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.