# Vacuum energy from the quantization of the electromagnetic field

I'm studying the quantization of the free electromagnetic field and at a certain point there is the Hamiltonian operator as: $$\hat{H}^f_{em}=\sum_\lambda\frac 1 2 (\hat{P}^2_\lambda+\omega_\lambda^2\hat{Q}^2_\lambda)$$ Since every contribute in the Hamiltonian of an harmonic oscillator I can write the eigenvalues of the free electromagnetic field as sum of the oscillators energy: $$H^f_{em}=\sum_\lambda (n_\lambda+\frac 1 2 )\hbar \omega_\lambda$$ Then my professor told us that fixing all the $$n_\lambda=0$$ the resulting energy represent the vacuum energy.

I don't understand this observation because I remember that in the quantum oscillator $$\frac 1 2 \hbar \omega$$ represents the minimum energy of the particle, and so, $$n_\lambda=0$$ state should be the state at which all the photons are at the lowest energy (instead of the state at which there are no photons).

• $n_\lambda=0$ means “no photons of any frequency”. Oct 29 '19 at 3:52
• Just want to comment that the vacuum energy coming from EM $\int \sqrt{-g}F^{\mu\nu}F_{\mu\nu}= \int \sqrt{-g}F_{\mu\nu}F_{\mu'\nu'}g^{\mu\mu'}g^{\nu\nu'}$ can NOT contribute to the cosmological constant term $\Lambda \int \sqrt{-g}$. Under a scale transformation of the metric, the EM vacuum behave differently than the cosmological constant, due to the extra $g^{\mu\mu'}g^{\nu\nu'}$ dependence. Oct 29 '19 at 21:18

The vacuum is, by definition, the lowest energy state. The lowest energy state is the one with every photon occupation number $$n_{\lambda}=0$$.