Meaning of $\phi(\vec{x})$ in scalar field canonical quantization

Question

Following this question: Why can the Klein-Gordon field be Fourier expanded in terms of ladder operators?

I can see that the when we Fourier transform the $\phi(\vec{x})$ operator and plug it into the KG equation it tells us that $$\bigg( \frac{\partial^2}{\partial t^2}+|\vec{p}|^2+m^2 \bigg)\bar{\Phi(\vec{p})}=0;$$ that means that $\Phi(\vec{p})$ operator is a solution of the harmonic oscillator equation with frequency $\omega_p=\sqrt{m^2+|\vec{p}|^2}$ and if we quantize this harmonic oscillator, $\Phi$ is promoted to the amplitude of the harmonic oscillator in time and we can build $\Phi(\vec{p})$ as a combination of ladder operators $\{a_p^\dagger,a_p\}$ that takes the energy of a state, one unit of $\omega_p$ up or down.

So it leads me to see the operator $\phi(\vec{x})$ as a infinite sum of amplitudes $\Phi(\vec{p})$ operators such that each one of them have $\omega_p$ as the value of the energy steps.

Is this interpretation good for understanding $\phi(x)$ ?

Since $\Phi$ is like the operator in point quantum mechanics, and there, it doesn't have non trivial eigenvalues, doesn't it means that $\phi$ will not have either? At the same time, there are mixed states in the point mechanics that give $\langle \hat q\rangle\neq0$ so I can't say that $\langle \phi(x)\rangle$ is $0$.

Answer 1

So it leads me to see the operator ϕ(x ) as a infinite sum of amplitudes Φ(p) operators such that each one of them have ωp as the value of the energy steps. Is this interpretation good for understanding ϕ(x) ?

Yes for the free KG field, you can look at it as an infinite number of harmonic oscillators.

Since Φ is like the operator in point quantum mechanics, and there, it doesn't have non trivial eigenvalues, doesn't it means that ϕ will not have either? At the same time, there are mixed states in the point mechanics that give ⟨q⟩≠0 so I can't say that ⟨ϕ(x)⟩ is 0.

The expectation value vanishes for the ground state, by definition of the ground state. One does not expect the expectation value to vanish in general states.

There do however exist coherent states, which are eigenvectors of the field operator.