HW #5

1. Harmonic Oscillator

(a)

We rewrite the Hamiltonian H = 
        ((p^2)/(2  m)) + (1/2) m (ω^2) (x^2)using a = (Sqrt[((m ω)/(2 ℏ))]) (x + 
              i (p/(m ω))), \(TraditionalForm\`a\^† = \(\@\(\(m\ ω\)\/\(2\ ℏ\)\)\) \((x - i\ p\/\(m\ ω\))\)\).  We first calculate
	\(TraditionalForm\`\(a\^†\) 
          a = \(\(\(m\ ω\)\/\(2\ ℏ\)\) \((x - 
                i\ p\/\(m\ ω\))\) \((x + 
                i\ p\/\(m\ ω\))\) = \(\(\(m\ ω\)\/\(2\ ℏ\)\) \((x\^2 - \(i\/\(m\ ω\)\)[p, x] + 
                  p\^2\/\(\(m\^2\) ω\^2\))\) = \(\(m\ ω\)\/\(2\ ℏ\)\) \((x\^2 - ℏ\/\(m\ ω\) + 
                  p\^2\/\(\(m\^2\) ω\^2\))\)\)\)\).  
Therefore, 
	ℏ ω (a^†) 
          a = (1/2) m (ω^2) (x^2) - (1/2) ℏ ω + 
          ((p^2)/(2  m)),

and hence H = ℏ ω((a^†) a + 
              (1/2)).

(b)

The ground state condition a | 0⟩ = 0 can be written in the position representation as
	\(TraditionalForm\`⟨x | a | 
            0⟩ = \(⟨x | \(\@\(\(m\ ω\)\/\(2\ ℏ\)\)\) \((x + i\ p\/\(m\ ω\))\) | 
              0⟩ = \(\(\@\(\(m\ ω\)\/\(2\ ℏ\)\)\) \((x + i\ \(1\/\(m\ ω\)\) \(ℏ\/i\) 
                    d\/\(d\ x\))\) ⟨x | 
                  0⟩ = 0\)\)\),
and hence
	(x +  (ℏ/(m ω)) 
                (d/(d x))) (Subscript[ψ, 0])(x) = 0.
This equation can be solved easily and we find
	(Subscript[ψ, 0])(x) = 
        N (e^(-m ω (x^2)/2 ℏ)).
To normalize the wave function, we compute
	\(TraditionalForm\`∫\_\(-∞\)\%∞\(\((e\^\(\(-m\)\ ω\ x\^2/2  ℏ\))\)\^2\) 
            d\ x = \@\(\(π\ ℏ\)\/\(m\ ω\)\)\).
Therefore, the correctly normalized ground state wave function is
	(Subscript[ψ, 0])(
          x) = (((((m ω)/(π ℏ))))^(1/4)) 
          (e^(-m ω (x^2)/2 ℏ)).
The shape of the wave function is

(Subscript[ψ, 0])[x_] := (((((m ω)/(π ℏ))))^(1/4)) 
        (E^(-m ω (x^2)/2 ℏ))

Plot[(Subscript[ψ, 0])[
            x] /. {m -> 1, ω -> 1, ℏ -> 
              1}, {x, -3, 3}];

(c)

The first excited state is given by |1⟩ = 
          (a^†) | 0⟩, and its position representation by
	\(TraditionalForm\`⟨x | 
            1⟩ = \(⟨x | a\^† | 
              0⟩ = \(\@\(\(m\ ω\)\/\(2\ ℏ\)\)\) \((x - i\ \(1\/\(m\ ω\)\) \(ℏ\/i\) 
                  d\/\(d\ x\))\) ⟨x | 
                0⟩\)\)
\(TraditionalForm\`\(\(=\)\(\(\@\(\(m\ ω\)\/\(2\ ℏ\)\)\) \((x + x)\) \(\((\(m\ ω\)\/\(π\ ℏ\))\)\^\(1/4\)\) 
            e\^\(\(-m\)\ ω\ x\^2/\((2  ℏ)\)\) = \(\((\(m\ ω\)\/\(π\ ℏ\))\)\^\(1/
                  4\)\) \(\@\(\(m\ ω\)\/\(2\ ℏ\)\)\) 2  
            x\ e\^\(\(-m\)\ ω\ x\^2/\((2  ℏ)\)\)\)\)\)
Its shape is

(Subscript[ψ, 1])[
        x_] := (((((m ω)/(π ℏ))))^(1/
              4)) (Sqrt[((m ω)/(2 ℏ))]) 2  
        x (E^(-m ω (x^2)/(2  ℏ)))

Plot[(Subscript[ψ, 1])[
            x] /. {m -> 1, ω -> 1, ℏ -> 
              1}, {x, -3, 3}];

Check that it is properly normalized:

Integrate[(Subscript[ψ, 1])[x]^2, {x, -∞, ∞}, 
      Assumptions -> Re[((m ω)/ℏ)] > 0]

