Expanding free scalar field in terms of ladder operators

Question

I'm having some difficulty with the finer points of expanding a field in terms of ladder operators. Note that this is not identical to the other related question I asked. From Peskin / Schroeder;

At any fixed time $t_0$ we can of course expand $\phi$ in terms of ladder operators $$\phi(\textbf{x},t_0)=\displaystyle\int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}}(a_pe^{i\textbf{p.x}}+a_p^\dagger e^{-i\textbf{p.x}}).$$ Then to obtain $\phi(\textbf{x},t)$ for $t\neq t_0$ we just switch to the Heisenberg picture $$\phi(\textbf{x},t)=e^{iH(t-t_0)}\phi(\textbf{x},t_0)e^{-iH(t-t_0)}.$$

I think the notation is a little dodgy. Define $\phi^S(\textbf{x})=\phi(\textbf{x},t_0)$. Then $\phi^S(\textbf{x})$ is time independent and corresponds to the Schrodinger picture of $\phi(x)$. We can then of course define the conjugate momenta $\pi^S(\textbf{x})=\pi(\textbf{x},t_0)$ in the Schrodinger picture. Together these satisfy the (equal time) commutation relations. We can then introduce the creation and annihilation operators and diagonalise the Hamilmtonian.

$$\phi(\textbf{x},t_0)=\displaystyle\int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}}(a_p(t_0)e^{i\textbf{p.x}}+a_p(t_0)^\dagger e^{-i\textbf{p.x}}).$$

My main issue is due to the "time dependence" of $a_p$ and $a^\dagger_p$. These operators are defined on the time slice $t=t_0$. In particular $a_p=a_p(t_0)$ and the vacuum state is defined to be the state annihilated by $a_p$. But $a_p$ is really only defined for $t=t_0$. So we could pick a difference time slice $t=t_1$ and expand the field in terms of new ladder operators

$$\phi(\textbf{x},t_1)=\displaystyle\int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}}(a_p(t_1)e^{i\textbf{p.x}}+a_p(t_1)^\dagger e^{-i\textbf{p.x}}).$$

Since the vacuum state must also be annihilated by the these annihilation operators does this imply the field is completely time independent?

Edit: Also Peskin / Schroeder make no mention of time slices, they merely say they are working in the Schrodinger picture. Is the way I've defined things above essentially what's going on - i.e. theyve defined their creation / annihilation operators at some reference time ($t=0$ by the looks of it) and just left this completely implicit? So their creation operators would create a particle with momentum $\textbf{p}$ at time $t=0$ whereas mine defined on the time-slice $t=t_0$ would create a particle with momentum $\textbf{p}$ at time $t=t_0$?

Answer 1

No.

If $$a_p(t_0)\left | 0\right>=0$$ then $$a_p(t)\left|0\right>=e^{iH(t-t_0)}a_p(t_0)e^{-iH(t-t_0)}\left|0\right>=0$$ since we usually think of the vacuum as a energy eigenstate with energy zero (even if it's not zero, the above equation is still true).