📂Quantum Mechanics

Ladder Operators for Angular Momentum

Definition

The Angular Momentum Operator corresponding to $L_{z}$ Ladder Operators are defined as follows.

$$L_{+} := L_{x} + \i L_{y} \\ L_{-} := L_{x} - \i L_{y}$$

$L_{+}$ is called the raising operator, and $L_{-}$ is called the lowering operator.

Explanation ^{1 2}

The names of the operators, raising/lowering, are due to the fact that $L_{\pm}$ raises or lowers the state of the simultaneous eigenfunction of the angular momentum operator $L_{z}$. Let the simultaneous eigenfunction of $L^{2}$ and $L_{z}$ be $\psi$, and assume that the eigenvalue equation for operator $L_{z}$ is as follows.

$$L_{z} \psi = \mu \psi$$

At this time, $L_{\pm}\psi$ can be shown to be an eigenfunction of $L_{z}$ with an eigenvalue greater than $\psi$ by $\pm \hbar$. By using the following property $(2)$,

$$\begin{align*} L_{z}(L_{\pm}\psi) &= (L_{\pm}L_{z} \pm \hbar L_{\pm})\psi \\ &= L_{\pm}L_{z}\psi \pm \hbar L_{\pm}\psi \\ &= L_{\pm}\mu\psi \pm \hbar L_{\pm}\psi \\ &= \mu L_{\pm}\psi \pm \hbar L_{\pm}\psi \\ &= (\mu \pm \hbar)L_{\pm}\psi\\ \end{align*}$$

On the other hand, by $(3)$, it can be seen that the eigenvalue for $L^{2}$ does not change. If the eigenvalue equation for $L^{2}$ is $L^{2}\psi = \lambda \psi$, then

$$\begin{align*} L^{2}(L_{\pm}\psi) &= (L_{\pm}L^{2})\psi \\ &= L_{\pm}(L^{2}\psi) \\ &= L_{\pm}l\lambda\psi \\ &= \lambda L_{\pm}\psi \end{align*}$$

Properties

The following relationships hold for the ladder operators.

$$\begin{align} L_{+}L_{-} &= L^{2} - L_{z}^{2} + \hbar L_{z} \nonumber \\ L_{-}L_{+} &= L^{2} - L_{z}^{2} - \hbar L_{z} \nonumber \\ L^{2} &= L_{+}L_{-} + L_{z}^{2} - \hbar L_{z} \nonumber \\ &= L_{-}L_{+} + L_{z}^{2} + \hbar L_{z} \nonumber \\ [L_{z}, L_{\pm}] &= \pm\hbar L_{\pm} \\ L_{z}L_{\pm} &= L_{\pm}L_{z} \pm \hbar L_{\pm} \\ [L^{2}, L_{\pm}] &= 0 \\ L_{x} &= \dfrac{1}{2}(L_{+} + L_{-}) \nonumber \\ L_{y} &= -\dfrac{\i}{2}(L_{+} - L_{-}) \nonumber \\ \end{align}$$

$(1)$ is the condition that $L_{\pm}$ becomes the Ladder Operator of $L_{z}$.

Proof

It can be shown by simple calculation. Note that the commutativity of multiplication generally does not hold for operators.

$$\begin{align*} L_{+}L_{-} &= (L_{x} + \i L_{y})(L_{x}-\i L_{y}) \\ &= {L_{x}}^{2} + \i L_{y}L_{x} - \i L_{x}L_{y} + {L_{y}}^{2} \\ &= {L_{x}}^{2}+{L_{y}}^{2} - \i[L_{x},L_{y}] \\ &= {L}^{2} -{L_{z}}^{2} +\hbar L_{z} \end{align*}$$

$$\begin{align*} L_{-}L_{+} &= (L_{x} - \i L_{y})(L_{x} + \i L_{y}) \\ &= {L_{x}}^{2} - \i L_{y}L_{x} + \i L_{x}L_{y} + {L_{y}}^{2} \\ &= {L_{x}}^{2} + {L_{y}}^{2} + \i[L_{x},L_{y}] \\ &= L^{2} - {L_{z}}^{2} -\hbar L_{z} \end{align*}$$

From the above two results, we obtain the following.

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

Commutation Relations of Angular Momentum Operator

$$\begin{align*} \left[ L_{y}, L_{z} \right] &= \i \hbar L_{x} \\ \left[ L_{z}, L_{x} \right] &= \i \hbar L_{y} \\ \left[ L^{2}, L_{x} \right] &= \left[ L^{2}, L_{y} \right] = 0 \end{align*}$$

The commutation relations of the angular momentum operator calculate as follows.

$$\begin{align*} [L_{z},L_{\pm}] &= [L_{z}, L_{x} \pm \i L_{y}] \\ &= [L_{z}, L_{x}] \pm \i [L_{z}, L_{y}] \\ &= \i\hbar L_{y} \pm \i (-\i\hbar L_{x}) \\ &= \pm \hbar(L_{x} \pm \i L_{y}) \\ &= \pm \hbar L_{\pm} \end{align*}$$

From the above result, we naturally obtain the following equation.

$$L_{z}L_{\pm} = L_{\pm}L_{z} \pm \hbar L_{\pm}$$

Likewise, the following equation is also obtained from the commutation relations of the angular momentum operator.

$$\begin{align*} [L^{2}, L_{\pm}] &= [L^{2}, L_{x} \pm \i L_{y}] \\ &= [L^{2},L_{x}] \pm \i [L^{2} , L_{y}] \\ &= 0 \end{align*}$$

By combining $L_{+}$ and $L_{-}$, we obtain the following.

$$\begin{align*} L_{x} &= \dfrac{1}{2} (L_{+} + L_{-}) \\ L_{y} &= -\dfrac{\i}{2}(L_{x} - L_{-}) \end{align*}$$

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