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Fixed amplitude of electric field operator in quantum optics

  1. Aug 26, 2015 #1
    Hi guys, I'm trying to understand why does the amplitude of the electric field operator in a cavity is fixed at

    [tex]\left ( \displaystyle\frac{\hbar\omega}{\epsilon_{0}V} \right )^\frac{1}{2} [/tex]

    Every book I read says it is a normalization factor... but, normalizing an operator?, what is the meaning of that in this context?, they also call it the "aplitude per photon"...

    I know the "V" (volume of the cavity) comes from the normalization of the cavity modes, but all the other quantities seem arbitrary to me.

    It does makes sense for the amplitude to be fixed that way in order to obtain a correct expresion for the field intensity... but in the quantization process I don't see a mathematical reason for that specific value... why can't I multiply this amplitude by an arbitrary constant?

    Thanks! :wink:
  2. jcsd
  3. Aug 26, 2015 #2


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    It's an operator. That is, in a sense, it is a projection.

    What do you get when you apply the same operator twice? That tells you what you need to put in for a normalization.
  4. Aug 26, 2015 #3
    Thanks for your answer, but I think your advice does not apply in this case, as the electric field operator is not a projection (its trace is zero).
  5. Aug 27, 2015 #4


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    Start with a classical single mode field that satisfies Maxwell's equation
    E_x(z,t)= ( \frac{2\omega^2}{V\epsilon_0}) q(t) \sin (kz)

    where q(t) is the canonical position. If you now quantize this equation in the usual way by introducing the annihilation (and creation) operators
    [itex] \hat{a}= (2 \hbar \omega)^{-1/2} (\omega \hat{q} +i \hat{p}) [/itex]
    you get that prefactor for the E field operator.

    Also, note that V is the effective (mode) volume, not the volume of the cavity.
  6. Aug 27, 2015 #5
    First of all, thanks for the answer f95toli.

    That's exactly my point, every quantum optics book starts assuming that form (with a fixed amplitude) for a classical single mode field, even though you can multiply that same solution by an arbitrary constant and still satisfy maxwell equations.
  7. Aug 27, 2015 #6


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    But then the fields won't satisfy the canonical commutation relations.
  8. Aug 31, 2015 #7
    Thanks Avodyne! That's the reason! :biggrin:
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