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

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 $$0$$.