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