Jonginn Yun

What The Shot-Noise Limit Means

The shot-noise limit asks the following reference question:

For an output mode, here the $2\omega$ output mode, whose average occupation is $\langle n_{2\omega}\rangle$, how large would the noise be if the detection events were independent, i.e. Poisson distributed?

The quantum state that represents this independent-event reference is the coherent state.

For a coherent state $|\beta\rangle$, photon number follows Poisson statistics:

$$\mathrm{Var}(n)=\langle n\rangle .$$

Therefore a coherent state supplies the conventional 0 dB shot-noise reference. This does not assume the real output is necessarily coherent; it defines the reference floor against which the measured noise is compared.

Time-Bin Picture

The note uses a time-bin or temporal-mode picture. Instead of starting directly with continuous arrival times, take a short time interval $\Delta t$ and treat the photon number detected in that interval as the observable. The temporal mode for that bin is denoted $\hat b$, and the measured number operator is

$$\hat N = \hat b^\dagger \hat b .$$

Thus continuous detection is approximated by chopping time into very small bins and counting photons in each bin.

Coherent State Gives Poisson Counts Per Bin

For an input coherent state,

$$\hat D(\alpha)|0\rangle,$$

the photon-number distribution in a time bin is

$$p(n) = e^{-|\alpha|^2}\frac{|\alpha|^{2n}}{n!}.$$

This is the Poisson distribution.

From Poisson Distribution To Poisson Process

A single Poisson-distributed random variable is not yet a Poisson process. To obtain a point process in time, one also needs:

1. In a short interval $\Delta t$, the probability of one click is proportional to $\Delta t$.
2. The probability of two or more clicks in the same short interval is $\mathcal O(\Delta t^2)$.
3. Different time bins are independent.

When these are satisfied, the total count $N(T)$ in the interval $[0,T]$ is Poisson distributed, and the counting process has independent increments. In the continuous-time limit this is a Poisson point process.

Photodetection Reason For Independence

In Glauber photodetection theory, the click rate at time $t$ is proportional to

$$\langle \hat a^\dagger(t)\hat a(t)\rangle .$$

Two-click correlations involve normally ordered moments such as

$$\langle \hat a^\dagger(t) \hat a^\dagger(t+\tau) \hat a(t+\tau) \hat a(t) \rangle .$$

For a coherent state,

$$\hat a|\alpha\rangle=\alpha|\alpha\rangle,$$

so all normally ordered moments factorize. For example,

$$\begin{aligned} &\langle \hat a^\dagger(t) \hat a^\dagger(t+\tau) \hat a(t+\tau) \hat a(t) \rangle\\ &\qquad = \langle \hat a^\dagger(t)\hat a(t)\rangle \langle \hat a^\dagger(t+\tau)\hat a(t+\tau)\rangle . \end{aligned}$$

This means that a click at one time does not change the probability of a click a little later. There is no bunching or antibunching, so the arrival events behave independently.

Nonclassical states can have time correlations and therefore can show bunching or antibunching.

Precise Meaning Of The Coherent-State Assumption

The statement coherent state assumption = Poisson process assumption should be read with the detection model included:

• coherent input field,
• Poisson photon counts in each time bin,
• factorized normally ordered correlations between bins,
• independent click events in the continuous-time limit,
• white shot-noise correlations.

Thus the statement that photon arrival times are modeled as a Poisson process is justified when the relevant input fields, such as a local oscillator or signal field, are coherent or treated as coherent mixtures.

What $S_{XX}=1/4$ Means

$S_{XX}(\omega)$ is the Fourier transform of the autocorrelation of the quadrature $X(t)$, up to whatever one-sided, two-sided, or symmetrized convention the note is using.

Vacuum fluctuations are white over the relevant band, so the spectrum is flat. In this normalization, the vacuum or coherent-state quadrature floor is

$$S_{XX}=\frac{1}{4}.$$

This flat value is what is called the shot-noise floor or shot-noise limit in this quadrature normalization.

Connection Between Poisson Shot Noise And $S_{XX}=1/4$

The photon-counting statement is

$$\mathrm{Var}(n)=\langle n\rangle .$$

To connect this to quadrature noise, use bright-beam linearization. Write

$$\hat a(t)=\alpha+\delta \hat a(t), \qquad \hat n=\hat a^\dagger\hat a \approx |\alpha|^2+\alpha(\delta\hat a+\delta\hat a^\dagger),$$

where $\alpha$ is chosen real. Then

$$\Delta \hat n(t) \approx 2\alpha\,\Delta \hat X(t).$$

The spectra obey

$$S_{nn}(\omega)\approx 4\alpha^2 S_{XX}(\omega).$$

For a coherent, shot-noise-limited field,

$$S_{XX}=\frac{1}{4},$$

so

$$S_{nn}(\omega) \approx 4\alpha^2\cdot\frac{1}{4} = \alpha^2.$$

Since

$$\langle n\rangle=|\alpha|^2=\alpha^2,$$

one obtains

$$S_{nn}\sim \langle n\rangle .$$

Thus the quadrature floor $S_{XX}=1/4$ and Poisson photon-counting shot noise are the same physical limit expressed in two different normalizations.