1. Dirac Reasoning
We start with the 4-dimensional Klein-Gordon wave equation as the basic description of a free oscillating field in 4-dimensions and its equivalent energy-momentum equation.
When converting this equation into a 4-dimensional linear differential equation, Dirac [1] showed that the result is a 4-dimensional linear equation with 4 matrices acting on a 4-vector field.
Following Dirac’s idea, of taking the square root of the wave operator, as suggested:
(1)
in order to force all cross-terms such as
to vanish, one must assume
(2)
(3)
(4)
Or, if
represent the A, B, C, D terms, then, in a compact form
(5)
These conditions are met if A, B, C and D are matrices, with the implication that the wave function has components, as the dimension of the matrices. This explained the appearance of two-component wave functions in Pauli’s theory of spin. The minimum would be to have A, B, C and D as 4 × 4 matrices. This is done in combinations of 2 × 2 Pauli matrices, in such a way that will obey the
constrains. This leads to the assertion, that Ψ is a 4-vector of complex wave functions.
In a 3 + 1 world,
is a 4 × 4 matrix, so A, B, C and D are 4 × 4 matrices. However, in an N-dimensional world, one needs N × N matrices4, with N ≥ 4, to set up a system with the properties required. In Dirac’s original work N= 4, so the wave function had four components, not two, as in the Pauli theory, or one, as in the Schrödinger equation. When looking into the initial development strategy of Dirac, his initial demand was to create an equation with the first derivatives:
(6)
Since it originated in the Klein-Gordon wave equation, which by itself originated from Schrödinger equation, it was assumed there that Ψ is complex and therefore can be presented as a 2-vector of two complex wave functions.
Instead of using the conventional 4 × 4
matrices (A, B, C and D) as introduced by Dirac, a set of four N × N matrices
, acting on a complex N-vector Ψ are introduced, instead of the conventional 4 × 4 gamma matrices in 4 dimensions.
These N × N χ matrices replace the 4 × 4 γ matrices in Dirac 4 × 4 equation to introduce the Dirac N × N equation:
(7)
where Ψ is now a complex N-vector.
1) Is the connection between the universe dimensions and the number of matrices D in the equation, defined and accepted? The answer is that since the KG wave equation is 4 dimensional, so is the number D of matrices.
2) Is the constraint
, a pure mathematical result of the required relations between the matrices in order to allow for the linearity, or is the
a must because of relativistic covariance?
3) Will the constraint depend, on space curvature due to gravitation?
We accept then that in a D dimensional universe there are exactly D matrices and they are of N × N dimensions – representing a N × N group representation.
As has been proven [2], and partly in [3], the connection between D and N must obey the relation
, provided the representation is irreducible.
Otherwise, if this condition is not met, the representation must be reducible and one can split it into two or more lower-rank representations.
In the case of d = 4, we are forced to have N = 4, or else there are two 2 × 2 matrices to represent the structure and thus 2-vectors instead of 4-vactor.
In the case of D = 3 and N = 4 one cannot have an irreducible representation.
2. N-Dimensional Representation of Dirac Equation
Any massive fermion must obey, by its definition, Dirac’s equation
(8)
This is a must because of Energy Momentum considerations for a massive particle, and Lorentz covariance.
Dirac’s
matrices obey the constraint
(9)
In a 3 + 1 world,
is a 4 × 4 matrix, so
are 4 × 4 matrices. However, in an N-dimensional world, one needs N × N matrices, with N > 4, to set up a system with the properties required.
To extend to other dimensions, the
notation is introduced instead of
.
These χ, N × N matrices, replace the 4 × 4 γ matrices in Dirac 4 × 4 equation. Thus, one extends the Dirac equation to a N × N equation:
(10)
Here Ψ is now a complex N-vector,
is the N × N unit matrix, and
are [N × N] matrices, satisfying the requirement:
(11)
3. Group Limitations on D
According to Dirac’s solution, the fermionic
matrices must satisfy the canonical anti-commutation relation
.
The fermionic matrices, create a Group representation. This group representation can be proven to be an irreducible representation if and only if, there exists a certain relationship between the size of the group (the number of elements d) and the size of the matrices N2.
The connection is
.
Consider a D-dimensional space-time, with a flat Minkowski metric
where
. (The original Dirac matrices correspond to taking D = N = 4).
