# Number of components of Dirac spinor in arbitrary dimensions.

1. Jul 10, 2008

### arroy_0205

How does one calculate the number of components a Dirac spinor in arbitrary dimensions? As far as I understand, the textbooks treat the four spacetime dimensions and here the spinor has four components because the gamma matrices must be 4x4 in nature to satisfy the required algebra. Now suppose I want to see what the case will be in five/six/arbitrarily higher dimensions. How should I proceed? Is the way to first identify dimensionality of gamma matrices which will satisfy the anticommutation algebra etc, and hence deduce the structure of the spinor? That will probably take a long time by trial and error. Is there any systematic approach?

Second question: In four dimensions, the "minimum" dimensionality of gamma matrices required to satisfy the anticommutation relations, etc is four but that means, I can choose higher dimensional matrices also, if I can find suitably in four dimensions and then the spinor structure will be higher than four and the physics will be more complicated, I guess. But no textbook deals with such picture, and I think there must be something wrong with this argument. Can anybody help me with this issue?

2. Jul 10, 2008

### CompuChip

As for the first question, it turns out that the number of components is generally $$2^{\lfloor d \rfloor}$$ with d the number of spacetime dimensions and the brackets denoting floor rounding. Nobody has ever explained to me why it is precisely this number though, and what "in general means" - just that it's true and complicated and it's a coincidence that there are 4 components in 4 dimensions. So if someone could derive this result, I'd like to know as well.

As for the second one, the gamma matrices are representations of an algebra. In four dimensions, the 4x4 matrices are an irreducible representation. But of course, if you have such 4x4 matrices $\gamma^\mu$ then you can always make, e.g. an 4nx4nrepresentation by setting
$$\Gamma^\mu = \begin{pmatrix} \gamma^\mu & 0 & \cdots & 0 \\ 0 & \gamma^\mu & \cdots & 0 \\ \vdots & 0 & \ddots & \vdots \\ 0 & 0 & & \gamma^\mu \end{pmatrix}$$
with 0 the 4x4 null matrix, so basically it's
$$\operatorname{diag}(\underbrace{\gamma^\mu, \cdots, \gamma^\mu}_{n \times})$$
Then of course the corresponding spinors will have more components, but since nothing has changed physically these are all obsolete. Probably you could decompose the whole thing into irreducible representations and consider all of them separately (and in the above case, you would get n copies of the original system, since all irreducible parts are the same).

3. Jul 10, 2008

### arroy_0205

Thanks for your response, but I have doubt about what you say. According to what you say, the no. of components of spinor in d=4 is 2^4=16, which is not the case. Probably you meant 2^(d/2) where the ceiling function(d/2) is to be understood. That would make sense. But we should understand why that is true.

4. Jul 10, 2008

### CompuChip

Yes, sorry, there was a /2 missing.

5. Jul 14, 2008

### per.sundqvist

I think the reason they are four is that you get a coupled set of two-component spinors, which represent the binary nature of spin (up-down), also known to reproduce Pauli's term etc. Even if you can use a 120x120 matrix in four dimensions, how should you interpret the spinor components? The four represents the 3D nature and the nature of spin as we know it. But what about say 5 dimensions, What is the corresponding nature of spin in that case? -Thats probably the problem...

6. Jul 16, 2008

### samalkhaiat

Dirac matrices $\Gamma_{a}$, a = 1,2,...,n, are defined to be irreducible representations of Clifford algebra

$$\{\Gamma_{a},\Gamma_{b}\} = 2 \eta_{ab} \mathbb{1} \ \ (1)$$

for arbitrary dimensions n and arbitrary signatures of the metric $\eta_{ab} = \pm \delta_{ab}$.
By adding a unit element, these matrices allow us to obtain $2^{n}$ numbers;

$$\mathbb{1}, \Gamma_{a}, \Gamma_{a}\Gamma_{b},...,\Gamma_{1}\Gamma_{2}...\Gamma_{n}$$

which are linearly independent and thus form a basis of a $2^{n}$-dimensional vector space.

The following two theorems (proved in some textbooks on Clifford algebra) answer all questions raised in this thread.

"For a given even dimension, n =2p, and a given signature, all irreducible representations of the Clifford algebra (1) are equivalent and are given by $2^{p}\times 2^{p}$ matrices"

This means that for any two representations $\{\Gamma \}$ and $\{\Gamma^{'} \}$, there is a non-singular matrix S such that

$$\Gamma^{'} = S \Gamma S^{-1}$$

"For a given odd dimension, n =2p+1, and a given signature, there exist two and only two mutually inequivalent irreducible representations, both being $2^{p}\times 2^{p}$ matrices"
This means that if $\{\Gamma \}$ is in one equivalence class then $\{- \Gamma \}$ is in the other.

So, in any dimension n the gamma matrices are $2^{p} \times 2^{p}$, where $p = 1,2,3,..,n/2 \ \mbox{or} \ (n-1)/2$.

Spinor representation of SO(n)

Any element of SO(n) can be written as

$$e^{\omega_{ab}M^{ab}}$$

where $\omega_{ab}$ are the n(n-1)/2 real parameters of SO(n) and $M^{ab} = -M^{ba}$ are a set of n(n-1)/2 linearly independent matrices (generators) satisfying the Lie algebra of SO(n);

$$[M_{ab},M_{cd}] = \delta_{ac}M_{bd} + \delta_{bd}M_{ac} - \delta_{ad}M_{bc} - \delta_{bc}M_{ad}$$

Let $\Psi_{A}, \ A = 1,2,..,m$ be a multi-component field that forms a representation of SO(n), i.e., transforms linearly as

$$\Psi_{A} \rightarrow \left( e^{\omega_{ab}M^{ab}} \right)_{AB} \Psi_{B}$$

where the generators M are represented by the $m \times m$ matrices

$$\left(M^{ab}\right)_{AB}$$

A representation of M in terms of Clifford numbers;

$$\left(M^{ab}\right)_{AB} \propto \left( \Gamma^{[a} \Gamma^{b]} \right)_{AB}$$

where $A,B = 1,2,3,...2^{p}$, $p \geq 1$, define the spinor representation of SO(2p) and SO(2p+1); Since the generators M are given by $2^{p} \times 2^{p}$ matrices in this representations, the multi-component field $\Psi_{A}$ has $2^{p}$ components and is called (multi)spinor. So in n = 4,5 Dirac bispinor has 4 components, in n = 6,7, it has 8 components, etc.

regards

sam

7. Oct 11, 2010

Hello
For an pseudo-Euclidean Space E (with an arbitrary number of time like dimension) with dim(E)=2n or dim(E)=2n+1, a spinor has 2^n components, according:
Abraham Paist, Journal of Mathematical Physics, Volume 3, Number 6, November-December 1962, p1135-1139, "on Spinors in n Dimensions"

8. Feb 22, 2012

### arkoroy

Can anyone please mention some textbooks which provides the complete proof of the above two theorems?

9. Feb 22, 2012

### Bill_K

"Representations of Groups" by Hermann Boerner. Chap 8 is "Spin Representations".