HW #4

1. 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.

To simplify the problem, define γ = 
        4  (π^2) (ν^2) (σ^2)/(c^2). Then the electric field is:

Ey[x_, t_] = 
      E0*Sin[((c (Sqrt[γ]))/σ) t - ((Sqrt[γ])/σ) x] (E^(-(x - c t)^2/(2  (σ^2))))

The form can be sketched as

Plot[
        Ey[x, t] /. {t -> 0, γ -> 400 (π^2), 
            E0 -> 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,"

norm = 
      Integrate[Ey[x, 0]^2, {x, -∞, ∞}, 
        Assumptions -> {γ > 0, σ > 0}]

We could set the overall normalization (Subscript[E, 0]) = 1 throughout since it will drop out after taking the norm correctly into account. But we can also leave it in as a check that everything is working correctly.

Next the expectation value

Integrate[x*Ey[x, 0]^2, {x, -∞, ∞}, 
      Assumptions -> {γ > 0, σ > 0}]

OK, this vanishes.  Finally the variance,

temp = 
      Integrate[x^2*Ey[x, 0]^2, {x, -∞, ∞}, 
        Assumptions -> {γ > 0, σ > 0}]

variance = temp/norm

Note that the E0 dependence did in fact drop out.

FullSimplify[variance]

One can write it as \(TraditionalForm\`\((Δ\ x)\)\^2 = \(1\/2\) \(\(σ\^2\)(
            1 + \(2  γ\)\/\(e\^γ - 1\))\)\), 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 as a function of the variable "f" is calculated below. We might as well pick a specific position like "x=0" to evaluate the Fourier transform.

ft[f_] = 
      Integrate[
        Ey[0, t]*
          (E^(I 2  π f t)), {t, -∞, ∞}, 
        Assumptions -> {γ > 0, σ > 0, c > 0,  
            f ∈ Reals}]

Again starting with the norm (noting that \(TraditionalForm\`\(ft[f]\^*\) 
          ft[f]\  = \ \(-\(\(ft[f]^2\)\(\ \)\)\)\)):

norm2 = 
      Integrate[-ft[f]^2, {f, 0, ∞}, 
        Assumptions -> {γ > 0, σ > 0, c > 0}]

Integrate[-ft[f]^2, {f, -∞, ∞}, 
      Assumptions -> {γ > 0, σ > 0, c > 0}]

Next, the average frequency,

exptf = 
      Integrate[f*-ft[f]^2, {f, 0, ∞}, 
          Assumptions -> {γ > 0, σ > 0, c > 0}]/
        norm2

exptf /. σ -> 
        c (Sqrt[ γ ])/(2  π ν)

Limit[Erf[(Sqrt[γ])], γ -> ∞]

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 variance in the frequency is (remember to normalize!)

exptf2 = 
      Integrate[(f^2)*-ft[f]^2, {f, 0, ∞}, 
          Assumptions -> {γ > 0, σ > 0, c > 0}]/
        norm2

FullSimplify[%]

% /. σ -> 
          c (Sqrt[ γ ])/(2  π ν) // 
      Simplify

Limit[%, γ -> ∞]

Then the square of the dispersion in the frequency is:

dispersion2 = exptf2 - exptf^2 // Simplify

dispersion2 /. σ -> 
          c (Sqrt[ γ ])/(2  π ν) // 
      Simplify

Limit[
      dispersion2, γ -> ∞]
% /. σ -> 
        c (Sqrt[ γ ])/(2  π ν)

Namely,

	\(\((Δ\ f)\)\^2 = \(\(ν\^2\)(\((1 - e\^\(-γ\))\) \((\((1 - e\^\(-γ\))\) + 2  γ)\) - 2  γ\ Erf[γ\^\(1/2\)]\^2)\)\/\(2  γ\ \((1 - e\^\(-γ\))\)\^2\)\)

which simplifies to\(\(
\)\(
\)\(	\)\(\((Δ\ f)\)\^2 = \(\(ν\^2\) 1\/\(\(2\)  \(γ\)\(\ \)\) = 
                c\^2\/\(8 \( π\^2\) σ\^2\)\)\)\)

whenγ ≫ 1.  Therefore,\(\(
\)\(
\)\(	\)\(\(\((Δ\ x)\)\^2\) \((Δ\ f)\)\^2 = 
              c\^2\/\(16  π\^2\)\)\)

Once interpreted as a photon,\(\(
\)\(
\)\(	\)\(\((Δ\ f)\)\^2 = \(\(c\^2\)(Δ\ p)\)\^2/h\^2\)\)

,and hence\(\(
\)\(	\)\(
\)\(	\)\(\(\((Δ\ x)\)\^2\) \((Δ\ p)\)\^2 = ℏ\^2\/4\)\)

as expected.

