Quantum orbital angular momentum ladder operators

The ladder operators of quantum orbital angular momentum are defined as:

$\small \hat{L}_+=\hat{L}_x+i\hat{L}_y\; \; \; \; \; \; \; \; 108$

$\small \hat{L}_-=\hat{L}_x-i\hat{L}_y\; \; \; \; \; \; \; \; 109$

where $\small \hat{L}_+$ is the raising operator and $\small \hat{L}_-$ is the lowering operator.

To demonstrate why the operators are named as such, we substitute eq108 in $\small \left [\hat{L}_+,\hat{L}_z\right ]$ :

$\small \left [\hat{L}_+,\hat{L}_z\right ]=\left [\hat{L}_x+i\hat{L}_y,\hat{L}_z\right ]=\left [\hat{L}_x,\hat{L}_z\right ]+i\left [\hat{L}_y,\hat{L}_z\right ]$

Substitute eq100 and eq101 in the above equation, noting that $\small \left [ \hat{L}_a,\hat{L}_b \right ]=-\left [ \hat{L}_b,\hat{L}_a \right ]$ ,

$\small \hat{L}_+\hat{L}_z-\hat{L}_z\hat{L}_+=-i\hbar\hat{L}_y-\hbar\hat{L}_x$

$\small \hat{L}_+\hat{L}_z=\hat{L}_z\hat{L}_+-\hbar\hat{L}_+\; \; \; \; \; \; \; \; 110$

Similarly, we find

$\small \hat{L}_-\hat{L}_z=\hat{L}_z\hat{L}_-+\hbar\hat{L}_-\; \; \; \; \; \; \; \; 111$

If $\small Y$ is an eigenfunction of $\small \hat{L}_z$ with eigenvalue $\small c$ ,

$\small \hat{L}_z Y=cY\; \; \; \; \; \; \; \; 112$

Operating on eq112 with $\small \hat{L}_+$ , we have $\small \hat{L}_+\hat{L}_z Y=c\hat{L}_+Y$ . Substitute eq110 in the this equation and rearranging, we have $\small \hat{L}_z\hat{L}_+ Y=(c+\hbar)\hat{L}_+Y$ . So, $\small \hat{L}_+Y$ is an eigenfunction of $\small \hat{L}_z$ with eigenvalue $\small (c+\hbar)$ . Operating on eq112 with $\small \hat{L}_+$ transforms $\small Y$ into another eigenfunction $\small \hat{L}_+ Y$ with an eigenvalue higher than c by $\small \hbar$ . If we operate on eq112 with $\small \hat{L}_+$ twice, we have

$\small \hat{L}_+\hat{L}_z\hat{L}_+ Y=(c+\hbar)\hat{L}_+^{\; 2}Y$

$\small \left ( \hat{L}_z\hat{L}_+-\hbar\hat{L}_+\right )\hat{L}_+ Y=(c+\hbar)\hat{L}_+^{\; 2}Y$

$\small \hat{L}_z\hat{L}_+^{\; 2}Y=(c+2\hbar)\hat{L}_+^{\; 2}Y$

By mathematical induction,

$\small \hat{L}_z\hat{L}_+^{\; k}Y=(c+k\hbar)\hat{L}_+^{\; k}Y \; \; \; \; \; k=0,1,2,\cdots\; \; \; \; \; \; \; \; 113$

Similarly, if we operate on eq112 $\small k$ times with $\small \hat{L}_-$ , we have

$\small \hat{L}_z\hat{L}_-^{\; k}Y=(c-k\hbar)\hat{L}_-^{\; k}Y \; \; \; \; \; k=0,1,2,\cdots\; \; \; \; \; \; \; \; 114$

In other words, the raising operator progressively raises the eigenvalue of $\small Y$ by $\small \hbar$ , while the lowering operator progressively lowers the eigenvalue of $\small Y$ by $\small \hbar$ , i.e. each operator generate a ladder of eigenvalues.

Question

Show that $\small \hat{L}_+\hat{L}_-=\hat{L}^{2}-\hat{L}_z^{\; 2}+\hbar\hat{L}_z$ .

Answer

Substituting eq108 and eq109 in $\small \hat{L}_+\hat{L}_-$ ,

$\small \hat{L}_+\hat{L}_-=\hat{L}_x^{\; 2}-i\hat{L}_x\hat{L}_y+i\hat{L}_y\hat{L}_x+\hat{L}_y^{\; 2}$

Substitute eq75 in the above equation and then eq99 in the resultant equation, noting that $\small \left [ \hat{L}_a,\hat{L}_b \right ]=-\left [ \hat{L}_b,\hat{L}_a \right ]$ ,

$\small \hat{L}_+\hat{L}_-=\hat{L}^{2}-\hat{L}_z^{\; 2}+\hbar\hat{L}_z\; \; \; \; \; \; \; \; 115$

Similarly,

$\small \hat{L}_-\hat{L}_+=\hat{L}^{2}-\hat{L}_z^{\; 2}-\hbar\hat{L}_z\; \; \; \; \; \; \; \; 116$

If $\small Y$ is simultaneously also an eigenfunction of $\small \hat{L}^{2}$ with eigenvalue $\small b$ ,

$\small \hat{L}^{2}Y=bY\; \; \; \; \; \; \; \; 117$

Question

Show that $\small \hat{L}^{2}$ commutes with $\small \hat{L}_\pm ^{\; k}$ .

