I have to give a quick talk at our University about the Dirac Equation and the Clifford Algebra. Since I am still at the beginning of my studies, my knowledge in this regard is still very limited. Therefore, it is difficult for me to work into the subject.

On Wikipedia I found that the Clifford algebra is defined as a vector space $V$ over a field $K$, where $V$ is equipped with the quadratic form $Q: V \to K$. All elements $u$, $v$ of $V$ should also fulfill the following equation: $\{u,v\}=2\langle u,v\rangle 1\text{, where }\langle u,v\rangle=\frac{1}{2}\left(Q(u+v)-Q(u)-Q(v)\right)$

In case of the Dirac equation, $V$ should be the Minkowski space and $Q$ the spacetime interval (at least I found this in another post here).

My problem is here to understand, what the connection between the gamma matrices (which fulfill the anti-commutation relation above) and the Minkowski space $V$ is, because since the Minkowski space is a 4 dimensional vector space its elements are 4-vectors and not 4$\times$4-matrices. So shouldn't $V$ be the space of the complex 4$\times$4 Matrices here? I also can't see why $\langle u,v\rangle = g_{\mu\nu}$

Note that I post here to get as simple answers as possible, because there is already a lot about it on the internet, but my mathematical knowledge is not sufficient.

  • $\begingroup$ Dirac matrices furnish a representation of the Clifford algebra. $\endgroup$ Jun 16, 2022 at 12:11
  • $\begingroup$ Yes I've read this also in other posts. But what is about the fact, that V is the Minkowski space, but not the vectors of the minkowski space follow the dirac algebra, the gamma matrices do. This seems to be a contradiction to me. $\endgroup$
    – Aralian
    Jun 16, 2022 at 13:33
  • $\begingroup$ Is $V$ Minkowski space? To me, it looks more like the vector space spanned by $w^\mu \gamma_\mu$ where $w$ is in Minkowski space. In any case, I wouldn't recommend spending much time deciphering this notation because it's something that physicists don't use. $\endgroup$ Jun 16, 2022 at 14:06

3 Answers 3


I think this is the most useful article:


You'll find the Dirac equation in this section.

Indeed the gamma matrices do correspond to the basis 4-vectors of Minkowski space, and this is the beauty of the approach. Let me explain how:

The Clifford algebra is also called the Geometric Algebra. In GA, you take the "geometric product" of two vectors, and you get the inner (dot) product plus the exterior (cross) product -- but the cross product is not a vector, but rather a "bivector", an oriented plane with a magnitude. Likewise, if you multiply 3 orthogonal vectors, you get a "trivector" and so on. Finally, you can add scalars, vectors, bivectors, etc. creating a composite object called a "multivector", though it's not immediately clear this should be useful.

So the product of a vector $v$ with itself is a scalar $v^2$ (its squared norm) whereas the product of two orthogonal vectors $u$ and $v$ is a pure bivector $uv$, and also, $uv = -vu$.

Thus, if you have any orthonormal basis of vectors called $\gamma_\mu$ for a metric space, you can quickly verify that $\gamma_\mu \gamma_\nu + \gamma_\nu \gamma_\mu = 2\eta_{\mu\nu}$ where $\eta_{\mu\nu}$ is the metric.

So you can see that the spacetime basis vectors, under the geometric product, obey the exact same commutation relations as the gamma matrices.

That's pretty neat already, but it doesn't guarantee the entire Dirac equation can be converted into multivectors. But it turns out it can, as shown by David Hestenes. Here is one of his books that discusses the subject. But the result is in that section referenced above. It turns out the Dirac wavefunction can be written as an "even" multivector (scalar + bivector + quad-vector). Not only that, but the derivative term becomes the "spacetime gradient", since each partial is multiplied by its corresponding basis vector (a.k.a. gamma matrix). The whole thing is quite elegant and thought-provoking, whether or not it ultimately leads to any deeper understanding of quantum mechanics.

And by the way, you might ask, what could be the significance of an even multivector? Well, as explained in the article, it can actually be decomposed into the product of a scalar, a Lorentz transformation, and a complex phase. This has to do with the fact that the "quad-vector" also called the "pseudo-scalar" actually has the same behavior as the imaginary number $i$. And in GA, $i$ acts as a "duality operator" -- it converts a geometric object into its Hodge dual, e.g. turns a vector into the orthogonal trivector. But it also turns a bivector into its dual bivector, hence it is an operator on bivectors at heart. Thus a complex phase is a "duality rotation", turning a bivector only partially into its dual.

Now, in order for this transformation to work, it turns out you have to both pre-multiply an object by the wavefunction and post-multiply by its conjugate. But this is exactly what we do in much of quantum theory, so that too suddenly takes on a concrete geometric meaning.

