Quantization from Clifford Algebra

Justin Dressel

Schmid College of Science and Technology

Chapman University


APS March Meeting, 2018

A bit of History

The canonical Hilbert-space representation of quantum mechanics was formalized by von Neumann in 1932, based on the unification of Schroedinger and Heisenberg by Dirac.

"I would like to make a confession which may seem immoral:

I do not believe absolutely in Hilbert space any more.


After all, Hilbert space (as far as quantum mechanical things are concerned) was obtained by generalizing Euclidean space, footing on the principle of ‘conserving the validity of all formal rules’. . . .


Now we begin to believe that it is not the vectors which matter, but the lattice of all linear (closed) subspaces. Because: 1) The vectors ought to represent the physical states, but they do it redundantly, up to a complex factor, only 2) and besides, the states are merely a derived notion, the primitive (phenomenologically given) notion being the qualities which correspond to the linear closed subspaces."


John von Neumann (1935),
as quoted in Stud. Hist. Phil. Mod. Phys. 21, 493 (1996).

However, only three years later in 1935 he believed his formalization was wrong.


He came to the conclusion that the algebra contained the physics, with Hilbert space being only a convenient representational space.

Key Correspondence

The quantum eigenvalue-eigenvector framework appears algebraically without a (Hilbert) representation space by looking at algebraic ideals generated by primitive idempotents

\epsilon^2 = \epsilon
ϵ2=ϵ\epsilon^2 = \epsilon
\epsilon A \epsilon = \lambda_A \epsilon
ϵAϵ=λAϵ\epsilon A \epsilon = \lambda_A \epsilon
A = \sum_\lambda \lambda \epsilon_\lambda
A=λλϵλA = \sum_\lambda \lambda \epsilon_\lambda

(More generally, any algebraic element has an expansion in terms of idempotents and nilpotents)

Dirac "bras" and "kets" appears as left and right ideals

Density "operators" appear as central ideals

|\psi\rangle \doteq \psi \epsilon
ψψϵ|\psi\rangle \doteq \psi \epsilon
\langle\psi| \doteq \epsilon \psi
ψϵψ\langle\psi| \doteq \epsilon \psi
|\psi\rangle \! \langle \phi | \doteq \psi \epsilon \phi
ψϕψϵϕ|\psi\rangle \! \langle \phi | \doteq \psi \epsilon \phi

This gives a prescription : find an algebra that encodes something physically relevant, identify its idempotents that generate ideals

(See work by Bohm and Hiley)

An Analogy

Representation of a complex number:

a + i b \doteq \begin{pmatrix} a & b \\ -b & a \end{pmatrix}
a+ib(abba)a + i b \doteq \begin{pmatrix} a & b \\ -b & a \end{pmatrix}
(a+ib)^2 = (a^2 - b^2) + 2abi \doteq \begin{pmatrix}a & b \\ -b & a\end{pmatrix}^2 = \begin{pmatrix}(a^2 - b^2) & 2ab \\ -2ab & (a^2 - b^2)\end{pmatrix}
(a+ib)2=(a2b2)+2abi(abba)2=((a2b2)2ab2ab(a2b2))(a+ib)^2 = (a^2 - b^2) + 2abi \doteq \begin{pmatrix}a & b \\ -b & a\end{pmatrix}^2 = \begin{pmatrix}(a^2 - b^2) & 2ab \\ -2ab & (a^2 - b^2)\end{pmatrix}

Technically contains same information, but less structurally clear and with redundancies

Note that i is represented as a "symplectic matrix"

Difficult to tell which features are artifacts of the representation, and which are essential

Geometric meaning: 

a + i b = re^{i\theta}
a+ib=reiθa + i b = re^{i\theta}

Encodes relationship between Cartesian and polar axes in a plane, with multiplication dilating radius and rotating the angle

Easy to derive and understand from algebraic properties of \(i\) alone.

Matrix representation obfuscates the relationships

Which algebraic features define the quantum theory?

