Uncertainty Restrictions on Orbital Angular Momentum

The general uncertainty relation of quantum mechanics forbids the simultaneous and completely accurate measurements of two observables whose operators do not commute. As will be shown here, the one-dimensional orbital angular momentum operators do not mutually commute. It follows then that, for example, one cannot suggest a system in which there is non-zero orbital angular momentum in only one dimension of orientation, as the knowledge of the orbital angular momentum in said dimension would imply that it is impossible to know that there is zero orbital angular momentum in the other two dimensions. Certain restrictions can thus be derived from the general uncertainty relation for orbital angular momentum.

Orbital Angular Momentum Operators

The orbital angular momentum operator $\hat{L}$ is given by:

\begin{equation}
\hat{L} = \hat{r} \times \hat{p}
\end{equation}

$\hat{r}$ and $\hat{p}$ are the position operator and the momentum operator respectively. The two can be separated into dimensional components like so: $\hat{r} = \mathbf{x}\hat{x} + \mathbf{y}\hat{y} + \mathbf{z}\hat{z}$ and $\hat{p} = \mathbf{x}\hat{p}_x + \mathbf{y}\hat{p}_y + \mathbf{z}\hat{p}_z$, where respectively $\mathbf{x}$, $\mathbf{y}$, and $\mathbf{z}$ are unit vectors in the $x$, $y$, and $z$ dimensions; $\hat{p}_x$, $\hat{p}_y$, and $\hat{p}_z$ are the $x$, $y$, and $z$ component momentum operators; and $\hat{x}$, $\hat{y}$, and $\hat{z}$ are the $x$, $y$, and $z$ component position operators.

\begin{equation}
\hat{L} =
\begin{vmatrix}
\mathbf{x} & \mathbf{y} & \mathbf{z} \\
\hat{x} & \hat{y} & \hat{z} \\
\hat{p}_x & \hat{p}_y & \hat{p}_z
\end{vmatrix}
\end{equation}
\begin{equation}
\hat{L} = \mathbf{x}\left( \hat{y}\hat{p}_z - \hat{z}\hat{p}_y \right) + \mathbf{y} \left( \hat{z}\hat{p}_x - \hat{x}\hat{p}_z \right) + \mathbf{z}\left( \hat{x}\hat{p}_y - \hat{y}\hat{p}_x \right)
\end{equation}

$\hat{L}$ can also be broken down into its $x$, $y$, and $z$ components, $\hat{L}_x$, $\hat{L}_y$, and $\hat{L}_z$ respectively.

\begin{equation}
\hat{L} = \mathbf{x}\hat{L}_x + \mathbf{y} \hat{L}_y + \mathbf{z}\hat{L}_z
\end{equation}
\begin{align}
\hat{L}_x &= \hat{y}\hat{p}_z - \hat{z}\hat{p}_y \\
\hat{L}_y &= \hat{z}\hat{p}_x - \hat{x}\hat{p}_z \\
\hat{L}_z &= \hat{x}\hat{p}_y - \hat{y}\hat{p}_x
\end{align}

Substituting in the one-dimensional position operators and momentum operators gives:

\begin{align}
\hat{L}_x &= -i\hbar \left( y\frac{\partial}{\partial z} - z\frac{\partial}{\partial y} \right) \\
\hat{L}_y &= -i\hbar \left( z\frac{\partial}{\partial x} - x\frac{\partial}{\partial z} \right) \\
\hat{L}_z &= -i\hbar \left( x\frac{\partial}{\partial y} - y\frac{\partial}{\partial x} \right)
\end{align}

Commutators

To utilize the general uncertainty relation, the commutators between the dimensional component orbital angular momentum operators must be first found. Let $\psi$ be an arbitrary function of $x$, $y$, and $z$.

\begin{equation}
\left[\hat{L}_x, \hat{L}_y\right]\psi = -\hbar^2 \left[\left( y\frac{\partial}{\partial z} - z\frac{\partial}{\partial y} \right) \left( z\frac{\partial}{\partial x} - x\frac{\partial}{\partial z} \right) - \left( z\frac{\partial}{\partial x} - x\frac{\partial}{\partial z} \right) \left( y\frac{\partial}{\partial z} - z\frac{\partial}{\partial y} \right) \right] \psi
\end{equation}
\begin{equation}
\left[\hat{L}_x, \hat{L}_y\right]\psi = -\hbar^2 \left( y\frac{\partial}{\partial x} - x\frac{\partial}{\partial y}\right) \left( \frac{\partial z \psi}{\partial z} - z\frac{\partial \psi}{\partial z} \right)
\end{equation}
\begin{equation}
\left[\hat{L}_x, \hat{L}_y\right] = \hbar^2 \left( x\frac{\partial}{\partial y} - y\frac{\partial}{\partial x}\right)
\end{equation}
\begin{equation}
\left[\hat{L}_x, \hat{L}_y\right] = i\hbar \hat{L}_z
\end{equation}

