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Now everything is a big mess again! Photons are not as simple as other quanta since they are massless and have spin 1. The first subtlety is that they have only 2 not three spin states. That's because there's no restframe for massless quanta. The full group theoretical foundation can be found in my lecture notes on QFT in appendix B:
http://th.physik.uni-frankfurt.de/~hees/publ/lect.pdf
The 2nd subtlety is that there's no position operator for photons. You can't treat photons in a first-quantization way to begin with. It's the very point where you really need QFT, if you want to get things right (although there are claims in the literature to the contrary).
The 3rd subtlety is that the naive canonical quantization doesn't work, because the classical Maxwell theory is a gauge theory. As is also shown in my QFT notes, that must be so from the group-theoretical point of view, if you don't want to introduce continuous spin-like degrees of freedom, which are not observed.
A quick hand-wavy way out (for the free field) is to completely fix the gauge. For the free em. field the most convenient choice is the radiation gauge, which consists of two gauge conditions for the vector field
$$A^0=0, \quad \vec{\nabla} \cdot \vec{A}=0.$$
The equation of motion is
$$\Box \vec{A}=(\partial_t^2-\Delta)\vec{A}=0.$$
Now we look for plane-wave modes and get
$$\vec{A}(t,\vec{x})=a(\vec{k},\lambda) \vec{\epsilon}_{\lambda} \exp(-\mathrm{i} \omega t + \mathrm{i} \vec{k} \cdot \vec{x})+c.c.$$
##\lambda## labels two polarization vectors, which can be chosen arbitrarily, but they must fulfill ##\vec{\epsilon}_{\lambda} \cdot \vec{k}=0## (i.e., you have two linearly independent choices). The most natural choice are the two circular polarized modes, which corresponding to helicity eigenmodes to the eigenvectors ##h \in \{-1,1 \}##. Then there's the dispersion relation ##\omega=|\vec{k}|##.
Now we skip all the subtleties of canonical quantization by observing that we have a set of harmonic oscillators there and thus we can make the ansatz
$$\hat{\vec{A}}(t,\vec{x}) = \sum_{\vec{k},\lambda} \frac{1}{\sqrt{2 \omega}} \hat{a}(\vec{k},\lambda) \vec{\epsilon}(\vec{k},\lambda) \exp(-mathrm{i} k \cdot x)|_{k^0=\omega}+\text{h.c.}$$
Here, I've assumed a large quantization volume as a cube of length ##L## with periodic boundary conditions for the field operator, so that the momenta are discrete, ##\vec{k} \in \frac{2 \pi}{L} \mathbb{Z}^3##.
The creation and annhiliation operators fulfill the Bose commutator relations
$$[\hat{a}(\vec{k},\lambda) ,\hat{a}(\vec{k}',\lambda')]=0, \quad [\hat{a}(\vec{k},\lambda),\hat{a}^{\dagger}(\vec{k}',\lambda')]=\delta_{\vec{k},\vec{k}'} \delta_{\lambda,\lambda'}.$$
The total four-momentum is given by
$$\hat{P}^{\mu} = \sum_{\vec{k},\lambda} \hat{N}(\vec{k},\lambda) k^{\mu}, \quad \hat{N}(\vec{k},\lambda)=\hat{a}^{\dagger}(\vec{k},\lambda) \hat{a}(\vec{k},\lambda).$$
The ground state is the vacuum ##|\Omega \rangle## defined by
$$\hat{a}(\vec{k},\lambda) |\Omega \rangle=0.$$
A single-mode single-photon state is given by
$$|\vec{k},\lambda \rangle=\hat{a}^{\dagger}(\vec{k},\lambda) |\Omega \rangle.$$
A single-mode coherent state that is most closely related to a classical em. field is given by the eigenvector of ##\hat{a}(\vec{k},\lambda)##,
$$\hat{a}(\vec{k},\lambda) |\alpha \rangle = \alpha |\alpha \rangle.$$
Since ##\hat{a}## is NOT self-adjoint ##\alpha \in \mathbb{C}##. It's not a state of definite photon number. The probability to find ##n## photons (of momentum ##\vec{k}## and with polarization ##\lambda## of course) is given by a Poisson distribution
$$P(n)=\exp(-n|\alpha|^2) \frac{|\alpha|^{2n}}{n!}$$
The mean photon number and four-momentum is
$$\langle N(\vec{k},\lambda) \rangle=|\alpha|^2, \quad \langle \vec{P} \rangle=k^{\mu} |\alpha|^2.$$
The operators for the electromagnetic field components are given by
$$\hat{\vec{E}}(t,\vec{x})=-\partial_t \hat{\vec{A}}(t,\vec{x}), \quad \hat{\vec{B}}(t,\vec{x})=\vec{\nabla} \times \hat{\vec{A}}(t,\vec{x}).$$
For a given state you can calculate expectation values as for any observable according to Born's rule.
http://th.physik.uni-frankfurt.de/~hees/publ/lect.pdf
The 2nd subtlety is that there's no position operator for photons. You can't treat photons in a first-quantization way to begin with. It's the very point where you really need QFT, if you want to get things right (although there are claims in the literature to the contrary).
