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

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

 

Show that \big[L_{z},L_{x}\big] = iħL_{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] = iħL_{y}, where L_{y} = \big(zp_{x} - xp_{z}\big), \big[y,p_{y}\big] = yp_{y} - p_{y}y = iħ (Canonical Commutator).

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