In 3+1 dimensions, Dirac spinors have four complex components. In 2+1 dimensions, the representation of the Clifford algebra by $\sigma^3$ and $-i\sigma^3\sigma^i$, with $i\in\{1,2\}$ is 2-dimensional, so the dimensional "Dirac spinors" have two complex components. Is there a physical reasons why in 3+1 dimensions there are 4 components, while in one dimension lower there are only two?

I think mathematically one might simply say that the Clifford algebra is always even-dimensional, but I would like to know if there is also some physics behind this.

  • $\begingroup$ Are you specifically asking why there are two fewer complex components, as opposed to just one fewer? $\endgroup$
    – gj255
    Commented Oct 25, 2016 at 23:46
  • $\begingroup$ I would like to say: however you interpret the question, I would like to hear an answer or argument. $\endgroup$ Commented Oct 25, 2016 at 23:55
  • $\begingroup$ You can translate to what they mean for the quanta as particle/antiparticle and left/right. But I'm not sure what precisely is a satisfactory answer. $\endgroup$
    – AHusain
    Commented Oct 25, 2016 at 23:56
  • $\begingroup$ There is an interesting fact that in certain dimensions (2+1, 3+1, 5+1, and 9+1) the real dimension of a minimal spinor (in 2+1 and 3+1 this is a Weyl spinor) is exactly twice the number of real degrees of freedom of a particle moving on the lightcone. If you can find a way to remove half of the spinor degrees of freedom, then there is potential for defining a model with symmetry between the spinor and "position" degrees of freedom. Not sure if this is a coincidence or if there is "physics behind it." $\endgroup$ Commented Oct 26, 2016 at 3:03
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/53318/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented Oct 26, 2016 at 6:02

2 Answers 2


A representation of the Clifford algebra could be obtained by creation and annihilation fermionic modes $c_k$ and $c_k^{\dagger}$, by the following definition: $$ \Gamma_{2k-1}=c_k+c_k^{\dagger}\,\,; \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Gamma_{2k}=i(c_k-c_k^{\dagger}) $$

For $k$ being an integer between $0 <k< (d+1)/2$. For directions $\mu$ with negative signature you should put a imaginary unit $i$ up front. The Lorentz generator of

$$ \delta [...]^{\rho}=\frac{1}{2}\theta_{\mu\nu}(\eta^{\rho\nu}[...]^{\mu}) $$

where $\theta_{\mu\nu}=-\theta_{\nu\mu}$ is given by:

$$ \Sigma(\theta)=-\frac{i}{2}\theta_{\mu\nu}\Gamma^{\mu}\Gamma^{\nu} $$

Note that for "rotations" in $(2k,2k-1)$ plane is given by

$$ \Sigma(\theta)=-\frac{i}{2}\theta_{2k,2k-1}(c_k+c_k^{\dagger})(i)(c_k-c_k^{\dagger})=\frac{1}{2}\theta_{2k,2k-1}(-c_kc_k^{\dagger}+c_k^{\dagger}c_k)= $$ $$ =\theta_{2k,2k-1}\left(c_k^{\dagger}c_k-\frac{1}{2}\right)=\theta_{2k,2k-1}\left(n_k-\frac{1}{2}\right) $$

This provides the interpretation of the modes. The modes are the eigenvalues the $\Sigma^{2k,2k-1}$ generator, being $+1/2$ for $n_k=1$ and $-1/2$ for $n_k=0$. The same as $J_3$ for the $d=3$ case.

For even dimensions, there is a chiral matrix commutes with the Lorentz generators (e.g. $\gamma_5$ for $d=4$) and can be written in terms of the modes as:

$$ \bar{\Gamma}=\prod_k(1-2n_k) $$

This means that the eigenvalues of $\bar{\Gamma}$, the chirality, is $+1$ if there is a even number of occupied modes and is $-1$ for odd. Since this matrix commutes with Lorentz generator, for even dimensions we may split a Dirac spinor into two spinors, a chiral spinor ($\bar\Gamma =+1$) and anti-chiral spinor ($\bar\Gamma =-1$).

For odd dimensions one of this gamma matrices will be left over. For the last $k$-mode we take just the $\Gamma_{2k-1}$ and the $\Gamma_{2k}$ is unused. You may note that now there are two matrices that commutes with Lorentz generator, the $\Gamma_{2k}$ and the $\bar\Gamma$. It is important to note that they not commute with each other, so we can use just one of them to split the representation but not both. Whatever matrix we use we will end up with a reduction of the number of modes by one. If there is $n$ modes for $d=2n$, for $d=2n-1$ we have $n-1$ modes.

There is also other way to obtain the odd dimension gamma matrices for $d=2n-1$ by starting with $d=2(n-1)$ and using the $\bar\Gamma$ matrix as the $\Gamma^{d}$. This have $n-1$ modes from the start.

Now you can see that there is a distinction between odd and even dimensions. So, is always expected to have this jumps on the numbers of components when you reduce an even dimension $d$ to $d-1$, or increase an odd dimension $d$ to $d+1$, because of the increasing number of $c_k$ modes.


As explained by @Nogueira, the irreducible representations of the Dirac matrices have different dimensions depending on whether d=2n or d=2n-1.

So in d=3+1 dimensions the massive particle Dirac spinor is a 4-component object. It allows for the Dirac equation to be invariant under parity ${\bf r} \to -{\bf r}$.

The massive particle Dirac Spinor in d=2+1 dimensions is a two-component object, but there are two inequivalent irreducible representations! So if you pile them up (see below) you have a 4 component spinor.

Why might you want to pile them up? Well parity in 2 space dimensions is defined differently than in 3 space because ${\bf r} \to -{\bf r}$ would simply be a rotation in the plane. So define the parity operation in 2 space by flipping only one coordinate: $ x \to -x$ keeping $y \to y$. Then you will find that your 2-component Dirac equation is not invariant under Parity.

Neither of the two different irreducible representations in d=2+1 is invariant under parity but if you combine them (into a reducible representation) you get a 4 component parity invariant structure.

But it turns out the parity violating 2-component structure is useful and has interesting physics and relations to other structures in the plane. See eg. https://www.sciencedirect.com/science/article/pii/0003491682901646?via%3Dihub


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