Answer

For $\small k=1$ , we have $\small \left [ \hat{L}^{2},\hat{L}_ \pm \right ]=\left [ \hat{L}^{2},\hat{L}_ x\pm i\hat{L}_ y\right ]=\left [ \hat{L}^{2},\hat{L}_ x\right ]\pm i\left [ \hat{L}^{2},\hat{L}_ y\right ]$ . Since $\small \left [ \hat{L}^{2},\hat{L}_ x \right ]=0$ and $\small \left [ \hat{L}^{2},\hat{L}_ y \right ]=0$ ,

$\small \left [ \hat{L}^{2},\hat{L}_ \pm \right ]=0$

For $\small k=2$ , we apply the identity $\small \left [ \hat{A},\hat{B}\hat{C} \right ]=\left [ \hat{A},\hat{B}\right ]\hat{C}+\hat{B}\left [ \hat{A},\hat{C} \right ]$ :

$\small \left [ \hat{L}^{2},\hat{L}_ \pm ^{\; 2}\right ]=\left [ \hat{L}^{2},\hat{L}_ \pm\right ]\hat{L}_ \pm+\hat{L}_ \pm\left [\hat{L}^{2},\hat{L}_ \pm\right]=0$

By mathematical induction,

$\small \left [ \hat{L}^{2},\hat{L}_\pm^{\; k} \right ]=0\; \; \; \; \; k=0,1,2,\cdots\; \; \; \; \; \; \; \; 118$

Operating on eq117 with $\small \hat{L}_\pm^{\; k}$ and using eq118

$\small \hat{L}_\pm^{\; k} \hat{L}^{2}Y=b\hat{L}_\pm^{\; k}Y$

$\small \hat{L}^{2}\hat{L}_\pm^{\; k}Y=b\hat{L}_\pm^{\; k}Y\; \; \; \; \; k=0,1,2,\cdots\; \; \; \; \; \; \; \; 119$

The raising and lowering operators also apply to the spin angular momentum $\small \boldsymbol{\mathit{S}}$ , because the spin angular momentum component operators are postulated to obey the form of commutation relations as described by eq99, eq100 and eq101 (see eq165, eq166 and eq167). Similarly, the raising and lowering operators apply to the total angular momentum $\small \boldsymbol{\mathit{J}}$ (see Q&A below).

Question

Show that $\small \left [ \hat{J}_x,\hat{J}_y \right ]=i\hbar\hat{J}_z$ .

Answer

Let

$\small \hat{J}_i=\hat{M}_{1i}+\hat{M}_{2i}\; \; \; \; \; \; \; \; 120$

where $\small i=x,y,z$ ; $\small \hat{M}_{1i}$ and $\small \hat{M}_{2i}$ are component operators of $\small \hat{M}^{(1)}$ and $\small \hat{M}^{(2)}$ respectively.

$\small \hat{M}^{(1)}$ and $\small \hat{M}^{(2)}$ are operators of two sources of angular momentum, e.g. $\small \hat{M}^{(1)}$ and $\small \hat{M}^{(2)}$ are the orbital angular momentum operator and spin angular momentum operator respectively of a particle.

$\small \left [ \hat{J}_x,\hat{J}_y \right ]=\left ( \hat{M}_{1x}+\hat{M}_{2x} \right )\left ( \hat{M}_{1y}+\hat{M}_{2y} \right )-\left ( \hat{M}_{1y}+\hat{M}_{2y} \right )\left ( \hat{M}_{1x}+\hat{M}_{2x} \right )$

Expanding and rearranging the RHS of the above equation,

$\small \left [ \hat{J}_x,\hat{J}_y \right ]=\left [ \hat{M}_{1x},\hat{M}_{1y} \right ]+\left [ \hat{M}_{2x},\hat{M}_{2y} \right ]+\left [ \hat{M}_{1x},\hat{M}_{2y} \right ]+\left [ \hat{M}_{2x},\hat{M}_{1y} \right ]$

Since the 3^rd term on RHS of the above equation involves operators acting on different vector spaces (e.g. spatial coordinates vs spin coordinates), they must commute. The same goes for the 4^th term. So,

$\small \left [ \hat{J}_x,\hat{J}_y \right ]=\left [ \hat{M}_{1x},\hat{M}_{1y} \right ]+\left [ \hat{M}_{2x},\hat{M}_{2y} \right ]$

According to eq99 and eq165, $\small \left [ \hat{M}_{1x},\hat{M}_{1y} \right ]=i\hbar\hat{M}_{1z}$ and $\small \left [ \hat{M}_{2x},\hat{M}_{2y} \right ]=i\hbar\hat{M}_{2z}$ . So,

$\small \left [ \hat{J}_{x},\hat{J}_{y} \right ]=i\hbar\hat{J}_{z}\; \; \; \; \; \; \; \; 121$

Similarly, we have $\small \left [ \hat{J}_{y},\hat{J}_{z} \right ]=i\hbar\hat{J}_{x}$ and $\small \left [ \hat{J}_{z},\hat{J}_{x} \right ]=i\hbar\hat{J}_{y}$ .

Quantum orbital angular momentum ladder operators

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