Time derivative of a real scalar field in the Schrödinger Picture In Schrödinger picture operators do not depend on time explicitly. Consider a free scalar field with Lagrangian density
$\mathcal{L}=\frac{1}{2} \partial_\mu \phi \partial^\mu \phi - \frac{1}{2}m^2\phi$
where $\phi = \phi(\mathbf{x})$ is not explicitly dependent on time. In QFT, this is taken to be an operator-valued function of all space. Hence, does this mean that the time derivatives of $\phi$ vanish? ie. is it true that 
$ \partial^0 \phi = 0 $ and $ \frac{ d \phi}{dt} = 0$

Minor aside:
Also, at this point I would also like to ask a notational question: when one writes $\dot{\phi}$ (in, for example, the question here) does it mean $ \partial^0 \phi $ or $ \frac{ d \phi}{dt} $ or even $ \frac{ \partial \phi}{\partial t} $? 

If this is true, however, it would mean that the (0,0) component of $T^{\mu \nu} = \partial^\mu \phi \partial^\nu \phi - \eta^{\mu \nu}\mathcal{L}$ will be given by 
$T^{00} = - \mathcal{L} = p_0 = \int d^3x T^{00} =  - \frac{1}{2} (\nabla  \phi) ^2 + \frac{1}{2}m^2 \phi^2 $
which will not give the correct energy 
$P^0 = \int d^3x T^{00} = \int d^3x \left( \frac{1}{2} \phi^2 + \frac{1}{2} (\nabla  \phi) ^2 + \frac{1}{2}m^2 \phi^2\right),$
where $P^\mu$ is the 4-momentum. What is wrong? 
 A: The operator $\phi$ in quantum field theory just generalizes the operator $x$ in ordinary quantum mechanics. So I'm going to specialize everything you said to quantum mechanics, and then hopefully it will be obvious where the misconceptions are.

In Schrödinger picture operators do not depend on time explicitly. Consider a free particle with Lagrangian
$\mathcal{L}=\frac{1}{2} m \dot{x}^2$
where $x$ is not explicitly dependent on time. In QM, this is an operator. Hence, does this mean that the time derivatives of $x$ vanish? ie. is it true that 
$ \partial^0 x= 0 $ and $ \frac{ d x}{dt} = 0$

Minor aside:
Also, at this point I would also like to ask a notational question: when one writes $\dot{x}$ does it mean $ \partial^0 x$ or $ \frac{ d x}{dt} $ or even $ \frac{ \partial x}{\partial t} $? 

If this is true, however, it would mean that the energy will be given by 
$E = \frac12 m \dot{x}^2 = 0$
which will not give the correct energy. What is wrong? 

Hopefully it's clear what the issue is. In quantum mechanics, the state of a particle isn't specified by a number $x$, it's specified by a wavefunction. In Schrodinger picture, motion of the particle is encoded in the time-evoultion of the wavefunction, even if the operator $\hat{x}$ doesn't change in time at all. It's just the same in QFT.
A: The operator formalism of quantum mechanics can only really be used after formulating the system in Hamiltonian mechanics: First, introduce a canonical momentum $\pi$ and define $\mathcal{H}(\phi, \pi)$ as what you get when you rewrite $\pi \dot{\phi} - \mathcal{L}(\phi, \dot{\phi})$ under the assumption that $\pi = \frac{\partial \mathcal{L}}{\partial\dot\phi}$ (here, $\pi = \dot\phi$). If you do this you will get
$$ \mathcal{H}(\phi, \pi) = \frac{1}{2}\pi^2 + \frac{1}{2}(\nabla\phi)^2 + \frac{1}{2}m^2\phi^2. $$
From here on, $\phi$ and $\pi$ should be regarded as independent fields.
If you now want to use the Schrödinger picture, you promote $\phi$ and $\pi$ to operators, and further define $[\phi(\vec{x}), \pi(\vec{y})] = i\delta^{(3)}(\vec{x} - \vec{y})$. If, after this point, you see $\dot\phi$, or any time derivative of an operator, you are strictly speaking doing something wrong. In particular, writing $\partial^\mu \phi \partial_\mu \phi$ or $\dot\phi = \pi$ would be meaningless. If you see a time derivative, it means that you have not completely translated the problem into Hamiltonian mechanics.
If you are instead in the Heisenberg picture, which is closer to classical Hamiltonian mechanics, time derivatives of operators make sense. For example, you can use the equal-time commutation relation $[\phi(\vec{x},t), \pi(\vec{y},t)] = i\delta^{(3)}(\vec{x} - \vec{y})$ to show that $\dot\phi(\vec{x},t) = i[H(t), \phi(\vec{x},t)] = \int d^3 y\ i [\mathcal{H}(\vec{y},t), \phi(\vec{x},t)] = \pi(\vec{x},t)$. Here, there is no problem writing $T^{\mu\nu} = \partial^\mu \phi \partial^\nu \phi - \eta^{\mu\nu} \mathcal{L}$, which will indeed give you $T^{00}(\vec{x},t) = \mathcal{H}(\vec{x},t)$. Although formally speaking, all operators should be in terms of $\phi$, $\pi$ and possibly $t$ (but not $\dot\phi$) when using Hamiltonian mechanics, even in the Heisenberg picture.

About the notational aside, in field theory, $t = x^0$ and $x^1$, $x^2$, $x^3$ are independent variables; the field $\phi(\vec{x},t)$. So $\vec{x}$ does not depend on $t$, which means that $\dot\phi$, $\partial^0\phi$, $\frac{d\phi}{dt}$ and $\frac{\partial\phi}{\partial t}$ all mean the same thing.
A: I guess the energy, according to my book, should actually be 
$P_0 = \frac{1}{2}\int  \pi^2 (x) + (\nabla\phi)^2 + m^2\phi^2 d^3 x$
where $\pi(x) = \frac{\partial \mathcal{L}}{\partial \dot{\phi}}$ is the conjugate momentum, which for the klein-gordon field is $\dot{\phi} = 0$, then they should be equal.
On the notation, $\partial^{0} = \frac{\partial \phi}{\partial x_0} = \partial_{0} = \frac{\partial \phi}{\partial x^0}$ then substitute $x_0 = x^0 = ct$ or just $t$ in natural units.