(A final note on bivectors: in 4 dimensions, a bivector in general can not always be written as the product of two orthogonal vectors [such a bivector is called "simple"] but it can be written as the sum of a simple bivector and its dual)


The vector space for the Dirac equation is not 4-dimensional Minkowski space. The quantities that satisfy the Dirac equation are bispinors which are spinor-like quantities that have 4 components. In reality, this is just a combination of two other representations: the "regular" spinor representation (with 2-component spinors) and its conjugate representation. Hence the vector space is going to be the space of such bispinors. However, it is not the spinors themselves that satisfy the Clifford algebra; it's the Dirac matrices. So indeed, the vector space $V$ of the Clifford algebra is the space of $4 \times 4$ complex matrices.

$Q$ is also not the spacetime interval. It is simply the quadratic form for the vector space elements. In a Clifford algebra, the square of an element $v$ is defined as:

\begin{equation} v^{2} = Q(v)\, I \end{equation}

where $I$ is the identity element of the algebra (the object which preserves other elements of the group under multiplication). The case of $Q(v) = v^{2}$ is the trivial case one encounters in "regular" algebra. It can however take the form $Q(v) = -v^{2}$ which for example holds true for Grassmann (anti-commuting) numbers. After finding an orthogonal basis in the vector space, if you have $p$ basis elements that satisfy the "regular" quadratic relation and $q$ which behave like Grassmann numbers, then you can write the Clifford algebra symbolically as $Cl_{p,q}(\mathbb{R})$.

The Clifford algebra of the Dirac gamma matrices is $Cl_{1,3}(\mathbb{R})$, which means if we identify those orthogonal elements of $V$, one of them satisfies $e_{0}^{2} = 1$ while the other three satisfy $e_{i}^{2} = -1$. If we perform the analogue of a Wick rotation (or a series of such) on $Cl_{1,3}(\mathbb{R})$, we may turn it into $Cl_{4}(\mathbb{C})$. The same can be done for $Cl_{3,1}(\mathbb{R})$ which satisfies the opposite relations ($e_{0}^{2} = -1$ and $e_{i}^{2} = 1$). Therefore we may choose either of the two with an analogous algebra still in place. It becomes apparent then why the anticommutator of the gamma matrices should be the Minkowski metric $\eta ^{\mu \nu}$: the basis elements match to each of the spacetime components (which of course form an orthogonal basis).

The reason why you are going to have four orthogonal basis elements is due to the dimensionality of the Clifford algebra and the way the Dirac representation of it is constructed. The dimensions of a Clifford algebra of $m$ total elements in $V$ is going to be $2^{m}$. As I mentioned in the first paragraph, the Dirac bispinor representation is the sum of the spinor representation and its conjugate. Since the spinor representation is 2-dimensional, the associated Clifford algebra is 4-dimensional, hence it will have 4 orthogonal basis elements.

You can think of it in terms of the associated group of the Clifford algebra we are examining, which is going to be the Lorentz group $SO(1,3)$. The matrices which shall satisfy the Clifford algebra will combine to form a set of other matrices which obey the Lie algebra of the Lorentz group. Said group can be seen as a manifold of dimensions equal to that of the associated algebra. Every point on the manifold maps to an element of the group, and every such element can be generated by a series of linear infinitesimal transformations, with the unit element of the group as the starting point:

\begin{equation} g(\alpha) = \mathbb{1} + i \alpha ^{a}T^{a} \, , \quad g \in G \end{equation}

where $\alpha^{a}$ is a vector of infinitesimal measure ($\alpha ^{2} \approx 0$) and $T^{a}$ are the generators of the algebra. This is how one associates Lie groups to their respective Lie algebras. The Lorentz group for our case that is generated by matrices abiding by the aforementioned Lorentz group Lie algebra, when embedded in 5-dimensional Minkowski space it is going to be a 4-dimensional hyperboloid:

(You may ignore the coordinate tags, I used a Latex graph of mine used in another context).

In that case, the four matrices of the Clifford algebra are the unit matrix and the 3 Pauli matrices $\sigma ^{i}$. These matrices will form the generators of the algebra as a spinor under a Lorentz transformation will transform as such:

\begin{equation} \Psi _{L} \xrightarrow{\text{$\Lambda$}} e^{-\frac{i}{2} \omega _{\mu \nu} S^{\mu \nu}_{(L)}} \Psi _{L} \end{equation}

\begin{equation} \Psi _{R} \xrightarrow{\text{$\Lambda$}} e^{-\frac{i}{2} \omega _{\mu \nu} S^{\mu \nu}_{(R)}} \Psi _{R} \end{equation}

where $\Psi_{L}$ are the left-handed spinors (spinors in the first of the spinor representations) and $\Psi _{R}$ the spinors in the conjugate representation. The tensorial quantity $\omega _{\mu \nu}$ is a constant while $S^{\mu \nu}$ are the generators of the Lorentz group:

\begin{equation} S^{ij} = \frac{1}{2} \epsilon ^{ijk} \, \sigma ^{k} \end{equation}

\begin{equation} S^{0i} = \begin{cases} -\frac{i}{2} \sigma ^{i} \quad , \quad (\text{L}) \\ \frac{i}{2} \sigma ^{i} \quad , \quad (\text{R}) \end{cases} \end{equation}

