References: Mark Srednicki, Quantum Field Theory, (Cambridge University Press, 2007) – Chapter 1, Problem 1.1.
The Klein-Gordon equation is an early attempt at a relativistic quantum theory, but it contains a second-order time derivative which leads to probability not being conserved over time. Dirac proposed another equation that attempts to solve this problem for particles of spin 1/2. The Dirac equation is essentially a modification of the Schrödinger equation:
Here, is now a vector in spin space with components . The objects and (for ) are square matrices (where the subscript indicates the component of the matrix being considered), also in spin space, and repeated indices are summed over spatial coordinates only. [We won’t worry about how Dirac arrived at this equation for now; we’ll just accept it and see where it leads.]
To make this equation formally equivalent to the Schrödinger equation, the hamiltonian operator on the RHS must now be a matrix. We can also use the definition of the momentum operator to get
This might not look much like the relativistic energy:
but if we square 2 (remembering that matrix products need not commute), we have
We can define the anticommutator as
We can write the first term on the RHS of 4 as
so we get
In order to make this equal to , we need the matrices and to satisfy the conditions:
The first condition requires the anticommutator of and to be zero unless , in which case the anticommutator gives the identity matrix. Remember that the superscripts and specify which matrix we’re talking about, while the subscripts indicate the component of the matrix. The conditions aren’t derived; rather they are imposed to make the energy come out right. With these conditions, we have
which gives the correct operator for the square of the energy.
The question arises as to what these matrices and are. One candidate is the set of three Pauli spin matrices
By direct calculation, we see that . For example
and so on. However, in order to satisfy 9, we need to find a single matrix that anticommutes with all 3 spin matrices. We get
We then get
So resulting in . Thus there is no non-zero matrix that anticommutes with all 3 of the Pauli spin matrices.
So what can we say about the Dirac matrices? From 10, we see that the eigenvalues of are all 1, so the eigenvalues of must be .
The trace (sum of the diagonal elements) of a matrix is equal to the sum of its eigenvalues (theorem from matrix algebra). To find the trace of , we can use the anticommutators 8 and 9, together with another theorem from matrix algebra which states that for any square matrices and of the same order.
Hence , so must have an equal number of and eigenvalues. In other words, must be even dimensional, so the smallest size is .
We can also find the trace of by starting with and following through the same steps as above (using ) to show that .