Coherent states are quantum states that are said to act "as classically as possible". You can define coherent states for the harmonic oscillator, or more generally for any collection of harmonic oscillators, such as a free quantum field. It is often said that a quantum field in a coherent state behaves like a classical field, e.g. that this is how classical electromagnetism emerges from QED.
I've only been taught about coherent states in optics, where you need to use special procedures and expensive equipment to create them. But if the world looks like it's made of classical fields, it must be easy to make a coherent state, so easy that it's inevitable.
How are coherent states of quantum fields made in the real world?
Best Answer
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.