The second excited state is given by (Sqrt[2]) | 2⟩ = 
          (a^†) | 1⟩, and its position representation by
	\(TraditionalForm\`⟨x | 
            2⟩ = \(\(1\/\@2\) ⟨x | 
                a\^† | 
                0⟩ = \(1\/\@2\) \(\@\(\(m\ ω\)\/\(2\ ℏ\)\)\) \((x - 
                i\ \(1\/\(m\ ω\)\) \(ℏ\/i\) 
                  d\/\(d\ x\))\) ⟨x | 
                1⟩\)\)
Its shape is

Simplify[(1/(Sqrt[2])) (Sqrt[((m ω)/(2 ℏ))]) (x (Subscript[ψ, 1])[x] - (ℏ/(m ω)) D[(Subscript[ψ, 1])[x], x])]

(Subscript[ψ, 2])[
        x_] := (((E^(-((m (x^2) ω)/(2 ℏ)))) (2 m (x^2) ω - ℏ) (((((m ω)/ℏ)))^(1/4)))/((Sqrt[2]) (π^(1/4)) ℏ))

Plot[(Subscript[ψ, 2])[
            x] /. {m -> 1, ω -> 1, ℏ -> 
              1}, {x, -3, 3}];

Check that it is properly normalized:

Integrate[(Subscript[ψ, 2])[x]^2, {x, -∞, ∞}, 
      Assumptions -> Re[((m ω)/ℏ)] > 0]

(d)

From the definitions of the annihilation and creation operators, we can solve for x,
	x = (Sqrt[(ℏ/(2  m ω))]) (a + 
                (a^†)).
Starting with the expectation values,
	⟨x⟩=\(⟨n | x | 
                    n⟩ = \(\(\@\(ℏ\/\(2  
                            m\ ω\)\)\) ⟨n | 
                      a + a\^† | 
                      n⟩

                \(\(=\)\(\(\@\(ℏ\/\(2  
                              m\ ω\)\)\) ⟨n | \((\@n | n - 1⟩\) + \@\(n + 1\) | 
                        n + 1⟩\)\)\)\)).
Because of the orthonormality of the Hamiltonian eigenstates ⟨n | 
            m⟩ = (Subscript[δ, n,  m]),

⟨x⟩  =  
        0.

Moving on to the variance,
 	\(TraditionalForm\`⟨x\^2⟩ = \(\(ℏ\/\(2  
                  m\ ω\)\) ⟨n | \((a + a\^†)\)\^2 | n⟩ = \(ℏ\/\(2  
                  m\ ω\)\) ⟨n | \(a\^†\) a + 
                  a\ a\^† | 
                n⟩\), 
\(\(=\)\(\(ℏ\/\(2  
                  m\ ω\)\) ⟨n | 
                2  N + \([a, a\^†]\) | 
                n⟩ = \(ℏ\/\(2  
                  m\ ω\)\) \(\((2  n + 1)\)\(.\)\)\)\)\)

\(TraditionalForm\`⟨\((Δ\ x)\)\^2⟩\  = \ \(ℏ\/\(2  m\ ω\)\) \((2  n + 
              1)\)\).

From the definitions of the annihilation and creation operators, we can solve for p, 
	\(TraditionalForm\`p = \(-i\) \(\@\(\(ℏ\ m\ ω\)\/2\)\) \(\((a - a\^†)\)\(.\)\)\)
Together with the expression for x fromthe previous problem, you can easily verify [x, p] = i ℏ.  Starting with the expectation values,
	\(TraditionalForm\`⟨p⟩ = \(⟨n | p | 
              n⟩ = \(-i\) \(\@\(\(ℏ\ m\ ω\)\/2\)\) ⟨n | a - a\^† | 
                n⟩\)\),

⟨p⟩  =  
        0 .

Moving on to the variance,
 	\(TraditionalForm\`⟨p\^2⟩ = \(\(-\(\(ℏ\ m\ ω\)\/2\)\) ⟨n | \((a - a\^†)\)\^2 | 
                n⟩ = \(\(ℏ\ m\ ω\)\/2\) ⟨n | \(a\^†\) a + a\ a\^† | 
                n⟩\), 
\(\(=\)\(\(\(ℏ\ m\ ω\)\/2\) ⟨n | 
                2  N + \([a, a\^†]\) | 
                n⟩ = \(\(ℏ\ m\ ω\)\/2\) \(\((2  n + 1)\)\(.\)\)\)\)\)

\(TraditionalForm\`⟨\((Δ\ p)\)\^2⟩\  = \ \(\(ℏ\ m\ ω\)\/2\) \((2  n + 
              1)\)\) .

Therefore,
 	\(TraditionalForm\`\((Δ\ x)\) \((Δ\ p)\) = \(ℏ\/2\) \((2  n + 1)\)\).
 The ground state n = 0 is a minimum uncertainty state, while the excited states have larger uncertainties.

