HW #5

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,
	\(TraditionalForm\`\(\(⟨x⟩ = \(⟨n | x | 
                  n⟩ = \(\(\@\(ℏ\/\(2  
                            m\ ω\)\)\) ⟨n | 
                      a + a\^† | 
                      n⟩
 = \(\@\(ℏ\/\(2  m\ ω\)\)\) ⟨n | \((\@n | 
                            n - 1⟩\) + \@\(n + 1\) | 
                      n + 1⟩\)\)\)\()\)\) = 0\)
because of the orthonormality of the Hamiltonian eigenstates ⟨n | 
            m⟩ = (Subscript[δ, n,  m]).  

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\/\(2  
                  m\ ω\)\) \(\((2  n + 1)\)\(.\)\)\)\)\)

(e)

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⟩ = 
            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\) \(\((2  n + 1)\)\(.\)\)\)\)\)
 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.

(f)

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\`\(\(\(\(\(|\)\(f\)\)\(⟩\)\) = ∑\+\(n = 0\)\%∞\( f\^n\/\(n!\)\) \((a\^†)\)\^n | 
            0\)\(⟩\)\)\), we can write it as \(TraditionalForm\`\(\(\(\(\(|\)\(f\)\)\(⟩\)\) = ∑\+\(n = 0\)\%∞ f\^n\/\@\(n!\) | 
            n\)\(⟩\)\)\).  Then
	 \(TraditionalForm\`\(\(\(\(\(\(a | f\)\(⟩\)\) = 
              a \(∑\+\(n = 0\)\%∞ f\^n\/\@\(n!\)\) | 
                n\)\(⟩\)\) = ∑\+\(n = 1\)\%∞\( f\^n\/\@\(n!\)\) \@n | 
            n - 1\)\(⟩\)\)\), 
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\)\(⟩\)\) = ∑\+\(m = 0\)\%∞ f\^\(m + 1\)\/\@\(m!\) | m\)\(⟩\)\)\),
where the dummy variable was chgned to m = n - 1.  Pulling one factor of f ouf of the sum,
	a | f⟩ = 
              f (∑_(m = 0)^∞) ((f^m)/(Sqrt[m!])) | 
                m⟩ = 
          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\`\(\(\(\(\(|\)\(f\)\)\(⟩\)\) = ∑\+\(n = 0\)\%∞\( f\^n\/\(n!\)\) \((a\^†)\)\^n | 
            0\)\(⟩\)\)\), 
	\(TraditionalForm\`\(\(\(\(\(\(a | f\)\(⟩\)\) = 
              a \(∑\+\(n = 0\)\%∞\( f\^n\/\(n!\)\) \((a\^†)\)\^n\) | 
                0\)\(⟩\)\) = ∑\+\(n = 0\)\%∞\( f\^n\/\(n!\)\)[a, \((a\^†)\)\^n] | 
            0\)\(⟩\)\)\), 
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\)\(⟩\)\) = ∑\+\(n = 1\)\%∞ f\^n\/\(\((n - 1)\)!\)\ \((a\^†)\)\^\(n - 1\) | 
            0\)\(⟩\)\)\),
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\)\(⟩\)\) = 
              f \(∑\+\(m = 0\)\%∞ f\^m\/\(m!\)\ \((a\^†)\)\^m\) | 0\)\(⟩\)\) = 
          f | f\)\(⟩\)\)\),

(g)

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
	(((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.

(h)

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\)\)
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))⟩.  The exectation value of the position operator is
	\(TraditionalForm\`⟨f, t | x | f, 
          t⟩ = \(\(e\^\(i\ ω\ t/
                    2\)\) ⟨f\ e\^\(\(-i\)\ ω\ t\) | 
                x | f\ e\^\(\(-i\)\ ω\ t\)⟩ 
            
            e\^\(\(-i\)\ ω\ t/2\) = ⟨f\ e\^\(\(-i\)\ ω\ t\) | \(\@\(ℏ\/\(2  m\ ω\)\)\) \((a + 
                    a\^†)\) | 
              f\ e\^\(\(-i\)\ ω\ t\)⟩\)\)
\(TraditionalForm\`\(\(=\)\(\(\@\(ℏ\/\(2  
                      m\ ω\)\)\) \((f\ e\^\(\(-i\)\ ω\ t\) + \(f\^*\) e\^\(i\ ω\ t\))\) ⟨f\ e\^\(\(-i\)\ ω\ t\) | f\ e\^\(\(-i\)\ ω\ t\)⟩ = \(\@\(ℏ\/\(2  m\ ω\)\)\) 
            2 \((\(Re(f)\) cos\ ω\ t + \(Im(f)\) 
                  sin\ ω\ t)\) ⟨f, t | f, 
              t⟩\)\)\)
Therefore, the expectation value is
\(TraditionalForm\`⟨x⟩ \((t)\) = \(⟨f, t | x | f, t⟩\/⟨f, t | f, t⟩ = \(\@\(ℏ\/\(2  m\ ω\)\)\) 
            2 \((\(Re(f)\) cos\ ω\ t + \(Im(f)\) 
                  sin\ ω\ t)\)\)\),
and shows the oscillatory behavior just like the classical solution.

(i)

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 as Daniel explained in the section.  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 value,
	\(TraditionalForm\`⟨x⟩ \((t)\) = \(⟨f | x(t) | f⟩\/⟨f | f⟩ = \(1\/⟨f | 
                    f⟩\) ⟨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\)\)\)
=(Sqrt[(ℏ/(2  m ω))]) 
          2 (Re(f) cos ω t + Im(f) 
                sin ω t),
which agrees with the calculation in the Schrödinger picture.