Coherence

Remarks

The Cauchy–Schwarz inequality gives $\abs{\gamma_{12}(\tau)} \le 1$. Temporal coherence is described by $\gamma_{11}(\tau)$ at one point, and spatial coherence by $\gamma_{12}(0)$. Put two equal pinholes at $\mathbf r_1$ and $\mathbf r_2$, as in Young's experiment, and look where the path delay between them is $\tau$. There the interference fringes have visibility $V = \frac{2\sqrt{I_1 I_2}}{I_1 + I_2} \abs{\gamma_{12}(\tau)}$. By the Wiener–Khinchin theorem, $\Gamma_{11}$ is the Fourier transform of the power spectrum of the light, so a narrow spectrum gives a long coherence time.

Example: a monochromatic plane wave $U = a \, e^{i(\mathbf k \cdot \mathbf r - \omega t + \Phi)}$, with $a > 0$ and a phase $\Phi$ uniform on $[0, 2\pi)$, has $$\Gamma_{12}(\tau) = a^2 e^{i \mathbf k \cdot (\mathbf r_1 - \mathbf r_2)} e^{-i\omega\tau}.$$ So $\abs{\gamma_{12}(\tau)} = 1$ for all points and delays: the wave is fully coherent.

Non-example: if $U(\mathbf r_1, \cdot)$ and $U(\mathbf r_2, \cdot)$ are independent with mean zero, then $\Gamma_{12} = 0$ and the two points are incoherent. Light from two separate lamps behaves this way, and it gives no stable fringes.