# Why is the definition of sin and cos in terms of exponentials is similar to the definition of $L_x$ & $L_y$ in terms of raising & lowering operators?

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

• – user856
Mar 15, 2019 at 10:04

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 $$L_+, L_-$$.
• 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. Mar 14, 2019 at 18:13
• @zabop: I corrected and expanded the post. It was wrong before, as I had suppressed what is now called "the matrix $W$". Mar 18, 2019 at 17:55