Angular Momentum: Show that [Lx,Ly] = iħLz | Quantum Science Philippines
Quantum Science Philippines

Angular Momentum: Show that [Lx,Ly] = iħLz

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

 

Show that \big[L_{x},L_{y}\big] = iħL_{z}

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_{x},L_{y}\big] = \big(L_{x}L_{y} - L_{y}L_{x}\big)

= \big(yp_{z} - zp_{y}\big) \big(zp_{x} - xp_{z}\big) - \big(zp_{x} - xp_{z}\big) \big(yp_{z} - zp_{y}\big)

= yzp_{z}p_{x} - yxp_{z}^{2} - z^{2}p_{y}p_{x} + zxp_{y}p_{z} - \big(zyp_{x}p_{z} - z^{2}p_{x}p_{y} - xyp_{z}^{2} + xzp_{z}p_{y}\big)

= yz\big(p_{z}p_{x} - p_{x}p_{z}\big) + yx \big(p_{z}^{2}-p_{z}^{2}\big) + z^{2} \big(p_{x}p_{y} -p_{y}p_{x}\big) +zx\big(p_{y}p_{z} - p_{z}p_{y}\big)

however, yx \big(p_{z}^{2} - p_{z}^{2}\big) = z^{2} \big(p_{x}p_{y} -p_{y}p_{x}\big) = 0. Now we are left with the terms,

= yp_{x} \big(p_{z}z - zp_{z}\big) + xp_{y} \big(zp_{z} - p_{z}z\big)

= -yp_{x} \big(zp_{z} - p_{z}z\big) + xp_{y}\big(zp_{z} - p_{z}z\big)

= -yp_{x} \big[z,p_{z}\big] + xp_{y} \big[z,p_{z}\big]

where \big[z,p_{z}\big] = \big(zp_{z} - p_{z}z\big)

= \big[z,p_{z}\big] \big(xp_{y} - yp_{x}\big)

= iħ\big(xp_{y} - yp_{x}\big)

Therefore,

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

Leave a Reply