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Basic Photon Question

  1. Jan 13, 2008 #1
    Hi everyone,

    I understand that a photon is interlinked packet of energy with oscillating electric and magnetic fields. My question is what is the electric field strength associated with a single photon? For a capacitor we of course use V/d and could use this to calculate the e-field of an atom (using the separation of the proton and electron for d), but what about the e-field of a photon so one can compare between the two?

    Thanks,

    L
     
    Last edited: Jan 13, 2008
  2. jcsd
  3. Jan 14, 2008 #2

    f95toli

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    The expectation value of the E field for a single photon (or, more generally, for ANY state with a well-defined number of photons; i.e number state) is zero.
    However, the variance of the E field is non-zero.
     
  4. Jan 14, 2008 #3

    Claude Bile

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    Photons are packets of energy and momentum, it doesn't really make sense to say that a photon has an intrinsic field.

    Claude.
     
  5. Jan 14, 2008 #4

    pam

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    The energy density of an EM wave = [tex]E^2/4\pi[/tex] in Gaussian notation.
    Multiplying this by a reasonable length of the wave packet times a reasonable area for the wave front gives the energy in the EM wave. This energy equals hf for a single photon.
    Using this, E for a single photon can be found.
     
  6. Jan 15, 2008 #5

    f95toli

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    No it can not. E (or, more correctly since it is QM, <E>) for a single photon is zero. It is NOT "undefined" or anything like that; it is just exactly zero; <n|E|n>=0
    Any text about basic quantum optics will cover this.

    Or, see e.g.
    http://people.seas.harvard.edu/~jones/ap216/lectures/ls_3/ls3_u3/ls3_unit_3.html

    Eq III-3A is the E-field operator; if you take the expectation value of this with respect to a number state (any state with a specfic number of photons, e.g. one) you always get zero.

    I know this isn't exactly a pedagogical explanation; but unfortunately it is impossible to avoid math when dealing with photons; I don't think there is an "intuitive" explanation.
     
  7. Jan 15, 2008 #6

    clem

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    I think Pam is referring to the amplitude of E. Since E is oscillating its expectation value is zero.
     
  8. Jan 15, 2008 #7

    Claude Bile

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    There is a flaw in this method.
    Having a wave-packet of finite length implies that the packet contains a spectrum of frequencies. It is therefore wrong to imply that you can define a specific energy for the photon. At best all you can do is define the width of the frequency spectrum.

    Claude.
     
  9. Jan 17, 2008 #8

    pam

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    That is the importance of the phrase "reasonable length".
    For instance, if a ruby laser of a given power is pulsed on for 1 nanosecond,
    the average amplitude of E per photon can be calculated.
     
  10. Jan 17, 2008 #9

    f95toli

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    That still doesn't make sense. An ordinary laser will not generate a Fock state (more likely a coherent state), so the number of photons is stricly speaking not fixed. Hence, there can't be an "E field per photon".
    Of course it is always possible to use a semi-classical calculation if the power is high enough, e.g. just divide the energy with [itex] \hbar \omega [/itex]. That will give you a handle on the average number of photons in the system and is useful for e.g. looking at the thermal occupancy of states. However, for a single photon system this expectation value will be very,very small (much smaller than the fluctations) meaning it still doesn't make sense to talk about this in terms of the field per photon.
     
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