Given a quadratic space $(V, q)$, and an orthogonal transformation $\sigma \in O(V,q)$, the spinor norm of $\sigma$ is nonzero if and only if $\sigma$ is not in the image of the canonical map $Pin(V,q) \to O(V,q)$. If $\sigma \in SO(V,q)$, we can say the same using the map $Spin(V,q) \to SO(V,q)$. It is a refined way to measure the failure of an element to come from the spin group, and you can view it as a boundary map in Galois cohomology (see, e.g., Wikipedia).

I don't know of a deep connection to physics, but it does come up in the following sense: if $(V,q)$ is an indefinite real space of dimension at least 3, like $\mathbb{R}^{3,1}$, then the pin group only has 2 connected components, while the orthogonal group has 4 - we can reflect in vectors of either positive or negative norm to get P or T type discrete symmetries. The pin group only maps to the subgroup of $O(3,1)$ generated by reflections in positive norm vectors, as these are precisely the transformations with positive spinor norm. The spin group is connected, and maps to the connected group $SO_0(3,1)$, while $SO(3,1)$ contains PT symmetries that reverse orientation of both space and time.

The spinor genus is a version of this that uses additional valuations - two lattices have the same spinor genus if for all completions you can transform the respective base changes by orthogonal transformations with trivial spinor genus.