Angular Momentum: Show that [Lz,Lx] = iħLy. | Quantum Science Philippines

## Angular Momentum: Show that [Lz,Lx] = iħLy.

By: Jenard B. Cerbanez, MS Physics I, MSU-IIT

Show that $\big[L_{z},L_{x}\big] = i$ħ $L_{y}$

Solution:

Given that, $L = L_{x} + L_{y} + L_{z}$,     where $L_{x} = yp_{z} - zp_{y}$ $L_{y}= zp_{x} - xp_{z}$, $L_{z} = xp_{y} - yp_{x}$

Thus, $\big[L_{z},L_{x}\big] = L_{z}L_{x} - L_{x}L_{z}$ $= \big(xp_{y} - yp_{x}\big) \big(yp_{z} - zp_{y}\big) - \big(yp_{z} - zp_{y}\big) \big(xp_{y} - yp_{x}\big)$ $=xyp_{y}p_{z} - xzp_{y}^{2} - y^{2}p_{x}p_{z} + yzp_{x}p_{y} - \big(yxp_{z}p_{y} - y^{2}p_{z}p_{x} - zxp_{y}^{2} + zyp_{y}p_{x}\big)$ $= xy\big(p_{y}p_{z} - p_{z}p_{y}\big) + xz \big(p_{y}^{2}-p_{y}^{2}\big) + y^{2} \big(p_{z}p_{x} -p_{x}p_{z}\big) +yz\big(p_{x}p_{y} - p_{y}p_{x}\big)$

however, $xz \big(p_{y}^{2} - p_{y}^{2}\big) = y^{2} \big(p_{z}p_{x} -p_{z}p_{x}\big) = 0$. Now we are left with the terms, $= xp_{z} \big(p_{y}y - yp_{y}\big) + zp_{x} \big(yp_{y} - p_{y}y\big)$ $= -xp_{z} \big(yp_{y} - p_{y}y\big) + zp_{x}\big(yp_{y} - p_{y}y\big)$ $= \big(yp_{y} - p_{y}y\big) \big(zp_{x} - xp_{z}\big)$ $= \big[y,p_{y}\big] \big(zp_{x} - xp_{z}\big)$

where $\big[y,p_{y}\big] = \big(yp_{y} - p_{y}y\big)$

Therefore, $\big[L_{z},L_{x}\big] = i$ħ $L_{y}$, where $L_{y} = \big(zp_{x} - xp_{z}\big), \big[y,p_{y}\big] = yp_{y} - p_{y}y = i$ħ (Canonical Commutator).