Quantum orbital angular momentum operators (Cartesian coordinates)

The quantum orbital angular momentum operators in Cartesian coordinates are derived from the classical angular momentum components

$L_x=r_yp_z-r_zp_y$

$L_y=r_zp_x-r_xp_z$

$L_z=r_xp_y-r_yp_x$

by replacing the position and linear momentum components with their corresponding operators. Since $r_x,r_y,r_z$ are position components with position operators $\hat{r}_x,\hat{r}_y,\hat{r}_z$ respectively, and $p_x,p_y,p_z$ are linear momentum components with linear momentum operators $\frac{\hbar}{i}\frac{\partial}{\partial r_x},\frac{\hbar}{i}\frac{\partial}{\partial r_y},\frac{\hbar}{i}\frac{\partial}{\partial r_z}$ respectively (see eq4), we have

$\hat{L}_x=\frac{\hbar}{i}\left (r_y \frac{\partial}{\partial r_z}-r_z\frac{\partial}{\partial r_y}\right )\; \; \; \; or\; \; \; \;\hat{L}_x=\frac{\hbar}{i}\left (y \frac{\partial}{\partial z}-z\frac{\partial}{\partial y}\right )\; \; \; \; \; \; \; \; 72$

$\hat{L}_y=\frac{\hbar}{i}\left (r_z \frac{\partial}{\partial r_x}-r_x\frac{\partial}{\partial r_z}\right )\; \; \; \; or\; \; \; \;\hat{L}_y=\frac{\hbar}{i}\left (z \frac{\partial}{\partial x}-x\frac{\partial}{\partial z}\right )\; \; \; \; \; \; \; \; 73$

$\hat{L}_z=\frac{\hbar}{i}\left (r_x \frac{\partial}{\partial r_y}-r_y\frac{\partial}{\partial r_x}\right )\; \; \; \; or\; \; \; \;\hat{L}_z=\frac{\hbar}{i}\left (x \frac{\partial}{\partial y}-y\frac{\partial}{\partial x}\right )\; \; \; \; \; \; \; \; 74$

From eq70, we have

$\hat{L}^{2}=\hat{L}_x^{\, \, 2}+\hat{L}_y^{\, \, 2}+\hat{L}_z^{\, \, 2}\; \; \; \; \; \; \; \; 75$

Since $\hat{L}^{ 2}$ is defined as the operator for the square of the magnitude of $\boldsymbol{\mathit{L}}$ , each of its eigenvalues is the square of the magnitude of the orbital angular momentum of an electron.

Question

Can we construct an angular momentum operator using $\boldsymbol{\mathit{L}}=\boldsymbol{\mathit{i}}L_x+\boldsymbol{\mathit{j}}L_y+\boldsymbol{\mathit{k}}L_z$ , such that

$\hat{\boldsymbol{\mathit{L}}}=\boldsymbol{\mathit{i}}\hat{L}_x+\boldsymbol{\mathit{j}}\hat{L}_y+\boldsymbol{\mathit{k}}\hat{L}_z\; \; \; \; \; \; \; \; 76$

which is then used to generate eigenvalues?

Answer

$\hat{L}^{2}$ , a scalar operator, is preferred over $\hat{\boldsymbol{\mathit{L}}}$ , a vector operator, because it easier to manipulate in quantum computations. $\hat{L}^{2}$ commutes with its component operators, allowing us to simultaneously determine the eigenvalues of $\hat{L}^{2}$ and say, $\hat{L}_z$ (which is useful, for example in the verification of singlet and triplet eigenstates). It also commutes with the time-independent Hamiltonian $\hat{H}$ , also a scalar operator, implying that we can select a common complete set of eigenfunctions for $\hat{L}^{2}$ and $\hat{H}$ . Note that $\hat{L}^{2}$ has a form that is consistent with the angular portion of $\hat{H}$ in spherical coordinates (compare eq49 with eq96). An eigenvalue of $\hat{H}_{angular}$ is the energy associated with the angular motion of an electron, while the square root of an eigenvalue of $\hat{L}^{2}$ is the magnitude of the orbital angular momentum of an electron in a particular state.