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Heisenberg versus Maxwell?

  1. May 26, 2010 #1

    nomadreid

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    There are two basic facts that I have difficulty reconciling:
    (a) the magnetic and electrical components make up a complementary pair of variables
    (b) Maxwell's equations describe the magnetic component in terms of the electrical one and vice-versa

    My question is definitely not original and assuredly not even very clever, and so should have a standard and elementary answer out there somewhere. I could formulate it as follows:

    Suppose one determined one of these components with a measurement. That should make the other one completely indeterminate, including its rate of change. But then what sense would the corresponding equation make? Otherwise put, given a precise measurement of one of the variables, Maxwell's equations should give us the other variable, which is impossible. What is wrong with this formulation?

    I assume that part of the answer is that the uncertainty principle is a statistical principle, but Maxwell's equations are not, so this makes it even more difficult to fit them both together, even though the Uncertainty Principle, if I understand my history (also not certain), was derived from Maxwell's equations.

    Thanks for any help.
     
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  3. May 27, 2010 #2

    Born2bwire

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    A lot of potential comments for this.

    The Heisenberg Uncertainty Principle is a consequence of the mathematics of quantum mechanics. It has nothing to do with Maxwell's equations. It also only makes remarks about the spread in measurements of a statistical set between non-commutable observables. This occurs because with non-commutable observables, the eigenspaces of the two observables differ. This means that when we make a measurement of observable A, we see the system in an eigenstate of A. But, when we then measure B, we can measure a spread of possible eigenstates of B since the states of B may be a superposition of states of A.

    Still, this does not prevent you from making a measurement to arbitrary precision, it just states that you will end up with a spread of measured results.
     
  4. May 27, 2010 #3

    nomadreid

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    Thank you for replying, Born2bwire, but I'm afraid it doesn't answer the question. Since the Heisenberg Uncertainty principle has to do with measurements of electric and magnetic fields, and Maxwell's equations also, there is a link between them. I understand that the uncertainties in the Uncertainty Principle are from the standard deviation of many measurements, and that one can make individual precise measurements -- in fact, one can make only precise measurements -- but nonetheless one cannot make measurements of two complementary variables simultaneously, even for a single piece of data from each. Maxwell's equations seem to imply that one can determine (by calculation) the magnetic field from a measurement of an electric field, and that the two fields would be determined simultaneously -- something forbidden by the Uncertainty Principle. Your answer leaves this difficulty unresolved.
     
  5. May 27, 2010 #4

    bapowell

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    Why do you say this? The HUP involves the measurements of conjugate operators -- observables that don't commute (eg x an p). E and B are not conjugate variables.

    EDIT: Although the HUP doesn't apply to E and B in the first place, Maxwell's Eqs are classical equations of motion. So, even if the HUP did apply, you wouldn't see it in Maxwell's Eqs. For example, consider a mass attached to a spring. Classically, we have:

    [tex]\frac{p^2}{2m} = \frac{kx^2}{2}[/tex]

    at maximum displacement. This expression suggests that simultaneous evaluation of x and p is possible, and, of course it is classically. The HUP supplements this classical equation of motion.
     
    Last edited: May 27, 2010
  6. May 27, 2010 #5

    nomadreid

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    "Why do you say this? The HUP involves the measurements of conjugate operators -- observables that don't commute (eg x an p). E and B are not conjugate variables."

    According to
    http://www.answers.com/topic/squeezed-quantum-states
    and other sources, they are.

    Your counterexample to the determination of p and x with a classical equation actually just says that this classical equation is only approximate. So is your answer that Maxwell's equations are only approximate, and that a full statement of the relationships between E and B would have to be altered a bit to take into account the HUP?
     
  7. May 27, 2010 #6

    bapowell

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    Thanks for the link. I'm confused about the statement that electric and magnetic field components are conjugate. :confused:

    Anyway, Maxwell's equations are indeed exact. The treatment of electromagnetism in quantum theory is really a field theory, called quantum electrodynamics. In QED, Maxwell's equations are the field equations of the photon, and the commutation relations are imposed on the photon field separately (this is where the quantum mechanics comes in).

    I know there is a limited formalism for understanding electric and magnetic fields in QM, but I'm not familiar with it. I have a hard time imagining it unless the electric and magnetic fields are treated classically -- then I suppose it makes sense to talk about a single particle state being an eigenstate of E or B. After all, the E and B fields are themselves particles (photons), so the E or B field of a particle is very different from a kinematical property, like x or p.
     
  8. May 27, 2010 #7
    Not necessarily. De Broglie clearly indicated that the particle is separate from an associated wave or field. Granted not everyone subscribes to that idea, but certain experiments clearly indicate it is so for the electron.

    If E and B fields are expressed in a spinor or tensor then they might have conjugate elements. But I think you were right from the beginning: E and B as typically considered are represented by real numbers only and the HUP doesn't apply.

    It might be Special Relativity that the OP was thinking was derived from Maxwell's Equations. Especially the mathematical side developed by Lorentz and Poincare.
     
  9. May 27, 2010 #8

    tom.stoer

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    No, the conjugate variables in QED are defined as follows: A° is a non-dynamical field as it comes w/o time derivative, i.e. A° has no conjugate momentum. Now fix the A°=0 temporal gauge and eliminate A° completely. The requirement that it's conjugate momentum remains zero as required by consistency leaves behind a constraint equation div E = j°, the Gauss law.

    For all other components of the vector potential A one can determine the conjugate momenta. One finds that the conjugate momentum of Ai is just Ei. So counting degrees of freedom we now have three photons, but we know that there are only two polarizations. That's OK because one degree of freedom is eliminated by solving the Gauss law constraint which results in two physical polarizations for the photon.
     
    Last edited: May 28, 2010
  10. May 27, 2010 #9

    Born2bwire

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    You can measure two non-commutable observables simultaneously. The uncertainty principle does not prevent that. It only states that in a statistical set of measurements you will find a variance in your measurements whose minimum variance is related to the uncertainty principle. It is my recollection that the electric and magnetic field operators are not commutable with each other for all possible combinations. However, this still does not prevent us from measuring the fields and achieving results that match calculations. The uncertainty principle talks about the variance in the measurements, the observables are the expectation (mean) of the measurements.

    The article that you linked, concerning squeezed states, is concerned with the inherent "noise" that arises from the vacuum state. The vacuum state has a zero mean but a non-zero fluctuation or variance. This manifests itself as noise when we do photon measurments and normally it is evenly distributed across the phase space. However, the uncertainty principle places a restriction between two non-commutable observables. Thus, we can increase the noise in one observable and thus decrease the minimum noise in the other observable. This results in a squeezed state and allows us to reduce the noise along one quadrature, thus improving the signal to noise ratio.
     
  11. May 30, 2010 #10

    nomadreid

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    Thanks to all those who responded to correct my original assumption about the conjugacy of E and B. I took a while to respond because I had to go back to find the source of this erroneous assumption. An example of learning by one's mistakes. So thanks again.
     
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