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).

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

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

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

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

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

$= i$ħ$\big(xp_{y} - yp_{x}\big)$

Therefore,

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

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## Angular Momentum: Show that [Ly,Lz] = iħLx.

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

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

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 $\big[x,p_{x}\big] = \big(xp_{x} - p_{x}x\big) =i$ħ (Canonical Commutator)

Therefore,

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

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## Matrix representation of the square of the spin angular momentum

By: Maria Christine L. Lugo, MS Physics I, MSU-IIT

Show that

a.) $(S^{2} = S_{1}^{2}+S_{2}^{2}+ 2S_{1z}S_{2z}+S_{1+}S_{2-}+S_{1-}S_{2+})$

b.) $S^{2} =\hbar^{2}\left( \begin{array}{cccc} 2& 0&0&0 \\ 0 &1&1&0 \\0&1&1&0\\0&0&0&2 \end{array} \right)$

Solution:

a.) We can expand $S^{2}$ by its components $S_{1}^{2}$ and S_{2}^{2},

$S^{2} = S_{1}^{2}+S_{2}^{2} + 2(S_{1}\cdot S_{2})$

But we know that,

$S_{1}\cdot S_{2} = \frac{1}{2} (S_{1+}S_{2-}+S_{1-}S_{2+}) +S_{1z}S_{2z}$

Therefore,

$S^{2} = S_{1}^{2} + S_{2}^{2} + 2[\frac{1}{2}(S_{1+}S_{2-} + S_{1-}S_{2+})+S_{1z}S_{2z}]$

$S^{2} = S_{1}^{2} + S_{2}^{2} +2S_{1z}S_{2z}+S_{1+}S_{2-}+S_{1-}S_{2+}$

b.) Using the eigenvalue equation of $S^{2}$ and substituting the basis $(|++>, |+->, |-+>, |- ->)$

$S^{2}|sm> = (S_{1}^{2}+S_{2}^{2} +2S_{1z}S_{2z}+ S_{1+}S_{2-}+S_{1-}S_{2+})|sm>$

We have,

$\begin{array}{lll} S^{2}|++>&=& S_{1}^{2}|++>+ S_{2}^{2}|++>+2S_{1z}S_{2z}|++> +S_{1+}S_{2-}|++>+S_{1-}S_{2+}|++>\\ \\&=&\frac{3}{4}\hbar^{2}|++>+\frac{3}{4}\hbar^{2}|++>+2(\frac{1}{2}\hbar)(\frac{1}{2}\hbar)|++>\\ \\&=&2\hbar^{2}|++>\end{array}$

$\begin{array}{lll} S^{2}|+->&=& S_{1}^{2}|+->+ S_{2}^{2}|+->+2S_{1z}S_{2z}|+-> +S_{1+}S_{2-}|+->+S_{1-}S_{2+}|+->\\ \\&=&\frac{3}{4}\hbar^{2}|+->+\frac{3}{4}\hbar^{2}|+->-\frac{1}{2}|+->+\hbar^{2}|-+>\\ \\&=&\hbar^{2}(|+->+|-+>)\end{array}$

$\begin{array}{lll} S^{2}|-+>&=& S_{1}^{2}|-+>+ S_{2}^{2}|-+>+2S_{1z}S_{2z}|-+> +S_{1+}S_{2-}|-+>+S_{1-}S_{2+}|-+>\\ \\&=&\frac{3}{4}\hbar^{2}|-+>+\frac{3}{4}\hbar^{2}|-+>-\frac{1}{2}|-+>+\hbar^{2}|+->\\ \\&=&\hbar^{2}(|-+>+|+->)\end{array}$

$\begin{array}{lll} S^{2}|->&=& S_{1}^{2}|- ->+ S_{2}^{2}|- ->+2S_{1z}S_{2z}|- -> +S_{1+}S_{2-}|- ->+S_{1-}S_{2+}|- ->\\ \\&=&\frac{3}{4}\hbar^{2}|- ->+\frac{3}{4}\hbar^{2}|- ->+2(\frac{1}{2}\hbar)(\frac{1}{2}\hbar)|- ->\\ \\&=&2\hbar^{2}|- ->\end{array}$

Therefore, the matrix representation will give us

$S^{2} =\hbar^{2}\left( \begin{array}{cccc} 2& 0&0&0 \\ 0 &1&1&0 \\0&1&1&0\\0&0&0&2 \end{array} \right)$

## Orthogonality of Two Eigenvectors

### Orthogonality of Two Eigenvectors

by: Mariel A. Escobal, MS Physics I, MSU-IIT

### Prove that two eigenvectors of a Hermitian operator corresponding to two different eigenvalues are orthogonal.

Consider two eigenvectors |ψ⟩ and |φ⟩ of the Hermitian operator Â,

Â|ψ⟩ = λ|ψ⟩     (1)

Â|φ⟩ = μ|φ⟩     (2)

Since Â is Hermitian, we can write (2) as

⟨φ|Â = μ⟨φ|     (3)

Multiplying (1) by ⟨φ| on the left and (3) by |ψ⟩ on the right, we get

⟨φ|Â|ψ⟩ = λ⟨φ|ψ⟩     (4)

⟨φ|Â|ψ⟩ = μ⟨φ|ψ⟩     (5)

Subtracting (4) and (5), we get

(λ-μ)⟨φ|ψ⟩ = 0

Consequently, the fact that λ≠μ, (λ-μ) ≠ 0 which then implies that ⟨φ|ψ⟩ = 0.

And as we know, this is the condition for two orthogonal eigenvectors.