Is this how spinors transform and is it the square root of a vector? I have an expression and the transformation rules, and I wonder if this qualifies as a spinor. Can the following expression written with complex Clifford algebra be seen as a spinor? In any case, it certainly reproduces quantum mechanics of spin wave vectors.
The expression for a single spin-1/2 particle given by Euler angles $\theta,\phi$ is
$$\Psi=\frac{1}{\sqrt{2}}\left((e_1+if_1)\cos\frac{\theta}{2}+(e_2+if_2)\sin\frac{\theta}{2}e^{i\phi}\right)$$
where $e_1,f_1,e_2,f_2$ are orthonormal basis vectors which all square to $e_i^2=f_i^2=+1$ and $i$ is the imaginary unit. This can be recognized as a way to write the usual spin wave vector $\psi=(\cos\frac{\theta}{2},\sin\frac{\theta}{2}e^{i\phi})$. For $n$-dimensional wave vectors it would be $\Psi=\frac{1}{\sqrt2}\sum_k(e_k+if_k)\psi_k$.
If I take the expression
$$\Omega=i(\Psi\Psi^\dagger-1)$$
where the multiplication is the clifford multiplication of vectors, then for a single particle I get
$$
\Omega=T+X\sin\theta\cos\phi+Y\sin\theta\sin\phi+Z\cos\theta 
$$
with
$$
\begin{align*}
X&=\frac{1}{2}\left(e_1f_2+e_2f_1\right)\\
Y&=\frac{1}{2}\left(e_1e_2+f_1f_2\right)\\
Z&=\frac{1}{2}(e_2f_2-e_1f_1)\\
T&=\frac{1}{2}(e_1f_1+e_2f_2)
\end{align*}
$$
which looks like the Bloch sphere.
The probability to measure a state $\Omega^{(A)}$ in an observable $\Omega^{(B)}$ is the inner product
$$
P(A\to B)=\langle \Omega^{(A)}\Omega^{(B)\dagger}\rangle
$$
which is conventionally expressed by the Born rule $P=|\langle \psi^{(A)}|\psi^{(B)}\rangle|^2$.
Spatial rotations around a normalized axis $(r_x, r_y, r_z)$ and an angle $\alpha$ can be performed with the equations
$$
\begin{align*}
\Omega'&=R\,\Omega\, R^\dagger\\
\Psi'&=R\Psi
\end{align*}
$$
with the rotor
$$
\begin{align*}
R&=\exp\left(\frac{\alpha}{2}(T+Xr_x+Yr_y+Zr_z)\right)\\
&=\cos\frac{\alpha}{2}+\sin\frac{\alpha}{2}(T+Xr_x+Yr_y+Zr_z)
\end{align*}
$$
With all that information, and assuming the algebra is correct (which it is), does the initial expression $\Psi$ qualify as a spinor? Is the relation between $\Omega$ and $\Psi$ what is informally called the "square root of a vector"?
(You could also do the whole calculation with real clifford algebra if you use $i=e_0f_0$. Also it works for a wave vector of any dimension)
...It's been a while and no answer :( With so many spinor questions in this forum, I thought checking if transformation rules apply should be easy. Maybe, someone can leave some thoughts in the comments. It's a new formulation and it surely reproduces QM and rotations, but I'm not familiar with spinor theory to tell how they relate. However, it looks suspiciously close the what I've seen about spinors and it looks like an explicit expression for spinors.
A lightning intro to clifford algebra: think of the basis vectors as matrices, because they similarly do not commute. All you need now is that each basis vector squared is $a^2=+1$, and if you have two different basis vectors they anti-commute under multiplication $ab=-ba$. $\dagger$ means reversing the order in a basis vector product (and taking the complex conjugate of the scalar coefficient).
Also note the relation: $\hat a^\dagger_\uparrow=e_1+if_1$ and  $\hat a^\dagger_\downarrow=e_2+if_2$ for a connection to creation operators.
 A: 
... what is called informally the square root of a vector

A spinor is often described by physicists as a double valued representation of the orthogonal group. Mathematicians, who don't like double valued anything rephrase this and say a k-valued representation of a group is actually a representation of a k-covering of the group. Now double coverings of any orthogonal groups exist and are called, unsuprisingly, spin groups. Thus, by the preceding, a spinor is a representation of a spin group and usually this is taken to be a fundamental representation and so the smallest rep.
So where are the 'square roots'?
Well, one informal way of thinking about it is to think of the square root in the complex plane. This organises itself into a double covering of the complex plane branched at the origin. But we can extend this to a double covering of the complex sphere branched at two points (the other branch point is the additional infinite point). And this rather looks like our double coverimg above, but isn't of course. Firstly because it is branched at two points, but also because its a double cover of the sphere rather than of its rotation group.
Now, not only can spinors be defined over flat space, they can be defined over  manifolds. But it turns out that not every manifold can support a spinor, those that do are called spin. And this in turn means they must support what is called a spin bundle, otherwise known as the bundle of spin frames or spin structures. This is the manifold equivalent of a spin group and we form a spinor bundle by using a spin representation. A spinor here, is more precisely a section of this bundle, and is more properly called a spinor field.
Now, it turns out that over a Kahler manifold, spin bundles are isomorphic to the square root of the top holomorphic cotangent bundle. Such bundles are vector bundles. So their sections are vector fields. So in some sense, spinors are not just "informally a square root of a vector field", they actually are one.
A: A couple of observations:

*

*Yes, $\Omega$ and $\Psi$ transform as vector and spinor. Note that the vector basis $e_1,f_1,e_2,f_2$ generates Clifford algebra $Cl(4,0)$. The spinor $\Psi$ can take a more general form such as the whole $Cl(4,0)$ space (16 components), or even part of $Cl(4,0)$ (8 components, as advocated by David Hestenes). OP's choice is limited to vector portion of $Cl(4,0)$ (4 components).


*For Clifford algebra $Cl(4,0)$, the symmetry group spanned by $Cl(4,0)$ bi-vectors is $SO(4)/Spin(4)$. The OP's $SU(2)$ (spanned by $X, Y, Z$) is a special case of Chris Doran's approach of embedding $SU(m)$ in $SO(2m)/Spin(2m)$ (with m=2). The OP's $T,X,Y,Z$ correspond to equations 6.1, 6.4a, 6.4b, and 6.5 in Doran's paper, respectively.


*The most general transformations on $\Omega$ and $\Psi$ are spanned by generators corresponding to all the $Cl(4,0)$ 6 bi-vectors and 4 tri-vectors, which constitute the de Sitter group $SO(4,1)$. In this case, the spinor $\Psi$ has to live in the whole Clifford algebra $Cl(4,0)$ space (8 odd + 8 even components), since the tri-vector-related operators would transform between Clifford-odd and Clifford-even sectors of the spinor $\Psi$.
