- #1
Mr-R
- 123
- 23
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.
$$\hat{H}=\int d^3p \sum^{2}_{\lambda=1} E_p \hat{a}^\dagger_{p\lambda} \hat{a}_{p\lambda}$$
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.