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Maxwell's equations, Orthogonality, electric and magnetic fields in EM

  1. Feb 17, 2008 #1
    Maxwell's equations give that the electric and magnetic fields in E-M radiation are orthogonal. This is a classic equation, but can it be related to the orthogonality of, for example, the momentum and position operators which lead to non-commutivity?
     
  2. jcsd
  3. Feb 17, 2008 #2
    what do u mean about "orthogonality" of q and p?????
     
  4. Feb 17, 2008 #3

    malawi_glenn

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    non-commutivity and orthonogality of states are not the same thing!

    Non- commutivity means that you get different results when measuring the quantities A then B of state S, from when doing B then A of state S.

    Ortonogality is about how states are related to each other. See the physical states as vectors in Hilbert space. A state n can be a superposition of several eigenstates to an operator O. Another state m can be a superposition of other eigenstates to the operator O. The different eigenstates o to O are mutual orthogonal. Compare with the cartesian unit(basis) vectors x,y,z. Every vector can be written as a superposition of x y z, for example A = 9x-5y+1z. The basis vectors are mutually orthogonal, but not every cartiesian vector are mutually orthogonal.

    But to answer your real question, there is no connection I would say.
     
  5. Feb 17, 2008 #4
    oops

    First, I apologize for the haste in which I posed my question, and give my thanks to malawi_glenn for the deserved slap-on-the-wrist. Strictly, of course, as justly pointed out, non-commutivity is not justification for claiming any sort of orthogonality.(Especially to Marco_84: To see what was going on in my head: if we define ACD(X) as "ability to completely determine the observable X", then ACD (momentum) = 1-ACD(position), thereby fitting one sort of orthogonality, but contrary to the more appropriate definition of orthogonality in Hilbert space.) So, if I don't try everyone's patience too much, let me ask more simply: is it possible to measure the magnetic field and the electric field of a particle simultaneously? (I suspect, from malawi_glenn's closing comment that it is.)
    Thanks again.
     
    Last edited: Feb 17, 2008
  6. Feb 17, 2008 #5

    malawi_glenn

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    The magnetic field and E-field are generated by the same mechanism, so I would say yes, you can measure them simultaneously. But this was just by plain reason, I cant give you a theoretical derivation of this (not at this moment).
     
  7. Feb 17, 2008 #6
    nomadreid - I think you are confusing orthogonality of vectors in our 3-D position space with the idea of orthogonality in Hilbert space (although I'm not sure I understand your use of "orthogonal" with respect to incompatible observables). I could talk about the x and y displacement vectors that give position of a particle in space, and they are orthogonal but are certainly not incompatible in the sense of not being simultaneously observable. Eigenstates of position in the x direction would be orthogonal to eigenstates in the y direction, but that's just because they are linearly independent, not be cause the 3-space vectors are orthogonal.
     
  8. Feb 17, 2008 #7
    OK, Thanks

    OK, thanks for these last two replies. Clear. I think my original confusion was partly caused by the experiments involving spin, whereby the measurement apparati (apparatuses?) for spin in one or the other direction means that one cannot measure for the spin in two directions at once.
     
  9. Feb 18, 2008 #8
    Ok i think i've got where is you're problem:

    I think ure reading Stern_Gerlach right???

    What this experiment proves is the algebraic relation beetween the angular momentum operators:

    [tex][L_{\mu},L_{\nu}]=i\epsilon_{\mu\nu\lambda}L_{\lambda}[/tex]

    This expression suggest us that we cannot simoultanesly diagonalize all of them at the same time. In other words they do not commute.
    A good "experimental way" to see this is rotating a box by 90° along two differebt axis and then switch the operations.

    Then.

    You're question about E and B fields require the tools of second quantization since they are dynamic variables with infinetly degrees of freedom.

    Usually the canonical quantization of the radiation field is made upon the 4-vector potential [tex]A_{\mu}[/tex]... see Itzykson_Zuber "Quantum field theory" chapter 3.

    Ciao

    Marco
     
  10. Feb 18, 2008 #9
    Itzykson_Zuber "Quantum field theory"

    Marco: thanks for the reference. Unfortunately, not living where I can easily get hold of this book from a library or even a local bookstore, I would have to order it (and pay, and wait, for its overseas delivery, not to mention other hassles and expenses involved). Hence: before I will go to the trouble to buy it, I will look online for an explanation of the points you made in the last two sentences (maybe you have a recommendation for a link?), but if I can't find it: would you say it is worth buying it? (I read that some people found it dated.) (This is a real question, not a rhetorical one.) Consider that my background is in mathematics, not physics (i.e., I do not know what sorts of assumptions the book makes about the reader.)
     
  11. Feb 18, 2008 #10
    Electricity and Magnetism ARE commuting variables, though - their relationship is not, as I understand, governed by the uncertainty principle, and therefore the analogy would not apply.

    The use of "state vectors" is simply a mathematically convenient way of speaking about states. Orthogonal states merely means mutually exlcusive base states. The orthogonality of the EM field, however, does have a real space aspect. The Hilbert space representing the states of a system has no direct physical analogue.
     
  12. Feb 18, 2008 #11

    malawi_glenn

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    yes, the magnetic moment and electric moment of a nucleus for instance are commuting operators.
     