Similarly, it can be derived that:

\begin{align}
\left[\hat{L}_y, \hat{L}_z\right] &= i\hbar \hat{L}_x\\
\left[\hat{L}_z, \hat{L}_x\right] &= i\hbar \hat{L}_y\\
\end{align}

Orbital Angular Momentum Uncertainty Relation

By the general uncertainty relation,

\begin{equation}
{\sigma_\hat{A}}^2{\sigma_\hat{B}}^2 \geq \frac{1}{4}\left|\left\langle \left[\hat{A}, \hat{B}\right]\right\rangle\right| ^2
\end{equation}

where $\sigma_\hat{A}$ and $\sigma_\hat{B}$ are the standard deviations of the measurements of $\hat{A}$ and $\hat{B}$ respectively. Applying this relation to $\hat{L}_x$ and $\hat{L}_y$ gives the following:

\begin{equation}
{\sigma_{\hat{L}_x}}^2{\sigma_{\hat{L}_y}}^2 \geq \frac{1}{4}\left|\left\langle \left[\hat{L}_x, \hat{L}_y\right]\right\rangle\right|^2
\end{equation}

Furthermore, substituting in the expressions for the standard deviations of $L_x$ and $L_y$ as well as their commutator gives:

\begin{equation}
\left\langle\left(\hat{L}_x - \left\langle\hat{L}_x\right\rangle\right)^2 \right\rangle \left\langle\left(\hat{L}_y - \left\langle\hat{L}_y\right\rangle\right)^2 \right\rangle \geq \frac{\hbar^2}{4} \left\langle \hat{L}_z \right\rangle^2 \\
\end{equation}

Set the expectation values of $\hat{L}_x$ and $\hat{L}_y$ to zero in order to create a scenario with the least possible eigenvalues of ${\hat{L}_x}^2$ and ${\hat{L}_y}^2$.

\begin{equation}
\left\langle {\hat{L}_x}^2 \right\rangle \left\langle {\hat{L}_y}^2 \right\rangle \geq \frac{\hbar^2}{4} \left\langle \hat{L}_z \right\rangle^2
\end{equation}

Consider the following: letting $a$, $b$, and $c$ be arbitrary positive numbers, if $ab\geq c$, then $a+b \geq a + c/a$ and $a+b \geq b + c/b$. Since

\begin{align*}
\partial (a + c/a)/{\partial a}|_{a=\sqrt{c}} &= 0 \\
\partial (a + c/a)/{\partial a}|_{0 < a < \sqrt{c}} &< 0 \\ \partial (a + c/a)/{\partial a}|_{a>\sqrt{c}} &> 0
\end{align*}

the absolute minimum of $a+c/a$ occurs when $a=\sqrt{c}$. Similarly, it can be found that the absolute minimum of $b+c/b$ occurs when $b=\sqrt{c}$. Hence, $a+b\geq 2\sqrt{c}$. Applying the same logic to relation $(20)$, the following can be written:

\begin{equation}
\left\langle {\hat{L}_x}^2 \right\rangle + \left\langle {\hat{L}_y}^2 \right\rangle \geq \hbar \left\langle \hat{L}_z \right\rangle
\end{equation}
\begin{equation}
{\hat{L}_x}^2 + {\hat{L}_y}^2 \geq \hbar \hat{L}_z
\end{equation}

By the same process, it can be also found that:

\begin{align}
{\hat{L}_y}^2 + {\hat{L}_z}^2 &\geq \hbar \hat{L}_x \\
{\hat{L}_z}^2 + {\hat{L}_x}^2 &\geq \hbar \hat{L}_y \\
\end{align}

Restrictions on Orbital Angular Momentum

Note that relations $(22)$ to $(24)$ restrict the dimensional components of orbital angular momentum such that the total orbital angular momentum vector can never be confined to one dimension. This dictates that given the square of the total angular momentum, $\hat{L}^2 = {\hat{L}_x}^2 + {\hat{L}_y}^2 + {\hat{L}_z}^2$, the values of the dimensional components of that orbital angular momentum are restricted to specific ranges.

\begin{align}
\hat{L}^2 &\geq \hbar \hat{L}_x + {\hat{L}_x}^2 \\
\hat{L}^2 &\geq \hbar \hat{L}_y + {\hat{L}_y}^2 \\
\hat{L}^2 &\geq \hbar \hat{L}_z + {\hat{L}_z}^2
\end{align}