phy-4600/solutions/chap7/prob5

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

    L̂𝓏❙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

    L̂𝓍 ≐
        ħ͟  ⎛ 0 1 0 ⎞
        √2 ⎜ 1 0 1 ⎟
           ⎝ 0 1 0 ⎠.

The general eigenvalue equation is 

    L̂𝓍❙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 

    L̂𝓍❙1  1❭𝓍 =  ħ❙1  1❭𝓍 ,
    L̂𝓍❙1  0❭𝓍 =  0❙1  1❭𝓍 , and
    L̂𝓍❙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