(e)

There are many ways to show this.  

First of all, using the relation (a^†) | 
              n⟩ = (Sqrt[n + 1]) | 
            n + 1⟩, we can show that 
	((((a^†)))^n) | 
              0⟩ = (Sqrt[n!]) | 
            n⟩ 
by recursion.  It obviously holds for n = 1.  If it holds for n, the n + 1-th one is
	\(TraditionalForm\`\(\(\(\(\(\(\(\(\(\(\((a\^†)\)\^\(n + 1\) | 
                        0\)\(⟩\)\) = \(\(a\^†\)(a\^†)\)\^n | 
                      0\)\(⟩\)\) = \(a\^†\) \@\(n!\) | n\)\(⟩\)\) = \(\@\(n!\)\) \@\(n + 1\) | 
              n + 1\)\(⟩\)\)\ 

        	\(\(=\)\(\@\(\((n + 1)\)!\) | 
            n + 1\)\)\)\(⟩\)\)\)
, and it holds again.  Therefore, it holds for all n.  

Using the definition 
	\(TraditionalForm\`\(\(\(\(\(|\)\(f\)\)\(⟩\)\) = \(\(\(e\^\(f\ a\^†\) | 
                    0\)\(⟩\)\)\ e\^\(-\(\(|\)\(f\)\( | \^2\)\(\(/\)\(2\)\)\)\) = ∑\+\(n = 0\)\%∞\( f\^n\/\(n!\)\) \((a\^†)\)\^n | 
                0\)\)\(⟩\)\)\ e\^\(-\(\(|\)\(f\)\( | \^2\)\(\(/\)\(2\)\)\)\)\), 
we can write it as 
	\(TraditionalForm\`\(\(\(\(\(|\)\(f\)\)\(⟩\)\) = ∑\+\(n = 0\)\%∞ f\^n\/\@\(n!\) | n\)\(⟩\)\) 
        e\^\(-\(\(|\)\(f\)\( | \^2\)\(\(/\)\(2\)\)\)\)\).  
Then
	 \(TraditionalForm\`\(\(\(\(\(\(a | f\)\(⟩\)\) = 
                  a \(∑\+\(n = 0\)\%∞ f\^n\/\@\(n!\)\) | 
                    n\)\(⟩\)\) 
              e\^\(-\(\(|\)\(f\)\( | \^2\)\(\(/\)\(2\)\)\)\) = ∑\+\(n = 1\)\%∞\( f\^n\/\@\(n!\)\) \@n | n - 1\)\(⟩\)\) 
        e\^\(-\(\(|\)\(f\)\( | \^2\)\(\(/\)\(2\)\)\)\)\), 
where the summation is now taken only from i = 1 because i = 0 term vanishes by a | 0⟩ = 0.  Continuing on, 
	\(TraditionalForm\`\(\(\(\(\(\(a | 
                      f\)\(⟩\)\) = ∑\+\(n = 1\)\%∞ f\^n\/\@\(\((n - 1)\)!\) | n - 1\)\(⟩\)\) 
              e\^\(-\(\(|\)\(f\)\( | \^2\)\(\(/\)\(2\)\)\)\) = ∑\+\(m = 0\)\%∞ f\^\(m + 1\)\/\@\(m!\) | m\)\(⟩\)\) 
        e\^\(-\(\(|\)\(f\)\( | \^2\)\(\(/\)\(2\)\)\)\)\),
where the dummy variable was chgned to m = n - 1.  Pulling one factor of f out of the sum,

\(TraditionalForm\`\(\(\(\(\(\(a | f\)\(⟩\)\) = 
                f \(∑\+\(m = 0\)\%∞ f\^m\/\@\(m!\)\) | 
                  m\)\(⟩\)\) 
            e\^\(-\(\(|\)\(f\)\( | \^2\)\(\(/\)\(2\)\)\)\) = f | f\)\(⟩\)\)\).

Another way to show the same result is by first showing the relation
	\(TraditionalForm\`\([a, \((a\^†)\)\^n]\) = \(n(a\^†)\)\^\(n - 1\)\).
It obviously holds for n = 1.  If it holds for n, the n + 1-th one is
	\(TraditionalForm\`\([a, \((a\^†)\)\^\(n + 1\)]\) = \(\([a, \(\(a\^†\)(a\^†)\)\^n]\) = \([a, 
                a\^†]\) \((a\^†)\)\^n + \(a\^†\)[
              a, \((a\^†)\)\^n]\)\)
\(TraditionalForm\`\(\(=\)\(\((a\^†)\)\^n + \(a\^†\) \(n(a\^†)\)\^n = \((n + 1)\) \((a\^†)\)\^\(n + 1\)\)\)\)
and hence it holds as well.  Therefore, it holds for any n.

