Is there a physical interpretation of the quadrature operators of the quantised EM field in a cavity? I am considering a cavity setup using two mirrors perpendicular to the $z$-axis separated by a distance $L$, as seen here

Assuming the electric field is polarised along the $x$-axis and uniform in value along the $y$-axis, the EM vector potential of a single mode of the cavity is given,
$$\textbf{A}(z,t)=\hat{\textbf{e}}_{x}A(t)\sin(k_{n}z)$$
whilst the electric and magnetic fields are respectively given by
$$\textbf{E}(z,t)=-\hat{\textbf{e}}_{x}\dot{A}(t)\sin(k_{n}z)$$
and
$$\textbf{B}(z,t)=\hat{\textbf{e}}_{y}k_{n}A_{k_{n}x}(t)\cos(k_{n}z)$$
where
$$k_{n}=\frac{n\pi}{L}$$
indicates the allowed modes of the mirror cavity of length $L$. Quantisation proceeds by defining the variables
$$A(t)=\sqrt{\frac{2}{\epsilon_{0}V}}q(t)$$
and
$$\dot{A}(t)=\sqrt{\frac{2}{\epsilon_{0}V}}p(t),$$ evaluating the energy of the EM radiation in a cavity and then applying canonical quantisation to the dynamical variables. Calculating the energy yields the Hamiltonian of a harmonic oscillator;
$$H=\frac{1}{2}(\omega_{n}^{2}q^{2}+p^{2})\rightarrow \hat{H}=\frac{1}{2}(\omega_{n}^{2}\hat{Q}^{2}+\hat{P}^{2})$$
where $\omega_{n}=ck_{n}$.
My question is, is there a physical interpretation of the canonically conjugate variables, $$[\hat{Q},\hat{P}]=i\hbar~?$$ Intuitively, if I imagine a coherent state of the mode, it would make sense for the variables to be associated with the momentum and position of the light beam, as I can imagine the beam bouncing backwards and forwards between the mirrors (like seen here, for example). However I can't see any physical reason why the amplitude of the vector potential $A$ and its time derivative $\dot{A}$ would be associated to position and momentum respectively.
 A: Thanks to a comment from KF Gauss, and by thinking of the comparison with the classical picture, I managed to work out what is happening.
$q(t)$ and $p(t)$ evolve according to a classical harmonic oscillator Hamiltonian, so will evolve sinusoidally, out of phase with each other by $\pi/2$. Assuming q(t) is initially at its maximum value,
$$q(t)=A\cos(\omega_{n}t),\;\;p(t)=-\omega_{n}A\cos(\omega_{n}t+\pi/2)$$
If we let
$$\textbf{E}(t)=E(t)\hat{\textbf{e}}_{x},\textbf{B}(t)=B(t)\hat{\textbf{e}}_{y}$$
Then we see that
$$E(t)=-\sqrt{\frac{2}{\epsilon_{0}V}}p(t)\sin(k_{n}z)$$
and
$$B(t)=\sqrt{\frac{2}{\epsilon_{0}V}}k_{n}q(t)\cos(k_{n}z)$$
We see that q(t) and p(t) represent time varying amplitudes that cause the amplitude of the electric and magnetic fields to vary with time, thus causing the electric field to form standing wave solutions, as demonstrated for $n=5$ here,

where $z$ is in units of $L$ the centre of the dark red bands are where $\sin(k_{n}z)=1$ hence $E(t)=-\frac{2}{\epsilon_{0}V}q(t)$ and the centre of the dark blue bands are where $\sin(k_{n}z)=-1$ hence $E(t)=\frac{2}{\epsilon_{0}V}q(t)$.
So after quantisation,
$$\hat{E}=-\sqrt{\frac{2}{\epsilon_{0}V}}\sin(k_{n}z)\hat{P}$$
and
$$\hat{B}=\sqrt{\frac{2}{\epsilon_{0}V}}k_{n}\cos(k_{n}z)\hat{Q}$$
that is, up to some $z$-dependant constant, the quadrature operators are equivalent to the field operators. Finally, returning to this gif, we realise that this isn't a wavepacket of light bouncing back and forth between the mirrors, because $\hat{Q}$ has no relation to position. Taking the expectation value of the electric field of a coherent state,
$$\langle\alpha e^{i\omega_{n} t}|\hat{E}|\alpha e^{i\omega _{n}t}\rangle=-\sqrt{\frac{2}{\epsilon_{0}V}}\sin(k_{n}z)\langle\alpha e^{i\omega_{n} t}|\hat{P}|\alpha e^{i\omega_{n} t}\rangle=\sqrt{\frac{4\omega_{n}\hbar}{\epsilon_{0}V}}|\alpha|\sin(k_{n}z)\cos(\omega_{n} t-\theta-\pi/2)$$
for $\alpha=re^{i\theta}$
we see that the amplitude of a coherent state oscillates in time, returning the classical standing wave image, where a real valued $\alpha$ corresponds to the assumption that $\langle E(t)\rangle$ is at the maximum value of its oscillation at $t=0$.
So in conclusion, my confusion was mistaking this oscillation in the gif as oscillation of a light wavepacket between the mirrors of the cavity, when it in fact (in the context i'm considering) represents the oscillation of the amplitude of the semi-classical electric field. Finally, the quadrature operators are equivalent to the field operators up to a constant.
A: This may not seem like a direct answer to your question...
Analytical mechanics is the study of describing the dynamics of a system represented by a given dynamical variable (canonical coordinates) and its canonically conjugate momentum.
In the case of analytical mechanics or quantum mechanics for "one-particle", the dynamical variables of the system are the coordinates associated with the particles.
On the other hand, as the name implies, analytical mechanics and quantum mechanics for “fields”, which describe electromagnetic fields or string oscillations etc., are the study of the dynamics of fields. In this case, the dynamical variable of the system is the field itself. In other words, in analytical mechanics and quantum mechanics for field, the field itself is concerned with its own time evolution, and there is no direct correspondence between this “field” and each individual particle.
In other words, the field operators are called "coordinates" simply because the canonical coordinates in quantum field theory are called "coordinates" by analogy with the quantum mechanics of a single particle, and not in the sense that the coordinates of each particle are represented by a field.
So, what do you do when you want to image the coordinates or momentum of a "specific particle" in quantum field theory?
In quantum mechanics, what determines the position and momentum of a particle was the wave function of the particle, in other words, its state.
In quantum field theory, this state of a particle is defined using a creation operator, which is defined by the mode expansion of the field operator:
$$\psi(x)=\sum_{p}(a_p^{\dagger} e^{ipx}+h.c.)$$
This "single particle state" corresponds to the wave function of the particle. Indeed, $\langle x|a_p^\dagger|0\rangle=e^{ipx}$ holds (see, for example, Schwartz’s QFT)
The same is true for two or more particles. In this sense, the field operator includes any number of particles in its definition, and you can construct states of particles at any position by letting the (interacting) field operator act on the vacuum, but the field operator itself never represents the "coordinates" of the particles.
Also note that field operators and particle wave functions are often confused because they use the same symbols, but they are completely independent concepts.
If we create these plane wave states, we can also create wave packet states, which are closer to the more classical particle picture. Such wave packets are often used when particle locality is important, such as in the derivation of the LSZ formula.
The coherent light is close to the wave packet "states”. Of course, it is possible to construct such states in the manner I mentioned above, using the creation-annihilation operators defined by the field operators. However, the "field operator" itself is not the same as the "wave function" of a particle. Even though they are related, but completely independent concept, so the field operator $A^\mu$ or $\phi$ above does not represent the coordinates of the particle.
