I agree with @Floris that the statement doesn't make sense at face value, but since I know what he is trying to say, I might be able to translate.
No signal has a single frequency. There is always a very small spread in a signal's frequency content. (Pure single frequency implies a signal that started in the infinite past and will continue into the infinite future.) But a nearly-single-frequency wave will be indistiguishable from a true-single-frequency wave if you look at it for a finite time interval.
So take two waves, and look at them during the same time interval. They both will look like waves with some frequency $f$, with some phase shift $\phi$ between them.
Now wait a while and look at the same two waves again. Again, they both will look like waves of frequency $f$. And they will have a phase shift between them, call it $\phi_\mathrm{new}$.
If you wait a short enough time between measurements, the two phase values will be the same. If you wait long enough, they will have been seen to have drifted apart.
Over time, $\phi_\mathrm{new}$ drifts away from $\phi$ because of those tiny fluctuations in the frequency. If $\phi_\mathrm{new} = \phi$ we could say that there is a constant phase shift; if not, then we can say that the phase shift is not constant.
Optical detectors average intensity over a long time, often on the order of seconds. If the phase shift is not constant during that interval (an incandescent source for example), interference patters will be washed out, and not visible. The light is incoherent over that interval. If the phase shift is constant over the (e.g. a laser), interference patterns are visible. The light is said to be coherent over the interval.
Coherent states appear in nature because they're what you get when you drive a harmonic oscillator through a dipole interaction.
Suppose we push on a harmonic oscillator with a time dependent force $F(t)$.
The Hamiltonian associated with that push is
$$\hat{H}_x = -\hat{x} F(t) \equiv -(\hat{a} + \hat{a}^\dagger) f_x(t)$$
where
we've used $\hat{x} = x_\text{zpf}(\hat{a}+\hat{a}^\dagger)$ where $x_\text{zpf}$ is the zero point fluctuation of the oscillator, i.e. $ x_\text{zpf}^2 = \langle 0 | \hat{x}^2 | 0 \rangle$.
we've defined $f_x(t) \equiv F(t) x_\text{zpf}$.
A drive proportional to $\hat{x}$ like this is very common in Nature because, for example, it's common to couple to one of the degrees of freedom of an oscillator and push on it.
From now on, we'll drop the operator hats.
Let's also add in a $y$ drive term so that the total drive is
\begin{align}
H_\text{drive}
&= -(a + a^\dagger) f_x (t) - (-i)(a - a^\dagger)f_y(t) \\
&= a(-f_x + i f_y) + a^\dagger(-f_x - i f_y) \, . \\
\end{align}
Define $f(t)\equiv -f_x(t)+i f_y(t)$ so that
$$H_\text{drive} = af(t) + a^\dagger f(t)^* \, .$$
The full Hamiltonian in the Heisenberg picture is then
\begin{align}
H(t)
&= H_\text{oscillator}(t) + H_\text{drive}(t) \\
&= \hbar \Omega \left(a^\dagger(t) a(t) + \frac{1}{2} \right) + (a(t) f(t) + \text{h.c.})
\end{align}
where $\Omega$ is the frequency of the oscillator.
Heisenberg's equation of motion works out to
$$\left( \frac{d}{dt} + i \Omega \right) a(t) = -\frac{i}{\hbar}f(t)^* \, \text{Id}$$
where $\text{Id}$ means the identity operator.
It's reasonably intuitive here (but you should check with explicit calculation) that $a(t)$ picks up only phase factors and scalar offsets.
In particular, the solution is
$$a(t) = a(0)e^{-i \Omega t} - \frac{i}{\hbar} \int_0^t e^{i \Omega t'} f(t')^* \, dt' \, . $$
Therefore, if we start in a coherent state and drive the system, the new state is still a coherent state:
\begin{align}
a(t) | \alpha \rangle
&= \left( a(0)e^{-i \Omega t} - \underbrace{\frac{i}{\hbar} \int_0^t e^{i \Omega t'} f(t')^* \, dt'}_\text{some complex number $c$} \right) | \alpha \rangle \\
&= \underbrace{\left( \alpha e^{-i \Omega t} - c \right)}_{\alpha'} |\alpha \rangle \\
&= \alpha' |\alpha \rangle \, .
\end{align}
Therefore, we still have a coherent state, it's just a different one than before the driving.
The ground state is a coherent state, so we've shown that if you start in the ground state, and then drive the system with a dipole interaction, you always get a coherent state.
Best Answer
Coherent states are eigenvectors for the (bosonic) annihilator,$$\hat a ~|\alpha\rangle = \alpha~|\alpha\rangle,$$and if we define the position and momentum quadratures as $\hat x = \hat a^\dagger + \hat a,$ $\hat p = i \hat a^\dagger - i \hat a,$ we have $[\hat x, \hat p] = 2i$ and the dimensionless Hamiltonian $\hbar\omega ~ \hat a^\dagger \hat a = \frac12 \hbar\omega~x^2 + \frac12 \hbar\omega~p^2 + \text{const.}$ to guide us. We can immediately see that in the coherent state we have $\langle x \rangle = \alpha^* + \alpha = 2 ~\Re~{\alpha}$ whereas $\langle p \rangle = i~\alpha^* - i ~ \alpha = 2 ~\Im~\alpha, $ so the position and momentum plane is basically just the complex plane $\mathbb C$ that $\alpha$ lives on.
Now this Hamiltonian of course has an eigenbasis $\hat a^\dagger \hat a ~ |n\rangle = n ~|n\rangle$ and in terms of that basis we see a recurrence that if $|\alpha\rangle = \sum_n c_n |n\rangle$ then we can work out that $\alpha |\alpha\rangle = \hat a |\alpha\rangle$ implies $$c_n \sqrt{n} = \alpha c_{n-1},\text { so } c_n = c_0 \frac{\alpha^n}{\sqrt{n!}}.$$Then working out $1 = \langle \alpha|\alpha\rangle = c_0\sum_n \big(|\alpha|^2\big)^n/n! = c_0 \exp\big(|\alpha|^2\big)$ gives the normalization constant $c_0.$
However note that under this Hamiltonian, $|n\rangle \mapsto e^{-in\omega t} |n\rangle$ and therefore, $$|\alpha\rangle \mapsto \exp\left(-|\alpha|^2\right) \sum_{n=0}^\infty \frac{\alpha^n}{\sqrt{n!}} ~ e^{-i~\omega t~n} |n\rangle,$$which we see on the right hand side combines by normal exponent rules to form $(\alpha e^{-i\omega t})^n.$ In other words the time evolution is that $|\alpha(t)\rangle = |\alpha_0 e^{-i\omega t}\rangle,$ and our coherent state simply makes a circle on the complex plane as it evolves.
It is in this precise sense that I understand the word "coherent," it is the meaning "it stays perfectly together as it goes along its journey." It's the same way I would say "lasers are a coherent phenomenon; light by its nature wants to spread out in a $1/r^2$ law but in a laser, the different wave packets are all arranged with just the right phase differences so that they destructively interfere for the waves that are trying to get out of the beam proper, and constructively interfere in the next position of the beam."