Is There a Hidden Factor in the Definition of [A,A]?

[Dox]

Hi everyone,

Homework Statement

I've been studying a paper in which there is a connection given by,

$$A = f(r)\sigma_1 dx+g(r)\sigma_2 dy,$$​

where $$\sigma$$'s are half the Pauli matrices. I need to calculate the field strength,

$$F = dA +[A,A].$$​

Homework Equations

$$A = f(r)\sigma_1 dx+g(r)\sigma_2 dy,$$

$$F = dA +[A,A]$$​

The Attempt at a Solution

I have computed it, but a factor is given me problems. I would say,

$$dA = f' \sigma_1 dr\wedge dx + g'\sigma_2 dr\wedge dy$$​

and

$$[A,A] = 2 f g \sigma_3 dx\wedge dy,$$​

with a factor 2 coming from the fact that there are two contributions... like a binomial.

Is it OK or there is a half factor hidden in the definition of $$[A,A]$$?

Thank you so much.

DOX​

 

[fzero]

If the $$\sigma_i$$ are equal to 1/2 the Pauli matrices, then their commutation relation is

$$[\sigma_i,\sigma_j] = i\epsilon_{ijk}\sigma_k.$$

If you recheck your calculation, you'll find that there's no factor of 2, $$[A,A] = f g \sigma_3 dx\wedge dy.$$

 

[Dox]

If the $$\sigma_i$$ are equal to 1/2 the Pauli matrices, then their commutation relation is

$$[\sigma_i,\sigma_j] = i\epsilon_{ijk}\sigma_k.$$

If you recheck your calculation, you'll find that there's no factor of 2,


$$[A,A] = f g \sigma_3 dx\wedge dy.$$

Thank you fzero. I've found out that people use to write a commutator for this factor but it is just a wedge product... that's why I was getting a different factor of 2.

THX.