Starting with the definition 
	\(TraditionalForm\`\(\(\(\(\(\(\(|\)\(f\)\)\(⟩\)\) = 
                e\^\(f\ a\^†\) | 
                  0\)\(⟩\)\)\ e\^\(-\(\(|\)\(f\)\( | \^2\)\(\(/\)\(2\)\)\)\) = ∑\+\(n = 0\)\%∞\( f\^n\/\(n!\)\) \((a\^†)\)\^n | 
            0\)\(⟩\)\)\)\(TraditionalForm\`e\^\(-\(\(|\)\(f\)\( | \^2\)\(\(/\)\(2\)\)\)\)\), 
	\(TraditionalForm\`\(\(\(\(\(\(a | f\)\(⟩\)\) = 
                  a \(∑\+\(n = 0\)\%∞\( f\^n\/\(n!\)\) \((a\^†)\)\^n\) | 0\)\(⟩\)\) 
              e\^\(-\(\(|\)\(f\)\( | \^2\)\(\(/\)\(2\)\)\)\) = ∑\+\(n = 0\)\%∞\( f\^n\/\(n!\)\)[
                  a, \((a\^†)\)\^n] | 0\)\(⟩\)\) 
        e\^\(-\(\(|\)\(f\)\( | \^2\)\(\(/\)\(2\)\)\)\)\), 
Here, we used the fact a | 0⟩ = 0.  Using the relation shown above,
	\(TraditionalForm\`\(\(\(\(\(\(a | 
                      f\)\(⟩\)\) = ∑\+\(n = 1\)\%∞\( f\^n\/\(n!\)\) n\ \((a\^†)\)\^\(n - 1\) | 
                    0\)\(⟩\)\) 
              e\^\(-\(\(|\)\(f\)\( | \^2\)\(\(/\)\(2\)\)\)\) = ∑\+\(n = 1\)\%∞ f\^n\/\(\((n - 1)\)!\)\ \((a\^†)\)\^\(n - 1\) | 0\)\(⟩\)\) 
        e\^\(-\(\(|\)\(f\)\( | \^2\)\(\(/\)\(2\)\)\)\)\),
where the summation is now taken only from i = 1 because i = 0 term vanishes by [a, 1] = 0.  Changing the dummy variable to m = n - 1, 
	\(TraditionalForm\`\(\(\(\(a | 
                f\)\(⟩\)\) = ∑\+\(m = 0\)\%∞\( f\^\(m + 1\)\/\(m!\)\) \((a\^†)\)\^m | 
              0\)\(⟩\)\) 
        e\^\(-\(\(|\)\(f\)\( | \^2\)\(\(/\)\(2\)\)\)\)\),

\(TraditionalForm\`\(\(\(\(\(\(a | f\)\(⟩\)\) = 
                f \(∑\+\(m = 0\)\%∞ f\^m\/\(m!\)\ \((a\^†)\)\^m\) | 0\)\(⟩\)\) 
            e\^\(-\(\(|\)\(f\)\( | \^2\)\(\(/\)\(2\)\)\)\) = f | f\)\(⟩\)\)\).

Finally, we verify the normalization,
	\(TraditionalForm\`⟨f | 
            f⟩\  = \ \(e\^\(-\(\(|\)\(f\)\( | \^2\)\)\)\) \(∑\+\(m = 0\)\%∞\(∑\+\(n = 0\)\%∞\(\((\(f\^*\))\)\^m\/\(m!\)\) \(f\^n\/\(n!\)\) ⟨0 | \(\(a\^m\)(a\^†)\)\^n | 0⟩\)\)\).
Consider a acting on the left and (a^†)acting on the right, again by orthonormality we must have m = n, so
	\(TraditionalForm\`⟨f | 
            f⟩\  = \ \(e\^\(-\(\(|\)\(f\)\( | \^2\)\)\)\) \(∑\+\(n = 0\)\%∞\(\(\(|\)\(f\)\( | \^2\)\)\/\(\(n!\) \(n!\)\)\) ⟨n | \(\@\(n!\)\) \@\(n!\) | n⟩\)\)
\(TraditionalForm\`	\(\(=\)\(\(e\^\(-\(\(|\)\(f\)\( | \^2\)\)\)\) \(∑\+\(n = 0\)\%∞\(\(\(|\)\(f\)\( | \^2\)\)\/\(n!\)\) ⟨n | n⟩\) = \ 
          e\^\(-\(\(|\)\(f\)\( | \^2\)\)\)\ e\^\(\(|\)\(f\)\( | \^2\)\)\)\)\)

⟨f | 
            f⟩  =  1.

