JimWhoKnew said:
For the other modes, the reality constraint couples the modes ##\vec{k}## and ##-\vec{k}## , so it is less obvious.
We can gain some more insight into the relation between the free field and the 1D harmonic oscillator:
For simplicity, I'll consider the field in one spatial dimension.
The field is real, so we can expand it with ##\cos kx## and ##\sin kx## . Define:$$b_k:=\frac{a_k+a_{-k}}{\sqrt{2}} \quad, \quad c_k:=\frac{a_k-a_{-k}}{\sqrt{2}} \quad.$$In this expansion the coupling between ##k## and ##(-k)## is removed and each "new" mode is truly normal (independent) to the others. This means that the amplitude of each mode (which is a real quantity) behaves like a 1D HO. Sounds abstract, but that is just what we would have expected from a guitar's string if it had only one polarization. We get$$[b_k,b_k^\dagger]=[c_k,c_k^\dagger]=1$$(see remark in #5) and$$[b_k,c_k]=[b_k,c_k^\dagger]=0 \quad .$$Define further$$\frac{b_k+b_k^\dagger}{\sqrt{2}} :=\alpha_k \hat X_{\cos,k} \quad, \quad \frac{c_k-c_k^\dagger}{i\sqrt{2}}:=\beta_k \hat P_{\sin,k}$$where ##\alpha_k## , ##\beta_k## are real proportionality factors. Absorbing ##\omega_k## and normalization factors into these proportionality factors, we have:$$\hat\phi(x)=\sum_{k\ge0} \left( \alpha_k \hat X_{\cos,k} \cos kx+\beta_k \hat P_{\sin,k} \sin kx \right)\quad.$$For a real function ##f(x)## expanded by$$f(x)=\sum_{k\ge0} \left(f_{\cos,k}\cos kx+f_{\sin.k}\sin kx \right) \quad,$$we have$$\hat\phi(x)|f\rangle=f(x)|f\rangle$$where$$|f\rangle=\prod_{k\ge0}|x_{\cos,k}=f_{\cos,k}/\alpha_k\rangle|p_{\sin,k}=f_{\sin,k}/\beta_k\rangle$$(which should be the same as the one discussed in post #4 when ##f(x)=\phi(x)## ). The 1D harmonic oscillator states ##|x\rangle## and ##|p\rangle## correspond to the "generalized" wavefunctions ##\delta(x-x')## and ##\exp[ix(p-p')]## respectively, so ##|f\rangle## should also be regarded as a "generalized" state ( ##\langle f|f'\rangle=\delta^\infty (f-f')## ).