Canonical commutation relations:

\displaystyle [\hat{p},\, \hat{x}] = -i\hbar \,\,\,\,\,\,\,\Rightarrow\,\,\,\,\,\ [\hat{p}_\mu, \,\hat{x}_\nu] = i \hbar\, \eta_{\mu\nu}
[p^,x^]=i [p^μ,x^ν]=iημν\displaystyle [\hat{p},\, \hat{x}] = -i\hbar \,\,\,\,\,\,\,\Rightarrow\,\,\,\,\,\ [\hat{p}_\mu, \,\hat{x}_\nu] = i \hbar\, \eta_{\mu\nu}
E = \hbar \omega, \,\,\,\,\,\, \vec{p} = \hbar \vec{k}
E=ω,p=kE = \hbar \omega, \,\,\,\,\,\, \vec{p} = \hbar \vec{k}

DeBroglie Relations:

(4D) Poisson bracket:

Classical symplectic structure more compactly encoded with an i

Consider classical fields on phase space:

\displaystyle [\hat{E}/c,\, c\hat{t}] = i\hbar
[E^/c,ct^]=i\displaystyle [\hat{E}/c,\, c\hat{t}] = i\hbar

Minkowski metric

f(x^\mu,p^\mu), \, g(x^\mu,p^\mu)
f(xμ,pμ),g(xμ,pμ)f(x^\mu,p^\mu), \, g(x^\mu,p^\mu)
\displaystyle \hat{\nabla} \equiv \sum_\mu \left( \hat{x}^\mu\frac{\partial}{\partial x^\mu} +i\frac{\hat{p}^\mu}{\hbar}\frac{\partial}{\partial p^\mu} \right)
^μ(x^μxμ+ip^μpμ)\displaystyle \hat{\nabla} \equiv \sum_\mu \left( \hat{x}^\mu\frac{\partial}{\partial x^\mu} +i\frac{\hat{p}^\mu}{\hbar}\frac{\partial}{\partial p^\mu} \right)

Define complex phase space derivative:

\displaystyle [\hat{\nabla}f,\,\hat{\nabla}g] = \sum_{\mu\nu} \left(\frac{\partial f}{\partial x^\mu}\eta^{\mu\nu}\frac{\partial g}{\partial p^\nu} - \frac{\partial g}{\partial x^\mu}\eta^{\mu\nu}\frac{\partial f}{\partial p^\nu}\right) = \{f,g\}_{P.B.}
[^f,^g]=μν(fxμημνgpνgxμημνfpν)={f,g}P.B.\displaystyle [\hat{\nabla}f,\,\hat{\nabla}g] = \sum_{\mu\nu} \left(\frac{\partial f}{\partial x^\mu}\eta^{\mu\nu}\frac{\partial g}{\partial p^\nu} - \frac{\partial g}{\partial x^\mu}\eta^{\mu\nu}\frac{\partial f}{\partial p^\nu}\right) = \{f,g\}_{P.B.}

Lie bracket of vector fields on manifold
(as a commutator)

Operators as Basis Vectors?

\displaystyle \hat{\nabla} \equiv \sum_\mu \left( \hat{x}^\mu\frac{\partial}{\partial x^\mu} +i\frac{\hat{p}^\mu}{\hbar}\frac{\partial}{\partial p^\mu} \right)
^μ(x^μxμ+ip^μpμ)\displaystyle \hat{\nabla} \equiv \sum_\mu \left( \hat{x}^\mu\frac{\partial}{\partial x^\mu} +i\frac{\hat{p}^\mu}{\hbar}\frac{\partial}{\partial p^\mu} \right)

Note the structure of this derivative. It would be a standard gradient if we could interpret \(\hat{x}^\mu\) and \(i\hat{p}^\mu/\hbar\) as unit basis elements of phase space.


\displaystyle \tilde{x}^\mu \equiv i\frac{\hat{p}^\mu}{\hbar}
x~μip^μ\displaystyle \tilde{x}^\mu \equiv i\frac{\hat{p}^\mu}{\hbar}

To make this interpretation, we need to define an associative algebra with non-commuting elements that embeds an antisymmetric form as a commutator.

This defines a symplectic Clifford algebra
(known as the Weyl algebra)

Such a Clifford algebra must be of even dimension, and factorable into complementary halves


The algebra is defined as the associative quotient algebra that preserves the symplectic relation:

\omega(\hat{x}^\mu,\tilde{x}^\nu) \equiv [\hat{x}^\mu,\tilde{x}^\nu] = \eta^{\mu\nu}
ω(x^μ,x~ν)[x^μ,x~ν]=ημν\omega(\hat{x}^\mu,\tilde{x}^\nu) \equiv [\hat{x}^\mu,\tilde{x}^\nu] = \eta^{\mu\nu}

where \(\omega\) is the symplectic form that connects the dual halves of the space to the metric

A general phase space covector is thus part of a 2x4 dimensional algebra, with the noncommutative algebraic product encoding the symplectic structure of the space