  13. Feb 18, 2008 #12
    Also many friends at my UNI think that is dated and they prefer something else...
    but i think that It is one of the most rigourus mathematically speaking...
    An if i got right you are mathematician!!

    Im currently studying those topics and using 4/5 books.
    Obviously Peskin.. then Lebellac is a really good one for connections with statistical mechanics...The bible (Weinberg) but i think it does not go so deep with the calculus, maybe im not that smart for those kind of books :)

    regards
    Marco
     
  14. Feb 19, 2008 #13
    "Devil's circles"

    Thanks to peter0301, malawi_glenn, and Marco_84 for the answers!
    Marco, I will follow up your recommendation and look into ordering those books. Yes, my academic specialty is in mathematics, and as this is in pure (as opposed to applied), maths/math (depending on your geographical position), various philosophical considerations involving Model Theory have brought me to looking into Quantum Theory; I have a number of books from different angles. I know that QM grew up as a potpourri of "well, this works, so use it"; nonetheless, except for the standard axiomatisations of Hilbert space, Measure Theory, and Projection Logics and a listing of about 20 "irreducible constants", I have not come across an axiomatisation of QM which is as compact as I feel is out there somewhere. For example, looking for a characterization of non-commuting variable pairs, I come across the explanation that two observables are incompatible if their associated Hamiltonians are not simultaneously diagonalizable, or more generally in terms of their spectral measures and associated Borel sets etc. However, since there seems to be no other criteria for assigning the details of the Hamiltonians, the characterization seems to be begging the question. p and q are incompatible because P and Q are non-commuting because p and q are incompatible…… I am obviously missing something that allows everyone to tell me that E and M are compatible (unless they just tell me that it is because the Hamiltonians commute, which again would be begging the question). Hence I must read further. But I am always grateful for guiding hints.
     
  15. Feb 19, 2008 #14
    minor correction

    oops, I meant "Hermitian operators" instead of "Hamiltonians" in that last diatribe, sorry. But that does not change my point.
     
  16. Feb 19, 2008 #15
    To answer this kinds of questions you should start reading an axiomatisation of QM and the problem of mesaure (not math) in physics in generally. I have a good book, but is in italian :(
    It start explaining what is the difference between an experiment of first kind and of the second kind... introducing then the postulates and stuff...
    I dont know... but im pretty sure that a lot of this work was done already by Von neumann and Gleason...

    If you cannot find anything that you can appreciate i think i would try to translate it...

    regards
    marco
     
  17. Feb 19, 2008 #16
    Non hò bisogno di traduzioni

    Grazie, Marco. Italian is a language that I read without difficulty (although I know longer use it actively, so speaking and writing it has dropped out of my capabilities, but I still read it). As far as the measure problem in general in Physics, I do have several references on that as well. (Most expositions are similar to Peeble's exposition, chapter 4 of his "Quantum Mechanics", although a somewhat different approach is taken by Nielsen and Chuang in their "Quantum Computation and Quantum Information".) All of these do put a few more steps, mainly working from Schrödinger's equation, in the "circolo vizioso" that I mentioned. So many steps, that one can lose sight of the vicious circle, but after I worked through these, the vicious circle seemed to still be there: there was always a step in which one was, essentially assuming compatibility or non-compatibility. I intend to work through them again to see if I can resolve the problem: that is, find some general criteria to find out, given any two observables without any hidden assumptions about their compatibility, to be able to test if they are compatible or not by any means besides experimental ones.
     
  18. Feb 19, 2008 #17
    You don't have to buy anything, it's here: http://people.seas.harvard.edu/~jones/ap216/lectures/lectures.html. EM-theory, QM and second-quantization (QED-theory). There, dE*dB>constant, is for example derived (similar to Heisenberg) and the Hamiltonian for the harmonic oscillator H=x^2+p^2 is an analogue to the EM-energy H~E^2+B^2, and operators could be thought of as: p~d/dx and B~dE/dx (not really but as a weak analogy it may help)

    Per
     
  19. Feb 19, 2008 #18
    Thanks. er...where?

    Thanks very much, per.sundqvist. I have bookmarked it, and will be looking through it carefully. At a first glance, I do not see dE*dB>const anywhere -- the closest I see is in III-4b of "Representation of Photon States" where dE*dH = ......; in that and the preceding section ("Quantization of the E-M Field" that deals with the canonical quantization of the fields), B is only mentioned explicitly in a formulation of Maxwell's Eq's, and implicitly in H. Do you mean that I should carry through the process to come up with that, or is this inequality derived elsewhere in the notes?
     
  20. Feb 19, 2008 #19
    I've never seen a simple or intuitive axiomizaiton of QM. I don't think there really is a "good one." Certainly you cannot boil it down to something as simple and intuitive as the 5 Euclidean postulates or Einstein's postulates of SR.

    The closest I've seen is many-worlds, but that nonetheless assumes a priori the validity of the Schrodinger Equation.
     
  21. Feb 19, 2008 #20

    reilly

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    Sad to say, in general the quantum E and B fields do not commute -- ultimately this can be seen by expressing the fields in terms of creation and destruction operators. All this stuff is discussed in detail in Mandel and Wolf, Optical Coherence and Quantum Optics, and in many other books. A Google on E&M field commutation rules be useful, no doubt.Regards,Reilly Atkinson
     
    Last edited: Feb 19, 2008
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