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I tried to make a group multiplication table for the Pauli matrices, but I keep getting multiples in front of the elements. What am I doing wrong? I thought the Pauli matrices formed a group that was closed under commutation? Here's what I have:

$\begin{array}{c|ccc} [,] & \sigma_1 & \sigma_2 & \sigma_3 \\ \hline \sigma_1 & 0& 2i\sigma_3 & -2i\sigma_2 \\ \sigma_2 & -2i\sigma_3 & 0 & 2i\sigma_1 \\ \sigma_3 & 2i\sigma_2 & 2i\sigma_1 & 0 \end{array}$

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Lie algebras are not a group w.r.t. to the commutator (the Lie bracket).

The first reason is that the commutator is not associative.

Another is that they almost always lack an identity element, since the identity matrix is, for example, not in $\mathfrak{su}(2)$, and Schur's lemma would, in the fundamental representation, guarantee that only multiples of the identity can be commuting with all elements of the algebra. From the lack of the identity, it follows that also the existence of an inverse is ill-defined, so they can't be a group.

They are closed under the Lie bracket operation though, but your table doesn't contradict that.

It is not possible to make the commutator into a group operation on finite-dimensional Lie algebras - just look at the trace of $[A,B] = 1$ to see that this cannot be true as long as the trace is defined for both sides of the equation.

The closest thing to providing a more "familiar" associative structure with a neutral element on a Lie algebra (aside from its vector addition) is to pass to the universal enveloping algebra. Even there, you still lack inverses for the algebra operation (which is good, since otherwise we'd have a (skewed) field, which is boring).

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  • $\begingroup$ Interesting! I guess I had assumed the commutator and Lie bracket operation were the same. Won't I still get the multiples of twice i, however? Is this okay to have and still be a closed group? $\endgroup$ – jjgoings Oct 15 '14 at 18:43
  • $\begingroup$ @jjgoings: I didn't say that the Lie bracket and the commutator are different things, they aren't. If the multiples of $\mathrm{i}$ are bothering you, they come from physicists insisting that the generators of Lie groups be Hermitian. If you take them to be anti-Hermitian (i.e. absorb the $\mathrm{i}$ into their definition), the $\mathrm{i}$ disappears. $\endgroup$ – ACuriousMind Oct 15 '14 at 18:44
  • $\begingroup$ Okay, my mistake. I see then if I redefine $\sigma$ as $i\sigma/2$ the multiples go away. I still have negative signs, however, but that's just the anti-commutativity. If I add the identity (and negative identity?) would this form a group? $\endgroup$ – jjgoings Oct 15 '14 at 18:59
  • $\begingroup$ @jjgoings: I expanded the answer. You seem to be awfully eager to turn this into a group. Why? There are many mathematical structures (like Lie algebras) which are perfectly valid and useful without being a group. $\endgroup$ – ACuriousMind Oct 15 '14 at 19:08
  • $\begingroup$ Thanks. I realized why it wouldn't work just before you responded. I wanted to make it a group because I'm interested in the symmetries of Hartree-Fock theory. You can derive many useful constraints on the wave function by examining how it transforms under spin rotations and time reversal symmetries. One problem I've run into, though, is that it is difficult to classify the different symmetry invariances into groups (and subgroups) using just the generators of the unitary transformations (in this case Pauli matrices). $\endgroup$ – jjgoings Oct 15 '14 at 19:23

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