Wikipedia claims Pauli Matrices with an $i$: $i \sigma_1, i \sigma_2, i \sigma_3$ form a basis of $\mathfrak{su}(2)$.

But what about the following relation?:

$$[\frac{1}{2} \sigma_i, \frac{1}{2} \sigma_j]=\frac{i}{2}\epsilon_{ijk}\sigma_k$$

Pauli cooked up the matrices set for spin..

There is no other sane way to get the $1/2$ angle.

I did try the Dirac Plate Trick, didn't manage. :D

Is the article wrong?

  • 3
    $\begingroup$ Sorry, I don't understand the question. What do you mean by "what about HALF"? $\endgroup$ – AccidentalFourierTransform Jul 10 '18 at 13:42
  • $\begingroup$ If a set of vectors is a basis, so is that set of vectors, divided by two. $\endgroup$ – knzhou Jul 10 '18 at 15:16
  • $\begingroup$ @AccidentalFourierTransform I meant the half angle when doing exponentiation to a group element. It is a spin double cover, just because of the Pauli matrices. It is a pretty important relation. $\endgroup$ – user192234 Jul 10 '18 at 15:32

You are right, one often uses $S_i = \sigma_i / 2$ as the generators of $\mathfrak{su}(2)$, they satisfy $$ [ S_i, S_j ] = \mathrm i\epsilon_{ijk}\, S_k . $$ However, $\mathfrak{su}(2)$ is a vector space, and the $\{ \sigma_i \}$ are a basis of it just as well as the $\{ S_i \}$.

This shows that the structure constants of an algebra are not uniquely defined, they depend on the choice of basis.

Edit: @doetoe is right in his answer that the convention in math is different by a factor of $\mathrm i$, because we talk about the real algebra $\mathfrak{su}(2)$. What I wrote above is the usual notation in Physics and technically applies only to the complexified $\mathbb C \otimes \mathfrak{su}(2)$.


$\mathfrak{su}(2)$ is a real (but not a complex) Lie algebra, so the commutator of two generators must be a real linear combination of generators, which is not the case for the set $\frac12\sigma_i$, but does hold for $i\sigma_i$.

Note that scalar factors (like the $\frac12$ here) will never change whether or not a set of elements forms a basis.


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