The special unitary group's elements fulfil the unitary condition, i.e. $aa^\dagger=1$, and have determinant 1 (special). I have read, that the Pauli matrices generate SU(2). Since all Pauli matrices have determinant -1, how can they generate the group, when they are not contained.

I'm referring to the common representation $\sigma_x=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ $\sigma_y=\begin{pmatrix}0&-i\\i&0\end{pmatrix}$ $\sigma_z=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$

  • $\begingroup$ Do you know about Lie groups and Lie algebras? $\endgroup$
    – themaker
    Apr 1, 2017 at 11:00
  • 1
    $\begingroup$ The Pauli matrices are not themselves part of $SU(2)$ (as a Lie group), but rather they are part of its generating Lie algebra, in the sense that they form a basis from which you can write down (using a map from the algebra to the group) any element of $SU(2)$. $\endgroup$
    – Demosthene
    Apr 1, 2017 at 11:19

1 Answer 1


Pauli matrices are not generators of the Lie group $SU(2)$, but after multiplication by$~\mathbf i$ they are so-called infinitesimal generators. These are elements of the Lie algebra $\mathfrak{su}(2)$ of $SU(2)$, and for such elements $X$ the exponential mapping $t\mapsto\exp(tX)$ gives a one-parameter subgroup of the Lie group. In view of this relation, the determinant of the Pauli matrices is not important; what is relevant is that they are Hermitian (so after multiplication by$~\mathbf i$ they become anti-Hermitian, and the exponential mapping gives unitary matrices) and have trace zero (so that exponential mapping gives matrices of determinant$~1$).

No countable set of group elements could generate the uncountable groups $SU(2)$, but the one-parameter subgroups for the Pauli matrices do generate $SU(2)$. Maybe more pertinent is the fact that the Pauli matrices generate the real vector space of traceless Hermitian matrices. Thus every element of $SU(2)$ is the image under the exponential mapping of a real linear combination of the multiples by $~\mathbf i$ of the Pauli matrices (although the fact that $SU(2)$ is connected and compact is instrumental for this fact).


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