\hat{F} = \sum_\mu \left(\hat{x}^\mu F^{(x)}_\mu + \tilde{x}^\mu F^{(x')}_\mu \right)
F^=μ(x^μFμ(x)+x~μFμ(x))\hat{F} = \sum_\mu \left(\hat{x}^\mu F^{(x)}_\mu + \tilde{x}^\mu F^{(x')}_\mu \right)

Hamilton's Equations of Motion

\displaystyle \hat{\nabla} \equiv \sum_\mu \left( \hat{x}^\mu\frac{\partial}{\partial x^\mu} +i\frac{\hat{p}^\mu}{\hbar}\frac{\partial}{\partial p^\mu} \right)
^μ(x^μxμ+ip^μpμ)\displaystyle \hat{\nabla} \equiv \sum_\mu \left( \hat{x}^\mu\frac{\partial}{\partial x^\mu} +i\frac{\hat{p}^\mu}{\hbar}\frac{\partial}{\partial p^\mu} \right)

Consider the classical equations of motion:

\displaystyle \frac{\partial f}{\partial t} = \{f, H\}_{P.B.}
ft={f,H}P.B.\displaystyle \frac{\partial f}{\partial t} = \{f, H\}_{P.B.}
\displaystyle 0 = [\hat{\nabla}f,\,\hat{\nabla}\mathcal{H}] = \{f,\mathcal{H}\}_{P.B.}
0=[^f,^H]={f,H}P.B.\displaystyle 0 = [\hat{\nabla}f,\,\hat{\nabla}\mathcal{H}] = \{f,\mathcal{H}\}_{P.B.}
\mathcal{H}(x^\mu,p^\mu) = E - H(x^k,p^k)
H(xμ,pμ)=EH(xk,pk)\mathcal{H}(x^\mu,p^\mu) = E - H(x^k,p^k)
0 = \{f,\mathcal{H}\}_{P.B.(4D)} = \partial_{ct}f\partial_{E/c}\mathcal{H} - \partial_{ct}\mathcal{H}\partial_{E/c}f - \{f,H\}_{P.B.(3D)}
0={f,H}P.B.(4D)=ctfE/cHctHE/cf{f,H}P.B.(3D)0 = \{f,\mathcal{H}\}_{P.B.(4D)} = \partial_{ct}f\partial_{E/c}\mathcal{H} - \partial_{ct}\mathcal{H}\partial_{E/c}f - \{f,H\}_{P.B.(3D)}

Can be rewritten as:

Immediate relation to the Poisson bracket appearing as the first-order term of the Moyal (deformation) quantization expansion.

Conclusion : The Symplectic Clifford algebra of phase space is a natural setting for discussing the quantum-classical correspondence through deformation quantization


The unit basis elements of phase space are the noncommuting quantum "operators"
One can interpret the "quantum uncertainty" as intrinsic to the phase space basis

1st-order derivative out of an infinite Taylor expansion that extrapolates from a definite phase space point to a displacement away

Spin : Pauli Operators form a 3D
Orthogonal Clifford Algebra (for rotation)

  • The orthogonal Clifford algebra of 3-space has 8 elements
  • The 4 "grades" of the algebra correspond geometrically to oriented subspaces of differing dimensions
\hat{\sigma}_y\hat{\sigma}_z \quad \hat{\sigma}_z\hat{\sigma}_x \quad \hat{\sigma}_x\hat{\sigma}_y
σ^yσ^zσ^zσ^xσ^xσ^y\hat{\sigma}_y\hat{\sigma}_z \quad \hat{\sigma}_z\hat{\sigma}_x \quad \hat{\sigma}_x\hat{\sigma}_y
\hat{\sigma}_x \quad \hat{\sigma}_y \quad \hat{\sigma}_z
σ^xσ^yσ^z\hat{\sigma}_x \quad \hat{\sigma}_y \quad \hat{\sigma}_z

Geometric meaning:







Line Segment

Plane Segment

Volume Segment

Matrix Representation of 3D Clifford Algebra

  • Simplification:
i\hat{\sigma}_x \quad i\hat{\sigma}_y \quad i\hat{\sigma}_z
iσ^xiσ^yiσ^zi\hat{\sigma}_x \quad i\hat{\sigma}_y \quad i\hat{\sigma}_z
\hat{\sigma}_x \quad \hat{\sigma}_y \quad \hat{\sigma}_z
σ^xσ^yσ^z\hat{\sigma}_x \quad \hat{\sigma}_y \quad \hat{\sigma}_z

