# Angular momentum in 3D. Quantum Algebra.

(Part of the Wolverhampton Lectures of Physics's Quantum Physics Course)

We have already discussed the WLP_XI/Ladders angular momentum in 2D to describe rotation in the plane, with $L_z=xp_y-yp_x$. Of course, angular momentum $\vec=\vec r\times\vec p$ is a 3D quantity, and just like the 2D harmonic oscillator was best described in terms of one component of angular momentum (in particular, whether clockwise or anticlockwise), motion in full space will also be found to be often best described in terms of rotation. For the hydrogen, for instance, the $l$ and $m$ degrees of freedom actually refer to rotation of the electron around its nucleus. We are therefore tempted to now consider the three components of the vector $\vec L$:

\begin{align} L_x&=yp_z-zp_y\,,\\ L_y&=zp_x-xp_z\,,\\ L_z&=xp_y-yp_x\,. \end{align}

These are operators thanks to the correspondence principle $p_x\rightarrow -i\hbar\partial_x$, $p_y\rightarrow -i\hbar\partial_y$ and $p_z\rightarrow -i\hbar\partial_z$. Unlike momentum $\vec p$, however, the components of angular momentum are not compatible observables, since we can check that they do not commute. For instance:

\begin{align} [L_x,L_y]&=L_xL_y-L_yL_x\\ &=-\hbar^2\left[(y\partial_z-z\partial_y)(z\partial_x-x\partial_z)- (z\partial_x-x\partial_z)(y\partial_z-z\partial_y)\right]\\ &=-\hbar^2[\color{red}{y\partial_zz\partial_x}-y\partial_zx\partial_z-z\partial_yz\partial_x+z\partial_yx\partial_z-z\partial_xy\partial_z+z\partial_xz\partial_y+x\partial_zy\partial_z\color{red}{-x\partial_zz\partial_y}]\\ &=-\hbar^2[\color{red}{y\partial_x+yz\partial_z\partial_x}-yx\partial^2_z-z^2\partial_y\partial_x+zx\partial_y\partial_z-zy\partial_x\partial_z+z^2\partial_x\partial_y+xy\partial_z^2\color{red}{-x\partial_y-xz\partial_z\partial_y}]\\ &=i\hbar(-i\hbar)[x\partial_y-y\partial_x]\\ &=i\hbar L_z\,. \end{align}

We have highlighted in red the parts where there is not a trivial commutation, and how this affects the computation, namely:

\begin{align} y\partial_z z\partial_x&=y\partial_x+yz\partial_z\partial_x\,,\\ x\partial_z z\partial_y&=x\partial_y+xz\partial_z\partial_y\,. \end{align}

Note that we have assumed commutation of the partial derivatives, which is satisfied in most cases of physical relevance and is the reason why projections of the linear momentum are compatible. One can similarly show that the other cases can be obtained by cyclic permutation of the indices:

\begin{align} [L_x,L_y]&=i\hbar L_y\,,\\ [L_y,L_z]&=i\hbar L_x\,,\\ [L_z,L_x]&=i\hbar L_y\,, \end{align}

which can be summarized in one expression only:

$$[L_k,L_l]=i\hbar\varepsilon_{klm}L_m\tag{1}$$

by introducing the Levi-Civita symbol:

$$\varepsilon _{klm}={\begin{cases}+1&{\text{if }}(k,l,m){\text{ is }}(1,2,3),(2,3,1),{\text{ or }}(3,1,2),\\-1&{\text{if }}(k,l,m){\text{ is }}(3,2,1),(1,3,2),{\text{ or }}(2,1,3),\\\;\;\,0&{\text{if }}k=l,{\text{ or }}k=m,{\text{ or }}l=m\,.\end{cases}}$$

Any two components of angular momentum are not compatible between each others, and invoking the generalized Heisenberg uncertainty,

$$\sigma_{L_x}^2\sigma_{L_y}^2\ge\left\langle{1\over 2i}[L_x,L_y]\right\rangle^2={\hbar^2\over 4}\langle L_z\rangle^2\,,$$

we find that the magnitude of angular momentum in one direction constrains the uncertainties on angular momenta in the other ones. This is unlike position and momentum, that do not commute but as a constant, while angular momentum components do not commute and follow a cyclic permutation of themselves. This peculiarity constitutes a so-called quantum algebra, whereby the non-commutativity of the operators emerges as a central feature of the mathematical relationships (addition and products) in the theory. In the formalism, this means that angular momentum as a vector—again, unlike linear momentum—cannot be known, since its components are not compatible, i.e., they cannot be all, in fact, even pairwise, be known with accuracy. Measuring part of the angular momentum disrupt the other parts. One cannot, therefore, label a quantum states through its components of angular momentum. However, one can know the magnitude of the angular momentum and one component, since, as we shall show

$$[L_x,L^2]=[L_y,L^2]=[L_z,L^2]=0$$

where, by definition, $L^2=\vec L^\cdot\vec L=L_x^2+L_y^2+L_z^2$. We prove the first one, starting with

$$[L_x,L^2]=[L_x,L_x^2]+[L_x,L_y^2]+[L_x,L_z^2]\,.$$

Note that we could express, e.g., $L_s^2$ in differential operator form and compute the commutators as we did above to derive Eq. (1), but that would be lengthier than turning to pure (quantum) algebra:

first $[L_x,L_x^2]=0$ since an operator commutes with its powers. For $[L_x, L_y]^2$, we use the fact that:

$$[A,BC]=[A,B]C+B[A,C]$$

to compute

$$[L_x,L_y^2]=[L_x,L_y]L_y+L_y[L_x,L_y]$$

and using the algebra of angular momentum:

$$[L_x,L_y^2]=i\hbar(L_zL_y+L_yL_z)$$

Note that we also have a notation $\{L_z,L_y\}$ for the anticommutator $L_zL_y+L_yL_z$. Computing $[L_x,L_z^2]$ similarly, we find:

$$[L_x,L_z^2]=-i\hbar\{L_z,L_y\}$$

$[L_x,L_y^2]$ and $[L_z,L_y^2]$ are not zero but their sum is, so finally:

$$[L_x,L^2]=0$$

and as announced, the $x$-component of angular momentum commutes with the magnitude (squared) of the angular momentum. The same holds for the $y$- and $z$-components. Usually, we use the $z$-component as the so-called quantization axis. This is a convention, as this $z$ axis can be aligned with any direction anyway.

Because these operators commute, we can find a common set of eigenfunctions, i.e., there exist some functions $\psi$ such that:

\begin{align} L_z\psi&=\mu\psi\\ L^2\psi&=\lambda\psi\,. \end{align}

We will turn to finding out explicitely what these eigenfunctions and eigenvalues are in the next Lecture. Interestingly, this will be best achieved by using pure quantum algebra, i.e., the central equation Eq. (1), without involving directly the differential equations.