If we now combine the left-handed and right-handed spinors into a Dirac (bi)spinor:

\begin{equation} \Psi _{\alpha} = \begin{pmatrix} \Psi_{L} \\ \Psi{R} \\ \end{pmatrix} \end{equation}

you can see that it will transform similarly under a Lorentz transformation, but its generators will now have to be $4 \times 4$ rather than $2 \times 2$ matrices:

\begin{equation} S^{\mu \nu} = \frac{i}{4} [\gamma ^{\mu} , \, \gamma ^{\nu}] \Rightarrow \begin{cases} &S^{ij} = \frac{1}{2} \epsilon ^{ijk} \, \begin{pmatrix} \sigma ^{k} & 0 \\ 0 & \sigma ^{k} \\ \end{pmatrix} \\ &S^{0i} = -\frac{i}{2} \begin{pmatrix} \sigma ^{i} & 0 \\ 0 & -\sigma ^{i} \\ \end{pmatrix} \end{cases} \end{equation}

  • 1
    $\begingroup$ Your claim that "$Cl_{1,3}(\mathbb{R})$ is isomorphic to $Cl_{3,1}(\mathbb{R})$" is not true, $Cl_{1,3}$ and $Cl_{3,1}$ are NOT isomorphic. $Cl_{1,3}$ is isomorphic to $M_2(H)$, 2*2 matrices of quaternions, while $Cl_{3,1}$ is isomorphic to $M_4(R)$, 4*4 matrices of real numbers. $\endgroup$
    – MadMax
    Jun 16, 2022 at 17:26
  • $\begingroup$ You are correct. I was thinking of performing the analogue of a Wick rotation to transform $Cl_{1,3}$ and $Cl_{3,1}$ into $Cl_{4}$, making them equivalent. I shall edit to clarify this point. $\endgroup$
    – rhomaios
    Jun 16, 2022 at 17:34

Thank you for your answers, they were really helpful. I also had a look at this post here: Relation between the Dirac Algebra and the Lorentz group

I will also talk about the lorentz covariance of the dirac equation. While proving this, it turns out that you have to show that $S\gamma^\nu S^{-1} = \gamma^\mu\Lambda_{\mu\nu}$, which is really complicated.

Following the post above, I came up with the following, which is of course not a prove, but should be ok to get the idea.

Like in the post, one can find (1): \begin{align*} 2\eta^{\mu\nu}1 \\=2\eta^{\mu\nu}S^{-1}S \\= \{S^{-1}\gamma^{\mu'}S,S^{-1}\gamma^{\nu'}S\} \\ =\{\gamma^{\mu'},\gamma^{\nu'}\} \\ \end{align*}

as well as (2): \begin{align*} 2\eta^{\mu\nu}1 \\ =\ ... \\ =\{\Lambda_{\mu\rho}\gamma^\rho,\Lambda_{\nu\sigma}\gamma^\sigma\} \\ =\{\gamma^{\mu'},\gamma^{\sigma'}\} \end{align*}

So you can see, that $\gamma^{\mu'}=S^{-1}\gamma^\mu S = \Lambda_{\mu\nu}\gamma^\nu$

which shows, that the codition holds and therefor the dirac equation is infact lorentz covariant.

So I'd interprete this as following. The fact, that the gamma matrices fulfill the clifford algebra is essential for the dirac equation to be invariant under lorentz transformations. One could say, that the clifford algebra is in some sense the connection between the dirac equation and the lorentz group.

In the process of showing the covariance of the dirac equation, you also assume that the gamma matrices do not transform or transform as scalars under lorentz transformations. Can (2) be interpreted in such way, that $\gamma^{\mu'}=\Lambda_{\nu\mu}\gamma^\nu$ is the lorentz transformation of the gamma matrix $\gamma^\nu$? So in (2) we see, that this transformed gamma matrix also fulfills the anti-commutation relation, so we can choose $\gamma^\mu=\gamma^{\mu'}$ and the assumption holds.

  • $\begingroup$ As you said, what you wrote is not a formal proof, so it's hard to draw any meaningful conclusions about the significance of the Clifford algebra just from this. After all, your proof tells you that gamma matrices transform as vectors under Lorentz transformations (not scalars as you mention later), but have this unique transformation form in the spinor representation. That is due to the Lorentz group generators appearing in the transformations, which in turn don't commute with the gamma matrices. $\endgroup$
    – rhomaios
    Jun 20, 2022 at 19:03
  • $\begingroup$ You could of course say that Lorentz covariance of the Dirac equation in the Dirac representation as it is formulated necessitates this particular algebra, but that is entirely by construction. In a way, we already "know" the formal properties of the Lorentz group, thus we can look for any number of different bases in a multitude of representations to express that algebra. It so happens to be that in the spinor representation, the basis elements satisfy this Clifford algebra. $\endgroup$
    – rhomaios
    Jun 20, 2022 at 19:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.