Angular momentum system is prepared in the state ❙Ψ❭ = 1͟ ❙11❭ - 2͟ ❙10❭ + ι͟2͟ ❙22❭ + ι͟1͟ ❙20❭ √10 √10 √10 √10 Possible results of 𝐋² measurement? The 𝐋̂² operator was developed in lecture and in the text. The related eigenvalue equation is 𝐋̂²❙lm❭ = l(l+1) ħ²❙lm❭ The possible measurements of 𝐋̂² for this system are those associated with the initial state vector, i.e. the values lm = 11, 10, 22, 20. For lm = 11 & 10, 𝐋̂² = 2ħ² For lm = 22 & 20, 𝐋̂² = 6ħ². Since the same 𝐋̂² is measures for two states, the sum of the probabilities of measuring those states is the probability of measuring that squared angular momentum. 𝓟(𝐋̂²=2ħ²) = │❬11❙Ψ❭│² + │❬10❙Ψ❭│². ❬11❙Ψ❭ = ❬11❙⎛ 1͟ ❙11❭ - 2͟ ❙10❭ + ι͟2͟ ❙22❭ + ι1͟ ❙20❭ ⎞ ⎝ √10 √10 √10 √10 ⎠. The eigenstates for this system are LaPlace's spherical harmonic functions, which comprise an orthogonal set, I.E.: ❬lm❙l′m′❭ = δₗₗ′ δₘₘ′. │❬11❙Ψ❭│² = │❬11❙ 1͟ ❙11❭│² = ¹/₁₀. │ √10 │ │❬10❙Ψ❭│² = │❬10❙ 2͟ ❙10❭│² = ⁴/₁₀. │ √10 │ (𝐚) 𝓟(𝐋̂²=2ħ²) = ½. No need to calculate the other set: since the vector is normalized, the probability of measuring 𝐋̂²=6ħ² is also 𝓟(𝐋̂²=6ħ²) = ½. For L̂𝓏 the eigenvalue equation is L̂𝓏❙lm❭ = mħ❙lm❭, so the expected measurements are, for lm = 11 & 21, L̂𝓏 = ħ. for lm = 10 & 20, L̂𝓏 = 0. In this case, the probabilities will be 𝓟(L̂𝓏=ħ) = │❬11❙Ψ❭│² + │❬21❙Ψ❭│² and 𝓟(L̂𝓏=0) = │❬10❙Ψ❭│² + │❬20❙Ψ❭│². The method has already been demonstrated, so taking the probability components and summing, (𝐛) 𝓟(L̂𝓏=ħ) = ²/₁₀. 𝓟(L̂𝓏=0) = ⁸/₁₀. Histogram of probabilities: 𝓟 𝓟 ╭─────────────────┬────────────────╮ 1 │ │ │ 1 │ │ │ .8 │ │ ▓ │ .8 │ │ ▓ │ │ │ ▓ │ .5 │ ▓        ▓    │           ▓    │ .5    │   ▓        ▓    │           ▓    │        │   ▓        ▓    │           ▓    │        │   ▓        ▓    │   ▓       ▓    │     .1 │   ▓        ▓    │   ▓       ▓    │ .1    ╰─────────────────┼────────────────╯ 2ħ² 6ħ² 0 ħ 𝐋̂² L̂𝓏