Quantum Science Philippines
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] = 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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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).

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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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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}

(See link https://www.quantumsciencephilippines.com/?p=5698 )

 

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)

 

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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.

 

 

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