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 ⎠.

Applying the operator to the states in Ψ,

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

    L̂𝓍❙1 0❭ ≐
                ħ͟  ⎛ 0 1 0 ⎞⎛0⎞ = ħ͟ ⎛1⎞ =  ħ͟ (❙1 1❭ + ❙1 -1❭), and
               √2  ⎜ 1 0 1 ⎟⎜1⎟  √2 ⎜0⎟   √2 
                   ⎝ 0 1 0 ⎠⎝0⎠     ⎝1⎠

    L̂𝓍❙1 -1❭ ≐
                ħ͟  ⎛ 0 1 0 ⎞⎛0⎞ = ħ͟ ⎛0⎞ =  ħ͟ ❙1 0❭.
               √2  ⎜ 1 0 1 ⎟⎜0⎟  √2 ⎜1⎟   √2 
                   ⎝ 0 1 0 ⎠⎝1⎠     ⎝0⎠


    L̂𝓍❙Ψ❭ =  ⎛  2͟ L̂𝓍❙1 1❭ + ι 3͟ L̂𝓍❙1 0❭ - 4͟ L̂𝓍❙1 -1❭ ⎞
             ⎝ √29           √29        √29         ⎠

     2͟ L̂𝓍❙1 1❭ =  2͟ ħ❙1 0❭, 
    √29          √58    

    ι 3͟ L̂𝓍❙1 0❭ = ι 3͟ ħ (❙1 1❭ + ❙1 -1❭), and
     √29           √58

     4͟ L̂𝓍❙1 -1❭ =  4͟ ħ❙1 0❭.
    √29           √58

Then,

    L̂𝓍❙Ψ❭ =  ħ⎛ -2͟  ❙1 0❭ + ι 3͟ (❙1 1❭ + ❙1 -1❭)⎞
              ⎝ √58         √58                 ⎠

Normalizing the function,

    C⎛⎛-2͟ ⎞² + ⎛ι 3͟ ⎞² + ⎛ι 3͟ ⎞²⎞ = 1.
     ⎝⎝√58⎠    ⎝ √58⎠    ⎝ √58⎠ ⎠


 STOPPED HERE 
 
    C = 58⎛⎛1͟⎞ - ⎛ι 3 ⎞⁻² + ⎛ι 3 ⎞⁻²⎞
          ⎝⎝4⎠   ⎝    ⎠     ⎝    ⎠  ⎠
So,

    \|❬1 1❙L̂𝓍❙Ψ❭\|^2 = 





    L̂𝓍❙Ψ❭ =  ⎛  + ι 3͟ L̂𝓍❙1 0❭ - 4͟ L̂𝓍❙1 -1❭ ⎞
             ⎝     √29        √29         ⎠