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86 lines
3.2 KiB
Plaintext
86 lines
3.2 KiB
Plaintext
Angular momentum system is prepared in the state
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❙Ψ❭ = 1͟ ❙11❭ - 2͟ ❙10❭ + ι͟2͟ ❙22❭ + ι͟1͟ ❙20❭
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√10 √10 √10 √10
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Possible results of 𝐋² measurement?
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The 𝐋̂² operator was developed in lecture and in the text. The related eigenvalue equation is
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𝐋̂²❙lm❭ = l(l+1) ħ²❙lm❭
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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.
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For lm = 11 & 10, 𝐋̂² = 2ħ²
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For lm = 22 & 20, 𝐋̂² = 6ħ².
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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❭ + ι͟2͟ ❙22❭ + ι1͟ ❙20❭ ⎞
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⎝ √10 √10 √10 √10 ⎠.
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The eigenstates for this system are LaPlace's spherical harmonic functions, which comprise an orthogonal set, I.E.:
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❬lm❙l′m′❭ = δₗₗ′ δₘₘ′.
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│❬11❙Ψ❭│² = │❬11❙ 1͟ ❙11❭│² = ¹/₁₀.
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│ √10 │
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│❬10❙Ψ❭│² = │❬10❙ 2͟ ❙10❭│² = ⁴/₁₀.
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│ √10 │
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(𝐚)
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𝓟(𝐋̂²=2ħ²) = ½.
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No need to calculate the other set: since the vector is normalized, the probability of measuring 𝐋̂²=6ħ² is also
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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,
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(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:
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𝓟 𝓟
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╭─────────────────┬────────────────╮
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1 │ │ │ 1
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│ │ │
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.8 │ │ ▓ │ .8
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│ │ ▓ │
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│ │ ▓ │
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.5 │ ▓ ▓ │ ▓ │ .5
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│ ▓ ▓ │ ▓ │
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│ ▓ ▓ │ ▓ │
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│ ▓ ▓ │ ▓ ▓ │
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.1 │ ▓ ▓ │ ▓ ▓ │ .1
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╰─────────────────┼────────────────╯
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2ħ² 6ħ² 0 ħ
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𝐋̂² L̂𝓏
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