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169 lines
4.8 KiB
Plaintext
169 lines
4.8 KiB
Plaintext
There is an angular momentum system with the state function
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❙Ψ❭ = 2͟ ❙1 1❭ + ι 3͟ ❙1 0❭ - 4͟ ❙1 -1❭
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√29 √29 √29
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In general the eigenvalue equation for the L̂𝓏 operator is
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L̂𝓏❙l m❭ = m ħ❙l m❭, where m ħ are the possible measurements.
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The possible measurements of this system, then, are, for m = {-1, 0, 1}:
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-ħ, 0, ħ.
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The probability for is given by
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│❬1 m′❙Ψ❭│², with m′ = {-1, 0, 1}.
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The eigenstates form an orthogonal set such that
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❬l′ m′❙l m❭ = δₗₗ′ δₘₘ′.
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Then,
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❬1 1❙Ψ❭ = ❬1 1❙⎛ 2͟ ❙1 1❭ + ι 3͟ ❙1 0❭ - 4͟ ❙1 -1❭ ⎞
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⎝√29 √29 √29 ⎠
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= ❬1 1❙ 2͟ ❙1 1❭ = 2͟ .
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√29 √29
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(𝐚)
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│❬1 1❙Ψ❭│² = 4͟ = ⁴/₂₉.
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29
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Similarly,
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│❬1 0❙Ψ❭│² = 9͟ = ⁹/₂₉ and
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29
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│❬1 -1❙Ψ❭│² = 1͟6͟ = ¹⁶/₂₉.
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29
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The eigenvalue equations for the L̂𝓏 operator are simplified because L̂𝓏 is diagonal in the z basis. The L̂𝓍 operator produces the same measurements, but the matrix representation of the L̂𝓍 operator must be applied. It is
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L̂𝓍 ≐
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ħ͟ ⎛ 0 1 0 ⎞
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√2 ⎜ 1 0 1 ⎟
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⎝ 0 1 0 ⎠.
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The general eigenvalue equation is
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L̂𝓍❙λ,mₗ❭ = λ❙λ,mₗ❭, where the eigenvalues λ are the possible measured values of L̂𝓍. The eigenvalues can be obtained from the secular equation
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det│L̂𝓍 - λ𝕀│ = 0
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ħ͟ ⎛ 0 1 0 ⎞ - ⎛ λ 0 0 ⎞ = ⎛ -λ ħ/√2 0 ⎞
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√2 ⎜ 1 0 1 ⎟ ⎜ 0 λ 0 ⎟ ⎜ ħ/√2 -λ ħ/√2 ⎟
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⎝ 0 1 0 ⎠ ⎝ 0 0 λ ⎠ ⎝ 0 ħ/√2 -λ ⎠.
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│⎛ -λ ħ/√2 0 ⎞│ = (-λ(λ² - ħ²/2) + (ħ²/2) λ) = -λ³ + ħ²λ.
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│⎜ ħ/√2 -λ ħ/√2 ⎟│
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│⎝ 0 ħ/√2 -λ ⎠│
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λ(-λ² + ħ²) = -λ(λ² - ħ²)) = 0.
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One eigenvalue is immediately obvious: λ = 0. The other two are given by
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λ² = ħ², so the eigenvalues are
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λ = 0,±ħ.
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These are exactly the expected measured values for a spin component.
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The eigenvalue equations are
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L̂𝓍❙1 1❭𝓍 = ħ❙1 1❭𝓍 ,
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L̂𝓍❙1 0❭𝓍 = 0❙1 1❭𝓍 , and
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L̂𝓍❙1 -1❭𝓍 = -ħ❙1 -1❭𝓍 .
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Matrix analysis can be used to find the eigenvectors for these eigenstates. The first one is
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ħ͟ ⎛ 0 1 0 ⎞ ⎛ a ⎞ = ħ ⎛ a ⎞, which gives the system
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√2 ⎜ 1 0 1 ⎟ ⎜ b ⎟ ⎜ b ⎟
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⎝ 0 1 0 ⎠ ⎝ c ⎠ ⎝ c ⎠
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⎧ b = √2 a
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⎨ (a + c) = √2 b
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⎩ b = √2 c
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Applying the operator to the states in Ψ,
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L̂𝓍❙1 1❭ ≐
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ħ͟ ⎛ 0 1 0 ⎞⎛1⎞ = ħ͟ ⎛0⎞ = ħ͟ ❙1 0❭.
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√2 ⎜ 1 0 1 ⎟⎜0⎟ √2 ⎜1⎟ √2
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⎝ 0 1 0 ⎠⎝0⎠ ⎝0⎠
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L̂𝓍❙1 0❭ ≐
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ħ͟ ⎛ 0 1 0 ⎞⎛0⎞ = ħ͟ ⎛1⎞ = ħ͟ (❙1 1❭ + ❙1 -1❭), and
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√2 ⎜ 1 0 1 ⎟⎜1⎟ √2 ⎜0⎟ √2
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⎝ 0 1 0 ⎠⎝0⎠ ⎝1⎠
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L̂𝓍❙1 -1❭ ≐
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ħ͟ ⎛ 0 1 0 ⎞⎛0⎞ = ħ͟ ⎛0⎞ = ħ͟ ❙1 0❭.
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√2 ⎜ 1 0 1 ⎟⎜0⎟ √2 ⎜1⎟ √2
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⎝ 0 1 0 ⎠⎝1⎠ ⎝0⎠
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L̂𝓍❙Ψ❭ = ⎛ 2͟ L̂𝓍❙1 1❭ + ι 3͟ L̂𝓍❙1 0❭ - 4͟ L̂𝓍❙1 -1❭ ⎞
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⎝ √29 √29 √29 ⎠
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2͟ L̂𝓍❙1 1❭ = 2͟ ħ❙1 0❭,
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√29 √58
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ι 3͟ L̂𝓍❙1 0❭ = ι 3͟ ħ (❙1 1❭ + ❙1 -1❭), and
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√29 √58
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4͟ L̂𝓍❙1 -1❭ = 4͟ ħ❙1 0❭.
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√29 √58
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Then,
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L̂𝓍❙Ψ❭ = ħ ⎛ -2͟ ❙1 0❭ + ι 3͟ (❙1 1❭ + ❙1 -1❭)⎞
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⎝ √58 √58 ⎠
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Normalizing the function,
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C⎛⎛-2͟ ⎞² + ⎛ι 3͟ ⎞² + ⎛ι 3͟ ⎞²⎞ = 1.
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⎝⎝√58⎠ ⎝ √58⎠ ⎝ √58⎠ ⎠
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C⎛4 - 9 - 9⎞ = 58 = C(14).
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⎝ ⎠
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C = 58/14 = 29/7.
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The probability of measuring one of the possible angular momenta will be given by
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│❬l m❙L̂𝓍❙Ψ❭│².
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❬1 1❙L̂𝓍❙Ψ❭ = ❬1 1❙2͟9͟⎛ -2͟ ❙1 0❭ + ι 3͟ (❙1 1❭ + ❙1 -1❭)⎞
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7 ⎝ √58 √58 ⎠
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= 2͟9͟ ι͟3͟ ❬1 1❙1 1❭ = ι 8͟7͟
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7 √58 7√58
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L̂𝓍❙Ψ❭ = ⎛ + ι 3͟ L̂𝓍❙1 0❭ - 4͟ L̂𝓍❙1 -1❭ ⎞
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⎝ √29 √29 ⎠
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