Quantum theory of optical coherence

Fix a detected position and polarization of waves. Write the electric field operator as $\hat{E}(t)={\hat{E}}^{(+)}(t)+{\hat{E}}^{(-)}(t)$, where the positive-frequency part contains annihilation operators and ${\hat{E}}^{(-)}=({\hat{E}}^{(+)}{)}^{†}$. An ideal absorbing photodetector responds towith density matrix $\rho$. The diagonal gives the single-count rate up to the detector coupling. Multiple absorptions require higher, normal-ordered field products. Roy J. Glauber used these correlations to develop the quantum account of optical coherence.^[ref]

The ordering follows from a detection amplitude. If absorption takes an initial pure state $|i⟩$ to an unobserved final state $|f⟩$, its field factor is $⟨f|{\hat{E}}^{(+)}|i⟩$. Squaring its complex modulus and summing over a complete orthonormal basis of final states gives $⟨i|{\hat{E}}^{(-)}{\hat{E}}^{(+)}|i⟩$. Averaging over initial preparations replaces this expression by a trace. Thus a detector that absorbs cannot count the Fock vacuum, although the vacuum has quantum fluctuations.

For one selected normal mode, ${\hat{E}}^{(+)}(t)=\mathcal{E}u(t)a$. All Fock states with nonzero occupation, coherent states, and thermal light in this same mode havewhere both denominators are nonzero. Its magnitude is one even when the expected value of the field vanishes. Perfect first-order optical coherence therefore does not identify a coherent state. Pair detection supplies additional information.

For a source whose counting correlations are stationary, define the dimensionless correlationAssume finite moments and nonzero denominators. The numerator is in normal ordering: two detections remove two photons. The denominator is the product of the two single-count factors, so independent counts have reference value one. This is a diagonal restriction of the four-point second-order field correlation, rather than its complete definition.^[ref]

For one selected mode with number operator $N=a^{†}a$ and $\mu =⟨N⟩>0$, evaluate both detections at the same time. The canonical commutation relation $a{a}^{†}=a^{†}a+1$ implies $N^2=(a^{†}{)}^2{a}^2+N$, henceThe factorial moment counts ordered pairs of distinct photons. Using $⟨N^2⟩$ would include pairing each photon with itself and would give the wrong coincidence prediction. At $\mu =0$ this normalized expression is undefined, rather than zero.