Can I apply the Born rule to a Dirac spinor? How does a Dirac spinor such as:
$$
\psi = \pmatrix{a_0+ib_0\\a_1+ib_1\\a_2+ib_2\\a_3+ib_3}
$$
Connect to a probability?
Can one apply the Born rule of this object?
 A: The following has been taken from Greiner' s Relativistic Quantum Mechanics where he  makes it clear the Dirac equation can be thought of as a relativistic wave equation.
This answer is a precursor to how does it connect to probability (Chapter $2$).
To shorten some of the arguments presented:
We define
$$ \hat \alpha_i = \begin{pmatrix}
0 & \hat \sigma_i \\
\hat \sigma_i & 0
\end{pmatrix} $$
And $\hat \beta$ as:
$$ \hat \beta= \begin{pmatrix}
I & 0 \\
0 & -I
\end{pmatrix} $$
where $\sigma_i$ is the $i$'th $2 \times 2$ Pauli matrix and $I$ is the identity matrix. We wish to construct a four-current density for the Dirac equation:
$$ i \hbar \frac{\partial \psi}{\partial t}= \frac{\hbar c}{i} \sum_{i =1}^3  \hat \alpha_k \frac{\partial}{\partial x^k} \psi + m_0 c^2 \hat \beta \psi $$
Multiplying the above with a Hermitian conjugate:
$$ i \hbar \psi^\dagger \frac{\partial \psi}{\partial t}= \frac{\hbar c}{i} \sum_{i =1}^3  \hat \psi^\dagger \alpha_k \frac{\partial}{\partial x^k} \psi + m_0 c^2 \hat \psi^\dagger \beta \psi $$
Taking the conjugate of the Dirac equation and multiplying $\psi$. We get $2$ equations:
$$ - i \hbar \frac{\partial \psi^\dagger}{\partial t} \psi = - \frac{\hbar c}{i} \sum_{i =1}^3  \hat \alpha_k \frac{\partial \psi^\dagger }{\partial x^k} \psi  + m_0 c^2 \hat \beta \psi^\dagger \psi $$
Substracting the above $2$ equations:
$$ i \hbar  \frac{\partial \psi^\dagger \psi }{\partial t} = \frac{\hbar c}{i } \sum_{i=1}^k \frac{\partial}{\partial x^k} (\psi^\dagger \hat \alpha_k \psi)$$
(Note $\alpha^\dagger = \alpha$ and $\beta^\dagger = \beta$)
Hence this is precisely of the form:
$$ \frac{\partial \rho}{\partial t} + \nabla j = 0$$
where $\rho = \psi^\dagger \psi $ and $j =  (\psi^\dagger \hat \alpha_k \psi) $
Since $ \rho $ is positive definite we accept $j$ as a probability density current.
A: As soon as you're dealing with relativistic equations, you're no longer in the realm of 1-particle quantum mechanics.
It is a basic preliminary result of quantum field theory that in order to have relativistic causality preserved, you must include both positive and negative frequencies in the amplitude.
There is a classic piece of literature where Feynman explains this concept at length: Elementary Particles and the Laws of Physics: The 1986 Dirac Memorial Lectures.
But the same material is covered in any book or lecture notes on QFT. Look for the first chapters where the authors explain the different propagators.
Fundamentally, this is because any relativistic equation worth its salt must satisfy Einstein's energy-momentum identity,
$$
E^{2}-c^{2}\boldsymbol{p}^{2}=m^{2}c^{4}
$$
which gives you two solutions for the dispersion relation,
$$
E=\pm\sqrt{m^{2}c^{4}+c^{2}\boldsymbol{p}^{2}}
$$
And Dirac's equation is no exception. It is essentially the "square root" of Klein-Gordon's equation.
As @josephh has said, a 4-spinor is not a wave function. Not even the more fundamental Weyl 2-spinors (of which Dirac 4-spinors are essentially direct sums) are wave functions. If you try to build a quadratic bilinear that carries statistical information about the state, you must give up positivity. This forces you to re-interpret the most natural quadratic bilinear as a number operator, instead of a probability density. This number becomes a charge density operator (which can be either positive or negative), the amplitudes become operators instead of states, and the 1-particle states are more fundamental objects on which these space-time dependent operators act.
A: The Born rule is the probability postulate of quantum mechanics.
