# Are spinors just wavefunctions in the dirac field?

1. Aug 24, 2007

### captain

are spinors just wavefunctions in the dirac field?

2. Aug 24, 2007

### jostpuur

I'm been mostly in a belief, that spinors are just vectors, and the name "spinor" is used to emphasize their transformation properties. Altough the Wikipedia seems to have lot more to say http://en.wikipedia.org/wiki/Spinor

I don't think that "function in a field" means anything, unless you can explain what you meant with it in more detail.

3. Aug 24, 2007

### captain

i was just guessing what it was based on what i had seen in textbooks.

4. Aug 24, 2007

### meopemuk

Spinors are 2-dimensional objects whose transformations with respect to rotations are described by the 2D irreducible representation of the rotation group generated by Pauli matrices. They are members in the following sequence of unitary irreducible representations of the rotation group: scalars (1D), spinors (2D), vectors (3D), ...

Electron's wavefunction is a spinor (i.e., a two-component) function in the momentum (or position) space. Dirac's quantum field is a 4-dimensional operator function on the Minkowski spacetime. Its rotational transformations are generated by commutators of Dirac's gamma-matrices. So, it is not exactly a spinor. Sometimes these 4D objects are called bi-spinors.

Eugene.

Last edited: Aug 24, 2007
5. Aug 24, 2007

### jostpuur

When you have a three vector $x\in\mathbb{R}^3$, the fact that it is defined with three numbers, is not yet everything about it. An equally important property is how it transforms in rotations. Its transformation properties also justify imagining it as an arrow in the space. An arrow in the space is something that you can rotate like you can rotate stuff in the physical world.

A spinor $\psi\in\mathbb{C}^2$ that you use to describe an internal state of an electron is basically two complex numbers, but the most important thing is how these numbers transform in rotations. The formula for rotations is defined with Pauli matrices, and the rotation operator is $\exp(-i\theta\cdot\sigma/2)$ (I don't remember if this was for passive of active rotations).

So if you just think it's a vector with certain transformation properties, you should get quite far.

Where you actually have these spinors is a different matter then. That probably can lead us to some depthful debate about content of QM again. Talking about the Dirac's field, isn't the Dirac's field itself a spinor-valued field? I'm not sure. I'm not sure what the Dirac's field is anymore... In non-relativistic theory of electrons you can at least simply replace the complex number of the wavefunction with a spinor, that means you replace $\Psi:\mathbb{R}^3\to\mathbb{C}$ with a $\Psi:\mathbb{R}^3\to\mathbb{C}^2$ and postulate transformation properties.

6. Aug 24, 2007

### genneth

If you know anything about groups, they can be understood as thus:

Consider the group of rotations in 3D, aka SO(3). We can look for representations of this group, which means finding some vector space in which we can map elements of SO(3) as operators (matrices) in that space. So obviously, there's the representation in R^3 where the elements are just what we usually think of as rotations. We call this the vector representation. Another representation is in 1D -- i.e. R. There, we map all the elements to 1 -- it's completely trivial and degenerate. This is the scalar representation. So we might wonder if there's a 2D one; as it happens, yes there is, but only in C^2. The mapping is a little more involved. These are the spinors.

7. Aug 25, 2007

### Anonym

No. The spinors are the mathematical objects similar to the scalars, vectors, tensors, spintensors, etc. of any dimensions. They are useful in the numerous physical applications. In particular, the solutions of the Dirac equation are four component spinors.The best presentation I know is the first hand E. Cartan “The Theory of Spinors”.

Regards, Dany.

Last edited: Aug 25, 2007