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Quantisation of EM field

  1. May 27, 2008 #1


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    Particle energy and momentum are obtained from the wave function as eigenfunctions of the spatial and time derivative operators.

    Is this true of ElectroMagnetic fields? In other words are E and B eigenfunctions of a differenial operator? I can see that E and B could be interpreted as a kind of generalised velocity (q) because that's how the flux of E and B are definded through a surface. But I can't find any websites or references that show how E and B corespond to generalised co-ordinates and their derivatives. Nor can I find what the EM wavefunction or what its Schrodinger equation might be

    Anybody help me?
  2. jcsd
  3. Jun 1, 2008 #2
    The dynamical quantity in quantum mechanics (in canonical quantization) is the wavefunction, [itex]\psi(\mathbf{x},t)[/itex], which must satisfy the (time-dependent) Schrodinger equation
    [tex]i\frac{\partial\psi}{\partial t}=\hat{H}\psi,[/tex]​
    and as you said, this yields the energy and momentum allowed by the system.

    On the other hand, the electric and magnetic fields, [itex]E(\mathbf{x},t)[/itex] and [itex]B(\mathbf{x},t)[/itex] are not wavefunctions -- they are classical objects which satisfy classical equations of motions, namely the Maxwell equations, whose solution merely give field configurations.

    It is important not to confuse a field (such as the electromagnetic or gravitational) for a wavefunction. I can see how this is tempting since they are both objects that assign a numeric value to every point in space and time.

    The correct way to obtain the electromagnetic wavefunction and its associated Schrodinger equation is to canonically quantize the electromagnetic field, a starting point for the study of quantum field theory. The end result of this gives a Schrodinger equation identical to that stated above, but with a drastically different Hamiltonian and with a wavefunctional assigning a (complex) number to every electromagnetic field configuration.

    Canonical quantization is a well established procedure and details can be found in any respectable quantum field theory text. Briefly, the procedure is analogous to canonically quantizing simple mechanical systems such as a free particle or a simple harmonic oscillator: write down the Lagrangian that describes the system (the Maxwell action in our case), identify the generalized coordinates and their canonical momentum conjugate, perform a Legendre transform to obtain the hamiltonian, upgrade the coordinates and momenta to operators, and impose (equal time) canonical commutation relations. Slight complications arise due to gauge invariance.
  4. Jun 1, 2008 #3


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    In quantizing the electromagnetic field, we don't use [itex]\vec E[/itex] and [itex]\vec B[/itex]. Instead, we use the electric potential [itex]\phi[/itex] and the magnetic vector potential [itex]\vec A[/itex], combined into the relativistic 4-vector potential [itex]A_{\mu}[/itex].
    Last edited: Jun 1, 2008
  5. Jun 2, 2008 #4


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    Many thanks for the replies. The quantisation of EM fields is still a puzzle to me. Speaking strategically, I get the idea of using a Hamiltonian in a wider variety of non-mechanical situations, but I still can't see what the goal is is. For example, there is no doubt even at the beginning of the analysis of, say, simple harmonic motion that the reader will obtain a wavefunction from which position, momentum and energy can be obtained as eignefunctions. But having converted the normal co-ordinates of the vector potential to operators, I just do not know what they operate on, nor what eigenfunctions they generate, nor can I see what quantities become probabalistic (e.g. an electroSTATIC field!). Would apreciate a reference or further help.
  6. Jun 2, 2008 #5
    I believe I know what you're trying to get at; you're trying to understand what a field operator really is and on what sort of Hilbert space (the bras and kets) they act. I believe this is best understood by thinking in analogy with elementary quantum mechanics, but I must make a few clarifications before I proceed.
    In elementary quantum mechanics, eigenstates of the position operator are states of definite position. Such states, in the position representation, are dirac delta functions located at positions. Likewise, eigenstates of the momentum operator are states of definite momentum, and such states are sinusoidal in the position representation.

