Angular Momentum: Show that [Ly,Lz] = iħLx. | Quantum Science Philippines
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Angular Momentum: Show that [Ly,Lz] = iħLx.

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

 

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

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

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

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

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

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

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

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

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

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

where \big[x,p_{x}\big] = \big(xp_{x} - p_{x}x\big) =iħ (Canonical Commutator)

Therefore,

\big[L_{y},L_{z}\big] = iħL_{x}, where L_{x} = \big(yp_{z} - zp_{y}\big).

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