Q:

Field operator in Schrödinger picture

In Schrödinger picture operators do not depend on time explicitly. Consider a free scalar field with Lagrangian density
$${\cal L} ~=~ \frac{1}{2}\partial_{\mu} \phi\partial^{\mu}\phi-\frac{m^2}{2}\phi^2,$$
where $\phi=\phi(t,x,y,z)$. In quantum field theory we take it as an operator. Is this operator Schrödinger or Heisenberg one? How can we express free field Lagrangian in both pictures?

A:

When the field $\phi$ is being written as a function of spacetime, with explicit dependence on time, then this field is said to be written in the Heisenberg picture.  You can see how this is consistent with the terminology used in non-relativistic quantum mechanics by recalling that (See e.g. Peskin and Schroeder eq. 2.43)
$$
  \phi(t,\mathbf x) = e^{i H(t-t_0)}\phi(\mathbf x, t_0)e^{-iH(t-t0)}.
$$
This shows that we can choose some time slice, say at time $t_0$, in Minkowski space at which be define our usual Shrodinger picture operators (the canonical degrees of freedom of the field theory) and at later times, these operator degrees of freedom are related to those at time $t_0$ via the usual Hamiltonian, unitary time evolution above.
In order to use notation that is more in line with that used in non-relativistic quantum mechanics (e.g. see Sakurai p. 82 bottom), you might be inclined to define
$$
  \phi^{(S)}(\mathbf x) = \phi(\mathbf x, 0), \qquad \phi^{(H)}(\mathbf x)(t)
$$
in which case the above time evolution relation would take the familiar form
$$
  \phi^{(H)}(\mathbf x)(t) = e^{iHt}\phi^{(S)}(\mathbf x)e^{-iHt}
$$
but this would be somewhat non-standard notation in field theory as far as I am aware.

