📂Quantum Mechanics

Matrix Representation of Angular Momentum Operator

Formula

The matrix representation of the angular momentum operator is as follows. When $\ell = 1$, $m = 1, 0, -1$ holds,

$$\underset{\normalsize L_{z} = }{\scriptsize} \begin{array}{cccc} \scriptstyle{m=1} & \scriptstyle m=0 & \scriptstyle m=-1 & \\ \hbar\left[ \begin{array}{c} 1 \\ 0 \\ 0 \end{array} \right. & \begin{array}{c} 0 \\ 0 \\ 0 \end{array} & \left. \begin{array}{c} 0 \\ 0 \\ -1 \end{array} \right] & \begin{array}{l} \scriptstyle m=1 \\ \scriptstyle m=0 \\ \scriptstyle m=-1 \end{array} \end{array}$$

$$\underset{\normalsize L_{x} = }{\scriptsize} \begin{array}{cccc} \scriptstyle{m=1} & \scriptstyle m=0 & \scriptstyle m=-1 & \\ \dfrac{\hbar}{\sqrt{2}}\left[ \begin{array}{c} 0 \\ 1 \\ 0 \end{array} \right. & \begin{array}{c} 1 \\ 0 \\ 1 \end{array} & \left. \begin{array}{c} 0 \\ 1 \\ 0 \end{array} \right] & \begin{array}{l} \scriptstyle m=1 \\ \scriptstyle m=0 \\ \scriptstyle m=-1 \end{array} \end{array}$$

$$\underset{\normalsize L_{y} = }{\scriptsize} \begin{array}{cccc} \scriptstyle{m=1} & \scriptstyle m=0 & \scriptstyle m=-1 & \\ \dfrac{\hbar}{\sqrt{2}}\left[ \begin{array}{c} 0 \\ -\i \\ 0 \end{array} \right. & \begin{array}{c} \i \\ 0 \\ -\i \end{array} & \left. \begin{array}{c} 0 \\ \i \\ 0 \end{array} \right] & \begin{array}{l} \scriptstyle m=1 \\ \scriptstyle m=0 \\ \scriptstyle m=-1 \end{array} \end{array}$$

The matrix representation of the ladder operator of angular momentum is as follows. When $\ell = 1$, $m = 1, 0, -1$ holds,

$$\underset{\normalsize L_{+} = }{\scriptsize} \begin{array}{cccc} \scriptstyle{m=1} & \scriptstyle m=0 & \scriptstyle m=-1 & \\ \left[ \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right. & \begin{array}{c} \sqrt{2} \\ 0 \\ 0 \end{array} & \left. \begin{array}{c} 0 \\ \sqrt{2} \\ 0 \end{array} \right] & \begin{array}{l} \scriptstyle m=1 \\ \scriptstyle m=0 \\ \scriptstyle m=-1 \end{array} \end{array}$$

$$\underset{\normalsize L_{-} = }{} \begin{array}{cccc} \scriptstyle{m=1} & \scriptstyle m=0 & \scriptstyle m=-1 & \\ \left[ \begin{array}{c} 0 \\ \sqrt{2} \\ 0 \end{array} \right. & \begin{array}{c} 0 \\ 0 \\ \sqrt{2} \end{array} & \left. \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right] & \begin{array}{l} \scriptstyle m=1 \\ \scriptstyle m=0 \\ \scriptstyle m=-1 \end{array} \end{array}$$

Proof

Ladder Operator

$$\begin{align*} L_{+}\ket{\ell, m} &= \sqrt{(\ell-m)(\ell + m + 1)}\hbar\ket{\ell, m+1} \\ L_{-}\ket{\ell, m} &= \sqrt{(\ell+m)(\ell - m + 1)}\hbar\ket{\ell, m-1} \end{align*}$$

Since the two ladder operators satisfy the above equation, for $L_{+}$ we obtain the following equation.

$$\begin{align*} \braket{\ell,m^{\prime} | L_{+}|\ell,m} &= \sqrt{(\ell - m)(\ell + m + 1)}\hbar\braket{\ell,m^{\prime} | \ell,m+1} \\ &= \sqrt{(\ell - m)(\ell + m + 1)}\hbar \delta_{m^{\prime}, m+1} \end{align*}$$

At this time $\delta$ is the Kronecker delta. In the same way, for $L_{-}$, we obtain the following equation.

$$\begin{align*} \braket{\ell,m^{\prime} | L_{-}|\ell,m} &= \sqrt{(\ell + m)(\ell - m + 1)}\hbar\braket{\ell,m^{\prime} | \ell,m-1} \\ &= \sqrt{(\ell + m)(\ell - m + 1)}\hbar \delta_{m^{\prime}, m-1} \end{align*}$$

When $\ell = 1$, possible $m$ is $1, 0, -1$, so let’s say that for convenience, the rows and columns of $L_{\pm}$ are $1, 0, -1$. Then, the coordinate vector of the eigenfunction is as follows.

$$\ket{1, 1} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \qquad \ket{1, 0} = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, \qquad \ket{1, -1} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$$

Thus the state of the eigenfunction increases intuitively as follows with the expression $L_{+} \ket{\ell, m} = \sqrt{(\ell-m)(\ell + m + 1)}\hbar\ket{\ell, m+1}$. If $\ell = 1$ is assumed,

$$L_{+} \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} = \sqrt{2}\hbar \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \\[1em] L_{+} \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = \sqrt{2}\hbar \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$$