Geometric meaning:







Line Segment

Plane Segment

Volume Segment

\hat{\sigma}_x\hat{\sigma}_y\hat{\sigma}_z = i\hat{1}
σ^xσ^yσ^z=i1^\hat{\sigma}_x\hat{\sigma}_y\hat{\sigma}_z = i\hat{1}

(Representation-independent definition of "imaginary unit", which refers in this case to a unit volume segment)

Dot and Wedge Products

The usual Pauli operators are a faithful matrix representation, so the matrix product is precisely the orthogonal Clifford product

\hat{A}\hat{B} = \frac{\hat{A}\hat{B}+\hat{B}\hat{A}}{2} + \frac{\hat{A}\hat{B}-\hat{B}\hat{A}}{2}
A^B^=A^B^+B^A^2+A^B^B^A^2\hat{A}\hat{B} = \frac{\hat{A}\hat{B}+\hat{B}\hat{A}}{2} + \frac{\hat{A}\hat{B}-\hat{B}\hat{A}}{2}

Symmetric part is dot product

Antisymmetric (noncommutative) part is wedge product

\frac{\hat{A}\hat{B}+\hat{B}\hat{A}}{2} = (\hat{A}\cdot\hat{B})\hat{1}
A^B^+B^A^2=(A^B^)1^\frac{\hat{A}\hat{B}+\hat{B}\hat{A}}{2} = (\hat{A}\cdot\hat{B})\hat{1}
\frac{\hat{A}\hat{B}-\hat{B}\hat{A}}{2} = \hat{A}\wedge\hat{B}
A^B^B^A^2=A^B^\frac{\hat{A}\hat{B}-\hat{B}\hat{A}}{2} = \hat{A}\wedge\hat{B}

Unit operator is an artifact of the matrix representation

\langle \hat{A} \rangle_0 \equiv \frac{1}{2}\text{Tr}[\hat{A}]
A^012Tr[A^]\langle \hat{A} \rangle_0 \equiv \frac{1}{2}\text{Tr}[\hat{A}]

Scalar projection removes representation:

\star \hat{A} \equiv -i\hat{A}
A^iA^\star \hat{A} \equiv -i\hat{A}

Hodge Star operation flips grade k to (3-k):

Cross Product is closed for grade 1 (vectors):

\hat{\sigma}_i\times\hat{\sigma}_j \equiv -i(\hat{\sigma}_i\wedge\hat{\sigma}_j) = \epsilon_{ijk}\hat{\sigma}_k
σ^i×σ^ji(σ^iσ^j)=ϵijkσ^k\hat{\sigma}_i\times\hat{\sigma}_j \equiv -i(\hat{\sigma}_i\wedge\hat{\sigma}_j) = \epsilon_{ijk}\hat{\sigma}_k

(Same "noncommutativity" in classical mechanics)

4D Clifford Algebra

A particular 3D geometry is embedded naturally a particular "spacetime split" of the 4D spacetime choosing a reference frame

Note: it is simple to "upgrade" the qubit representation to this 4D "spacetime"

3D Clifford algebra is a subalgebra of the 4D Clifford algebra of spacetime

Removing matrix representation changes no physics, but clarifies correspondence


Euclidean 3D

Minkowski 4D (+,-,-,-)

\sigma_1^2 = \sigma_2^2 = \sigma_3^2 = 1
σ12=σ22=σ32=1\sigma_1^2 = \sigma_2^2 = \sigma_3^2 = 1
\gamma_0^2 = 1, \; \gamma_1^2 = \gamma_2^2 = \gamma_3^2 = -1
γ02=1,γ12=γ22=γ32=1\gamma_0^2 = 1, \; \gamma_1^2 = \gamma_2^2 = \gamma_3^2 = -1
\sigma_1 \equiv \gamma_1\gamma_0 \quad \sigma_2 \equiv \gamma_2\gamma_0 \quad \sigma_3 \equiv \gamma_3\gamma_0
σ1γ1γ0σ2γ2γ0σ3γ3γ0\sigma_1 \equiv \gamma_1\gamma_0 \quad \sigma_2 \equiv \gamma_2\gamma_0 \quad \sigma_3 \equiv \gamma_3\gamma_0

Apparent 3D vectors are timelike planes in 4D

Could represent 4D basis as Dirac "gamma matrices" if desired,
but matrix representation is again not strictly needed

