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Intensity in Electromagnetism versus probability density in Quantum Mechanics

  1. Apr 9, 2012 #1
    In the book Introduction to Quantum Optics: From the Semi-classical Approach to Quantized Light by Grynberg, Aspect and Fabre I came across the following statement on page 385:

    "By the end of the nineteenth century, classical electromagnetism,.........., provided a wave description of almost all known optical phenomena (adding the postulate that the quantity measured in optics, called the light intensity, is proportional to the average of the squared electric field of the Maxwellian wave)."

    An analogous statement for quantum mechanics might read: the Schrodinger equation describes all known non-relativistic electronic phenomena (adding the postulate that the quantity measured, called electron probability density, is proportional to the squared modulus of the Schrodinger wavefunction.)

    Now I knew that the postulate concerning the squared modulus of the wavefunction is fundamental to interpreting measurements in quantum mechanics but I didn't know that you need a postulate concerning the square of the electric field to interpret measurements in optics. Can't you always (in principle at least) measure the electric field itself? Is this additional postulate concerning the intensity a fundamental part of classical electromagnetism?
     
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  3. Apr 10, 2012 #2

    mfb

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    You can measure the electric field. But if you talk about "optics", the usual way to detect light does not measure the fast oscillation of the electromagnetic fields in a direct way, but only the light intensity on a surface.

    In QM, this is a bit different - there is no way to measure the wave function directly and in an objective way. The amplitude squared is a physical value, the phase is not. Only phase differences are measurable.
     
  4. Apr 13, 2012 #3

    A. Neumaier

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    My lecture on ''Optical models for quantum mechanics''
    http://www.mat.univie.ac.at/~neum/ms/optslides.pdf
    shows that the analogy is in fact quite close.
     
  5. Apr 13, 2012 #4

    Hans de Vries

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    The two are basically the 2 seperate steps from the (spin 1/2) field
    to the (spin 2) gravitation field.

    - spin 1/2 electron field
    - spin 1 vector field
    - spin 2 tensor field

    spin 1/2 electron field components add linear.
    [tex]\psi=\psi_1+\psi_2[/tex]
    The electron field "squared" gives the electric current
    [tex]j^\mu = \bar{\psi}\gamma^\mu\psi[/tex]
    So the currents of the electron field components don't add linear.
    [tex]j^\mu \neq j^\mu_1+j^\mu_2[/tex]
    The current gives rise to the electromagnetic potential and Fields.
    [tex]j^\mu \longrightarrow A^\mu,~\vec{E},~\vec{B}[/tex]
    The electromagnetic potential and Fields add linear.
    [tex]A^\mu =A^\mu_1+A^\mu_2,~~~\vec{E}=\vec{E}_1+\vec{E}_2,~~~\vec{B}=\vec{B}_1+\vec{B}_2[/tex]
    The electric fields square gives the energy density (and the rest of the SE Tensor)
    [tex]{\cal E}=\frac12\Big(E^2+B^2\Big)[/tex]
    So the energy densities from two electromagnetic fields do not add linear
    [tex]{\cal E}\neq{\cal E}_1+{\cal E}_2[/tex]
    The energy densities (and the other [itex]T^{\mu\nu}[/itex] tensor components) give rise to gravity.
    [tex]T^{\mu\nu} \longrightarrow curvature[/tex]
    Gravitational fields then add linear again (1st aprox)

    Hans.
     
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