If ${L}$ is a line bundle over a complex manifold, what does the square root line bundle $L^{\frac{1}{2}}$ mean?
After some google, I got to know that there are certain conditions for the existence of square root line bundle. In particular,I've following questions :


1. What is the square root of a line bundle, what are the conditions for its existence.
2. More importantly, how to think of the square root line bundle. Intuitively, It seems that a square root bundle $L^{\frac{1}{2}}$ is a line bundle s.t. the tensor bundle obtained by taking the tensor product of $L^{\frac{1}{2}}$ with itself gives the line bundle $L$.(correct if I am wrong).


I am particularly interested in square root of the Canonical Line bundle over a Riemann Sphere and the relation of square root bundle to Spinors in QFT.
Please provide some references to look at. 



What is a square root of a line bundle?
If $L$ is any line bundle over a compex manifold $X$, a square root of $L$ is a line bundle $M$ such that $M^{\otimes2}=L$. So your guess in part (2) is correct.


This square root (if it exists) is not unique in general, and two of them will differ by a $2$-torsion line bundle, that is a line bundle $\eta$ such that $\eta^{\otimes 2}$ is trivial.
In particular, if $\textrm{Pic}(X)$ is torsion free, then there is at most one square root. 


In some cases no square root exists. Some general results are:


1. A line bundle of degree $0$ has always at least one square root. This because $\textrm{Pic}^0(X)$ is a complex torus, hence a divisible group (in fact, there are roots of any order).
2. A line bundle over a Riemann surface of genus $g$ has a square root if and only if it has even degree. The number of different square roots equals in this case $2^{2g}$, the number of $2$-torsion points in $\textrm{Pic}^0(X) \cong \textrm{Jac}(X)$.
3. If $L$ is effective, that is $H^0(X, L) \neq 0$, and $Z \subset X$ is the zero locus of a
holomorphic section of $L$, then the existence of a square root of $L$ is equivalent to the existence of a double cover $Y \to X$ branched over $Z$. In particular, non-trivial square roots of the trivial bundle correspond to non-trivial *unramified* double covers of $X$.


The square root of the canonical bundle of the Riemann Sphere $S$ is unique, since $\textrm{Pic}(S)=\mathbb{Z}$, and it is isomorphic to $\mathcal{O}(-1)$, the dual of the hyperplane bundle (the unique line bundle of degree $1$, whose transition function is $z \to 1/z$).


A readable introduction to spinor bundles is provided in the book of MOORE "Lectures on Seiberg-Witten invariants". 


