3
$\begingroup$

I noted a similarity outlined below:

The angular momentum operators in $x$ and $y$ direction can be written:

$$L_x=\frac{1}{2}(L_++L_-)$$ $$L_y=\frac{1}{2i}(L_+-L_-)$$

$cos(x)$ and $sin(x)$ can be written:

$$\cos(x)=\frac{1}{2}(e^{ix}+e^{-ix})$$ $$\sin(x)=\frac{1}{2i}(e^{ix}-e^{-ix})$$

Is there a particular reason for this manifest similarity, or it is just a mere coincidence, without much depth in it?


Note: this question has been asked here as well: https://physics.stackexchange.com/q/466480/212053

$\endgroup$
4
$\begingroup$

Consider the inverse relation; $$ L_x+iL_y=L_+, \qquad L_x-iL_y=L_-, $$ and compare it with $$ x+iy=z, \qquad x-iy=\overline z.$$ With $x=\cos \theta, y=\sin \theta$, you have the analogy you spotted.


You ask why such relations are useful. The reason is that they massively simplify the description of plane rotations, because they reduce them to multiplication by $e^{i\theta}$.

More precisely, if we let $$ W:=\begin{bmatrix} 1 & i \\ 1 & -i\end{bmatrix},$$ then we can compactly write the above formulas in matrix form $$ \begin{bmatrix}L_+ \\ L_-\end{bmatrix} = W \begin{bmatrix} L_x \\ L_y\end{bmatrix}. $$ And now we notice that, for $\theta\in\mathbb R$, $$ \begin{bmatrix}e^{i\theta}L_+ \\ e^{-i\theta}L_-\end{bmatrix} = W \begin{bmatrix} \cos \theta & -\sin\theta \\ \sin \theta & \cos \theta\end{bmatrix} \begin{bmatrix} L_x \\ L_y\end{bmatrix}.$$ This is the simplification we were talking about: the linear change of variable induced by $W$ conjugates a rotation matrix to multiplication by $e^{\pm i \theta}$. This also explains the subscripts $\pm$ in $L_+, L_-$.

$\endgroup$
  • $\begingroup$ Yes, this is a bit better than my original formulas, though still leaves me wondering if there is a deeper relation between the two (either considered in the form of the OP or this answer), or just simply algebraic accident. $\endgroup$ – zabop Mar 14 at 18:10
  • 1
    $\begingroup$ You will always find something like this; given two real quantities, such as $x$ and $y$, you form a single complex one. You will then have to consider both this quantity and its conjugate as independent. Here's a good answer describing a rather general appearance of this trick. $\endgroup$ – Giuseppe Negro Mar 14 at 18:13
  • $\begingroup$ @zabop: I corrected and expanded the post. It was wrong before, as I had suppressed what is now called "the matrix $W$". $\endgroup$ – Giuseppe Negro Mar 18 at 17:55

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.