    Interpretation of such operators in elementary quantum mechanics carry over into the language of quantum field theory quite directly. Classical fields, such as the electromagnetic field, are viewed as the continuum limit of a large collection of (harmonic) oscillators located at every point in spacetime. All oscillators are dynamical and their displacement is interpreted as nonvanishing field values. In other words, if an oscillator at spacetime point [itex](\mathbf{x},\,t)[/itex] is displaced from its equilibrium value, one interprets this as a nonvanishing field value at [itex](\mathbf{x},\,t)[/itex].

    When moving to the quantum world, the displacement of each oscillator becomes uncertain in spirit of Heisenberg's principle. Furthermore, we may now write down a position operator for each oscillator [itex]\hat{X}_i[/itex], where the kernel symbol [itex]\hat{X}[/itex] measures oscillator [itex]i[/itex]'s displacement. Likewise, we can write down a "momentum" operator [itex]\hat{P}_i[/itex] which is associated with the rate at which the displacement of oscillator [itex]i[/itex] is changing (i.e. velocity, or more properly, conjugate momentum).

    Our notation is nonstandard: in literature the subscript [itex]i[/itex] is traded out for a more descriptive space argument [itex]\mathbf{x}[/itex], and the kernel symbol denoting the oscillator displacement is usually written [itex]\hat\phi[/itex] or [itex]\hat{A}[/itex], depending on the field one is talking about. One usually writes [itex]\hat{A}(\mathbf{x})[/itex] to refer to the operator that is associated with displacement of the quantum oscillator at point [itex]\mathbf{x}[/itex], and is called the field operator.
    Our picture now is a large collection of quantum oscillators whose collective displacements are uncertain. However, the interpretation that oscillator displacements correspond to nonvanishing field strengths remains unchanged in moving to quantum mechanics. Therefore we must conclude that field strength values at every point in space is subject to quantum uncertainty. This means, for example, that the value of the electrostatic field in the vicinity of an electron can no longer be specified with complete uncertainty, as suggested by merely using Coulomb's law to infer it (of course, one could do this at the expense of completely loosing information of the field's velocity).
    Eigenstates of the field operator [itex]\hat{A}(\mathbf{x})[/itex] can be understood to be eigenstates of the displacement operators for all the oscillators: they are states of definite field configurations.
  7. Jun 5, 2008 #6
    Errata to the above post:

    2nd paragraph: "...located at positions." should read "located at various positions."
    6th paragraph: "...can no longer be specified with complete uncertainty," should read "...can no longer be specified with complete certainty,". (This error is critical)
  8. Jun 12, 2008 #7
    Actually, in the so called "second quantization" (it's just the quantization of the fields), we have to lift all fields to operators.
    So, there is no longer the probabilistic interpretation for the fields.
    However, the norm squared of the physical states in the Hilbert space are still interpreted as probabilities.

    Moreover, in the quantization of fields, what we care about are usually not the eigenstates of the fields, but the physical states composed the Fock space which are constructed by operating on the vacuum creation operators.

    So, E and B fields cannot be thought of as eigenstates of some operators. Actually, we need gauge boson in the symmetry describing various interactions. The E and B fields are components of the kinetic term of gauge boson. It's owing to the fact the E and B are components of a tensor, so in some particular frame, we can observe that, for example, B field vanishes.
  9. Jun 14, 2008 #8
    Sure there are...
    ..these states can certainly be of definite field configurations, for which, like you said, can carry a probabilistic interpretation. ;)

    Agreed! Eigenstates of field operators are states of definite field configurations, and because their applicability in research problems are very rare, their mention is dropped entirely from texts.

    Although the gauge symmetry exhibited by the electromagnetic fields complicates the whole quantization proceedure, you can make contact with classical electrodynamics by going to the Coulomb gauge, and write mode expansions for the electric and magnetic field operators in that gauge. And, as I mentioned above, states of definite electric and magnetic field configurations should be eigenstates of their respective operators.
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