What has "quantisation" to do with associated graded algebras?

Question

I was currently reading an introduction into spin geometry by José Figueroa-O’Farrill. The first chapter handles Clifford algebras. When discussing the connection of the Clifford algebra to the exterior algebra, the author states:

Filtered algebras whose associated graded algebras are commutative (or supercommutative) can be interpreted as quantisations of their associated graded algebra, which inherits a Poisson bracket from the (super)commutator in the filtered algebra. This is precisely what happens for the Clifford algebra as we will now see.

Later on, this is made more precise: (the "$\mod F^{p+q-4}C$" just means "forget all terms of degree less than $p+q-4$)

We define a bracket $$[\cdot, \cdot]: \Lambda^pV \times \Lambda^qV\rightarrow \Lambda^{p+q-2}V$$ by $$[\alpha, \beta] := \alpha\beta - (-1)^{\vert\alpha\vert\vert\beta\vert}\beta\alpha \mod F^{p+q-4}C$$ It is an exercise to show that this is a Poisson bracket making $\Lambda V$ into a Poisson superalgebra. It is in this sense that $Cl(V,Q)$ is a quantisation of $\Lambda V$. We can think of $\Lambda V$ as the functions on the “phase space” for a finite number of fermionic degrees of freedom and $Cl(V,Q)$ as the corresponding quantum operator algebra. The Hilbert space of the quantum theory is then an irreducible representation of $Cl(V,Q)$.

(for more in-depth information see pages 8-10 of the above linked document.)

My question is therefore as follows: How is a "quantisation" of an algebra to be understood from a physical point of view? Can the usual way of quantization (i.e. replacing the position and momenta scalars by their corresponding operators) be understood in this picture?

Answer 1

The mathematical term of "quantization" of algebras is much broader than the physical notion of quantization of a classical theory, however, a very prominent quantization in the mathematical sense occurs during the physical process of quantization, which possibly is the origin of the mathematical terminology:

During (canonical) quantization, we turn the commutative algebra of classical observables (functions on phase space with the Poisson/Moyal bracket) into an anti-commutative algebra of operators on phase space. It is known (by the Groenewold-van Howe theorem) that's there's no proper quantization map with all desired properties if we insist on mapping the Poisson bracket to the quantum commutator, but there is if we use the Moyal bracket, which crucially has a parameter suggestively called $\hbar$ in it that controls how much it deviates from the classical Poisson bracket. Therefore, the algebras of quantum observables are quantizations in the mathematical sense of the algebra of classical observables, and the latter is obtained from the former by letting the quantization/deformation parameter go to zero, i.e. $\hbar\to 0$.

So this is the archetypical "quantization". In the case you're asking about, the two algebras in play are the "classical" exterior algebra with $v\wedge v = 0$ and the "quantum" Clifford algebra with $v\wedge v = Q(v)$ where $Q$ is a quadratic form, usually the Minkowski or Euclidean metric. Just like the quantum algebras of observables give the classical algebra of observables in the limit $\hbar\to 0$, the Clifford algebra gives the exterior algebra in the limit $Q\to 0$.

However, the text you quote suggest the similarity is more than just formal. If we allow for classical fermionic degrees of freedom (as they are needed, for example, for the BRST formalism), then the classical algebra of observables on a purely fermionic phase space of dimension $n$ is just the exterior algebra in $n$ dimensions. Introducing an $\hbar$ by which the quantum (anti-)commutator may deviate from the classical exterior algebra is akin to introducing the $Q$ from earlier, so for such a purely fermionic system, the Clifford algebra is indeed the quantum algebra to a system which has the exterior algebra as its classical algebra of observables.

However, it must be noted that this is not the main physical importance of the Clifford algebra. To physicists, the Clifford algebra is most useful because it defines the Dirac spinor representation by its irreducible representation.

Answer 2

This is just the mathematicians terminology of quantization of fermions. In the physicists terms the exterior algebra is a Grassmann algebra and just like the Poisson algebra is quantized into the Weyl algebra in the case of bosons:

$$\{x, p\}_{P.B.} = 1 \rightarrow [x, p] = i$$

the Grassmann algebra, in the case of fermions is quantized into a Clifford algebra

$$\{\psi_i, \psi_j\}_{P.B.} = \delta_{ij} \rightarrow \{\hat{\psi}_i, \hat{\psi}_j\} = \delta_{ij}$$

May be, the most unfamiliar relation of the above is the Poisson brackets between the Grassmann variables, but they can be easily obtained from canonical analysis of classiacal Lagrangians with Grassmann degrees of freedom such as:

$$ L = p_i \dot{q_i} + \dot{\psi_i}\psi_i$$

(The Grassmann part is just a zero dimensional free Dirac Lagrangian)

Most of the physical texts emphasize the role of Grassmann variables in the path integral formulation, but there is an equally rich theory of Hamiltonian quantization where they play a major role. Sometimes it is called pseudoclassical mechanics.

The physical interpretation of Grassmann variables is spin, since the spin generators can be constructed as:

$$ S_i = i \epsilon_{ijk} \psi_j\psi_k$$

You can check that these variables satisfy the spin algebra both classically (with Grassmann variables) and quantum mechanically with the Clifford variables.

Please see one of the original works describing this interpretation by Berezin and Marinov.

In fact, the above model is one of the simplest models exhibiting supersymmetry.This is why Clifford algebras occur in the quantization of supersymmetric models.

Also, the quantization equations above, together describe a simple example of quantization of a supermanifold.