #3

    There only one physical meaning I have been able to assign to the wavefunction.

    This is that the wavefunction is a description of the property distribution of the quantum system associated with the wavefunction.

    I see it like this. The properties of the quantum system are distributed over all the states defined and described by the superposed wave function. When an interaction occurs, a localization of the property distribution of the system takes place and the wave function appears to collapse. This apparent collapse is just a reorganization of the property distribution of the system, due to the exact nature of the interaction. Since we due not know the exact nature of the interaction, it appears probablistic, but is actually totally determined.

    The distribution interacts with the measuring system, and localizes in a neighborhood. The neighborhood is dependent on both the strengths and the asymmetries of the interactions of various parts of the distribution with the elements of the measuring system.

  5. Jul 24, 2005 #4


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    This was Einstein's view of things. But EPR situations and Bell's theorem give that view a hard time !

  6. Jul 24, 2005 #5


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    There are many different possible answers to your question and the least that one can say is that there is no consensus on the issue :smile:

    Probably there are 3 categories of views:

    1) the wave function IS physical

    2) the wavefunction is only a mathematical tool to help us describe links between observations, and it is a hollow exercise to try to say what is "physical"

    3) something is physical, and we're missing the point. By some coincidence, the wavefunction gives good results, but we haven't found yet the true underlying mechanism that will explain things to us.

    My personal view is 1), until 3) is proven, which I hope for :smile:

  7. Jul 24, 2005 #6


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    But if you accept (1) then you have to erect a category of things that are physical but not observable or measurable, and that have no consequences in spacetime except indirectly by calculating a probability. In your favored MWI, we don't even have that! My preference is to be agnostic favoring (3). And I protest that the way QM works with the wave function is not "by coincidence" but is based soundly on physics experiments and discoveries of the early 20th century. The physicsts got a first glimpse with Bohr's old QM of the Hydrogen atom, which enabled a rough calculation of its spectrum. Then deBroglie's "matter waves" and Heisenberg and Schroedinger, and much better calculations. And so on. The quantum world shows us this behavior pretty obviously, and 50 years of improved models gave an excellent account of it. But nothing exists to tell us what the quantum world really is.
  8. Jul 24, 2005 #7
    Surely one description of the reality with its onw domain of validity like any physical theory. :biggrin:
    The rest is pure speculation until prooved by tests.


    PS. Before this post, I thought you were a proponent of the espistemic view of physics.
  9. Jul 24, 2005 #8
    The whole idea of "physical but not directly measureable" isn't a huge problem. There is no way of measuring the electric or magnetic fields directly, but nobody would argue that they aren't physical. Their measurement requires tracking the motion of charges, then inferring from the equations of motion what the fields are.
  10. Jul 24, 2005 #9


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    The wave functions of electrons can also be put into x-y form just like photons. There are technical differences that have to do with the photon being a massless particle of spin 1 and the electron being massive and spin 1/2, but ignoring these details, there is no doubt that it is possible to polarize electrons vertically and horizontally in a manner at least somewhat similar to how one polarizes photons.

    The usual way photons can be polarized is vertically or horizontally. The corresponding usual polarization for electrons is up or down.

    There is an alternative polarization for photons, which is left or right circularly polarized. Similarly, there are other alternative polarizations for electrons.

  11. Jul 26, 2005 #10


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    I thought I expressed my agnosticism also :smile:. Only, I cannot wait until I'm dead and rotten to even try to make sense of what we have now today, so I prefer to have a picture - even though I try to make clear that it is only going hand in hand with the CURRENT formalism, and that this picture is bound to change when the formalism changes. I'm even convinced that EVERY generation of physicists after us will have exactly the same problem (dare to give a picture of what they know, or be agnostic and hope that their collegues of the future will settle the issue), so if you delay in having a picture until you have the final answer, you'll never have a picture.

    I find it extremely helpful to have a clear picture (even if it is weird and you don't want to believe it for real) of a given theory.

    Ah, it was said in the same way as I could call Newtonian mechanics to "work by coincidence". It also shows us that if you're hoping for the future to give us a better view, sometimes you have to wait a LONG time !
    Imagine that Newton said the same thing: that "action at a distance" was too weird to be true, so let's not attach a too profound meaning to "points" and "space" and things like that. Just stuff on paper. That would have left at least 15 generations of physicists with NO picture at all !
  12. Jul 28, 2005 #11
    Not sure I understand your statement. Newton was describing gravitational
    behavior, and chose not to hypothesize wrt its cause(s) -- which left the
    door open for competing models to 'explain' what Newton had generally

    I think that a good rule of thumb is that if an explanatory model (ie., a
    'picture' of an underlying 'reality') is 'weird', then it probably isn't true.

    Anyway, gravitational behavior, like the global behavior of any system,
    isn't *due to* action-at-a-distance. Action-at-a-distance just refers
    to the observation of simultaneous changes in the behavior of
    spacelike separated objects. Correlated actions-at-a-distance are a
    feature of all gravitational systems. The macroscopically visible objects
    in a gravitational system are the most intensly interacting regions of the
    individual wave complexes that make up the system, the invisible extensions
    of which are locally interacting with the extended wave complexes of the
    other bodies of the gravitational system.

    This is not exactly the same as what is happening to produce
    the correlations in Bell tests, where the global variable (changes
    in the Theta of the analyzing polarizers) is part of the system.
    The motion (some property or properties) of two disturbances
    produced by the same oscillator is (as long as no external influences
    are introduced on the way to the analyzers) related. The
    action-at-a-distance in Bell tests has to do with simultaneous
    changes in the global variable, Theta, as the setting of one
    or the other polarizer is varied, which also simultaneously affects
    the probability of joint detection.
    Last edited: Jul 28, 2005
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