phy-4600/solutions/chap7/prob5
2016-03-26 18:02:07 -04:00

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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
𝓏❙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
𝓍
ħ͟ ⎛ 0 1 0 ⎞
√2 ⎜ 1 0 1 ⎟
⎝ 0 1 0 ⎠.
The general eigenvalue equation is
𝓍❙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
𝓍❙1 1❭𝓍 = ħ❙1 1❭𝓍 ,
𝓍❙1 0❭𝓍 = 0❙1 1❭𝓍 , and
𝓍❙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
Following this to conclusion just like with spin operators will provide the eigenstates, and then from that the wave function can be expressed using the x basis, and probabilities obtained.
I need to stop here, but I will produce at least sthe histogram from part a:
(𝐜)
𝓟(L̂𝓏)
╭─────────────────────────╮
│ │
│ │
│ │
│ │
│ │
                     │ 
   │                         │    
¹⁶/₂₉ ├   ▓                     │    
 │   ▓                     │    
 │   ▓                     │ 
 │   ▓                     │ 
 │   ▓                     │ 
 │   ▓                     │ 
│   ▓                     │ 
⁹/₂₉ ├   ▓        ▓            │ 
 │   ▓        ▓            │ 
 │   ▓        ▓            │ 
 │   ▓        ▓            │ 
 │   ▓        ▓            │ 
⁴/₂₉ ├   ▓        ▓        ▓   │ 
 │   ▓        ▓        ▓   │ 
 │   ▓        ▓        ▓   │ 
 │   ▓        ▓        ▓   │ 
   ╰─────────────────────────╯
-ħ 0 ħ