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108 lines
3.1 KiB
Plaintext
108 lines
3.1 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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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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STOPPED HERE
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C = 58⎛⎛1͟⎞ - ⎛ι 3 ⎞⁻² + ⎛ι 3 ⎞⁻²⎞
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⎝⎝4⎠ ⎝ ⎠ ⎝ ⎠ ⎠
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So,
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\|❬1 1❙L̂𝓍❙Ψ❭\|^2 =
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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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