📂Quantum Mechanics

Simultaneous Eigenfunctions of Angular Momentum

Summary

Angular momentum operator $L^{2}$ and $L_{z}$’s normalized simultaneous eigenfunctions are denoted as $\ket{l, m}$. The eigenvalue equations are given below.

$$\begin{align*} L^{2} \ket{\ell, m} &= \ell(\ell+1)\hbar^{2}\ket{\ell, m} \\ L_{z}\ket{\ell, m} &= m\hbar\ket{\ell, m} \end{align*}$$

In this case, $\ell$ can only be integers or half-integers. For a given $\ell$, the minimum value of $m$ is $-\ell$, and the maximum value is $\ell$.

$$\begin{align*} \ell &= 0, \frac{1}{2}, 1, \frac{3}{2}, 2, \cdots \\ m &= -\ell, -\ell+1, -\ell+2, \cdots , \ell-2, \ell-1, \ell \quad \text{ for given } \ell \end{align*}$$

Explanation

When $\ell$ is an integer, it represents the eigenvalue equation for orbital angular momentum. Orbital angular momentum is the same as the angular momentum we know classically. When it is a half-integer, it is called spin angular momentum and is denoted by $\mathbf S$ instead of $\mathbf L$. This is a unique physical quantity that only appears in quantum phenomena, without a classical counterpart. The simultaneous eigenfunctions of the two operators $L^{2}, L_{z}$ are distinguished by $\ell, m$ and are denoted as $\ket{\ell, m}$.

Derivation

The angular momentum operators $L^{2}$ and $L_{z}$ commute.

$$\left[ L^{2}, L_{z} \right] = 0$$

Two commuting operators have simultaneous eigenfunctions, thus the normalized simultaneous eigenfunctions of $L^{2}$ and $L_{z}$ are denoted as $\ket{\psi}$, with eigenvalues $\lambda$ and $\mu$, respectively. The eigenvalue equation is given below.

$$\begin{align*} L^{2} \ket{\psi} &= \lambda \ket{\psi} \\ L_{z}\ket{\psi} &= \mu\ket{\psi} \end{align*}$$

Ladder operator of the angular momentum operator $L_{z}$

$$\begin{align*} L_{\pm} &= L_{x} \pm \i L_{y} \\ L^{2} &= L_{-}L_{+} + {L_{z}}^{2} + \hbar L_{z} \\ &= L_{+}L_{-} + {L_{z}}^{2} - \hbar L_{z} \\ \end{align*}$$

When the ladder operator $L_{\pm}$ is applied to the simultaneous eigenfunctions of $L_{z}$ and $L^{2}$, it changes the eigenvalue of $L_{z}$ by $\pm \hbar$, but does not change the eigenvalue of $L^{2}$.

$$\begin{align*} L_{z} (L_\pm \ket{\psi}) &= (\mu \pm \hbar)L_\pm \ket{\psi} \\ L^{2} (L_\pm \ket{\psi}) &= \lambda L_\pm \ket{\psi} \end{align*}$$

This is significant because, since $L_{\pm}$ does not change the eigenvalue of $L^{2}$, the eigenvalue of $L_{z}$ cannot infinitely increase. Given that $L^{2} = {L_{x}}^{2} + {L_{y}}^{2} + {L_{z}}^{2}$ and the eigenfunctions are normalized, when calculating the expectation value,

$$\braket{L^{2}} = \braket{{L_{x}}^{2}} + \braket{{L_{y}}^{2}} + \braket{{L_{z}}^{2}}$$

$$\implies \lambda = \braket{{L_{x}}^{2}} + \braket{{L_{y}}^{2}} + \mu^{2} \ge \mu^{2}$$

Therefore, the eigenvalue of $L_{z}$ cannot grow beyond a certain size. Let the largest eigenvalue be $\ell \hbar$, and the corresponding eigenfunction be $\ket{\psi_{\text{max}}}$. Then, the following two eigenvalue equations are obtained.

$$\begin{align*} L_{z}\ket{\psi_{\text{max}}} &= \ell \hbar \ket{\psi_{\text{max}}} \\ L^{2}\ket{\psi_{\text{max}}} &= \lambda \ket{\psi_{\text{max}}} \end{align*}$$

Since the maximum eigenvalue of $L_{z}$ is equal to the total angular momentum, the $x$ component value and $y$ component value of the angular momentum are $0$.

$$L_{x} \ket{\psi_{\text{max}}} = L_{y} \ket{\psi_{\text{max}}} = 0$$

The following equation can be derived.

