Is single photon monochromatic or not?

A single photon need not be monochromatic in the same way as an electron need not have a fixed energy. A coherent state is a superposition of states with different numbers of photons so that the uncertainty of electric and magnetic fields is minimal. There are no corresponding states for an electron, as there exists a super selection rule for the electron number (but not for photon number).

 

There can't be superpositions between states containing different number of electrons. This is easiest to see for states differing by one electron. As the electron is a fermion, any matrix element ##\left n| V | n+1\right must vanish, as it gets multiplied by -1 upon a rotation by 360 degrees (i.e. the identity) as the electron is a fermion.

 

There are many types of single photons. Some types are "monochromatic" and other types are not.

 

The uncertainty relation in the case of electron wavepacket is between position and momentum. Localized electron wavefunction in free space is supposed to be realized by introducing uncertainty in momentum, I guess it is this momentum spread which you referred to as "multi-frequency".

I think your English skill obscures your intention about what you actually wanted to say. The state of a quantum system is something one can control, for the states of photon, it can be number state, coherent state, squeezed state, etc. It's not like when the (expectation value of) number of photons increases, the state of light becomes more and more coherent. You can have number states with arbitrarily larrge number of photons, yet the electric field corresponding to such state is not coherent at all.
However, for the special case of coherent state, the observed electric field does exhibit certain dependency between the coherency, which may be defined as the uncertainty in phase at a given time, and the expected number of photons ##\langle \hat{n} \rangle##. For sufficiently large value of ##\langle \hat{n} \rangle##, the uncertainty in the phase of the electric field turns out to have the form ##\Delta \phi = 1/(2\sqrt{\langle \hat{n} \rangle})##. Therefore, if the number of photons is very large, the electric field becomes more and more coherent (i.e. it gets closer to being a classical field oscillation), in the sense that its phase becomes more definite.

 

Yes, correct, and the operator if the field is non-diagonal in particle number, hence it's expectation value vanishes.

 

Yes, that's the right expression. Now express the field operator in terms of the lowering and raising operators (similar to those in harmonic oscillator) and also remember that ##|\textrm{number of photons}\rangle## behaves exactly like the eigenstates of harmonic oscillator.