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).  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).

(c) 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.)

3. Classical Uncertainty Principle

(a) The Maxwell's equations in vacuum are given by 
	\(TraditionalForm\`\(∇\& -> \)\(\(·\)\(E\& -> \)\) = 0\)
\(TraditionalForm\`\(∇\& -> \)\(\(·\)\(B\& -> \)\) = 0\)
\(TraditionalForm\`\(∇\& -> \)\(\(×\)\(E\& -> \)\) + ∂\_t\( B\& -> \) = 0\)
\(TraditionalForm\`\(c\^2\) \(∇\& -> \)\(\(×\)\(B\& -> \)\) - ∂\_t\( E\& -> \) = 0\)
In this problem, there are only x and t dependence, and the only non-vanishing components are (Subscript[E, y])and (Subscript[B, z]).  Then the Maxwell's equations reduce to
	(Subscript[∇, x]) (Subscript[E, y]) + (∂_t) (Subscript[B, z]) = 0
-(c^2) (Subscript[∇, x]) (Subscript[B, z]) - (∂_t) (Subscript[E, y]) = 
        0
Putting them together, they reduce to a simple one-dimensional equation,
	\(TraditionalForm\`\(c\^2\) ∇\_x\^2 E\_y - ∂\_t\^2 E\_y = 0\).
Any function of the combination c t - x satisfies this equation, namely
	\(TraditionalForm\`\((\(c\^2\) ∇\_x\^2\(-∂\_t\^2\))\) \(f(c\ t - x)\) = 0\).
Because the form of (Subscript[E, y]) given in the problem is a function of c t  - x only, it solves the Maxwell's equations automatically.  The form can be sketched as

Plot[
      Sin[2  π ν (t - (x/c))] 
          (E^(-(((x  -  c t))^2)/2/(σ^2))) /. {t -> 
            0, ν -> 1, 
          c -> 1, σ -> 10}, {x, -30, 30}, 
      PlotRange -> {-1, 1}]

It oscillates just like the plane waves, but is localized.  The "uncertainty" is defined using the formula analogous to the quantum mechanical wave function.  First the "norm,"

Integrate[(((Sin[2  π ν (t - (x/c))] (E^(-(((x  -  c t))^2)/2/(σ^2)))))^2) /. {t -> 
            0}, {x, -∞, ∞}]

norm = 
      Simplify[PowerExpand[
          (1/2) (1 - (E^(-((4 (π^2) (ν^2) (σ^2))/(c^2))))) (Sqrt[π]) (Sqrt[(σ^2)])]]

We set the overall normalization (Subscript[E, 0]) = 1 throughout as it drops out after taking the norm correctly into account.

Next the expectation value

Integrate[(((Sin[2  π ν (t - (x/c))] 
                  (E^(-(((x  -  c t))^2)/2/(σ^2)))))^2) 
          x /. {t -> 0}, {x, -∞, ∞}]

OK, this vanishes.  Finally the variance,

Integrate[(((Sin[2  π ν (t - (x/c))] 
                  (E^(-(((x  -  c t))^2)/2/(σ^2)))))^2) 
          (x^2) /. {t -> 
            0}, {x, -∞, ∞}]

Simplify[
      PowerExpand[(((E^(-((4 (π^2) (ν^2) (σ^2))/(c^2)))) (Sqrt[π]) ((((σ^2)))^(3/2)) ((c^2) (-1 + (E^(((4 (π^2) (ν^2) (σ^2))/(c^2))))) + 
                    8 (π^2) (ν^2) (σ^2)))/(4 (c^2)))/
          norm]]

One can write it as \(TraditionalForm\`\((Δ\ x)\)\^2 = \(1\/2\) \(σ\^2\) \(\((1 - e\^\(-γ\))\) + 2  γ\ e\^\(-γ\)\)\/\(1 - e\^\(-γ\)\)\), where γ = 
        4  (π^2) (ν^2) (σ^2)/(c^2).  It is especially simple when γ ≫ 1, when \(TraditionalForm\`\((Δ\ x)\)\^2 = \(1\/2\) σ\^2\).

