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Angular momentum system is prepared in the state
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❙Ψ❭ = 1͟ ❙11❭ - 2͟ ❙10❭ + ι ͟2͟ ❙22❭ + ι ͟1͟ ❙20❭
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√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
𝐋 ̂²❙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.
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𝓟 (𝐋̂²=2ħ²) = │❬11❙Ψ❭│² + │❬10❙Ψ❭│².
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❬11❙Ψ❭ = ❬11❙⎛ 1͟ ❙11❭ - 2͟ ❙10❭ + ι ͟2͟ ❙22❭ + ι 1͟ ❙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.
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For L̂𝓏 the eigenvalue equation is
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L̂𝓏 ❙lm❭ = mħ❙lm❭,
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so the expected measurements are,
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for lm = 11 & 21, L̂𝓏 = ħ.
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for lm = 10 & 20, L̂𝓏 = 0.
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In this case, the probabilities will be
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𝓟 (L̂𝓏 =ħ) = │❬11❙Ψ❭│² + │❬21❙Ψ❭│² and
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𝓟 (L̂𝓏 =0) = │❬10❙Ψ❭│² + │❬20❙Ψ❭│².
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The method has already been demonstrated, so taking the probability components and summing,
(𝐛 )
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𝓟 (L̂𝓏 =ħ) = ²/₁₀.
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𝓟 (L̂𝓏 =0) = ⁸/₁₀.
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Histogram of probabilities:
𝓟 𝓟
╭─────────────────┬────────────────╮
1 │ │ │ 1
│ │ │
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.8 │ │ ▓ │ .8
│ │ ▓ │
│ │ ▓ │
.5 │ ▓ ▓ │ ▓ │ .5
│ ▓ ▓ │ ▓ │
│ ▓ ▓ │ ▓ │
│ ▓ ▓ │ ▓ ▓ │
.1 │ ▓ ▓ │ ▓ ▓ │ .1
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╰─────────────────┼────────────────╯
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2ħ² 6ħ² 0 ħ
𝐋̂² L̂𝓏