Graduate Canonical quantisation of the EM field

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SUMMARY

The discussion focuses on the canonical quantization of the electromagnetic (EM) field as presented in "Quantum Field Theory for the Gifted Amateur" by Lancaster and Blundell. The Hamiltonian derived is expressed as $$\hat{H}=\int d^3p \sum^{2}_{\lambda=1} E_p \hat{a}^\dagger_{p\lambda} \hat{a}_{p\lambda}$$ with ##E_p=|p|##. The conversation highlights the concept of vacuum energy and its relation to normal ordering, clarifying that the term $$\frac{1}{2}$$ in the Hamiltonian is removed through normal ordering, addressing the divergence issue in the second Hamiltonian expression $$\hat{H}=\sum_{p\lambda}\hbar \omega_p (\hat{a}^\dagger_{p\lambda}\hat{a}_{p\lambda}+\frac{1}{2})$$.

PREREQUISITES
  • Understanding of quantum field theory (QFT) principles
  • Familiarity with Hamiltonian mechanics in quantum systems
  • Knowledge of normal ordering in quantum operators
  • Basic concepts of vacuum energy and its implications in physics
NEXT STEPS
  • Study the derivation of the Hamiltonian for the electromagnetic field in QFT
  • Explore the concept of normal ordering and its role in renormalization
  • Investigate vacuum energy and its implications in special relativity
  • Review the differences between integration and summation in quantum Hamiltonians
USEFUL FOR

Physicists, particularly those specializing in quantum field theory, researchers in theoretical physics, and students seeking to understand the quantization of fields and the implications of vacuum energy.

Mr-R
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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.
 
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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.
 
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Yes that makes sense. Now I remember going through that infinite energy problem. Cheers!
 

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