Are Electromagnetic Fields Considered Spinors in Geometric Algebra?

[Tomsk]

Hi, I'm trying to teach myself a bit about spinors, mainly from reading about geometric algebra. There is something that I can't figure out though. According to GA, spinors are elements of the even graded subalgebra, so scalars, bivectors and so on. But the electromagnetic field is a bivector, but surely it's not a spinor... so how does that make sense?

Also: When they talk about even graded elements being spinors, are they talking about spin 1/2, 1, 3/2... or something else?

The other thing I'd like to ask about is this unclear section on wikipedia:

Consequences

There are many far-reaching consequences of the Clebsch-Gordan decompositions of the spinor spaces. The most fundamental of these pertain to Dirac's theory of the electron, among whose basic requirements are

* A manner of regarding the product of two spinors $\bar{\phi}\psi$ as a scalar. In physical terms, a spinor should determine a probability amplitude for the quantum state.
* A manner of regarding the product $\bar{\phi}\psi$ as a vector. This is an essential feature of Dirac's theory, which ties the spinor formalism to the geometry of physical space.

http://en.wikipedia.org/wiki/Spinor#Consequences
I didn't really understand the bit about Clebsch-Gordon decomposition but I really don't get how $\bar{\phi}\psi$ can be considered as a scalar and/or a vector. I can see why it would be a scalar but not a vector as well. (BTW the latex \bars don't seem to be showing up)

These are probably fairly basic things... I'm pretty new to it all!

 

[eok20]

The literature on spinors can be very confusing since every person uses different notation. The way I have always understood things is that spinors in n dimensions make up the representation space of a representation of the even subalgebra of the clifford algebra. That is, even elemenets of the Clifford algebra (like bivectors) act on spinors.

In regards to the product of spinors, I am not sure. I mean $$\bar \phi \psi$$ should be a scalar. Maybe for the other thing they meant the Kronecker product $$\psi \bar\phi$$ which is in the endomorphism space of spinors and therefore can be seen as an element of the even Clifford algebra.

The clearest reference on spinors I have found is Ian Porteous's book "Clifford algebras and the classical groups."

 

[ameliaswank]

ahhhhhh ... these are the things which really makes me confused also.. in my school life i never get good marks in geometry just because i never get it correctly... nd even now after passing school i can not make out differences in this field

 

[Tomsk]

That is, even elemenets of the Clifford algebra (like bivectors) act on spinors.

That seems quite different to saying that spinors ARE elements of the even graded subalgebra. It seems to make more sense based on what I've read about representing vectors and matrices and spinors as vectors though. If even elements of the Clifford algebra act on spinors, what do they actually DO to the spinors when they act?

The other thing why I'm not sure why spinors can be even graded elements- in 4d spacetime they have 4=2^(4/2) components, but there are 8=1/2*2^4 components in the even graded subalgebra (1 scalar, 1 pseudoscalar, 6 bivectors). So I'm kind of confused about how to connect this all up.

 

[blue_raver22]

Hello, it's great that you are trying to teach yourself about spinors and geometric algebra. Spinors can be a complex and abstract concept, so it's understandable that you may have some confusion.

Firstly, let's clarify the definition of spinors. In geometric algebra, spinors are defined as elements of the even-graded subalgebra, which includes scalars, bivectors, and higher grade elements. This means that spinors are not just restricted to being spin 1/2, 1, 3/2, etc. They can be any element in the even-graded subalgebra.

Now, to address your question about the electromagnetic field being a bivector, but not a spinor. In geometric algebra, the electromagnetic field is represented as a bivector because it has both a magnitude and direction. However, it is not considered a spinor because it does not satisfy the requirements of being an element of the even-graded subalgebra. Spinors must also satisfy certain algebraic properties, such as being closed under multiplication and addition, which the electromagnetic field does not fulfill.

Moving on to the Wikipedia section about Clebsch-Gordan decomposition, this is a mathematical technique used to decompose a higher dimensional representation into a combination of lower dimensional representations. In the context of spinors, this is used to describe how the spinor spaces can be decomposed into a combination of different spin states. The consequences of this decomposition are related to the properties of spinors in Dirac's theory of the electron.

Now, to address your confusion about how \bar{\phi}\psi can be considered as both a scalar and a vector. In Dirac's theory, spinors represent the quantum state of an electron. The product of two spinors, \bar{\phi}\psi, is known as a bilinear form and can be interpreted as both a scalar and a vector. The scalar interpretation comes from the fact that it represents the probability amplitude of the quantum state. The vector interpretation comes from the fact that it is related to the geometry of physical space, as described in Dirac's theory.

I hope this helps clarify some of your confusion about spinors and their properties. Keep exploring and learning, and don't be afraid to ask questions when things are unclear. Good luck on your journey to understanding spinors!