(f)

Using the result from (d), 
	\(TraditionalForm\`⟨f | x | 
            f⟩ = \(⟨f | \(\@\(ℏ\/\(2  m\ ω\)\)\) \((a + a\^†)\) | 
              f⟩ = \(\(\@\(ℏ\/\(2  
                        m\ ω\)\)\) \((f + \(f\^*\))\) = \(\@\(ℏ\/\(2  m\ ω\)\)\) 2 \( Re(f)\)\)\)\).  
Here, we used the fact \(TraditionalForm\`⟨f | 
            a\^† = ⟨f | \(f\^*\)\), obtained by taking the hermitian conjugate of a | f⟩ = 
          f | f⟩.  Similarly, using the result from (e),
	\(TraditionalForm\`⟨f | p | 
            f⟩ = \(⟨f | \(-i\) \(\@\(\(ℏ\ m\ ω\)\/2\)\) \((a - a\^†)\) | 
              f⟩ = \(\(-i\) \(\@\(\(ℏ\ m\ ω\)\/2\)\) \((f - \(f\^*\))\) = \(\@\(\(ℏ\ m\ ω\)\/2\)\) 
              2\ \(Im(f)\)\)\)\).
Now on the variance,
	⟨f | (x^2) | 
            f⟩ = ⟨f | (ℏ/(2  
                      m ω)) (((a + (a^†)))^2) | 
              f⟩ = (ℏ/(2  
                  m ω)) ⟨f | 
                (a^2) + a (a^†) + (a^†) 
                    a + ((((a^†)))^2) | f⟩
\(TraditionalForm\`\(\(=\)\(\(ℏ\/\(2  
                  m\ ω\)\) ⟨f | 
                a\^2 + \([a, \ a\^†]\) + 
                  2 \( a\^†\) a + \((a\^†)\)\^2 | 
                f⟩ = \(ℏ\/\(2  
                  m\ ω\)\) \((f\^2 + 1 + 
                2 \( f\^*\) f + \((\(f\^*\))\)\^2)\)\)\)\)
\(TraditionalForm\`\(\(=\)\(\(ℏ\/\(2  
                m\ ω\)\) \((\((f + \(f\^*\))\)\^2 + 1)\)\)\)\)
and hence

\(TraditionalForm\`⟨\((Δ\ x)\)\^2⟩ = \(\(ℏ\/\(2  m\ ω\)\)\(.\)\)\)

Similarly,
	\(TraditionalForm\`⟨f | p\^2 | 
            f⟩ = \(⟨f | \(-\(\(ℏ\ m\ ω\)\/2\)\) \((a - a\^†)\)\^2 | 
              f⟩ = \(\(ℏ\ m\ ω\)\/2\) ⟨f | \(-a\^2\) + 
                  a\ a\^† + \(a\^†\) 
                    a - \((a\^†)\)\^2 | f⟩\)\)
\(TraditionalForm\`\(\(=\)\(\(\(ℏ\ m\ ω\)\/2\) ⟨f | \(-a\^2\) + \([a, \ a\^†]\) + 
                  2 \( a\^†\) a - \((a\^†)\)\^2 | 
                f⟩ = \(\(ℏ\ m\ ω\)\/2\) \((\(-f\^2\) + 1 + 2 \( f\^*\) f - \((\(f\^*\))\)\^2)\)\)\)\)
\(TraditionalForm\`\(\(=\)\(\(\(ℏ\ m\ ω\)\/2\) \((\(-\((f - \(f\^*\))\)\^2\) + 1)\)\)\)\)
and hence

\(TraditionalForm\`⟨\((Δ\ p)\)\^2⟩ = \(\(\(ℏ\ m\ ω\)\/2\)\(.\)\)\)

Therefore,
	\(TraditionalForm\`\((Δ\ x)\) \((Δ\ p)\) = ℏ\/2\)
and hence the coherent state is a minimum uncertainty state for any f.

(g) [optional]

We take the solution from Eq. 2

\(\(ψf[x_] = \(\((\(m\ ω\)\/\(π\ \))\)\^\(1/4\)\) 
          Exp[\(-\((\(\@\(\(m\ ω\)\/\(2  \)\)\) x - f)\)\^2\) + \(1\/2\) \((f\^2 - Abs[f]\^2)\)];\)\)

and apply the operator a as in part (b)

\(\(resa = \(\@\(\(m\ ω\)\/\(2  \)\)\) \((x\ ψf[x] + 
              i\ \(1\/\(m\ ω\)\) \(\/i\) ∂\_x\ ψf[
                    x])\);\)\)

.  The desired result is

resd = f ψf[x];

.  We see that they are indeed the same

Simplify[resa - resd]

(h)

