Let's assume we have a carrier at $f_{c}$, and a message at $f_{m}$ ($f_{c} \gg f_{m}$). Amplitude modulation is used to modulate the message on the carrier. Let's call the modulated carrier $\text{rf}(t)$. At the receiver, we first have the antenna, followed by a polyphase filter, which creates an I and Q channel: $$ \begin{align} I_{pp}(t) &= \text{rf}(t) \\ Q_{pp}(t) &= \text{rf}\left( t-\frac{1}{4f_{c}} \right) \end{align} $$
After the polyphase filter there is a Zero-IF I/Q mixer circuit, which brings the carrier to DC and the message back into baseband. Note that the phase between the polyphase filter output and the local oscillator of the mixer is random.
$$ \begin{align} I_{bb}(t) &= \Re{\left((I_{pp}(t)+iQ_{pp}(t)) e^{-(i\omega_{lo}\,t+\theta_{rand}\,)}\right)} \\ Q_{bb}(t) &= \Im{\left((I_{pp}(t)+iQ_{pp}(t)) e^{-(i\omega_{lo}\,t+\theta_{rand}\,)}\right)} \end{align} $$
Now what I am wondering about is what is the phase between the baseband I and Q channel ($I_{bb}$ and $Q_{bb}$), after the mixer? First intuition for me was $90°$, however that cannot be the case, since the shift introduced by the polyphase filter is $90°$ with respect to $f_c$, and not $f_m$. So are I and Q going to be in-phase? Or might the phase also be 180°? Does it depend on the phase between polyphase filter output and the local oscillator?