Is time an observable in Relativistic Quantum Mechanics? Relativistic Quantum Mechanic is based, as far as I know, in the Dirac Equation. Now, the Schrödinger equation, in the abstract state space takes the form:
$$i\hbar \dfrac{d|\psi(t)\rangle}{dt}=H|\psi(t)\rangle.$$
If $|\mathbf{r}\rangle$ is the standard position representation, projecting the equation onto $|\mathbf{r}\rangle$ we get:
$$i\hbar \dfrac{\partial}{\partial t}\langle \mathbf{r}|\psi(t)\rangle = \langle \mathbf{r}|H|\psi(t)\rangle,$$
now, if we write $H = P^2/2m + V$ and $\Psi(\mathbf{r},t)=\langle \mathbf{r}|\psi(t)\rangle$ we get, as long as $P = -i\hbar\nabla$ the usual equation
$$i\hbar\dfrac{\partial \Psi}{\partial t}=-\dfrac{\hbar^2}{2m}\nabla^2\Psi+V\Psi.$$
The physical interpretation of $\Psi$ is obvious. We have $|\Psi(\mathbf{r},t)|^2=|\langle \mathbf{r}|\psi(t)\rangle|^2$ so that from the postulates of Quantum Mechanics, $|\Psi|^2$ is the probability density at time  $t$ for the position.
Now, I've heard that Relativistic Quantum Mechanics keeps the postulates and the only difference is that we change the Hamiltonian operator and we pick one specific state space.
But there's something wrong. After all, in Special Relativity space and time turn into a single thing: spacetime. Quantum Mechanics, on the other hand, treats quite differently space and time, where time is a parameter and indeed has not even one observable corresponding to it.
In that sense, in Relativistic Quantum Mechanics based on Dirac's Equation, is time made into one observable? How this assymetry between space and time that exists in Quantum Mechanics is dealt with in the context of Dirac's Equation?
 A: 
Is time made into one observable?

No. It is known that an operator $T$ that satisfies $[H,T]=i\hbar$ is either self-adjoing and $H$ unbounded below or anti-self-adjoint. Therefore, the theory is either intrinsically flawed (arbitrary negative energy) or $T$ is not observable (anti-self-adjoint $\Rightarrow$ imaginary eigenvalues).
A: 
In that sense, in Relativistic Quantum Mechanics based on Dirac's Equation, is time made into one observable?

No, but it is treated on equal footing to the space coordinates. Dirac's equation is
$$(i \hbar \gamma^\mu \partial_\mu - mc )\psi$$
where $\psi$ is not the classical wavefunction, but a four-component spinor. The components of the spinor are function of the four-dimensional spacetime position $\vec s = (\vec r, ct)$, whose absolute value squared $s^2 = (ct)^2-\vec r^2$ is Lorentz-invariant. So in a certain way it is not time that is promoted to an observable: the space coordinates are "downgraded" to labels.
Update:
With a brief Internet search, I found out that it is actually possible to introduce a time operator and obtain a consistent theory ("Pauli’s objection is shown to be resolved or circumvented", cit.), but apparently it is far easier to simply downgrade the space coordinates and treat them as labels. I've also found out that it is possible to introduce in certain sense a time operator even in classical quantum mechanics.
A: The theorem that time is not an observable is quite general, Unruh W., Wald R. prove this in "Time and the interpretation of canonical quantum gravity, Physical Review D Volume 40 issue 8 1989" in the following form: 
"... in the context of ordinary Schrodinger quantum mechanics, no dynamical variable in a system with Hamiltonian bounded from below can act as a perfect clock in the sense that there is always a nonvanishing amplitude for any realistic dynamical variable to 'run backwards'".
Quantum field theory circumvents the problem by reducing the spatial coordinates to parameters which enumerate the field operators $\varphi(x,t)$ too. 
In general, do not expect that there is anything quantum which is Lorentz covariant on the fundamental level. What people care about is Lorentz covariance on the level of observables, not more. 
A: The question is whether time is an operator in the sense of ${\hat T}|t\rangle~=~t|t\rangle$. This at first glance would seem to make sense because we do have a position operator ${\hat X}|x\rangle~=~x|x\rangle$. However, this does not work. This is a subtle question in many ways.
Quantum mechanics is unitary. Consider a state vector $|\psi(t)\rangle$ evolve into a small increment of time so $|\psi(t)\rangle~\rightarrow~|\psi(t + \delta t)\rangle$. A Taylor expansion this gives
$$
\psi(t + \delta t)\rangle~=~|\psi(t)\rangle + \frac{\partial|\psi(t)\rangle}{\partial t}\delta t + O(\delta t^2).
$$
Now write $|\psi(t)\rangle~=~e^{-i\omega t}|\psi(t_0)\rangle$, and use de Broglie-like relation $\omega~=~E/\hbar$, so that the energy $E$ is an eigenvalue of the Hamiltonian ${\hat H}|E\rangle~=~E|E\rangle$. We can see that the Hamiltonian operator is the Hermitian generator of the unitary time development operator $U(t)~=~e^{i{\hat H}t/\hbar}$. The Hamiltonian is then the generator that tells how a state evolves as $t~\rightarrow~t' > t$, and ${\hat H}~=~i\hbar\partial/\partial t$.
