In my classes the dirac equation is always presented as the "square root" of the Klein Gordon equation, then from this you can demand certain properties from the Matrices (anticommutation relations, square to one etc) and it turns out the four gamma matrices will satisfy all these relations.

However as I've been delving into group theory, specifically representation theory of the Lorentz group, it would seem the gamma matrices have much more physical significance and aren't just purely mathematical requirements, which is discussed here at into level: https://en.wikipedia.org/wiki/Gamma_matrices#Physical_structure

Can anyone lend themselves to the task to help myself understand how to think of these matrices physically?


2 Answers 2


I'll start in the context of three-dimensional space, and then I'll extend the context to four-dimensional space-time.

An observable should be invariant under a $2\pi$ rotation. A model is usually constructed in terms of field operators, and observables are expressed in terms of field operators, but the field operators themselves do not need to be invariant under a $2\pi$ rotation. This is important because of the spin-statistics theorem, which says that in relativistic QFT, a fermion field (whose corresponding particle obeys the Pauli exclusion principle) must change sign under a $2\pi$ rotation.

So, if we want to be able to handle the Pauli exclusion principle in a relativistic QFT, we need a way to construct fields that change sign under a $2\pi$ rotation. Representations of the rotation group $O(3)$ don't do this. We need something else. Clifford algebra gives us a nice way to construct that something else.

Still working in the context of three-dimensional space, suppose we have three matrices $\gamma_1,\gamma_2,\gamma_3$ that satisfy $$ \gamma_j\gamma_k+\gamma_k\gamma_j=2\delta_{jk}. $$ We can represent an ordinary vector as $\mathbf{v}=\sum_k v^k\gamma_k$. Familiar manipulations of vectors can be expressed using this representation. In the following equations, $\mathbf{v}\mathbf{u}$ means the matrix product of the matrix representations of $\mathbf{v}$ and $\mathbf{u}$. (As an abstract product, apart from any matrix representation, this would be called the Clifford product.) The dot product of two vectors $\mathbf{v}$ and $\mathbf{u}$ is $$ \frac{\mathbf{v}\mathbf{u}+\mathbf{u}\mathbf{v}}{2} =(\mathbf{v}\cdot\mathbf{u})I, $$ where $I$ is the identity matrix, and the more-natural replacement for the "cross product" is $$ \mathbf{v}\wedge\mathbf{u}\equiv \frac{\mathbf{v}\mathbf{u}-\mathbf{u}\mathbf{v}}{2}, $$ which is a linear combination of the basis bivectors $\gamma_j\gamma_k$. (This is called the wedge product, and it produces a bivector — as it should — rather than a vector.) A rotation through angle $\theta$ in the $1$-$2$ plane (for example) is given by $$ \mathbf{v}\mapsto \exp\left(\frac{\theta}{2}\gamma_1\gamma_2\right) \mathbf{v} \exp\left(-\frac{\theta}{2}\gamma_1\gamma_2\right). $$ This is an ordinary rotation of the vector $\mathbf{v}$ through angle $\theta$ (not $\theta/2$) in the $1$-$2$ plane (aka "about the $3$ axis"). A spinor is a single-column matrix $\psi$ that transforms under rotations according to $$ \psi\mapsto \exp\left(\frac{\theta}{2}\gamma_1\gamma_2\right)\psi. $$ To motivate this, notice that the product $\mathbf{v}\,\psi$ again transforms like spinor: $$ \mathbf{v}\,\psi\mapsto \exp\left(\frac{\theta}{2}\gamma_1\gamma_2\right) \mathbf{v} \exp\left(-\frac{\theta}{2}\gamma_1\gamma_2\right) \exp\left(\frac{\theta}{2}\gamma_1\gamma_2\right)\psi = \exp\left(\frac{\theta}{2}\gamma_1\gamma_2\right)\mathbf{v}\,\psi. $$ Even more, if we choose the matrix representation of the $\gamma$-matrices so that $\gamma_k^\dagger=\gamma_k$, then the quantity $\psi^\dagger\mathbf{v}\psi$ is invariant under all rotations. And if $\theta=2\pi$, then we can use $(\gamma_1\gamma_2)^2=-1$ to prove that the transformation reduces to $\psi\mapsto-\psi$, which is what we want. This means that $\psi$ by itself can't be an observable, but something involving a product of two $\psi$s can still be an observable because the minus signs cancel.

The smallest matrices that satisfy the first equation are $2\times 2$, so we can represent $\psi$ as a column matrix with two (complex) components. These correspond to the "spin up" and "spin down" components of an electron, for example. The preceding equations show how these two components mix with each other under a rotation.

In summary, regarding the physical significance of the $\gamma$-matrices in three-dimensional space: we can use them to describe ordinary vectors, including ordinary rotations, and they also provide a nice way to describe things that change sign under $2\pi$ rotations, as fermions should. So we get everything we need, all in one package.