Postulates are extra axioms used by physics theories in order to pick those mathematical solutions that will describe data and predict future data.
It does not apply on the wavefunctions, it defines the way wavefunctions predict the behavior of data by giving the probability distribution of the data by $Ψ^*Ψ$.
A: Sure you can! This is like the whole point of spinors!
For simplicity sake let's think about particles with spin $1/2$ (like electrons): Spinors are vectors in the Hilbert space of spin, you can write them as column vectors
$$\chi = \begin{bmatrix} a\\b\\\end{bmatrix} \ \ \ \ \ \ a,b \in \mathbb{C}$$
or equivalently you can write them as a linear combination of basis vectors
$$ \chi = a |+\rangle + b |-\rangle $$
And this last object is exactly one for which Born's rule applies! The probabilities of measuring the spin along one direction, up or down, are simply derived with the Born's rule (assuming of course normalized basis vectors): just take the module squared of the component and you are done!
A: The Born rule, in its simplest form, tells us that the probability (density) of finding a particle at a given point is the square of the particle's wave function at that point. That is, $$\rho = \mid \psi (x,t)\mid^2$$ which is always a positive real number.
As such, the Dirac spinor is not a wave function. They do not even represent traditional states in a Hilbert space. By convention, spinors are taken to be the elements of a representation space of the Dirac algebra.
But anyway, from the fermion 4-current $$j^{\mu}=\bar \psi\gamma^{\mu}\psi$$ and if we take the $\mu=0$ component, we define the probability density as $$j^0=\rho=\bar \psi\gamma^{0}\psi = \psi^{\dagger} \psi$$ and this quantity is not  always positive definite.
A: When treating the Dirac equations as a classical equation, one can find plane wave solutions with internal degrees of freedom that are represented as Dirac spinors. Let's denote them as:
$$ \psi = u_s \exp(-ipx) - \text{fermion} $$
$$ \phi = v_s \exp(ipx) - \text{anti-fermion} , $$
where $s$ represents the spin index. These solutions are combined with ladder (annihilation and creation) operators in the quantization of the fermion field to define field operators. These ladder operators are often denoted as $\hat{a}_s(p)$ and $\hat{a}_s^{\dagger}(p)$ for the fermions and $\hat{b}_s(p)$ and $\hat{b}_s^{\dagger}(p)$ for the anti-fermions. When the creation operator is applied to the vacuum it produces a state with the properties associated with the Dirac spinor solution
$$ |p_s\rangle = \hat{a}_s^{\dagger} | vac\rangle . $$
However, note that $|p_s\rangle$ is not a well-defined state in the infinite dimensional case, because it is not normalizable.
One can now use these ladder operators to define states. An arbitrary single particle fermion state would be created by an operator
$$ \hat{a}_F^{\dagger} = \sum_s \int \hat{a}_s^{\dagger}(p) F_s(p)\ \text{d}p , $$
where $F_s$ is an arbitrary normalized spectral coefficient function. The state that is produced by this operator is simply produced by
$$ |\psi\rangle = \hat{a}_F^{\dagger} | vac\rangle . $$
Although the concept of a wave function was introduced with the Schroedinger equation, the wave function is today understood in a broader sense as the function that is produced from a state when projected onto a coordinate basis. In other words, the wave function is given by
$$ \psi(x) = \langle x|\psi\rangle . $$
The nature of the coordinate basis is such that when they are used with the individual spinor states, the result is the original Dirac spinor solutions
$$ \langle x|p_s\rangle = u_s \exp(-ipx) . $$
In that sense, the Dirac spinor solution is the (unnormalizable) wave function of the individual plane wave states. Because it is not normalizable, it does not represent a physical wave function. However, all physical wave functions can be represented as superpositions of these Dirac spinor solutions:
$$ \psi(x) = \sum_s \int u_s \exp(-ipx) F_s(p)\ \text{d}p . $$
What about a measurement? A measurement is represented by
$$ \langle \hat{A}\rangle = \langle\psi| \hat{A}|\psi\rangle , $$
where $\hat{A}$ is the observable represented by a Hermitian operator. It can be the number operator defined in terms of the fermion field operators
$$\hat{n}(x) = \Psi^{\dagger}(x)\Psi(x) . $$
When you substitute this operator into the expression for the measurement and apply it to the single particle fermion state you'll obtain an expression for the probability distribution of observing a single electron at a given location $x$.