Suppose that time is an operator. We can now examine the energy development of a state $|\psi(E)\rangle~\rightarrow~|\psi(E + \delta E)\rangle$ and in the same manner we can see that the time operator is ${\hat T}~=~-i\hbar\partial/\partial E$. So far things seem alright. We can compute the commutator of the two operators acting on $|\psi(t)\rangle$ and $|\psi(E)\rangle$
$$
[{\hat T}, {\hat E}]|\psi(t)\rangle = [{\hat T}, i\hbar\frac{\partial}{\partial t}]|\psi(t)\rangle
$$
$$
= -i\hbar\left({\hat T}\frac{\partial}{\partial t}|\psi(t)\rangle~-~\frac{\partial}{\partial t}\big({\hat T}|\psi(t)\rangle\big)\right)~=~i\hbar|\psi(t)\rangle.
$$
Much the same works if we consider $|\psi(E)\rangle$ with ${\hat T}~=~-i\hbar\partial/\partial E$. 
Consider a Hermitian time operator ${\hat T}$ such that $[{\hat T},~H]~=~i\hbar$, so a unitary operator  $U_\epsilon~=~exp(-i\epsilon{\hat T})$ exists.  This is an energy development operator, where $\epsilon$ is in the set of reals.  The state $\psi$ in the eigenbasis of a Hamiltonian $H\psi~=~E\psi$, with commutator 
$$
[U_\epsilon,~H]~=~\sum_{n=0}^\infty{{(-i\epsilon)^n}\over{n!}}[{\hat T}^n,~H]~=~-\epsilon U_\epsilon.
$$
defines the composite operator $HU_\epsilon$
$$
HU_\epsilon\psi~=~(U_\epsilon H~-~[U_\epsilon,~H])\psi~=~(E~+~\epsilon)U_\epsilon\psi.
$$
$U_\epsilon\psi$ is an eigenstate of the Hamiltonian with eigenvalue $E~+~\epsilon$.  The Hamiltonian $HU_\epsilon$ is not discrete or bounded below, since $\epsilon$ has a continuum of values on the reals.  If the Hamiltonian $H$ is discrete and bounded below the $U_\epsilon$ maps these eigenvalues on the entire set of reals, and the time operator does not exist
Define the time operator 
$$
{\hat T}~=~i\hbar\sum_{j\ne k}{{|E_j\rangle\langle E_k|}\over{E_j~-~ E_k }}. 
$$
that acts upon a ket $|t\rangle~=~N^{1/2}\sum_nexp(iE_nt/\hbar)$ as
$$
{\hat T}|t\rangle~=~i\hbar\sum_{j\ne k}{{|E_j\rangle\langle E_k|}\over{E_j~-~E_k }}|t\rangle~=~iN^{-1/2}\hbar\sum_{j\ne k}(E_j - E_k)^{-1}|E_j\rangle e^{-iE_kt/ħ}.  
$$
This is a Fourier summation, which in the continuum limit gives $t|t\rangle$ by the Cauchy integral formula.  Now compute matrix elements of $[T,~H]$ for $|\psi\rangle~=~\sum_ja_j|E_j\rangle$
$$
\sum_ja_j\langle E_i|[T,~H]|E_j\rangle~=~i\hbar\sum_{j,k,l}a_j\langle E_i|(E_k~-~E_l)^{-1}|E_k\rangle\langle E_l |E_j\rangle
$$
$$
 =~i\hbar\sum_{j,k}a_j\langle E_i| (E_k~-~E_j)^{-1}|E_k\rangle,
$$
where the matrix element $\langle E_i|(E_k~-~E_j)^{-1}|E_k\rangle~=~\delta_{ik}(E_k~-~E_j)^{-1}$.  $|E_i\rangle$ is not in the projective summation of the time operator for $[{\hat T},~H]~=~i\hbar$, and generally $[{\hat T},~ H]~=~0$.
A Cauchy sequence of states will converge to a bounded state $|\psi\rangle~=$ $\sum_{j=0}^Na_j(N)|E_j\rangle$, for $N$ the bound on the complete set.  For the coefficient $\sim~(1/j)$ as $N~\rightarrow~\infty$ the accumulation point contains a dense set of points with $E~+~\delta E$ energy eigenvalues which satisfy $\sum_j\langle E_j|\psi\rangle~=~0$.  This means that the commutator $[T,~H]~=~i\hbar$ holds on a measure zero set, and for the function $\psi(t)$ an almost periodic function.