Now transition to four-dimensional space-time. We have basically the same story but with the rotation group $O(3)$ replaced by the Lorentz group. For consistency with the spin-statistics connection, we need a way to construct representations that change sign under a $2\pi$ rotation. Representations of the Lorentz group itself don't do this, but again we can use Clifford algebra. By the way, this all generalizes nicely to an arbitrary number of space-time dimensions, but I'll just show the 4-d case here.

Suppose we have 4 matrices $\gamma_\mu$ that satisfy $$ \gamma_\mu\gamma_\nu+\gamma_\nu\gamma_\mu=2\eta_{\mu\nu}, $$ where $\eta_{\mu\nu}$ are the components of the Minkowski metric. We can represent an ordinary four-vector as $\mathbf{v}=\sum_\mu v^\mu\gamma_\mu$. The preceding comments about dot products and the wedge-product apply here, too. (The "cross product", which pretends to construct a vector from the two input vectors, does not generalize to four-dimensional space-time, but the wedge product does.) A Lorentz transformation (boost or rotation) in the $\mu$-$\nu$ plane is given by $$ \mathbf{v}\mapsto \exp\left(\frac{\theta}{2}\gamma_\mu\gamma_\nu\right) \mathbf{v} \exp\left(-\frac{\theta}{2}\gamma_\mu\gamma_\nu\right). $$ The effect of the same Lorentz transformation on a Dirac spinor $\psi$ is $$ \psi \mapsto \exp\left(\frac{\theta}{2}\gamma_\mu\gamma_\nu\right)\psi. $$ Again, this changes sign under a $2\pi$ rotation, so we can use this for a fermion field. It can't be an observable by itself, but we can use it to construct observables because any product of an even number of these things is invariant under a $2\pi$ rotation. The smallest matrices that satisfy the defining relationship have size $4\times 4$. (In $2n$-dimensional space-time, they have size $2^n\times 2^n$, and they have this same size in $2n+1$-dimensional space-time.)

In summary, regarding the physical significance of the $\gamma$-matrices in four-dimensional space-time: we can use them to describe Lorentz boosts of things like ordinary vectors, and they also provide a nice way to describe things that change sign under $2\pi$ rotations, as fermions should. So we get everything we need, all in one package — without ever mentioning anything about square-roots of Klein-Gordon equations.

By the way, saying that a fermion field must change sign under a $2\pi$ rotation might seem problematic, because it says that ordinary spin-1/2 particles — like electrons, protons, and neutrons — must also have this property. They do, and that's not a problem. It's not a problem in, say, a single-electron state, because the sign-change in that case is just a change in the overall coefficient of the state-vector, which has no observable consequences. It would cause a problem in a state like $|\text{even}\rangle+|\text{odd}\rangle$ that is a superposition of states with even and odd number of fermions, and such superpositions are not allowed in QFT. States with even and odd numbers of fermions belong to different superselection sectors. What we can do is consider a superposition of two different locations of a single fermion, and then the sign-change under a $2\pi$ rotation does have indirect observable consequences. This has been demonstrated in neutron interference experiments, basically two-slit experiments with a macroscopic distance between the two paths in the interferometer. (Diffraction in a crystal was used as a substitute for "slits".) Magnets were used to cause precession of any neutron that passes through one of the paths, and the effect on the resulting two-slit interference pattern displays the effect of the sign-change under $2\pi$ rotations. This is reviewed in "Theoretical and conceptual analysis of the celebrated $4\pi$-symmetry neutron interferometry experiments", https://arxiv.org/abs/1601.07053.

  • $\begingroup$ Thank you this was very helpful, I'm leaving it open for a bit in case anyone else wants to chime in as well. One question, what does $$\mathbf{v}\mathbf{u}$$ mean here? If it's a bivector that makes sense for the wedge product but not so much for the scaler product. Thanks so much $\endgroup$
    – Craig
    Oct 27, 2018 at 21:32
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    $\begingroup$ @Craig Good question. $\mathbf{vu}$ is the matrix product of the two vectors' matrix representations, also called the Clifford product. In the scalar product case, the result is the dot product times the identity matrix. This follows from $\gamma_j\gamma_k+\gamma_k\gamma_j=2\delta_{ij}$. I added these words to the post. Thanks for catching this oversight. I also added an appendix about observables consequences of the sign-change under $2\pi$ rotations. $\endgroup$ Oct 27, 2018 at 22:30
  • $\begingroup$ can I say that the gamma matrices are the generators of the underlying Lie algebra which specifies how bi-spinors transform? $\endgroup$
    – Craig
    Mar 30, 2019 at 4:19
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    $\begingroup$ @Craig The products $\gamma^a\gamma^b$ with $a\neq b$ can be regarded as those generators, yes. Also, in 4-d spacetime, those products can be block-diagonalized, and the blocks are the generators that specify how Weyl spinors transform. $\endgroup$ Mar 30, 2019 at 5:07
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    $\begingroup$ @Craig The vector-transformation rule that I wrote is consistent with the one that you wrote, just expressed differently. The way you wrote it, the vector $v$ is represented as a column matrix (a matrix with a single column). The way I wrote it, the vector $v$ is represented as a square matrix (a linear combination of the $\gamma$-matrices: $\sum_k v^k \gamma_k$). These are two different ways to represent the same vector, both equally valid. The effect of a Lorentz transform looks different in these two representations because the representations are different, but the effect is the same. $\endgroup$ Dec 8, 2019 at 20:44