(b) The Fourier transform to the frequency domain is given by

Integrate[
      Sin[2  π ν (t - 
                  (x/c))] (E^(-(((x  -  c t))^2)/
                  2/(σ^2))) 
          (E^(I 2  π f t )) /. {x -> 0}, {t, -∞, ∞}, 
      Assumptions -> {Re[((c^2)/(σ^2))] > 0}]

Simplify[
      PowerExpand[-((I (E^(-((4 (π^2) ((f^2) + (ν^2)) (σ^2))/(c^2)))) ((E^(((2 (π^2) (((f - ν))^2) (σ^2))/(c^2)))) - (E^(((2 (π^2) (((f + ν))^2) (σ^2))/(c^2))))) (Sqrt[(π/2)]))/(Sqrt[((c^2)/(σ^2))]))]]

Again starting with the norm(dropping the overall f-independent factors),

Integrate[((((E^(-((4 (π^2) ((f^2) + (ν^2)) (σ^2))/(c^2)))) ((E^(((2 (π^2) (((f - ν))^2) (σ^2))/(c^2)))) - (E^(((2 (π^2) (((f + ν))^2) (σ^2))/(c^2)))))))^2), {f, 0, ∞}]

norm = 
      Simplify[PowerExpand[(((E^(-((4 (π^2) (ν^2) (σ^2))/(c^2)))) (-1 + (E^(((4 (π^2) (ν^2) (σ^2))/(c^2))))))/(2 (Sqrt[π]) (Sqrt[((σ^2)/(c^2))])))]]

Next, the average frequency,

Integrate[((((E^(-((4 (π^2) ((f^2) + (ν^2)) (σ^2))/(c^2)))) ((E^(((2 (π^2) (((f - ν))^2) (σ^2))/(c^2)))) - (E^(((2 (π^2) (((f + ν))^2) (σ^2))/(c^2)))))))^2) f, {f, 
        0, ∞}]

Simplify[
      PowerExpand[(((ν^2) Erf[2 π (Sqrt[(((ν^2) (σ^2))/(c^2))])])/(2 (Sqrt[π]) (Sqrt[(((ν^2) (σ^2))/(c^2))])))]/norm]

One can write it as \(TraditionalForm\`ν\ Erf[γ\^\(1/2\)]\/\(1 - E\^\(-γ\)\)\), where γ = 
        4  (π^2) (ν^2) (σ^2)/(c^2).  It is especially simple when γ ≫ 1, when it reduces to nothing but ν.  Finally the dispersion in the frequency is

Integrate[((((E^(-((4 (π^2) ((f^2) + (ν^2)) (σ^2))/(c^2)))) ((E^(((2 (π^2) (((f - ν))^2) (σ^2))/(c^2)))) - (E^(((2 (π^2) (((f + ν))^2) (σ^2))/(c^2)))))))^2) (f^2), {f, 
        0, ∞}]

Simplify[
      PowerExpand[(((E^(-((4 (π^2) (ν^2) (σ^2))/(c^2)))) (-(c^2) + 
                      (c^2) (E^(((4 (π^2) (ν^2) (σ^2))/(c^2)))) + 
                      8 (E^(((4 (π^2) (ν^2) (σ^2))/(c^2)))) (π^2) (ν^2) (σ^2)))/(16 (π^(5/2)) (σ^2) (Sqrt[((σ^2)/(c^2))])))/
            norm - (((((ν Erf[((2 π ν σ)/c)])/(1 - (E^(-((4 (π^2) (ν^2) (σ^2))/(c^2))))))))^2)]]

Namely, \(TraditionalForm\`\((Δ\ f)\)\^2 = \(\(ν\^2\)(\((1 - e\^\(-γ\))\) \((\((1 - e\^\(-γ\))\) + 2  γ)\) - 2  γ\ Erf[γ\^\(1/2\)])\)\/\(2  γ\ \((1 - e\^\(-γ\))\)\^2\)\) which simplifies to (((Δ f))^2) = (ν^2) 
            (1/(2  γ )) = 
          ((c^2)/(8  (π^2) (δ^2)))when γ ≫ 1.  Therefore, \(TraditionalForm\`\(\((Δ\ x)\)\^2\) \((Δ\ f)\)\^2 = c\^2\/\(16  π\^2\)\).

Once interpreted as a photon, \(TraditionalForm\`\((Δ\ f)\)\^2 = \(\(c\^2\)(Δ\ p)\)\^2/h\^2\), and hence \(TraditionalForm\`\(\((Δ\ x)\)\^2\) \((Δ\ p)\)\^2 = ℏ\^2\/4\), as expected.