HW #3

1.  Spin Matrices

We use the spin operators represented in the bases where (Subscript[S, z])is diagonal:

(Subscript[S, x]) = (ℏ/2) {{0, 1}, {1, 0}}; 
    (Subscript[S, y]) = (ℏ/2) {{0, -I}, {I, 0}}; 
    (Subscript[S, z]) = (ℏ/2) {{1, 0}, {0, -1}};

(Subscript[S, x]) // MatrixForm

(Subscript[S, y]) // MatrixForm

(Subscript[S, z]) // MatrixForm

(a) Obviously two matrices commute when they are the same: i = j.  Also, it is obvious that [(Subscript[S, i]), (Subscript[S, j])] is anti-symmetric in i⟷j because [(Subscript[S, j]), (Subscript[S, i])] = -[(Subscript[S, i]), (Subscript[S, j])].  Therefore, it only reamains to verify

(Subscript[S, x]) . (Subscript[S, y]) - (Subscript[S, y]) . (Subscript[S, x]) - I ℏ (Subscript[S, z])

(Subscript[S, y]) . (Subscript[S, z]) - (Subscript[S, z]) . (Subscript[S, y]) - I ℏ (Subscript[S, x])

(Subscript[S, z]) . (Subscript[S, x]) - (Subscript[S, x]) . (Subscript[S, z]) - I ℏ (Subscript[S, y])

(b) We define (Overscript[n,  ->])  = (sinθ cosϕ,  
          sinθ sinϕ,  cosθ)

(Subscript[n, x]) = Sin[θ] Cos[ϕ];  (Subscript[n, y]) = Sin[θ] Sin[ϕ]; 
    (Subscript[n, z]) = Cos[θ]

(Subscript[S, n]) = 
      Simplify[(Subscript[n, x]) (Subscript[S, x]) + (Subscript[n, y]) (Subscript[S, y]) + (Subscript[n, z]) (Subscript[S, z])]

Eigensystem[(Subscript[S, n])]

PowerExpand[%]

Simplify[%]

Therefore, one can take the the normalized eigenstates to be (({{cos (θ/2)}, {sin (θ/2) (e^i)ϕ}}))with eigenvalue +(ℏ/2)and  (({{-sin (θ/2) (e^(-iϕ))}, {cos (θ/2)}}))with eigenvalue -(ℏ/2).

(c) The state with spin along the (Overscript[n,  ->])  direction is the former, and its probability to have the positive (Subscript[S, z]) when measured is simply given by \(\(|\)\(⟨S\_z = \(\(+\(ℏ\/2\)\) | 
                    S\_n = \(+\(ℏ\/2\)\)\)⟩\)\( | \^2\)\)=|(1, 0)(({{cos (θ/2)}, {sin (θ/2) (e^i)ϕ}}))\( | \^2\)=(cos^2) (θ/2) .

(d) Between (Overscript[n,  ->])  = (sinθ cosϕ,  
          sinθ sinϕ,  cosθ)and (Overscript[n,  ->]) ' = (sinθ' cosϕ',  
          sinθ' sinϕ',  cosθ'), the probability is \(\(|\)\(⟨S\_\(n'\) = \(\(+\(ℏ\/2\)\) | 
                    S\_n = \(+\(ℏ\/2\)\)\)⟩\)\( | \^2\)\)=|\((cos\ θ'\/2, 
                  sin\ \(θ'\/2\) e\^\(-iϕ'\))\)(({{cos (θ/2)}, {sin (θ/2) (e^i)ϕ}}))\( | \^2\)=\(\(\(|\)\(cos\ \(θ'\/2\) cos\ θ\/2 + 
                  sin\ \(θ'\/2\) \(e\^\(-iϕ'\)\) 
                    sin\ \(θ\/2\) e\^iϕ\)\( | \^2\)\) = 
              cos\^2\ \(θ'\/2\) cos\^2\ θ\/2 + 
                sin\^2\ \(θ'\/2\) sin\^2\ θ\/2 + 
                2  cos\ \(θ'\/2\) cos\ \(θ\/2\) 
                  sin\ \(θ'\/2\) 
                  sin\ \(θ\/2\) \(cos(ϕ - ϕ')\)\)

TrigExpand[Cos[((Subscript[θ, 1])/2)]^2 
          Cos[((Subscript[θ, 2])/2)]^2 + Sin[((Subscript[θ, 1])/2)]^2 
          Sin[((Subscript[θ, 2])/2)]^2 + 
        2  Cos[((Subscript[θ, 1])/2)] Cos[((Subscript[θ, 2])/2)] Sin[((Subscript[θ, 1])/2)] 
          Sin[((Subscript[θ, 2])/2)] Cos[(Subscript[ϕ, 1]) - (Subscript[ϕ, 2])]]

Simplify[%]

This is nothing but \(TraditionalForm\`\(1\/2\) \((1 + \(n\& -> \)·\(n\& -> \)')\) = \(\(1\/2\) \((1 + cosη)\) = \(cos\^2\) η\/2\)\), where η is the angle between two vectors, as expected from the rotational invariance.

2. Sloppy Hydrogen Atom

According to the problem,

Energy  =  (1/(2  
                m)) ((((ℏ/d)))^2) - ((Z (e^2))/d)

Solve[D[Energy, d] == 0, d]

Simplify[Energy /. %[[1]]]

This actually agrees with the exact result.  (One should be cautioned, however, that the agreement with the exact result is a coincidence for this particular example.)