The construction can be generalized to an $n$-dimensional $\mathbb{F}$-vector space $V$ with a $\mathbb{F}$-bilinear symmetric non-degenerate form $g: V\times V\to \mathbb{F}$. Let $(e_k)_{k=1, \ldots, n}$ be a basis for $V$ and $g_{jk}:=g(e_j,e_k)$.

  1. The (possibly indefinite) orthogonal group $$O(V)~:=~\{M\in GL(V) ~|~\forall v,w\in V:~~ g(M(v), M(w))~=~g(v,w) \}$$ $$ ~\stackrel{\text{polarization}}{=}~\{M\in GL(V) ~|~\forall v\in V:~~ g(M(v), M(v))~=~g(v,v) \} \tag{1}$$ with corresponding Lie algebra $$so(V)~:=~\{m\in {\rm End}(V) ~|~\forall v,w\in V:~~ g(m(v),w)+g(v,m(w))~=~0 \}$$ $$~\stackrel{\text{polarization}}{=}\{m\in {\rm End}(V) ~|~\forall v\in V:~~ g(m(v),v)+g(v,m(v))~=~0 \}~. \tag{2}$$ There is a vector space isomorphism $$\bigwedge{}^2 V~\ni~\omega ~=~\frac{1}{2}\sum_{j,k=1}^n\omega^{jk}e_j\wedge e_k ~~\mapsto~~ -i((\cdot)^{\flat})\omega ~=~\sum_{j,k,\ell=1}^n e_j \omega^{jk}{}g_{k\ell} e^{\ast\ell} \in~so(V), \tag{3}$$ where $i:V^{\ast}\times \bigwedge V \to \bigwedge V$ denotes the interior product and $\flat:V\to V^{\ast}$ is the musical isomorphism $v\mapsto v^{\flat}:=g(v,\cdot)$.

  2. The Clifford algebra is defined as $$Cl(V)~:=~T(V)/I(V), \qquad T(V)~:=~\bigoplus_{n=0}^{\infty} T^n(V), $$ $$T^n(V)~:=~ \underbrace{V\otimes \ldots\otimes V}_{n\text{ factors}},\qquad T^0(V)~:=~\mathbb{F}, \tag{4} $$ where $I(V)$ is the 2-sided ideal in $T(V)$ generated by $$\{v\otimes v - g(v,v){\bf 1} \in T(V)~|~ v\in V\}.\tag{5}$$ The linear map $\Phi: V\to {\rm End}(\bigwedge V)$ given by a sum of exterior and interior multiplication $ v\mapsto e(v)+i(v^{\flat})$ can be extended to an algebra homomorphism $$\Phi: T(V)~\to~ {\rm End}(\bigwedge V)\tag{6}$$ so that $$ \Phi(v\otimes v)~=~ \Phi(v)\circ \Phi(v)~=~\ldots~=~g(v,v)\Phi({\bf 1}) \tag{7}$$ with kernel $ {\rm Ker}(\Phi)=I(V)$. In other words, there is an algebra homomorphism $$\widetilde{\Phi}: Cl(V)~\to~ {\rm End}(\bigwedge V)\tag{8}$$ Then $$Cl(V)~\ni~ c~~\mapsto~~ \widetilde{\Phi}(c)(1)~\in~ \bigwedge V\tag{9}$$ is a vector space isomorphism. In particular $$Cl(V)^{\rm even}~\ni~ c~=~\frac{1}{4}\sum_{j,k=1}^n\omega^{jk}(e_j\otimes e_k-e_k\otimes e_j) $$ $$~~\mapsto~~ \widetilde{\Phi}(c)(1) ~=~\frac{1}{2}\sum_{j,k=1}^n\omega^{jk}e_j\wedge e_k ~=~\bigwedge{}^2 V.\tag{10} $$

  3. The maps (3) & (10) can be combined to yield an imbedding $so(V) \hookrightarrow Cl(V)^{\rm even}$. In plain English: The generators of the Lie algebra (2) can be can be identified with anticommutators of gamma matrices (up to normalization). See also Refs. 1 & 2.


  1. S. Sternberg, Lie algebras, 2004; Chapter 9.

  2. W. Fulton & J. Harris, Representation theory, 1991; Lecture 20.


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