# Daily Archives: July 4, 2008

## Electroweak Unification, Leptons

Quarks will require 1-circulant and 2-circulant matrices. Before adding that complication, let’s unify the lepton quantum numbers. Two quantum numbers (other than generation) distinguish quarks and leptons, weak hypercharge $t_0$ and weak isospin $t_3$. Weak hypercharge is a U(1) quantum number; it has only one generator and therefore is commutative. Weak isospin arrives in two representations, singlets and doublets. The singlets have weak isospin quantum number of 0 and so we can represent them with any sort of 0. The doublets have spin-1/2, which we represent with the Pauli spin matrices:

There are three Pauli spin matrices, and they are linearly independent, so complex multilpes of them give 3 complex degrees of freedom. Since complex 2×2 matrices have 4 complex degrees of freedom, there is 1 complex degree of freedom left, the unit matrix:

The unit matrix is a natural basis for a U(1) symmetry, so we can combine weak hypercharge with weak isospin into 2×2 matrices. The number of degrees of freedom in the 2×2 matrices is just sufficient to support an SU(2) spin-1/2 and a U(1). Given a representation of an SU(2) operator, and a U(1) operator, the 2×2 matrix representation is simply the sum of the Pauli matrix representation of the SU(2) operator, and the U(1) value times the unit matrix. Similarly, given an arbitrary 2×2 matrix, we can split it into a U(1) portion and the SU(2) portions:

The above allows the weak hypercharge and weak isospin operators to share a 2×2 matrix representation. To unify the states, we have to pass to a density matrix representation.

Given a normalized quantum state vector (a,b), the density matrix representation of the state is a 2×2 matrix:

Similarly, given a U(1) state $\psi$, we can convert it into a 2×2 matrix. For qubits, such a (pure) density matrix would be boring, it would only be the unit matrix. But for wave functions that depend on position, the density matrix is not trivial and contains the relative phase information of the quantum state (which is the only phase information that is physical). But for this post, simply note that 2×2 matrices are rich enough to contain both types of quantum numbers. A density matrix is partially characterized by the fact that it is idempotent, that is, $\rho^2 = \rho$. This characterization is not complete in that the equation has other solutions, in this case $0^2 = 0$, and $1^2 = 1$. These other solutions have trace 0 and 2, the usual pure density matrix has trace 1. It turns out we need these other solutions so their having the wrong trace is an issue. Further down we will show how to convert the traces to 1, but for now let us postpone the discussion.