You are always free to define $$ a_{\boldsymbol k}\equiv \int\mathrm d^3 x\ \mathrm e^{ikx}(\omega_{\boldsymbol k}\phi(x)+i\pi(x)) \tag{1} $$ where $\pi=\dot\phi(x)$. If you take the time derivative of this definition, you get $$ \dot a_{\boldsymbol k}= \int\mathrm d^3 x\ \mathrm e^{ikx}(\partial^2+m^2)\phi(x) \tag{2} $$ which is non-zero for an interacting field. Therefore, $a=a(t)$, where $t$ is the time slice you chose in $(1)$; in other words, our definition of $a$ is not in general independent of the time slice, so $a$ depends parametrically on the value of $t$ at that slice.

Now, inverting the Fourier transform in $(1)$, you get $$ \phi(x)=\int\frac{\mathrm d^3 k}{(2\pi)^32\omega_{\boldsymbol k}}\ \mathrm e^{-ikx}a_{\boldsymbol k}(t)+\text{h.c.} $$ which is essentially P&S's statement. Note that, in general, this statemenet is mostly devoid of any practical meaning; its just a trivial consequence of the inversion theorem of the Fourier transform.

You are always free to define $$a_{\boldsymbol k}\equiv \int\mathrm d^3 x\ \mathrm e^{ikx}(\omega_{\boldsymbol k}\phi(x)+i\pi(x)) \tag{1}$$ where $\pi=\dot\phi(x)$. If you take the time derivative of this definition, you get $$\dot a_{\boldsymbol k}= \int\mathrm d^3 x\ \mathrm e^{ikx}(\partial^2+m^2)\phi(x) \tag{2}$$ which is non-zero for an interacting field. Therefore, $a=a(t)$, where $t$ is the time slice you chose in $(1)$; in other words, our definition of $a$ is not in general independent of the time slice, so $a$ depends parametrically on the value of $t$ at that slice.

Now, inverting the Fourier transform in $(1)$, you get $$\phi(x)=\int\frac{\mathrm d^3 k}{(2\pi)^32\omega_{\boldsymbol k}}\ \mathrm e^{-ikx}a_{\boldsymbol k}(t)+\text{h.c.}$$

which is essentially P&S's statement. Note that, in general, this statement is mostly devoid of any practical meaning; its just a trivial consequence of the inversion theorem of the Fourier transform.

You are always free to define $$ a_{\boldsymbol k}\equiv \int\mathrm d\boldsymbol x\ \mathrm e^{ikx}(\omega_{\boldsymbol k}\phi(x)+i\pi(x)) \tag{1} $$ where $\pi=\dot\phi(x)$. If you take the time derivative of this definition, you get $$ \dot a_{\boldsymbol k}= \int\mathrm d x\ \mathrm e^{ikx}(\partial^2+m^2)\phi(x) \tag{2} $$ which is non-zero for an interacting field. Therefore, $a=a(t)$, where $t$ is the time slice you chose in $(1)$; in other words, our definition of $a$ is not in general independent of the time slice, so $a$ depends parametrically on the value of $t$ at that slice.

Now, inverting the Fourier transform in $(1)$, you get $$ \phi(x)=\int\frac{\mathrm d\boldsymbol k}{(2\pi)^32\omega_{\boldsymbol k}}\ \mathrm e^{-ikx}a_{\boldsymbol k}(t)+\text{h.c.} $$ which is essentially P&S's statement. Note that, in general, this statemenet is mostly devoid of any practical meaning; its just a trivial consequence of the inversion theorem of the Fourier transform.

You are always free to define $$ a_{\boldsymbol k}\equiv \int\mathrm d^3 x\ \mathrm e^{ikx}(\omega_{\boldsymbol k}\phi(x)+i\pi(x)) \tag{1} $$ where $\pi=\dot\phi(x)$. If you take the time derivative of this definition, you get $$ \dot a_{\boldsymbol k}= \int\mathrm d^3 x\ \mathrm e^{ikx}(\partial^2+m^2)\phi(x) \tag{2} $$ which is non-zero for an interacting field. Therefore, $a=a(t)$, where $t$ is the time slice you chose in $(1)$; in other words, our definition of $a$ is not in general independent of the time slice, so $a$ depends parametrically on the value of $t$ at that slice.

Now, inverting the Fourier transform in $(1)$, you get $$ \phi(x)=\int\frac{\mathrm d^3 k}{(2\pi)^32\omega_{\boldsymbol k}}\ \mathrm e^{-ikx}a_{\boldsymbol k}(t)+\text{h.c.} $$ which is essentially P&S's statement. Note that, in general, this statemenet is mostly devoid of any practical meaning; its just a trivial consequence of the inversion theorem of the Fourier transform.