I'm supposed to find out all irreducible representations of Pauli group, that is, the group generated by Pauli matrices $\sigma_k(k=1,2,3)$.

It has 16 elements: $\pm 1, \pm i, \pm \sigma_k, \pm i \sigma_k$, and 10 classes: $\{1\}, \{-1\}, \{i\}, \{-i\}, \{\pm \sigma_k\}, \{\pm i \sigma_k\}$. Since the sum of square of dimension of representations equals the order of the group, and the number of irreducible representations equals the number of classes, the only possible combination is 8 1-D representations, and 2 2-D representations.

But the character of the 2-D representation must be $\chi=(2,-2,2i,-2i,0,\cdots,0)$ or the character will have a norm larger than 1, and that means reducible. Thus there can only be one 2-D irreducible representation of Pauli group. That contradicts above.

Then what goes wrong in my deduction?

  • $\begingroup$ if $\chi=(2,-2,2i,-2i,zeroes)$ is a character then so is $(2,-2,-2i,2i,zeroes)$ $\endgroup$
    – user8268
    May 10, 2012 at 13:40
  • $\begingroup$ @user8268: So you mean that it constitutes another representation if $i$ is represented by $-i I$, where I is the identity matrix? $\endgroup$
    – Siyuan Ren
    May 10, 2012 at 13:57
  • $\begingroup$ See vixra.org/abs/1507.0089 for a table of irreducible representations. $\endgroup$ Jan 7, 2022 at 17:52

1 Answer 1


The norm of the character is given by $$\frac{1}{|G|}\sum_g \chi(g)\chi^*(g)$$ Here $|G|=16$ and the sum is $4+4+4+4$, so the norm is actually $1$, not $2$. So there isn't a contradiction as that 2D representation (I assume you are thinking of the standard representation in terms of Pauli matrices) is irreducible.


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