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).

  • $\begingroup$ $n_\lambda=0$ means “no photons of any frequency”. $\endgroup$
    – G. Smith
    Oct 29, 2019 at 3:52
  • $\begingroup$ 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. $\endgroup$
    – MadMax
    Oct 29, 2019 at 21:18

1 Answer 1


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

  • $\begingroup$ I edited the question to be more clear about my problem, what you say is what I don't understand $\endgroup$
    – SimoBartz
    Oct 29, 2019 at 6:01
  • $\begingroup$ To me it looks like this is the lowest energy state of the em field, If I remove the em field from the space I remove this extra energy. I know I'm wrong but I don't understand why $\endgroup$
    – SimoBartz
    Oct 29, 2019 at 6:22
  • $\begingroup$ @SimoBartz Actually, when there were no electromagnetic field, it would not make sense to talk about its lowest energy state, i.e. characterizing something that would not exist. There concept of an existing quantum EM-field is that if there are no quanta(photons), the EM-field still manifests through its vacuum state which can still interact with matter under given circumstances. $\endgroup$ Oct 29, 2019 at 9:41

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