The 3rd subtlety is that the naive canonical quantization doesn't work, because the classical Maxwell theory is a gauge theory. As is also shown in my QFT notes, that must be so from the group-theoretical point of view, if you don't want to introduce continuous spin-like degrees of freedom, which are not observed.
A quick hand-wavy way out (for the free field) is to completely fix the gauge. For the free em. field the most convenient choice is the radiation gauge, which consists of two gauge conditions for the vector field
$$A^0=0, \quad \vec{\nabla} \cdot \vec{A}=0.$$
The equation of motion is
$$\Box \vec{A}=(\partial_t^2-\Delta)\vec{A}=0.$$
Now we look for plane-wave modes and get
$$\vec{A}(t,\vec{x})=a(\vec{k},\lambda) \vec{\epsilon}_{\lambda} \exp(-\mathrm{i} \omega t + \mathrm{i} \vec{k} \cdot \vec{x})+c.c.$$
##\lambda## labels two polarization vectors, which can be chosen arbitrarily, but they must fulfill ##\vec{\epsilon}_{\lambda} \cdot \vec{k}=0## (i.e., you have two linearly independent choices). The most natural choice are the two circular polarized modes, which corresponding to helicity eigenmodes to the eigenvectors ##h \in \{-1,1 \}##. Then there's the dispersion relation ##\omega=|\vec{k}|##.
Now we skip all the subtleties of canonical quantization by observing that we have a set of harmonic oscillators there and thus we can make the ansatz
$$\hat{\vec{A}}(t,\vec{x}) = \sum_{\vec{k},\lambda} \frac{1}{\sqrt{2 \omega}} \hat{a}(\vec{k},\lambda) \vec{\epsilon}(\vec{k},\lambda) \exp(-mathrm{i} k \cdot x)|_{k^0=\omega}+\text{h.c.}$$
Here, I've assumed a large quantization volume as a cube of length ##L## with periodic boundary conditions for the field operator, so that the momenta are discrete, ##\vec{k} \in \frac{2 \pi}{L} \mathbb{Z}^3##.
The creation and annhiliation operators fulfill the Bose commutator relations
$$[\hat{a}(\vec{k},\lambda) ,\hat{a}(\vec{k}',\lambda')]=0, \quad [\hat{a}(\vec{k},\lambda),\hat{a}^{\dagger}(\vec{k}',\lambda')]=\delta_{\vec{k},\vec{k}'} \delta_{\lambda,\lambda'}.$$
The total four-momentum is given by
$$\hat{P}^{\mu} = \sum_{\vec{k},\lambda} \hat{N}(\vec{k},\lambda) k^{\mu}, \quad \hat{N}(\vec{k},\lambda)=\hat{a}^{\dagger}(\vec{k},\lambda) \hat{a}(\vec{k},\lambda).$$
The ground state is the vacuum ##|\Omega \rangle## defined by
$$\hat{a}(\vec{k},\lambda) |\Omega \rangle=0.$$
A single-mode single-photon state is given by
$$|\vec{k},\lambda \rangle=\hat{a}^{\dagger}(\vec{k},\lambda) |\Omega \rangle.$$
A single-mode coherent state that is most closely related to a classical em. field is given by the eigenvector of ##\hat{a}(\vec{k},\lambda)##,
$$\hat{a}(\vec{k},\lambda) |\alpha \rangle = \alpha |\alpha \rangle.$$
Since ##\hat{a}## is NOT self-adjoint ##\alpha \in \mathbb{C}##. It's not a state of definite photon number. The probability to find ##n## photons (of momentum ##\vec{k}## and with polarization ##\lambda## of course) is given by a Poisson distribution
$$P(n)=\exp(-n|\alpha|^2) \frac{|\alpha|^{2n}}{n!}$$
The mean photon number and four-momentum is
$$\langle N(\vec{k},\lambda) \rangle=|\alpha|^2, \quad \langle \vec{P} \rangle=k^{\mu} |\alpha|^2.$$
The operators for the electromagnetic field components are given by
$$\hat{\vec{E}}(t,\vec{x})=-\partial_t \hat{\vec{A}}(t,\vec{x}), \quad \hat{\vec{B}}(t,\vec{x})=\vec{\nabla} \times \hat{\vec{A}}(t,\vec{x}).$$
For a given state you can calculate expectation values as for any observable according to Born's rule.
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