Importance of local conservation of probability In almost every textbook of quantum mechanics we can find the derivation of the local conservation of probability.
$$\nabla\cdot\vec{J}+\partial_t (\psi^*\psi)=0$$ where $\vec{J}$ is probabilty current.
It seems like an important aspect of quantum physics.
Furthermore, this topic is a must in the discussion of finding a relativisic wave equation. We can see how this is done for the Klein-Gordon equation and Dirac equation.
But then, suddenly, nobody talks about that anymore. I dont see it in quantum electrodynamics or in string theory or any other quantum theory.
Is local conservation of probability perhaps not as important as it seems? Do you know any reference(book or paper) deriving the local conservation of probability in the context of QFT, QED or String Theory?
 A: One may think it is maybe possible to reproduce the reasoning that leads to this equation from the Schrödinger one in any quantum system where the Schrödinger equation holds (i.e. surely QFT and QED, I don't know about string theory). However, this is actually rather difficult, and it is not possible to obtain the nice form of the continuity equation (in my opinion) apart from some special situation. In addition, I would like to point out that the local conservation equation is a direct consequence of the Schrödinger equation (and of the particular form of "standard" Hamiltonians in non-relativistic QM), and does not add any further insight than the Schrödinger equation.
One problem in QFT is that we often do not know the precise form of the interacting Hamiltonian, however we may suppose that we know it and it is of the form $H=H_0(\Pi)+V(\Phi)$; where $\Phi$ is the quantum field operator and $\Pi$ its conjugate momentum.
Another problem is that $\Pi$ does not necessarily behave like a derivation operator on the wavefunction $\Psi$, and $V(\Phi)$ as a multiplicative operator. However, this can be made possible using a "trick" at least in a special case, i.e. when the Hilbert space of the QFT is a Fock space $\Gamma(\mathscr{H})$, where $\mathscr{H}$ is the one-particle (separable) Hibert space. In fact there is a construction, called Q-space, that unitarily identifies the Fock space with an $L^2(\Omega,d\mu)$ space of functionals $\Psi(\phi)$ on $\Omega$ with Gaussian measure $d\mu(\phi)$. In this space, the Fock space field $\Phi(x)$ acts as the multiplication by the function $\phi(x)$, and the momentum $\Pi(x)$ as the functional derivative $-i\partial_{\phi(x)}$.
Now in the Q-space form the Hamiltonian becomes $H=H_0(\partial_\phi)+V(\phi)$, and this is analogous to the usual $L^2(\mathbb{R}^d)$ form of QM, and therefore the conservation equation
$$\partial_{\phi(t)} J\Bigl(\Psi(t,\phi),\partial_{\phi(t)}\Psi(t,\phi(t))\Bigr)+\partial_t\Bigl(\Psi(t,\phi)^*\Psi(t,\phi)\Bigr)=0\; ;$$
may be recovered (with a suitable current $J$, that in the case of $H_0(\Pi)\simeq \Pi^2$ has the usual form).
A: I'd say the answer to your question "Is local conservation of probability perhaps not as important as it seems?" should be positive, at least for relativistic quantum theory. Indeed, is probability really conserved? I'd say not necessarily. For example, particle-antiparticle pairs can be created in some processes. Is probability conserved in such processes? I doubt it, unless you say that a positron can be regarded as an electron with "negative probability distribution". What is important in relativistic quantum theory is current conservation, not probability conservation, whereas it is pretty much the same for non-relativistic quantum theory, as the energy is too low there for pair creation. 
A: I think the answer to your question is because it is difficult to work in the Schrodinger picture in QFT whilst maintaining a reasonable standard of logic. Let $|0\rangle_{H}$ be the vacuum state (ground state of the interacting Hamiltonian) in the Heisenberg picture. In general, the Schrodinger picture does not exist, so there is no Schrodinger picture time evolution,
\begin{equation}
|0(t)\rangle_{S} = e^{-itH_{S}}|0\rangle_{H}
\end{equation}
where $H_{S}$ is the Hamiltonian of the interacting theory. However, the Heisenberg picture is good so that operators make sense in the Heisenberg picture.
\begin{equation}
K(t)_{H}=e^{itH_{S}}K_{S}e^{-itH_{S}}
\end{equation}
At any time $t$, there is, presumably, a state $|0\rangle_{t}$ (which we can call the vacuum at $t$) which is annihilated by the Heisenberg picture annihilation operator evaluated at $t$,
\begin{equation}
a(t)_{H}|0\rangle_{t}=0
\end{equation}
but, since the Schrodinger picture does not exist, the vacuums $|0\rangle_{t}$ and $|0\rangle_{t'}$ for two different times $t,t'$ are not related by a unitary operator. So, at time $t$, the states that make sense are of the form $K(t)_{H}|0\rangle_{t}$ and all physical questions that can be asked must be answered by these states. In order to answer physical questions, we need to get c-numbers and so the operator $K(t)_{H}$ has to be put into normally ordered form with the annihilation operators to the right. Now, any terms in the normally ordered $K(t)_{H}$ that contain annihilation operators will not contribute in the physical state $K(t)_{H}|0\rangle_{t}$; Dirac, Lectures on Quantum Field Theory, Belfer Graduate School of Science, 1966, page 148 calls them latent terms. Now, these latent terms are still in $K(t)_{H}$, so they may evolve to other terms at $t'$ which are no longer latent in $K(t')_{H}$ and so contribute to the physical state $K(t')_{H}|0\rangle_{t'}$. This means that we can normalize a physical state $K(t)_{H}|0\rangle_{t}$ at one time $t$ but $K(t')_{H}|0\rangle_{t'}$ won't necessarily stay normalized and so it's not possible to logically get a notion of conserved probability.
A: I would say that the answer to your question relies in the construction of the $S$-matrix. When constructing the $S$-matrix, one requires that it is a unitary operator, hence one that conserves probability. Now, $S$-matrix is a typical object of the Interaction Picture, in which QFT and QED are mainly developed by physicists, so it is in some way "folklore". Indeed, on a formal level Haag's theorem states that, in four-spacetime dimensions, the transformation carrying from the Schroedinger to the interaction picture is not unitary, hence the two pictures are not equivalent. Obviously, one can't keep Haag's theorem on the some foot of a "true" mathematical theorem, since its hypotesis are based on our approximated description of the physical world and are subjected to our current knowledge. As far as I know, people working in QFT are inclined to consider it as the consequence of an over-formalization of the objects describing the theory.
I think that a formal presentation of such topics could be found in [1]. An informal outilne can be found in [2] and [3].

[1] Bogoliubov et alii, "Axiomatic Quantum Field Theory"
[2] Mandl, Shaw, "Quantum Field Theory"
[3] Landau, Lifschitz, "Course of theoretical physics IV"