The Heisenberg equation of motion is
	\(TraditionalForm\`i\ ℏ\ \(d\/\(d\ t\)\) 
          x = \(\([x, H]\) = \(\([x, p\^2\/\(2  m\)]\) = 
            i\ ℏ\ p\/m\)\), 
i\ ℏ\ \(d\/\(d\ t\)\) p = \(\([p, 
            H]\) = \(\([p, \(1\/2\) 
                m\ ω\^2\ x\^2]\) = \(-i\)\ ℏ\ m\ ω\^2\ \(\(x\)\(.\)\)\)\)\)
There are many ways to solve these coupled equations.  One way is to use the exponential of a matrix.  Write the equations in the matrix form,
	(d/(d t))(({{m ω x}, {p}}))=(({{ω p}, {-m (ω^2) x}}))=ω(({{0, 1}, {-1, 0}}))(({{m ω x}, {p}})).
Therefore,
	(({{m ω x}, {p}}))(t)=exp((({{0, -i}, {i, 0}}))i ω t)(({{m ω x}, {p}}))(0)=exp((Subscript[σ, 2]) i ω t)(({{m ω x}, {p}}))(0).
Here, (Subscript[σ, 2]) is one of the Pauli matrices.  The exponential factor can be worked out using its Taylor expansion,
	exp((Subscript[σ, 2]) 
            i ω t) = (∑_(n = 0)^∞) 
                (1/(n!)) ((((Subscript[σ, 2])^n)(i ω t))^n).
It is easy to check that ((Subscript[σ, 2])^2) = 1, and hence ((Subscript[σ, 2])^even) = 1, ((Subscript[σ, 2])^odd) = (Subscript[σ, 2]).  Therefore,
	\(exp(\(σ\_2\) 
              i\ ω\ t) = ∑\+\(n = 0, \ even\)\%∞\( 
                    1\/\(n!\)\) 
                1 \((i\ ω\ t)\)\^n + ∑\+\(n = 0, \ odd\)\%∞\( 1\/\(n!\)\) \(\(σ\_2\)(i\ ω\ t)\)\^n\)
=1 cos ω t + 
              i (Subscript[σ, 2]) sin ω t=(({{cos ω t, sin ω t}, {-sin ω t, cos ω t}})).
We find the solution
	(({{m ω x}, {p}}))(t)=(({{cos ω t, sin ω t}, {-sin ω t, cos ω t}}))(({{m ω x}, {p}}))(0)
=(({{m ω x(0) cos ω t + p(0) 
                        sin ω t}, {-m ω x(0) sin ω t + p(0) 
                        cos ω t}})).	
Using this solution, we calculate the expectation values,
	\(TraditionalForm\`⟨x⟩ \((t)\) = \(⟨f | x(t) | 
              f⟩ = ⟨f | \(x(0)\) 
                  cos\ ω\ t + \(\(p(0)\)\/\(m\ ω\)\) 
                  sin\ ω\ t | f⟩\)\)
\(TraditionalForm\`\(\(=\)\(\(\@\(ℏ\/\(2  m\ ω\)\)\) 
            2 \( Re(f)\) 
            cos\ ω\ t + \(1\/\(m\ ω\)\) \(\@\(\(ℏ\ m\ ω\)\/2\)\) 2\ \(Im(f)\) sin\ ω\ t\)\)\)

\(TraditionalForm\`⟨x⟩ \((t)\)\  = \(\@\(ℏ\/\(2  m\ ω\)\)\) \(\(2\)\([\)\((\(Re(f)\) 
                  cos\ ω\ t + \(Im(f)\) sin\ ω\ t]\)\)\).

\(TraditionalForm\`⟨p⟩ \((t)\) = \(⟨f | p(t) | 
              f⟩ = ⟨f | \(-m\)\ ω\ \(x(0)\) sin\ ω\ t + \(p(0)\) cos\ ω\ t | 
              f⟩\)\)
= -m ω (Sqrt[ (ℏ/(2  
                      m ω))]) 2  Re(f) 
            sinω t + (Sqrt[((ℏ m ω)/2)]) 
            2 Im(f) cos ω t

\(TraditionalForm\`⟨p⟩ \((t)\) = \ \(\@\(\(ℏ\ m\ ω\)\/2\)\) 
          2[\(-\(Re(f)\)\) sinω\ t + \ \(Im(f)\) 
                cos\ ω\ t]\).

Indeed, these solutions correspond to the classical ones.

(i) [optional]

The Schrödinger equation gives 	
	|n, 
                    t⟩ = 
                  (e^(-i H t/ℏ)) | 
                    n⟩ = 
              (e^(-i ℏ ω(n + 1/2) t/ℏ)) | 
                n⟩ = 
          (e^(-i ω(n + 1/2) t)) | 
            n⟩.  
Using what we showed above, 
	\(TraditionalForm\`\(\(\(\(\(|\)\(f\)\)\(⟩\)\) = ∑\+\(n = 0\)\%∞ f\^n\/\@\(n!\) | 
            n\)\(⟩\)\)\), 
its time evolution is
	\(TraditionalForm\`\(\(\(\(\(\(\(\(\(\(\(\(|\)\(f\)\), 
                            t\)\(⟩\)\) = \(e\^\(\(-i\)\ H\ t/ℏ\)\) \(∑\+\(n = 0\)\%∞ f\^n\/\@\(n!\)\) | 
                            n\)\(⟩\)\) = ∑\+\(n = 0\)\%∞ f\^n\/\@\(n!\) | n\)\(⟩\)\) 
                  e\^\(\(-i\)\ \(ω(n + 1/2)\) t\)

