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Photons from strings, from expert to layman.

  1. Apr 15, 2014 #1
    Don't know how I bumped into this but thought some of you might like this,


    https://groups.google.com/forum/#!msg/sci.physics.research/5w2S-j0Vyfw/A95XAYUGaWgJ

    By, Urs Schreiber

    4/11/03

    photon as strings for peasants (was: Meaning of dilaton field)

    Oz schrieb:
    > I presume photons are represented as strings.
    >
    > I would be quite grateful for a 'photon as strings for peasants'
    > description, if such a thing is possible.
    Yes, there is a state of the open string that corresponds to a photon.

    Consider first the open bosonic string: It can oscillate. Classically
    the oscillation can be described by an amplitude A. This amplitude is of
    course a spacetime vector with components A = A^mu e_mu, where e_mu are
    ordinary orthonormal basis vectors of spacetime. Every direction A^mu of
    oscillation behaves pretty much like a harmonic oscillator.

    Quantizing these harmonic oscillators sends the amplitude A^mu to the
    operator a^mu. This has an adjoint: c^mu = (a^dagger)^mu. Acting with
    this operator on the quantum state of the harmonic oscillator makes the
    oscillation energy in the direction e_mu go up by one quantum. There is
    a ground state, denoted |0>, which belongs to a string which oscillates
    as little as the fuzzyness of quantum mechanics will allow.

    Due to some details that I omit, it turns out that this ground state has
    a negative ground state energy. When we act on it with the operator c^mu
    to create a state in which the string oscillates a little bit along the
    e_mu direction, its energy goes up by one unit and becomes exactly zero.
    This state describes a massless photon with polarization along e_mu.

    So classically a bosonic string in a photon state with polarization in
    the e_1 direction roughly looks like this:


     c^1|0> =
     
                 |              |                 |
              |  |  |        |  |  |           |  |  |          |  
           |  |  |  |  |     |  |  |  |     |  |  |  |  |    |  |
     ...|  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | |  |  |......
           |  |  |  |  |     |  |  |  |     |  |  |  |  |    |  |
              |  |  |        |  |  |           |  |  |          |
                 |              |                 |
     ^
     | e_1 direction
     |
     |
     |
     |-----------> e_0 direction (time)


    (The photon must move at the speed of light. So you must really imagine
    a third axis perpendicular to the e_1-e_0 plane along which the string
    is moving while performing the above oscillations.)

    To get a more vivid impression of this classical approximation to the
    photon state go to
    http://viswiz.imk.fraunhofer.de/~nikitin/course_practicum/applet.html [Broken]
    and draw a very narrow ellise into the applet window. The major axis of
    this ellipse corresponds to the polarization vector of the "photon
    state" of the bosonic string that the applet will animate for you.

    This appeals to intuition, doesn't it? But there is a problem. If there
    are strings, then they are not bosonic, but supersymmetric. I suggest
    picturing an open superstring as a bosonic string which has at every
    point a little arrow attached. Like this:


       |
       |
       \                       <--- a generic piece of open bosonic string
        \
         \     _
         |    / \___
        /   _/      \
        \__/
         


       <
       <
       v                       <--- a generic piece of open superstring
        >
         ^     ^
         ^    < >>^<
        >   ^<      <
        v<<^


    See the little arrows at every point of the string? This are the
    fermionic superpartners of the oscillation degrees of freedom of the
    string. Such a little arrow in the direction e_mu is created by an
    operator called b^mu (er, roughly, imagine one or two technicalities
    omitted). It turns out the the photon state for the superstring looks
    formally just like that of the bosonic string, but with the operator
    a^mu replaced by the operator b^mu:

     superstring photon state polarized along e_mu  =  b^mu |0> .

    So when the superstring tries to mimic a photon it doesn't oscillate at
    all (except for the constant little ground state fluctuations demanded
    by quantum mechanics) - but it has a little arrow pointing along e_mu.
    Hence b^1|0> simply looks like this:


     c^1|0> =
     
     ^
     | e_1 direction
     |
     |   ....^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^....
     |
     |-----------> e_0 direction


    That's much quicker to draw, one of the many advantages of
    supersymmetry! :-)

    Of course the superstring can also perform honest oscillations like the
    bosonic string. But none of these correspond to massless states.


     
    > In some ways I find the stringy (particularly the loop ones - not
    > confused with LQG) idea quite nice except I am told that the string that
    > generates a photon is unimaginably small. This being in marked contrast
    > to the photon wavelength which is typically rather large.
    Be warned that the oscillation of the string does not "explain" the
    wavelike nature of the photon. The center-of-mass of that line that
    makes up the string is a point and does have quantum wave properties allκ
    by itself.

    > But there is the odd thing that a photon does seem to oscillate in and
    > out of our spacetime
    Does it seem so? I hadn't noticed! :-)
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. May 4, 2014 #2
    I'm sorry you are not generating any responses at the moment. Is there any additional information you can share with us? Any new findings?
     
  4. May 5, 2014 #3
    Was not expecting any responses. I just bumped into a short work by an expert in the field trying to make simple that which is complicated. Just thought some here would like it.
     
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