4D Clifford Algebra

  • Simplification:
\sigma_1 \quad \sigma_2 \quad \sigma_3 \quad | \quad i\sigma_1 \quad i\sigma_2 \quad i\sigma_3
σ1σ2σ3iσ1iσ2iσ3\sigma_1 \quad \sigma_2 \quad \sigma_3 \quad | \quad i\sigma_1 \quad i\sigma_2 \quad i\sigma_3
\gamma_0 \quad | \quad \gamma_1 \quad \gamma_2 \quad \gamma_3
γ0γ1γ2γ3\gamma_0 \quad | \quad \gamma_1 \quad \gamma_2 \quad \gamma_3






i \equiv \sigma_1\sigma_2\sigma_3 = \gamma_0\gamma_1\gamma_2\gamma_3
iσ1σ2σ3=γ0γ1γ2γ3i \equiv \sigma_1\sigma_2\sigma_3 = \gamma_0\gamma_1\gamma_2\gamma_3

(Representation-independent definition of "imaginary unit" is the 4-volume)

(but, commutes with even grade, anti-commutes with odd grade)


i\gamma_0 \quad | \quad i\gamma_1 \quad i\gamma_2 \quad i\gamma_3
iγ0iγ1iγ2iγ3i\gamma_0 \quad | \quad i\gamma_1 \quad i\gamma_2 \quad i\gamma_3
2^4 = 16\;\text{elements}
24=16elements2^4 = 16\;\text{elements}

Relative 3D space embedded as even-graded subspace

Planes (of rotation):

3 hyperbolic | 3 elliptic

Example : Maxwell's Equation

\displaystyle \nabla \equiv \sum_\mu \gamma^\mu \frac{\partial}{\partial x^\mu}
μγμxμ\displaystyle \nabla \equiv \sum_\mu \gamma^\mu \frac{\partial}{\partial x^\mu}

"quantum Dirac operator" is a classical gradient in 4-space with basis unit vectors of orthogonal Clifford algebra of spacetime

F = \sum_{k=1}^3 E^k\gamma_k\gamma_0 + \sum_{k=1}^3 B^k (i\gamma_k\gamma_0) = \vec{E}+i\vec{B}
F=k=13Ekγkγ0+k=13Bk(iγkγ0)=E+iBF = \sum_{k=1}^3 E^k\gamma_k\gamma_0 + \sum_{k=1}^3 B^k (i\gamma_k\gamma_0) = \vec{E}+i\vec{B}

Total electromagnetic field "tensor" is a bivector field on spacetime, which decomposes in a particular reference frame to a pair of 3-vectors

\nabla F = j
F=j\nabla F = j
j = \sum_\mu j^\mu \gamma_k
j=μjμγkj = \sum_\mu j^\mu \gamma_k

An electromagnetic source is a 4-vector consisting of the usual charge and current densities

This is Maxwell's Equation

\nabla F = \gamma_0(\vec{\nabla}\cdot\vec{E} + \partial_{ct}\vec{E}- \vec{\nabla}\times\vec{B}) + \gamma_0(\vec{\nabla}\cdot\vec{B} + \partial_{ct}\vec{B} + \vec{\nabla}\times\vec{E})i = \gamma_0(j^0 + \vec{j}) + \gamma_0 (0)i
F=γ0(E+ctE×B)+γ0(B+ctB+×E)i=γ0(j0+j)+γ0(0)i\nabla F = \gamma_0(\vec{\nabla}\cdot\vec{E} + \partial_{ct}\vec{E}- \vec{\nabla}\times\vec{B}) + \gamma_0(\vec{\nabla}\cdot\vec{B} + \partial_{ct}\vec{B} + \vec{\nabla}\times\vec{E})i = \gamma_0(j^0 + \vec{j}) + \gamma_0 (0)i

JD, et al. Physics Reports 589 1-71 (2015)


  • It is possible to think quantum mechanically without ever introducing a Hilbert space, using algebra directly
  • Both symplectic Clifford algebra and orthogonal Clifford algebra naturally appear as the essential structural algebras for both classical and quantum formalisms : the former encodes phase space structure, while the latter encodes spacetime geometric structure
  • The algebraic approach is naturally connected to deformation quantization, which is one of the most general ways to "quantize" a classical system : the quantum uncertainty appears in the basis vectors of phase space

Thank you!

Quantization from Clifford Algebra

By Justin Dressel

Quantization from Clifford Algebra

APS March Meeting 2018

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