                \(\(=\)\(\ \)\(∑\+\(n = 0\)\%∞\((f\ e\^\(\(-i\)\ ω\ t\))\)\^n\/\@\(n!\) | 
                    n\)\)\)\(⟩\)\) 
              e\^\(\(-i\)\ ω\ t/2\) = \(\(|\)\(f\ e\^\(\(-i\)\ ω\ t\)\)\)\)\(⟩\)\) 
        e\^\(\(-i\)\ ω\ t/2\)\)

\(TraditionalForm\`\(\(\(\(\(\(|\)\(f\)\), 
              t\)\(⟩\)\)\  = \(\(|\)\(f\ e\^\(\(-i\)\ ω\ t\)\)\)\)\(⟩\)\) 
        e\^\(\(-i\)\ ω\ t/2\)\).

where the coherent state in the last expression has the eigenvalue a | 
              f (e^(-i ω t))⟩ = 
          f (e^(-i ω t)) | 
            f (e^(-i ω t))⟩.

(j)

The probability density on x is \(TraditionalForm\`\(\(|\)\(⟨x | 
            f\ e\^\(\(-i\)\ ω\ t\)⟩\)\( | \^2\)\)\), i.e.

\(\(ρf[x_, t_] = 
        Conjugate[ψf[x]] ψf[
              x] /. {f -> \(\@\(\(m\ ω\)\/\(2  \)\)\) 
                x0\ Exp[\(-I\)\ ω\ t]};\)\)

.  For the convenience in plotting, we set all the constants to 1

\(\(\(\ \)\(ρfc[x_, t_] = 
      Simplify[ComplexExpand[ρf[x, t]] /. {m -> 1,  -> 
              1, \ ω -> 1, x0 -> 1}]\)\)\)

.  Now we animate over time interval t = (0,  2  π) with gaps Δ t  =  π/4:

<< Graphics`Animation`;

MoviePlot[ρfc[x, t], {x, -4,  4}, {t, 0, 
          2  π,  π/4},  AspectRatio ->  .5,  
        PlotRange -> {{-4, 4}, {0, 0.6}}];

Perfectly oscillatory Gaussian!  Here \(TraditionalForm\`m\  ∼ \ \); however, as one would expect by the Correspondence Principle, as \(TraditionalForm\`m\  ≫ \ \), the Gaussian approaches a delta distribution.

2. Time Evolution of Gaussian Wave Packet 	[optional]

(a)

The result for \(TraditionalForm\`⟨q | ψ⟩ = ϕ(q)\) from Hwk 4 is

\(\(ϕ[q_]\  = \ \(\((\(2\ d\^2\)\/\(π\ \^2\))\)\^\(1/4\)\) 
          Exp[\(I\ \((p - q)\)\ x0\)\/] 
          Exp[\(\(-\((p - q)\)\^2\) d\^2\)\/\^2];\)\)

.  It is given that
	\(TraditionalForm\`\(\(\(\(|\)\(ψ\)\), \(\(t\)\(>\)\)\  = \ ∫e\^\(\(-i\)\ \(q\^2\) t/2  m\ \) | 
              q\)\(⟩\)\) ⟨q | ψ⟩ dq\)
so
	\(TraditionalForm\`ψ(x, 
          t)\  = \ \(⟨x | ψ, 
            t⟩\  = \ ∫\(e\^\(\(-i\)\ \(q\^2\) 
                    t/2  m\ \)\) ⟨x | 
                  q⟩ ⟨q | ψ⟩ dq\)\)
\(TraditionalForm\`	\(\(=\)\(\ \)\(∫\(e\^\(\(-i\)\ \(q\^2\) 
                  t/2  m\ \)\) \(e\^\(i\ q\ x/\)\/\@\(2\ π\ \)\) \(ϕ(q)\) \(\(dq\)\(\ \ \)\(.\)\)\)\)\)
This gives

\(\(myassumptions = {m > 0,  > 0, d > 0, 
          x0 > 0, \ {p, q, t} ∈ \ 
            Reals};\)\)
\(ψ[x_, t_] = 
      Integrate[
        Exp[\(-\(\(\(\ \)\(I\ q\^2\ t\)\)\/\(2\ m\ \)\)\)] \(1\/\@\(2\ π\ \)\) 
          Exp[\(I\ q\ x\)\/] ϕ[
            q], {q, \(-∞\), ∞}, 
        Assumptions -> myassumptions]\)

(Note that I defined 'myassumptions' for use later so Mathematica understands the nature of the constants here.)

