Given a $2\times 2$ unitary, unimodular matrix $U\in {\rm SU}(2)$, the (elements of the) corresponding $3\times 3$ rotation matrix $R\in {\rm SO}(3)$ can be obtained from the map $$R_{jk}(U)=\frac{1}{2}{\rm Tr}(\sigma_j U \sigma_kU^\dagger)\tag{1}$$ where $\sigma_j,\sigma_k$ represent the Pauli matrices. What is the inverse map that determines (the elements of) $U$ from $R$?

  • 2
    $\begingroup$ See (3.23) to (3.25) here, and (4.59) to (4.61) for the Lorentz analog. $\endgroup$
    – bolbteppa
    May 31 at 22:18
  1. OP's formula (1) follows from the relation [1] $$U \sigma_k U^{-1}~=~ \sum_{j=1}^3\sigma_j R^j{}_k.\tag{A} $$ See also e.g. this & this related Phys.SE posts.

  2. The map
    $$SU(2)~\ni~ U~\mapsto~ R~\in~ SO(3)\tag{B}$$ is 2:1, so the inverse map does strictly speaking not exist. However, there exists a double-valued map, cf. e.g. my Phys.SE answer here.

  3. There is an injective map from rotation vectors $$\vec{\alpha}~\in~B(\vec{0},\pi)~:=~\{\vec{\alpha} \in\mathbb{R}^3 \mid |\vec{\alpha}|< \pi\} \tag{C} $$ (belonging to an open neighborhood) to $3\times 3$ rotation matrices $$R(\vec{\alpha})~=~\exp(i\vec{\alpha}\cdot \vec{L})~\in~ SO(3)~ \subseteq ~{\rm Mat}_{3\times 3}(\mathbb{R}). \tag{D}$$ Then we can pick a continuous branch $$U(\vec{\alpha})~=~\exp(\frac{i}{2}\vec{\alpha}\cdot \vec{\sigma})~\in~ SU(2)~ \subseteq ~{\rm Mat}_{2\times 2}(\mathbb{C}), \tag{E}$$ of the double-valued map in an open neighborhood. This establishes a continuous inverse map in an open neighborhood.


  1. G 't Hooft, Introduction to Lie Groups in Physics, lecture notes; chapters 3 + 6. The pdf file is available here.
  • 1
    $\begingroup$ I was looking for an inverse map i.e. finding (the elements of) $U$ knowing the elements of $R$, maybe up to a sign. With this, can we make an inversion? $\endgroup$
    – SRS
    May 31 at 16:22

I am addressing the comment of the OP to the complete answer of @Qmechanic. Given the real orthogonal 3×3 matrix (D), $$ R(\vec \alpha)= e^{i\vec \alpha\cdot \vec L} \equiv e^{\theta \mathbb L},\\ \vec \alpha\equiv \theta \hat n, ~~~|\hat n|=1, ~~~{\mathbb L} =\begin{pmatrix} 0&-n_z& n_y\\ n_z&0 &-n_x\\-n_y& n_x&0\end{pmatrix} , $$ real antisymmetric, where the three real coefficients $\vec \alpha$ of the Lie algebra element are broken down to an angle $\theta$ and a unit axis of rotation $\hat n$, so that ${\mathbb L}^3 =-{\mathbb L}~~\leadsto$, $$ R(\theta \hat n) = {\mathbb 1}+ \sin\theta ~{\mathbb L}+ (1-\cos \theta)~{\mathbb L}^2. $$

This is the celebrated matrix form of Rodrigues' rotation formula, with three terms, the middle one being real antisymmetric, and the extremal ones being real symmetric. So, manifestly, $$ (\operatorname{Tr}~R-1)/2= \cos{\theta}, $$ to solve for θ, whence $$ R-R^T= 2\sin \theta ~{\mathbb L}~~\leadsto \\ \hat n= (R_{32}-R_{23}, R_{13}-R_{31}, R_{21}-R_{12})^T/2\sin\theta, $$ so you have determined $\vec \alpha$, which specifies $U=\cos\theta/2 +i\sin\theta/2 ~~~\hat n\cdot \vec\sigma $, as per the answer commented on. Choose your quadrants and half-angle values to your convenience.


Let $g_{rs}$ be a matrix in $\text{SO(3)}$, while $U$ be a matrix in $\text{SU(2)}$. We can find as in the OP:

$$g_{rs} = \frac{1}{2}\text{Tr}\left(\sigma^r U \sigma^{s} U^{\dagger}\right) \tag{1}$$

The inverse of this $2:1$ mapping is

$$ U = \mp \frac{1+\sigma^r \sigma^s g_{rs}}{2\left(1+\text{Tr}~g\right)^{\frac{1}{2}}} \tag{2}$$

Bibliography: Carmeli, M., Malin, S. "Theory of spinors. An introduction", WS, 2000, page 7, eqns. 1.14 and 1.15.


For this problem the completness relation of the pauli matrices is here very useful:

$\sigma^j_{\alpha \beta}\sigma^j_{\gamma \delta}=2\delta_{\alpha \delta}\delta_{\beta \gamma}-\delta_{\alpha\beta}\delta_{\gamma \delta},$ where the superindices denote the typ of the pauli matrix and the subindices denotes the matrix index and we sum over repeated indices.

Given a matrix $R_{jk}(U)$, we can get the product of the matrix elements of $U$ and $U^\dagger$ back (so the map can not be uniquely inverted as mentioned above.) as shown in the following:

We take $R_{jk}(U)$ and multiply it with $\sigma^j_{\alpha \beta}$ and $\sigma^k_{\gamma \delta}$ where we in the following assume summation over all repeated indices. We get after writing the trace in $R_{jk}(U)$ also in index notation (using latin indices for the trace and greek letters for the new free indices):

$\sigma^{j}_{\alpha\beta}R_{jk}(U)\sigma^k_{\gamma \delta}=\frac{1}{2} \sigma^{j}_{\alpha\beta}\sigma^j_{ab} U_{bc} \sigma^k_{cd}U^\dagger_{da}\sigma^k_{\gamma \delta}$

Using the above completeness relation twice for the summations over j and k, then resolving all appearing kronecker deltas and using that $UU^\dagger=1$, we get in the end:

$\sigma^{j}_{\alpha\beta}R_{jk}(U)\sigma^k_{\gamma \delta}=U_{\alpha\delta}U^\dagger_{\gamma \delta}-\frac{1}{2}\delta_{\alpha \beta}\delta_{\gamma \delta}$

(if I do not messed something up....)

From that formula, one can extra the matrix elements of $U$ up to the an factor of $-1$ by probing different index combinations of $\alpha,\beta,\gamma,\delta$.


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