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.
Best Answer
I think this is the most useful article:
https://en.wikipedia.org/wiki/Spacetime_algebra
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)