2. Gaussian Wave Packet

(a) The norm is

	\(⟨ψ | ψ⟩ = \(∫⟨ψ | 
                  x⟩ ⟨x | ψ⟩ dx = ∫\(\([ψ \((x)\)]\)\^*\) ψ \((x)\) dx\)\) 

\(\(\(.\)\(\ \ \)\(Specify\)\)\ unnormalized\ state\ on\ spatial\ \(\(basis\)\(:\)\)\)

\(\(ψun[
          x_] = \ \ E\^\(I\ p\ x\ /\ \)\ \ E\^\(\(-\((x - x0)\)\^2\)/\((4  d\^2)\)\);\)\)

We now integrate:

norm3 = 
      Integrate[
        ComplexExpand[
          Conjugate[ψun[x]] ψun[
              x]], {x, -∞, ∞}, 
        Assumptions -> {d > 0}]

Therefore N is

1/Sqrt[norm3]

Verify:

ψ[x_] = ψun[x]/Sqrt[norm3];

Integrate[
      ComplexExpand[
        Conjugate[ψ[x]] ψ[
            x]],  {x, -∞,  ∞}, 
      Assumptions -> {d > 0}]

(b) Now calculate

	⟨x⟩ = \(⟨ψ | 
            x | ψ⟩ = \(∫⟨ψ | x⟩ 
              x ⟨x | ψ⟩ dx = ∫\(\([ψ \((x)\)]\)\^*\) x\ ψ \((x)\) \(\(dx\)\(\ \)\)\)\)
	
:

Integrate[
      ComplexExpand[
        Conjugate[ψ[x]] 
          x ψ[x]],  {x, -∞,  ∞}, 
      Assumptions -> {d > 0}]

(c) Calculate 2nd moment

	⟨(x^2)⟩ = \(⟨ψ | 
            x\^2 | ψ⟩ = \(∫⟨ψ | 
                  x⟩ \(x\^2\) ⟨x | ψ⟩ dx = ∫\(\([ψ \((x)\)]\)\^*\) 
              x\^2\ ψ \((x)\) \(\(dx\)\(\ \)\)\)\)

:

Integrate[
      ComplexExpand[
        Conjugate[ψ[x]] 
          (x^2) ψ[x]],  {x, -∞,  ∞}, 
      Assumptions -> {d > 0}]

So our dispersion Δx=(Sqrt[⟨(x^2)⟩-⟨x⟩^2])is

dispx = 
      Simplify[Sqrt[% - (x0^2)], Assumptions -> {d > 0}]

(d) We utilize the unity operator, noting that the momentum eigenvalue p in ψ(x) is not the same as the momentum coordinate q in ϕ(q),

	ϕ(q) = ⟨q|ψ> = \(TraditionalForm\`∫⟨q | 
              x⟩ ⟨x | ψ⟩ dx\) = \(TraditionalForm\`∫\(e\^\(\(-i\)\ q\ x/\)\/\@\(2  π\)\) \(ψ(x)\) dx\)
:

ϕ[q_]=TrigToExp[Simplify[ComplexExpand[Integrate[((\(Exp[\(-I\)\ q\ x/]\))/(\(\@\(2  π\ \)\))) ψ[x],{x, -∞, ∞},\(Assumptions -> {d > 0,  > 0}\)]],\(Assumptions -> {d > 0,  > 0}\)]]

Check norm:

\(Integrate[
      ComplexExpand[
        Conjugate[ϕ[q]] ϕ[
            q]], {q, \(-∞\), ∞}, \ 
      Assumptions -> {d > 0,  > 0}]\)

(e) Similar to ⟨x⟩ :

\(Integrate[
      ComplexExpand[
        Conjugate[ϕ[q]] 
          q\ ϕ[q]], {q, \(-∞\), ∞}, \ 
      Assumptions -> {d > 0,  > 0}]\)

(f) Let us now calculate Δp similarly to Δx:

\(Integrate[
      ComplexExpand[
        Conjugate[ϕ[q]] 
          q\^2\ ϕ[q]], {q, \(-∞\), ∞}, \ 
      Assumptions -> {d > 0,  > 0}]\)

\(dispp = 
      Simplify[Sqrt[% - p\^2], 
        Assumptions -> {d > 0,  > 0}]\)

Finally, compute the combined uncertainty ΔxΔp :

dispx * dispp

We obtain the minimum combined uncertainty as expected!

