Phase noise and linewidth of laser I learned that the linewidth of a single frequency laser is closely related to the phase noise. But this sentence from a book completely confuses me:
"A quantum-limited beam in a so-called coherent state also has some intensity and phase noise but can have a zero linewidth: there is no unbounded drift of the optical phase."
Why does coherent state have a zero linewidth, but with specific quantum phase noise?
 A: Warning: I'm not an optics expert. I am answering this question based on my own reasoning so it is very likely that I will miss something which is important in practical laser physics.
One notion of the linewidth is the width of frequencies present in the laser beam.
Suppose I have a laser which produces photons only in one exact frequency.
In other words, the laser excites only one mode of the electromagnetic field.
Then the line width is zero because there isn't a range of excited frequencies (modes).
Now, the interesting part (and answer to your question) is that if you consider even a single mode of the electromagnetic field, when you excite it with a laser the phase of that excitation is not a single well defined number.
Lasers put the electromagnetic field mode into a coherent state.
A coherent state, usually denoted $|\alpha\rangle$, has the special property that
$$ \hat{a}|\alpha\rangle = \alpha | \alpha \rangle \, .$$
In other words, a coherent state is an eigenstate of the lowering operator.
Note that for the coherent state
$$\bar{n} \equiv
\langle \alpha | \hat{n} | \alpha \rangle =
\langle \alpha | \hat{a}^\dagger \hat{a} | \alpha \rangle =
|\alpha|^2 \, .
$$
Anyway, the interesting thing is that if we define the quadratures of the electromagnetic field as
$$
\hat{X} = \frac{1}{2}(\hat{a} + \hat{a}^\dagger)\qquad
\hat{Y} = \frac{-i}{2}(\hat{a} - \hat{a}^\dagger)
$$
you can calculate that the mean square deviation of either quadrature for a coherent state is:
$$\sigma_x^2 \equiv
\langle \alpha | \hat{X}^2 | \alpha \rangle -
\langle \alpha | \hat{X} | \alpha \rangle^2 =
1/4 \neq 0\, .$$
It turns out that $\sigma_y = \sigma_x$ as well.
So, if you think of the electromagnetic field mode as living in a phase/amplitude plane, then we've shown that the mode occupies a nonzero amount of space in that plane.
Therefore, it doesn't have one single well defined phase (or amplitude).
This is illustrated in the diagram.

As you can see, the phase of the coherent state is random because the state has non-zero size in the $XY$ plane.
In fact, you can see that the standard deviation of the phase is $^{[a]}$
$$\sigma_{\phi} = \frac{\text{arc length}}{\text{radius}} = \frac{1}{2 \sqrt{\bar{n}}} \, .$$
We have now seen that the laser can have an arbitrarily small range of frequencies, but still has phase noise.
This is very different from usual case with classical signals.
Usually, any phase noise can be thought of as the accumulation of phase coming from a frequency noise.
In other words, the phase noise $\delta \phi$ can be written
$$\delta \phi = \delta \omega t \, .$$
In that case, a fluctuation in the frequency causes the phase to accumulate indefinitely.
If you wait long enough the phase will drift by an arbitrarily large amount.
With the coherent state the situation is quite different.
As you can see in the diagram, the phase noise is constant in time; it's just determined by the size of the blue ball and the length of the radial line, as explained in the equation for $\sigma_{\phi}$ above.
With the coherent state it is truly the phase (and amplitude) which is noisy; you can't really think of it as a frequency noise.
Practically, this means that when you measure the phase of the laser signal, you will find that it is composed of three parts:


*

*The frequency of the laser: $\phi = \omega t$.

*The technical frequency noise: $\delta \phi_{\text{technical}} = \int_0^t \delta \omega(t')\,dt'$.

*The quantum phase noise: randomly distributed with $\sigma_{\phi} = 1/2\sqrt{\bar{n}}$.


$[a]$ Someone should check this because I just did this computation myself and I'm not an optics expert.
