# Difference between spinor and vector field [duplicate]

How do we distinguish spinors and vector fields? I want to know it in terms of physics with mathematical argument.

-

## marked as duplicate by Qmechanic♦Feb 7 '13 at 1:38

Possible duplicate: physics.stackexchange.com/q/41211/2451 –  Qmechanic Feb 6 '13 at 21:45

The answer to your question depends on the context, but the basic unifying theme distinguishing different kinds of fields (like vector fields, scalar fields, etc.) is how these fields transform when they are acted on by Lie Groups (and or Lie Algebras) which falls under the mathematical subject of representation theory of Lie groups and Lie algebras. Here are some examples to illustrate what I mean:

Consider a real field $V^\mu(x)$ defined on 4-dimensional Minkowski space $\mathbb R^{3,1}$. We say that this field is a Lorentz 4-vector field provided when it is acted on by a Lorentz transformation $\Lambda = (\Lambda^{\mu}_{\phantom\mu\nu})$ (an element of the Lorentz Group $\mathrm{SO}(3,1)$), it transforms as follows: $$V^\mu(x) \to V'^\mu(x) = \Lambda^{\mu}_{\phantom\mu\nu} V^\nu(\Lambda^{-1}x)$$

Now Consider instead a field $\Psi^a(x)$ defined on 4-dimensional Minkowski space $\mathbb R^{3,1}$. We say that such a field is a Dirac Spinor provided when it is acted on by Lorentz transformations, it transforms as follows: $$\Psi^a(x) \to \Psi'^a(x) = R_{\mathrm{dir}}(\Lambda)^a_{\phantom a b}\Psi^b(\Lambda^{-1}x)$$ where $R_{\mathrm{dir}}$ is called the Dirac spinor representation of the Lorentz Group.

There are also other kinds of spinors, all of whom transform according to different representations of different Lie groups.

-
Nice answer, maybe an extenstion of this to a blog post would be nice too ...? –  Dilaton Feb 6 '13 at 20:40
@Dilaton. Thanks; that's a great idea. I'll do my next post on this and extend the concepts with more detail! –  joshphysics Feb 6 '13 at 21:34
It might be good to expand a bit on the spinor representation itself, since a transformation law of the form $\Psi^a \rightarrow M(\Lambda)^a_b \Psi^b$ is completely general. –  Vibert Feb 7 '13 at 9:31

Often we use spinors to rotate things. Using clifford algebra, you get a rotation formula looking like this:

$$a' = \psi a \psi^{-1}$$

where $a$ is some vector. This shows how vectors are rotated in a double-sided way.

If you chain two rotations together, you might get something like this:

$$a'' = \phi \psi a \psi^{-1} \phi^{-1} = \psi' a {\psi'}^{-1}$$

So $\psi' = \phi \psi$. Spinors transform in a single-sided way.

Geometrically, vectors are the oriented lines that you're used to, with a weight equal to the vector's magnitude. Spinors represent linear combinations of scalars and bivectors, oriented planes. Any two vectors can define a spinor, and in 3d, that spinor has components related to both the dot product and cross product. It measures, in a redundant way, how the two vectors are both aligned and not-aligned (including what plane they lie in). These vectors are related geometrically to the rotation as the two reflection vectors you can build up a rotation from--any rotation can be built up from two reflections.

-
+1, thanks for the beginnings of my geometric intuition about spinors :) Please could you give a reference or two where I could read a bit more about this geometric interpretation of spinors? –  Mark Mitchison Feb 6 '13 at 23:50
By double-sided you mean the inverse transformation matrix included? –  pkjag Aug 3 '14 at 19:57

Some mathematical representation has been given by others. If you like some discussion on the physics difference between these two objects you can consider the following. The difference is in the rotational symmetry of these fields. The wave function of a vector particle needs to be rotated once (360 degrees) in order to return to its original configuration (to be identical). The wave function of a spinor particle, must be rotated twice (720 degrees) to return to its original configuration. This is because of the spin difference of these two objects being $S = 1\hbar$ and $S = 1/2\hbar$ respectively. For a rotation by an arbitrary angle 'a' the rotation matrix operator is of the general form

$R(a)=\exp (iS.a/\hbar)$.

So you can see that if $S = 1\hbar$ then an angle $a = 2\pi$ returns the field to its original configuration, while if $S = (1/2)\hbar$ then you need $a = 4\pi$ to return the "object" to its original configuration. I hope this helps see the difference in the physics of these two fields.

-