$$L_{+} \ket{\psi_{\text{max}}} = L_{+}\ket{\psi_{\text{max}}} + \i L_{y}\ket{\psi_{\text{max}}} = 0\ket{\psi_{\text{max}}} + \i 0\ket{\psi_{\text{max}}} = 0$$

Assuming that $L_{+} \ket{\psi_{\text{max}}} = 0$ is physically meaningless, it is rational to set it as such. Using this property, the following calculations can be made.

$$\begin{equation} \begin{aligned} L^{2} \ket{\psi_{\text{max}}} &= (L_{-}L_{+} + {L_{z}}^{2} + \hbar L_{z})\ket{\psi_{\text{max}}} \\ &= (0 + \ell^{2}\hbar^{2} + \ell\hbar^{2})\ket{\psi_{\text{max}}} \\ &= \ell(\ell + 1)\hbar^{2} \ket{\psi_{\text{max}}} \\ &= \lambda \ket{\psi_{\text{max}}} \end{aligned} \end{equation}$$

Thus, $\lambda = \ell(\ell + 1)\hbar^{2}$ is obtained. Similarly, the lowest eigenvalue state $\ket{\psi_{\text{min}}}$ of $L_{z}$ can be known to exist. The value of $L_{-}$ in this state is physically meaningless, so it is rational to set it as $0$.

$$L_{-}\ket{\psi_{\text{min}}} = 0$$

Let the lowest eigenvalue be $\ell^{\prime} \hbar$, then the eigenvalue equations are as follows.

$$\begin{align*} L_{z}\ket{\psi_{\text{min}}} &= \ell^{\prime} \hbar \ket{\psi_{\text{min}}} \\ L^{2}\ket{\psi_{\text{min}}} &= \lambda \ket{\psi_{\text{min}}} \end{align*}$$

Similarly, the following equation is derived.

$$\begin{equation} \begin{aligned} L^{2} \ket{\psi_{\text{min}}} &= (L_{+}L_{-} + {L_{z}}^{2} - \hbar L_{z})\ket{\psi_{\text{min}}} \\ &= (0 + {\ell^{\prime}}^{2}\hbar^{2} - \ell^{\prime}\hbar^{2})\ket{\psi_{\text{min}}} \\ &= \ell^{\prime}(\ell^{\prime}-1)\hbar^{2} \ket{\psi_{\text{min}}} \\ &= \lambda \ket{\psi_{\text{min}}} \end{aligned} \end{equation}$$

From $(1)$ and $(2)$, the following equation is derived.

$$\begin{align*} \lambda &= \ell(\ell+1)\hbar^{2} \\ \lambda &= \ell^{\prime}(\ell^{\prime}-1)\hbar^{2} \\ \end{align*}$$

Thus, the following equation holds.

$$\begin{align*} && \ell(\ell+1)\hbar^{2} &= \ell^{\prime}(\ell^{\prime}-1)\hbar^{2} \\ \implies&& \ell(\ell+1) &= \ell^{\prime}(\ell^{\prime}-1) \\ \implies&& \ell(\ell+1) - \ell^{\prime}(\ell^{\prime}-1) &= 0 \\ \implies&& ( \ell^{\prime} + \ell )(\ell^{\prime} -(\ell+1) ) &= 0 \\ \end{align*}$$

Then $\ell^{\prime} = \ell + 1$ or $\ell^{\prime} = -\ell$ is obtained. However, since $\ell^{\prime}\hbar$ is the smallest eigenvalue and $\ell \hbar$ is the largest, it cannot be $\ell^{\prime} = \ell + 1$. Hence, the following is obtained.

$$\ell^{\prime} = -\ell$$

In other words, the largest eigenvalue of $L_{z}$ is $\ell\hbar$, and the smallest is $-\ell\hbar$.

$$-\ell\hbar \le m \le \ell \hbar$$

Thus, applying $L_{-}$ to the simultaneous eigenfunction each time changes the corresponding eigenvalues sequentially to $\ell \hbar$, $(\ell - 1)\hbar$, $(\ell - 2)\hbar$, $\dots$, $(-\ell + 1)\hbar$, $-\ell \hbar$. If the number of all possible states is $n+1$, then $\ell - n = -\ell$ and $\ell = \dfrac{n}{2}$ follow. Therefore, the possible values of $\ell$ are integers or half-integers (half of an integer). Additionally, the possible range of $m$ is $-\ell, -\ell+1, -\ell+2, \cdots , \ell-2, \ell-1, \ell$. Thus, the number of possible values of $m$ is $2\ell+1$.

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