I found this problem and need some help. It is given:

$$ \sigma_1 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} $$

$$ \sigma_2 = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} $$

$$ \sigma_3 = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} $$

These are the Pauli matrices.

Furthermore I have:

$$h \in \mathbb{R}^3$$ $$h \sigma := h_1 \sigma_1 + h_2 \sigma_2 + h_3 \sigma_3$$


$$\mathcal{su}(2) := \lbrace A \in M(2,\mathbb{C}) | A = h \cdot \sigma, h \in \mathbb{R}^3\rbrace$$

is a vector space.

Now I have a scalar product given by

$$\langle \cdot , \cdot \rangle_{\mathcal{su}(2)}: \mathcal{su}(2) \times \mathcal{su}(2) \rightarrow \mathbb{R} $$ $$\langle A, B\rangle_{\mathcal{su}(2)} = \frac{1}{2} trace (AB)$$

I got this and now I want to show, that

$$\phi: \mathbb{R}^3 \rightarrow \mathcal{su}(2)$$ $$h \mapsto h \cdot \sigma$$

is an isometric isomorphism

An isometric isomorphism has to be bijectiv, continous, the inverse has to be continous and the norm must be retained.

But how can I show it? I need some help!

  • $\begingroup$ What do you mean exactly by direct calculation. How do I start? $\endgroup$
    – Gilfoyle
    Jun 14, 2014 at 19:58
  • $\begingroup$ This can hardly be true, as $\dim U(2)=4$. Don't you mean $SU(2)$ (or something related)? $\endgroup$ Jun 14, 2014 at 20:08
  • $\begingroup$ @ Peter Franek: You are right! I fixed it. $\endgroup$
    – Gilfoyle
    Jun 14, 2014 at 20:22
  • $\begingroup$ Now that it became the Lie algebra, shouldn't you multiply all those matrices by $i$? It seems to me that $su(2)$ should be skew-hermitian matrices.. $\endgroup$ Jun 14, 2014 at 20:29
  • $\begingroup$ I am still trying to solve that problem. Is there anybody who can help me? $\endgroup$
    – Gilfoyle
    Jun 15, 2014 at 7:40

1 Answer 1


If you define $su(2)$ like this, in other words, the span of those Pauli matrices, you get the vector space of all hermitian matrices with trace $0$. (It is more common to define $su(2)$ as the set of all anti-hermitian matrices with zero trace, but let's skip this for the moment.) The map $\phi$ is clearly linear and its inverse $\phi^{-1}$ is $$\begin{pmatrix}a & b+ci \\ b-ci & -a\end{pmatrix}\mapsto (b,-c,a),$$ so $\phi$ is a linear isomorphism. It is an isometry because the three Pauli matrices form an orthonormal basis under the given scalar product (check that $(1/2) Tr(\sigma_i\,\sigma_j)=\delta_{ij}$) and if a linear map maps an orthonormal basis $\{e_1, e_2, e_3\}$ to an orthonormal basis $\{\sigma_1, \sigma_2, \sigma_3\}$, it is an isometry.

  • $\begingroup$ Thank you for your answer! But I still did not get the point. Up to know I consulted several books to understand your answer. I need to know where $\phi$ and $\phi^{-1}$ is coming from. I do not see it from what I have given. And how can I check that $(1/2) tr (\sigma_i \sigma_j) = \delta_{ij}$. Is it the same as $(1/2) tr (AB) = \delta_{ij}$?. Well I calculated $(1/2) tr (\sigma_i \sigma_j) = \delta_{ij}$ and my results are always zero. Is this right? Thank you a lot! $\endgroup$
    – Gilfoyle
    Jun 15, 2014 at 17:12
  • $\begingroup$ You defined $\phi$ in the title, right? And you defined $\sigma_1, \sigma_2, \sigma_3$. If you know how to multiply matrices and compute the trace, you can easily check that $(1/2) Tr(\sigma_1 \,\sigma_1)=1$, $(1/2) Tr(\sigma_1\,\sigma_2)=0$, $(1/2) Tr(\sigma_1\,\sigma_3)=0$, $(1/2) Tr(\sigma_2\,\sigma_2)=1$ etc, so $\{\sigma_1, \sigma_2, \sigma_3\}$ is an orthonormal basis of the space $su(2)$ that you defined how you defined it. Then you only need to know that if a linear map maps $\{(1,0,0), (0,1,0), (0,0,1)\}$ in $R^3$ to an orthonormal basis of the target space, the map is an isometry. $\endgroup$ Jun 15, 2014 at 17:51

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