3. Proving the Baker-Hausdorff Lemma

(a)  (d/(d t)) B(t)  =  (e^(t A)) 
          A B (e^(-t A))+ (e^(t A)) B (-A) (e^(-t A))

= (e^(t A)) [A,  
          B] (e^(-t A))

Noting that:

	(e^(t A)) [A,  
            B] (e^(-t A))   =  [A,  B(t)]

(b)  B(1)  =  
        B  +  (∫_0^1) (d/(d t)) B(
                t) dt
B (1)  =  
        B  +  (∫_0^1)[A,  B(t)] dt
(∑_(n = 0)^∞) (Subscript[B, n])  =  
        (Subscript[B, 0]) + (∑_(n = 0)^∞)[A, 
                (Subscript[B, n])] (∫_0^1) (t^n) dt
(∑_(n = 0)^∞) (Subscript[B, n])  =  
        (Subscript[B, 0]) + (∑_(n = 0)^∞)[A, (Subscript[B, n])] (1/(n + 1))
\(TraditionalForm\`∑\+\(n = 1\)\%∞ B\_n\  = \ ∑\+\(n = 0\)\%∞\([A, B\_n]\) 1\/\(n + 1\)\)
\(TraditionalForm\`∑\+\(n = 1\)\%∞ B\_n\  = \ ∑\+\(n = 1\)\%∞\(\([\)\(A, B\_\(n - 1\)\)\(]\)\) 1\/n\)

(Subscript[B, n])  =  (1/n)[A, (Subscript[B, n - 1])]

(c)  (Subscript[B, n])  =  (1/n)[A, 
                  (Subscript[B, n - 1])] = (1/n)[
                  A,  (1/(n - 1))[A,  (Subscript[B, n - 2])]] = (1/(n(n - 1)))[
              A,  [A,  
                (Subscript[B, n - 2])]]   =  ...

              = (1/(n(n - 1) ...  (n - n  +  1)))[
                  A,  [A,  ...  [A,  
                        (Subscript[B, n - n])] ... ]]

(Subscript[B, n])  =  (1/(n!))[
          A,  [A,  ...  [A,  
                B] ... ]]	(w/ n nested commutators)

(d)  (e^A) 
          B (e^(-A))  =  B(
            1)  =  (∑_(n = 
                  0)^∞)(Subscript[B, n])

\(TraditionalForm\`\(\(	\)\(\(e\^A\) 
        B\ e\^\(-A\)\  = ∑\+\(n = 0\)\%∞\ \(1\/\(n!\)\)[
          A, \ \([A, \ \(\(...\)\(\ \ \)\(\([\)\(A, \ 
              B\)\(]\)\)\(\ \)\(...\)\)\ ]\)]\)\)\)

(e)  \(TraditionalForm\`\(e\^\(i\ p\ a\ /\ \)\) 
          x\ e\^\(\(-i\)\ p\ a\ /\ \)\  = \ ∑\+\(n = 0\)\%∞\ \(1\/\(n!\)\)[
            A, \ \([A, \ \(\(...\)\(\ \ \)\([A, \ x]\)\(\ \)\(...\)\)\ ]\)]\)
\(TraditionalForm\`	\(\(=\)\(\ \)\(∑\+\(n = 0\)\%∞\ \(a\^n\/\(n!\)\) \(\((\(\(i\)\(\ \)\)\/)\)\^n\)\ [
              p, \ \([p, \ \(\(...\)\(\ \ \)\([p, \ 
                    x]\)\(\ \)\(...\)\)\ ]\)]\)\)\)

Only the first and second terms survive:

\(TraditionalForm\`\(\(		\)\(\(=\)\(\ \ \)\(x\  + \ 
          a\ \((i\/)\)\ \((\(-i\)\ )\)\)\)\)\)

\(TraditionalForm\`\(\(	\)\(\(e\^\(i\ p\ a\ /\ \)\) 
        x\ e\^\(\(-i\)\ p\ a\ /\ \)\  = \ x\  + \ a\)\)\)