Understanding the rotation operator $D\left(R\right)=\exp\left(-i\frac{\phi\hat{n}\cdot\vec{S}}{\hbar}\right)$ in spin-space

Question

In quantum mechanics, the rotating operator in the spin space is $ D\left(R\right)=\exp\left(-i\frac{\phi\hat{n}\cdot\vec{S}}{\hbar}\right)$ where $\phi$ is the angle of rotation and $\hat{n}$ the axis of rotation.

Can I treat $ D\left(R\right)=\exp\left(-i\frac{\phi\hat{n}\cdot\vec{S}}{\hbar}\right)$ as a sequence of rotations around each axis?

By this I mean:

$D\left(R\right)=\exp\left(-i\frac{\phi\hat{n}\cdot\vec{S}}{\hbar}\right)=\exp\left(-i\frac{\phi\cdot n_x\cdot S_x}{\hbar}\right)\exp\left(-i\frac{\phi\cdot n_y\cdot S_y}{\hbar}\right)\exp\left(-i\frac{\phi\cdot n_z\cdot S_z}{\hbar}\right)$, so $D(R)|\alpha\rangle$ yields the ket after rotation by $n_z\phi$ in the $z$ axis, then by $n_y\phi$ in the $y$ axis, and finally by $n_x\phi$ in the $x$ axis?

Isn't there a problem with the fact rotations don't commute, but addition is commutative? So if I write $\hat n \cdot \vec S=n_yS_y+n_xS_x+n_zS_z$, don't I get a whole new ket then the one I got when I wrote $\hat n \cdot \vec S=n_xS_x+n_yS_y+n_zS_z$?

Answer 1

Yes, there is a problem. If $[A,B]\neq 0$, then $$ \mathrm e^{A+B}\neq \mathrm e^A\mathrm e^ B $$ (the correct expression is given by the Baker–Campbell–Hausdorff formula)

Therefore, in general $$ \mathrm e^{i\boldsymbol S\cdot\boldsymbol n}\neq \mathrm e^{iS_xn_x}\mathrm e^{iS_yn_y}\mathrm e^{iS_zn_z} $$

If you want to write the general rotation as a product of three rotations, you must use Euler Angles.