3
$\begingroup$

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 $\hat n \cdot \vec S=n_xS_x+n_yS_y+n_zS_z$?

$\endgroup$
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.