How do I solve for $\beta_k$ in:
$e^{\alpha_1 G_1 + \alpha_2 G_2 +\alpha_3 G_3 } =e^{\beta_1 G_1} e^{\beta_2 G_2} e^{\beta_3 G_3} e^{\beta_4 G_4}$ ?
Note there is no $\alpha_4$ term.
(Also, do solutions even exist for this problem? Referring to the answer by MoisheKohan at Disentangling and reordering operator exponentials from Lie groups)
Here $G_k$ form $\mathfrak{gl}_2(\mathbb{C})=\mathfrak{sl}_2(\mathbb{C})\oplus\mathbb{C}$ Lie algebra:
$[G_1,G_2]=0,\\ [G_1,G_3]=[G_3,G_2]=G_4,\\ [G_1,G_4]= [G_4,G_2]=G_3,\\ [G_3,G_4]=-2G_1+2G_2$
These have the representations:
\begin{equation}\begin{aligned}
G_1 &= \begin{pmatrix}1&0\\0&0\end{pmatrix}\\
G_2 &= \begin{pmatrix}0&0\\0&1\end{pmatrix}\\
G_3 &= \begin{pmatrix}0&1\\1&0\end{pmatrix}\\
G_4 &= \begin{pmatrix}0&1\\-1&0\end{pmatrix}
\end{aligned}\end{equation}
Using these representations I end up with a matrix equation:
\begin{equation}\begin{aligned}
\begin{pmatrix}e^{\frac{\alpha_1+\alpha_2}{2}}\left[\cosh\left(\frac{1}{2}\sqrt{(\alpha_1-\alpha_2)^2+4\alpha_3^2}\right)+\frac{(\alpha_1-\alpha_2)}{\sqrt{(\alpha_1-\alpha_2)^2+4\alpha^2_3}}\sinh\left(\frac{1}{2}\sqrt{(\alpha_1-\alpha_2)^2+4\alpha_3^2}\right)\right]&\frac{2e^{\frac{\alpha_1+\alpha_2}{2}}\alpha_3\sinh\left(\frac{1}{2}\sqrt{(\alpha_1-\alpha_2)^2+4\alpha_3^2}\right)}{\sqrt{(\alpha_1-\alpha_2)^2+4\alpha^2_3}}\\\frac{2e^{\frac{\alpha_1+\alpha_2}{2}}\alpha_3\sinh\left(\frac{1}{2}\sqrt{(\alpha_1-\alpha_2)^2+4\alpha_3^2}\right)}{\sqrt{(\alpha_1-\alpha_2)^2+4\alpha^2_3}}&e^{\frac{\alpha_1+\alpha_2}{2}}\left[\cosh\left(\frac{1}{2}\sqrt{(\alpha_1-\alpha_2)^2+4\alpha_3^2}\right)-\frac{(\alpha_1-\alpha_2)}{\sqrt{(\alpha_1-\alpha_2)^2+4\alpha^2_3}}\sinh\left(\frac{1}{2}\sqrt{(\alpha_1-\alpha_2)^2+4\alpha_3^2}\right)\right]\end{pmatrix} &= LHS
\end{aligned}\end{equation}
and
\begin{equation}\begin{aligned}
RHS &= \begin{pmatrix}e^{\beta_1}\left(\cos\beta_4\cosh\beta_3-\sin\beta_4\sinh\beta_3\right)&e^{\beta_1}\left(\sin\beta_4\cosh\beta_3+\cos\beta_4\sinh\beta_3\right)\\e^{\beta_2}\left(-\sin\beta_4\cosh\beta_3+\cos\beta_4\sinh\beta_3\right)&e^{\beta_2}\left(\cos\beta_4\cosh\beta_3+\sin\beta_4\sinh\beta_3\right)\end{pmatrix}
\end{aligned}\end{equation}
Best Answer
I'm only writing this to avoid a garland of thrust-and-parry comments, and to remind you of the standard method. The standard drill you may have covered in the physics of spin 1/2 through Pauli matrices is the following.
First clean up your formulas and parameters that seem to completely overwhelm you. $$ G_3=\sigma_1; ~~G_4=i\sigma_2; ~~2G_1=\sigma_3+ I; ~~2G_2=I-\sigma_3; $$ It is thus obvious that $G_1+G_2$ is in the center of your Lie algebra, the 2x2 identity matrix, and factors out of the problem: it should be eliminated with extreme prejudice.
The remaining three Lie Algebra elements are traceless, and so the group elements of $sl(2)$ now map to an exponential of a traceless 2x2 matrix. That is, $$ e^{(\alpha_1 + \alpha_2)/2} e^{(\alpha_1-\alpha_2) \sigma_3/2 + \alpha_3 \sigma_1 } =e^{(\beta_1 +\beta_2)/2} e^{(\beta_1 -\beta_2)\sigma_3/2} e^{\beta_3 \sigma_1} e^{i\beta_4 \sigma_2} . $$ That is to say, after you appreciate that $\alpha_1+\alpha_2=\beta_1+\beta_2$, one α and one β is redundant, and may be eliminated. Do that, introducing primed variables for the half differences, to solve $$ e^{\alpha' \sigma_3 + \alpha_3 \sigma_1 } = e^{\beta' \sigma_3} e^{\beta_3 \sigma_1} e^{i\beta_4 \sigma_2} . $$ Now, given the cornerstone expansion of the Pauli vector adduced in the WP link provided, perform the multiplication on the RHS and equate it to the expansion of the LHS. One combination of the of the 3 remaining βs will be constrained to zero: In particular the coefficient of the $\sigma_2$, on the RHS, which is absent on the LHS--do you see why? So there are only two βs to solve for two αs.
If I were you, I'd take my remaining two αs to be pure imaginary, so the LHS is a group element of su(2); and $\beta_4$ real, while $\beta_3$ and $\beta'$ pure imaginary, so you merely compose three elements of su(2) on the right, three unitary 2x2 matrices, to a restricted unitary matrix on the LHS.