Embedding of $\mathrm{SU}(2)$ in $\mathrm{SU}(3)$ There is an embedding of $\mathfrak{su}(2)$ in $\mathfrak{su}(3)$ given by sending the standard basis
$$
\begin{pmatrix}i/2&0\\0&-i/2\end{pmatrix},\quad \begin{pmatrix}0&-1/2\\1/2&0\end{pmatrix},\quad\begin{pmatrix}0&-i/2\\-i/2&0\end{pmatrix}
$$
of $\mathfrak{su}(2)$ to
$$\begin{pmatrix}i\\&0\\&&-i\end{pmatrix},\quad\frac{1}{\sqrt{2}}\begin{pmatrix}&-1\\1&&-1\\&1\end{pmatrix},\quad-\frac{i}{\sqrt{2}}\begin{pmatrix}&1\\1&&1\\&1\end{pmatrix}.
$$
in $\mathfrak{su}(3)$. 
Since $\mathrm{SU}(2)$ is simply connected, this lifts to a Lie group homomorphism
$$
\varphi:\mathrm{SU}(2)\longrightarrow\mathrm{SU}(3).
$$

Question. Is there an explicit formula for this map? 

We can express any matrix in $\mathrm{SU}(2)$ as
$$
\begin{pmatrix}\alpha&-\bar{\beta}\\ \beta&\bar{\alpha}\end{pmatrix}
$$
where $\alpha,\beta\in\mathbb{C}$ and $|\alpha|^2+|\beta|^2=1$. Are there explicit expressions for the coordinates of $\varphi(\begin{smallmatrix}\alpha&-\bar{\beta}\\ \beta&\bar{\alpha}\end{smallmatrix})$ in terms of $\alpha,\beta$?
Using the exponential map, we can find special cases, such as
$$
\varphi\begin{pmatrix}\alpha&0\\0&\bar{\alpha}\end{pmatrix}=\begin{pmatrix}\alpha^2\\&1\\&&\bar{\alpha}^{2}\end{pmatrix}
$$
but I am having some difficulties finding the general formula.
 A: I don't know where you got the Lie algebra homomorphism from, but there is just one reasonable possibility what it can be. If you look at your formula, you see that three matrics in $\mathfrak{su}(3)$ that you are writing out don't have a joint eigenvektor, so they defined a complex irreducible representation of $\mathfrak{su}(2)$ and there is just one such representation in complex dimension 3. The simplest realization of this is via the adjoint representation on $\mathfrak{sl}(2,\mathbb C)$. This can be endowed with the Hermitian inner product defined by $(X,Y)\mapsto \frac12tr(X^*Y)$ and an orthonormal basis for this is given by $\left \{\begin{pmatrix} 0 & \sqrt2\\ 0 & 0 \end{pmatrix},\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix},\begin{pmatrix} 0 & 0\\ \sqrt2 & 0 \end{pmatrix}\right\}$. If I got it right, your matrices just describe the adjoint action of $\mathfrak{su}(2)$ in this basis.
The corresponding group homomorphism sends $A\in SU(2)$ to conjugation by $A$ as a linear map on $\mathfrak{sl}(2,\mathbb C)$. So one just has to determine the matrix representation of this with respect to the above basis. Working this out, you see that (if I got everything right)
$$\phi\begin{pmatrix} \alpha & -\bar\beta \\ \beta & \bar\alpha\end{pmatrix}= \begin{pmatrix}\alpha^2 & \sqrt2 \alpha\bar\beta & -\bar\beta^2 \\ -\sqrt2 \alpha\beta & |\alpha|^2-|\beta|^2 & -\sqrt2\bar\alpha\bar\beta \\ -\beta^2 & \sqrt2 \bar\alpha\beta & \bar\alpha^2  \end{pmatrix}.$$