This result looks a little odd with the i's floating about and t in the denominators, but ψ(x, 0) is indeed just the ψ(x) given in Hwk 4:

TrigToExp[
      Simplify[ComplexExpand[ψ[x, 0]], 
        Assumptions -> myassumptions]]

(b)

As we know 
	\(TraditionalForm\`⟨x⟩ \((t)\)\  = \ \(⟨ψ, 
            t | x | ψ, 
            t⟩\  = \ ∫\(\(ψ(x, t)\)\^*\) 
              x\ \(ψ(x, t)\) dx\)\)
which gives

meanx = 
      Integrate[
        ComplexExpand[
          Conjugate[ψ[x, t]] 
            x ψ[x, t]], {x, -∞, ∞}, 
        Assumptions -> myassumptions]

This is the expected result!

(c)

The second moment
	\(TraditionalForm\`⟨x\^2⟩ \((t)\)\  = \ \(⟨ψ, 
            t | x\^2 | ψ, 
            t⟩\  = \ ∫\(\(ψ(x, 
                    t)\)\^*\) \(\(\(x\)\(\ \)\)\^2\) \(ψ(x, 
                t)\) dx\)\)
which gives

meanx2 = 
      Integrate[
        ComplexExpand[
          Conjugate[ψ[x, t]] 
            (x^2) ψ[x, t]], {x, -∞, ∞}, 
        Assumptions -> myassumptions]

So the dispersion-square \(TraditionalForm\`⟨\((Δ\ x)\)\^2⟩ = ⟨x\^2⟩ - ⟨x⟩\^2\)is

dispx2 = 
      Simplify[meanx2 - (meanx^2), 
        Assumptions -> myassumptions]

Again, as expected, the dispersion increases with time; specifically, the disperson-square increases quadratically, as shown below with all constants set to 1.

\(\(Plot[
        dispx2 /. {d -> 1,  -> 1, m -> 1}, {t, 0, 
          10}];\)\)

(d)

There is only one Heisenberg equation of motion since [p, H] = 0, so like in part 1(h)
	\(TraditionalForm\`i\  \( d\/\(d\ t\)\) x\  = \ \ \(\(\([\)\(x, 
          H\)\(]\)\)\  = \ \(\(\([\)\(x, p\^2\/\(2  m\)\)\(]\)\)\  = \ 
            i\ \ p\/m\)\)\),
which clearly gives

x(t)  =  x(0) + (p/m) t
or
	\(TraditionalForm\`x(t)\  = \ 
        x\  + \ \(t\/m\) \(\/i\) ∂\/∂x\)
on the x basis.

For the second moment,
	(d/(d t)) 
          (x^2)  =  x ((d x)/(d t)) +  ((d x)/(d t)) 
              x =  (x(0) + (p/m) t) (p/m) + 
            (p/m) (x(0) + (p/m) t)
so

(x^2)(
          t) = ((((p/m)))^2) 
            (t^2) +  (x(0) (p/m) + (p/m) x(0)) t + (x^2)(
            0)
or
	(((x^2)(t) = \((\(t\/m\) \/i\)))^2)((∂^2)/∂)(x^2)+ \(\(t\/m\) \(\/i\) \((2  x ∂\/∂x + 
                1)\)\) + (x^2).

(e)

Again like in part 1(h),
	\(TraditionalForm\`⟨x⟩ \((t)\)\  = \ \(⟨ψ | 
              x(t) | ψ⟩\  = \ ∫\(\(ψ(x, 0)\)\^*\) \(x(t)\) \(ψ(x, 
                0)\) dx\)\)
which gives

\(meanxh = 
      Integrate[
        ComplexExpand[
          Conjugate[ψ[x, 
                0]] \((x\ ψ[x, 
                    0] + \(t\/m\) \(\/I\) ∂\_x\ ψ[x, 
                      0])\)], {x, \(-∞\), ∞}, 
        Assumptions -> myassumptions]\)

Similarly, for \(TraditionalForm\`⟨x\^2⟩ \((t)\)\)

\(meanx2h = 
      Integrate[
        ComplexExpand[
          Conjugate[ψ[x, 
                0]] \((\(\((\(t\/m\) \/I)\)\^2\) ∂\_x\ \(∂\_x\ ψ[x, 0]\) + \(t\/m\) \(\/I\) \((2  
                        x ∂\_x\ ψ[x, 0] + ψ[x, 
                        0])\) + \(x\^2\) ψ[x, 
                    0])\)], {x, \(-∞\), ∞}, 
        Assumptions -> myassumptions]\)

.  Thus, for \(TraditionalForm\`⟨\((Δ\ x)\)\^2⟩ \((t)\)\) we find

dispx2h = 
      Simplify[meanx2h - (meanxh^2), 
        Assumptions -> myassumptions]

These results in the Heisenberg picture are just as we calculated in parts (b) and (c) for the Schroedinger picture!

To check:

meanx - meanxh
dispx2 - dispx2h