The matrix $L_{+}$ is as follows.

$$\begin{align*} L_{+} &= \begin{bmatrix} \braket{1, 1 | L_{+}| 1, 1} & \braket{1, 1 | L_{+}| 1, 0} & \braket{1, 1 | L_{+}| 1,-1} \\ \braket{1, 0 | L_{+}| 1, 1} & \braket{1, 0 | L_{+}| 1, 0} & \braket{1, 0 | L_{+}| 1,-1} \\ \braket{1,-1 | L_{+}| 1, 1} & \braket{1,-1 | L_{+}| 1, 0} & \braket{1,-1 | L_{+}| 1,-1} \end{bmatrix} \\ &= \begin{bmatrix} 0\hbar \delta_{1, 2} & \sqrt{2}\hbar \delta_{1 , 1} & \sqrt{2}\hbar \delta_{ 1, 0} \\ 0\hbar \delta_{0, 2} & \sqrt{2}\hbar \delta_{0 , 1} & \sqrt{2}\hbar \delta_{ 0, 0} \\ 0\hbar \delta_{1, 2} & \sqrt{2}\hbar \delta_{-1, 1} & \sqrt{2}\hbar \delta_{-1, 0} \end{bmatrix} \\ &= \hbar \begin{bmatrix} 0 & \sqrt{2} & 0 \\ 0 & 0 & \sqrt{2} \\ 0 & 0 & 0 \end{bmatrix} \end{align*}$$

Similarly, for $L_{-}$, we have the following.

$$L_{-} \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} = \sqrt{2}\hbar \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \\[1em] L_{-} \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = \sqrt{2}\hbar \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$$

$$L_{-} = \hbar \begin{bmatrix} 0 & 0 & 0 \\ \sqrt{2} & 0 & 0 \\ 0 & \sqrt{2} & 0 \end{bmatrix}$$

■

Angular Momentum Operator

The simultaneous eigenfunction of $L_{z}$ and $L^2$ is $\ket{\ell, m}$, and the eigenvalue equation is

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

Thus,

$$\braket{\ell,m^{\prime} | L_{z}|\ell,m}=m\hbar \braket{\ell,m^{\prime} | \ell,m}=m\hbar \delta_{mm^{\prime}}$$

Therefore, it is $[(L_{z})_{mn}] = m \hbar \delta_{mn}$. At this time $\delta$ is the Kronecker delta. When $\ell = 1$, possible $m$ is $1, 0, -1$, so let’s say $L_{z}$ has rows $1$, $0$, and $-1$ for convenience. Let’s assume that the coordinate vector of each eigenfunction is as follows.

$$\ket{1, 1} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \qquad \ket{1, 0} = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, \qquad \ket{1, -1} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$$

Then, since it is $[(L_{z})_{mn}] = m \hbar \delta_{mn}$,

$$\begin{align*} L_{z} &= \begin{bmatrix} (L_{z})_{11} & (L_{z})_{10} & (L_{z})_{1-1} \\ (L_{z})_{01} & (L_{z})_{00} & (L_{z})_{0-1} \\ (L_{z})_{-11} & (L_{z})_{-10} & (L_{z})_{-1-1} \end{bmatrix} \\ &= \begin{bmatrix} 1\hbar\delta_{1,1} & 1\hbar\delta_{1,0} & 1\hbar\delta_{1,-1} \\ 0\hbar\delta_{0,1} & 0\hbar\delta_{0,0} & 0\hbar\delta_{0,-1} \\ -1\hbar\delta_{-1,1} & -1\hbar\delta_{-1,0} & -1\hbar\delta_{-1,-1} \end{bmatrix} \\ &= \hbar \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{bmatrix} \end{align*}$$

Also, since $L_{x} = \dfrac{1}{2} (L_{+} + L_{-})$ and $L_{y} = -\dfrac{\i}{2}(L_{+} - L_{-})$, $L_{x}$ and $L_{y}$ are as follows.

$$\begin{align*} L_{x} &= \dfrac{1}{2} (L_{+} + L_{-}) \\ &= \dfrac{1}{2}\left( \hbar\begin{bmatrix} 0 & \sqrt{2} & 0 \\ 0 & 0 & \sqrt{2} \\ 0 & 0 & 0 \end{bmatrix} + \hbar\begin{bmatrix} 0 & 0 & 0 \\ \sqrt{2} & 0 & 0 \\ 0 & \sqrt{2} & 0 \end{bmatrix} \right) \\ &= \dfrac{\hbar}{\sqrt{2}} \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} \\ \end{align*}$$

$$\begin{align*} L_{y} &= -\dfrac{\i}{2} (L_{+} - L_{-}) \\ &= -\dfrac{\i}{2}\left( \hbar\begin{bmatrix} 0 & \sqrt{2} & 0 \\ 0 & 0 & \sqrt{2} \\ 0 & 0 & 0 \end{bmatrix} - \hbar\begin{bmatrix} 0 & 0 & 0 \\ \sqrt{2} & 0 & 0 \\ 0 & \sqrt{2} & 0 \end{bmatrix} \right) \\ &= \dfrac{\hbar}{\sqrt{2}} \begin{bmatrix} 0 & -\i & 0 \\ \i & 0 & -\i \\ 0 & \i & 0 \end{bmatrix} \\ \end{align*}$$