Why Does the Intensity of Thermal Light Follow an Exponential Distribution?
Haifei1. An Elegant Argument
Nobel laureate Ketterle gave a very elegant argument for the \(g^{(2)}\) of a classical thermal light field in Open Course 8.422:
According to the central limit theorem, the electric field \(E\) follows a Gaussian distribution, and thus the light intensity \(I\) follows an exponential distribution.
The probability density function (pdf) of the exponential distribution is: \(f(x) = \gamma e^{-\gamma x}\) .
The moments of the exponential distribution have the following property:
\[\langle I^n \rangle = n! \langle I\rangle^n\]Thus, the classical second-order correlation function \(g^{(2)}(0)\) of a thermal light field is
\[g^{(2)} := \langle I^2\rangle/\langle I\rangle^2 = 2\]This argument is very elegant because it makes full use of the “chaotic” nature of thermal light: according to the central limit theorem, regardless of the probability distributions of the electric fields in the individual frequency modes, their superposition must follow a Gaussian distribution. Therefore, the square of the electric field follows an exponential distribution.
2. A Review of Probability and Statistics
Wait, is it really this simple? Let us analyze this using the probability and statistics knowledge we learned as undergraduates. If a random variable follows a Gaussian distribution, does its square really follow an exponential distribution?
Let us write down the cdf (cumulative distribution function) of the Gaussian distribution:
\(p(E\lt x) = \Phi(x) = \int_{-\infty}^{x} \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{t^2}{2\sigma^2}} \mathrm{d}t\)
Then the cdf of the light intensity is
\(p(I\lt y) = p(E^2 \lt y) = p(-\sqrt{y}\lt E\lt \sqrt{y}) = \Phi(\sqrt{y}) - \Phi(-\sqrt{y}) = 2\Phi(\sqrt{y}) - 1\)
Therefore, the pdf of the light intensity should be
\(f(y) = p^\prime(I \lt y) = 2 \Phi^\prime(\sqrt{y}) = \frac{1}{\sigma\sqrt{2\pi y}} e^{-\frac{y}{2\sigma^2}}\)
This is not an exponential distribution! Compared with a true exponential distribution, it has an extra \(y^{-1/2}\) factor.
Where is the problem?
We can take a different approach and ask the reverse question: if a random variable follows an exponential distribution, what distribution does its square root follow?
The cdf of the exponential distribution is:
\(p(I\lt y) = 1 - e^{-\gamma y}\)
Then the cdf of the electric field is:
\(p(E \lt x) = \frac{1 + p(-x\lt E\lt x)}{2} = \frac{1 + p(I\lt x^2)}{2} = \frac{2 - e^{-\gamma x^2}}{2}\)
Therefore, the pdf of the electric field should be:
\(f(x) = p^\prime(E\lt x) = \gamma x e^{-\gamma x^2}\)
This distribution is called the Rayleigh distribution, and compared with the Gaussian distribution it has an extra \(x\) factor.
Students with keen intuition may realize that the Rayleigh distribution actually corresponds to the magnitude of a two-dimensional random vector following a Gaussian distribution:
\[f(r) = \int_{0}^{2\pi} \frac{1}{2\pi} e^{-\frac{r^2}{2\sigma^2}} r\mathrm{d}\theta\]Wait, is the electric field actually a two-dimensional random variable? Exactly! In addition to its amplitude, an electric field also has a phase—it is in fact a complex number! If a complex random variable follows a Gaussian distribution, then its squared modulus does indeed follow an exponential distribution.
At this point, the problem is perfectly resolved.
3. From Classical to Quantum
However, all of the above can only show that the classical \(g^{(2)}\) is 2; it cannot show that the quantum \(g^{(2)}\) is also 2.
The only difference between quantum \(g^{(2)}\) and classical \(g^{(2)}\) is that the former takes into account the particle nature of light. If you detect one photon, then the light field has one fewer photon. Therefore, quantum \(g^{(2)}\) is not \(\langle n ^2\rangle/\langle n\rangle^2\), but rather \(\langle n (n-1)\rangle/\langle n\rangle^2\).
Could it be that quantum \(g^{(2)}\) is not actually equal to 2, but rather 1.9999? Fortunately, although quantum \(g^{(2)}\) and classical \(g^{(2)}\) differ in form, quantum \(g^{(2)}\) is still equal to 2. Let us derive this below.
According to the Boltzmann law, at thermal equilibrium we have:
\[p_n = p_0 e^{-n\omega/T} = (1-e^{-\omega/T}) e^{n\omega/T}\]where \(p_n\) is the probability of having n photons in a given mode.
This is a geometric distribution. A geometric distribution is the discrete version of an exponential distribution, while an exponential distribution is the continuous version of a geometric distribution.
Thus, the quantum \(g^{(2)}\) is
\[\begin{aligned} g^{(2)} := \frac{\langle n(n-1) \rangle}{\langle n\rangle^2} = \frac{\sum n(n-1) p_n }{(\sum n p_n)^2} = 2 \end{aligned}\]The calculation is left to the reader as an interesting exercise. The key is to calculate the moments of the exponential distribution, which are much more complicated than the moments of the exponential distribution.
We can see that although the form of the quantum \(g^{(2)}\) has changed, the corresponding probability distribution has also changed from the exponential distribution followed by the light intensity to the geometric distribution followed by the particle number. These two effects cancel each other perfectly, so that the quantum \(g^{(2)}\) is still equal to 2.
4. Quantumness
What makes quantum \(g^{(2)}\) special compared with classical \(g^{(2)}\)? Its special feature is that quantum \(g^{(2)}\) can be less than 1, whereas classical \(g^{(2)}\) cannot be less than 1.
For classical \(g^{(2)}\), \(g^{(2)} = \langle I^2 \rangle / \langle I\rangle^2 \ge [\cancel{\text{Var}(I)} + \langle I\rangle^2] / \langle I\rangle^2 = 1\); that is, the minimum value is attained if and only if the light intensity is constant, namely when the electric field is a sinusoidal wave with constant amplitude.
Quantum \(g^{(2)}\) is different. Detecting one photon means that the light field has one fewer photon, and a single-photon state contains only one photon: once it is detected, it is gone. Thus, for a single-photon state, quantum \(g^{(2)}\) is 0.
More generally, light fields with quantum \(g^{(2)}\) less than 1 are called sub-Poissonian light fields, while light fields with quantum \(g^{(2)}\) greater than 1 are called super-Poissonian light fields.
When a light field is in a coherent state (with definite or indefinite phase), quantum \(g^{(2)}\) equals 1, and the photon number then follows a Poisson distribution. Its classical counterpart is a sinusoidal wave with definite or indefinite phase (but definite amplitude).
Quantum \(g^{(2)}\) has another special feature: for a thermal light field, even if there is only one frequency mode, that mode can exhibit quantum \(g^{(2)}\) equal to 2. For a classical thermal light field, however, the preceding argument may involve superposition among different modes. This once caused confusion in the 8.422 Open Course.
Of course, if one also uses the Boltzmann law for the classical case, one can immediately obtain the exponential distribution. It is just that the argument using the central limit theorem is remarkably elegant. Perhaps there is a universal connection between the central limit theorem and the Boltzmann law.