Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A Canonical quantisation of the EM field

  1. Aug 3, 2016 #1
    I have just gone through chapter 14 on the QFT for the gifted amateur by Lancaster and Blundell. Quantising the electromagnetic field results in the Hamiltonian:
    $$\hat{H}=\int d^3p \sum^{2}_{\lambda=1} E_p \hat{a}^\dagger_{p\lambda} \hat{a}_{p\lambda}$$
    with ##E_p=|p|##. In this post ##p## represents the momentum 3-vector.
    My question is; how does the concept vacuum energy apply here? I think what is puzzling me is the fact that I see many authors arrive at this result:
    $$\hat{H}=\sum_{p\lambda}\hbar \omega_p (\hat{a}^\dagger_{p\lambda}\hat{a}_{p\lambda}+\frac{1}{2})$$.
    Also, the previous expression has an integration over ##p## as opposed to a sum.
    Maybe I am comparing it to the wrong Hamiltonian, but I think that after applying normal ordering I get rid of the 1/2 term.
     
  2. jcsd
  3. Aug 3, 2016 #2

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    The vacuum energy is subtracted in the first expression, and that's the correct one, because the 2nd is obviously divergent. In the usual formalism you call this description "normal ordering". The socalled "vacuum energy" is anyway unobservable within special relativity since there you can measure only energy differences and not the absolute energy of a system. This is a first (almost trivial) example for "renormalizing" an observable quantity like the energy of the electromagnetic field, whose associated operator is given by your first expression.
     
  4. Aug 3, 2016 #3
    Yes that makes sense. Now I remember going through that infinite energy problem. Cheers!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Canonical quantisation of the EM field
  1. Quantising Fields (Replies: 1)

Loading...