For a D-dimensional space ( D–1 spatial + 1 temporal) dimension, there must be D such matrices
(
), of size N × N each, adhering to the anticommutator relation,
.
Using the matrices
and
(a total of d + 1 matrices) we can construct a set of 2d N × N matrices as follows:
Over all combinations of indices, where
etc.
there are
such matrix combinations.
Denote these possible matrix combinations by
, where
.
This creates a set
, of size 2D.
Because of the anticommutation relation between the gammas, the product of any
by any
is, up to a sign, a third member
of the set.
Define a set
whose elements are the same members of
, but with negative sign, namely:
Create now a union set
, so G becomes the set of matrices
.
Re-assign notations for the
members of the set
as follows:
(
,
,
,
,
,
) with
terms, so that
.
All pair products
. and, for any
, there exists an inverse
, such that
.
Thus
, is a group of size
.
Assume now that the matrices in
create an irreducible representation.
Define a matrix
where Z is an arbitrary N × N matrix (the summation runs over all
).
It follows that
Since S commutes with all matrices in
, and since we assumed that
is an irreducible group of matrices, then, by Schur’s first lemma S must be proportional to the identity matrix:
Taking the trace on both sides gives
Hence
and so:
Taking the ijth element of both sides and summing over i andj, we arrive after Setting l = i, m = j
and so,
But, since
are made out of
multiplications
for all
, while for
or
Therefore:
The starting point was the assumption that G is an irreducible representation. So, if G is irreducible, then
4. Is
an Irreducible Representation Group?
We create all the possible classes of
and look at their dimensions.
Denoting the class of an element
by
we have:
For the group
it is easy to see that the conjugate classes are
Therefore, according to the Decomposition Theorem:
If and only if
is irreducible.
is the trace of the class
.
is the number of elements in the class.
All traces of the conjugate classes are null, except for
and
. For the group
to be irreducible one must have
In other words, for
to be irreducible, one must have
.
But, as we saw above, if G is irreducible, then
.
Thus, if
the group is irreducible, whereas if G is irreducible,
.
Therefore, the group is irreducible if, and only if,
.
Any other representation with N and D that do not satisfy the above, the representation must be reducible. If the relationship of N and D satisfies
, then the representation must be irreducible.
The above result was obtained independent of D and it shows that D, for an irreducible representation, must be even.
5. An Upper Limit on D
So far, it was shown that as long as the representation is irreducible, D must be an even number. But is there an upper limit on D?
For each dimension, there must be a single N × N Dirac matrice
.
These matrices obey the restriction
with
and where the metric signature of
is
.
For a D dimensional universe, the matrices, construct the following set
of 2D N × N matrices:
Over all combinations of indices, where
…etc.
Create next the set
Define the group
Thus,
with
.
Evidently,
is a group of order
and
contains a set of sub-groups:
They are described in the following:
etc.
All
(except for i = 0.1) are
or products of
’s (
). Hence, when two new matrices
and
are introduced in a 6-dimensional universe, we define
, and it is a straightforward procedure to show that
for all
.
Hence, by Schur’s Lemma, if A (
), commutes with all matrices
of the group, then the representation is necessarily reducible.
Therefore, the group
must be a reducible representation.
This forces us to conclude, that no irreducible representation
may be found for d> 4 and therefore, any fermionic universe must have d = 4.
6. Weyl Equation and the Neutrino
In the case of a massless fermion, m = 0 and Dirac equation becomes
(12)
This is Weyl’s equation. It has a solution given by N = 2 matrices
(13)
where
are Pauli’s 2 × 2 matrices, satisfying
.
In the case of massless neutrino, the Dirac equation reduces to D = 3 and the N = 4. The condition
is violated and the solution must be reducible.
Indeed, the massless neutrino is not described by a single Dirac equation. Rather, there are two uncoupled equations.
7. Conclusions
Dirac equation, describing a massive fermion, can be extended from 4-dimensions to higher D-dimensions, where D stands for the number of matrices and also the dimension of the universe in which the fermions exist.
Based on the canonical anti-commutation relation
, it was shown that the
matrices can be used as generators of a group G, which dimension is
and since N must be an integer, so must be d.
Therefore, no irreducible fermionic theory can have an odd dimensional d. It was further shown, that an upper limit on the dimensions is D < 5.
Any representation of an odd order or which does not satisfy
must be reducible to a lower order.
The final conclusion is then, that for a fermionic universe, D = 4.