I am having troubles at distinguishing between the different states of the quantized electromagnetic field.

  • Consider one single mode of the free quantized e.m. field characterized by its wavevector $\vec{k}=\frac{w}{c}\vec{n}$ and its polarization $\vec{\varepsilon}$ which we assume to be linear.

If I want to compute the expected value for the energy if this mode contains zero photons, what should I use? The Fock state $| 0 \rangle$ or a coherent state $| \alpha \rangle=e^{-|\alpha|/2}$?

Thanks in advance.

  • 1
    $\begingroup$ Note that the Fock state $|n=0\rangle$ is equal to the coherent state $|\alpha=0\rangle$. $\endgroup$
    – Noiralef
    Jan 15, 2017 at 13:18

1 Answer 1


A state labeled by the precise occupation numbers of all modes is a Fock state. Since you know the occupation number, you're dealing with a Fock state. This isn't the only way to label quantum states since you are completely leaving out the phase of the photon out of the description. To talk about both numbers and phases, you can decompose the annihilation operator like this: $\hat a = (\hat N+1)^{1/2}\hat{ e^{i\phi}}$, and similarly $\hat a^\dagger$. In fact, if you work out the commutation relations you get a (perhaps surprising) number-phase uncertainty, $$\Delta\hat N \Delta \hat\cos\phi \gtrsim \frac{1}{2}|<\hat\sin\phi>|,$$ which means that Fock states lack any information about phase whatsoever. Coherent states, which are basically Gaussian wave-packets, minimize this uncertainty. But this means that they don't have well defined occupation numbers.

  • $\begingroup$ Correction: I meant whether the occupation number of a given mode is known. Of course the modes themselves are known, but the number of photons in a given mode is what we're interested in. $\endgroup$
    – franz g.
    Jan 15, 2017 at 10:43
  • 2
    $\begingroup$ Maybe you should edit your answer? "A single mode is a Fock state" is very misleading. $\endgroup$
    – Noiralef
    Jan 15, 2017 at 13:17

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