HW #11

1. Bogoliubov Transformation

(a)

This is a straight-forward algebra
[b, 
          (b^†)] = [a cosh η + (a^†) 
              sinh η,  (a^†) c osh η + 
            a sinh η]
=[a, 
              (a^†)] (cosh^2) η + [(a^†), 
              a] (sinh^2) η
\(TraditionalForm\`\(\(=\)\(\(cosh\^2\) η - \(sinh\^2\) η\)\)\)
=1

(b)

Using b, (b^†) defined in (a), we can solve for a, (a^†) as
a = 
        b cosh η - (b^†) sinh η
(a^†) = (b^†) cosh η - 
          b sinh η
Substituing them into the Hamiltonian ,we find
H  = ℏ ω (a^†) 
            a + (1/2) V(a a + (a^†) (a^†))
=ℏ ω((b^†) 
                cosh η - 
              b sinh η) (b cosh η - (b^†) 
                sinh η)
\(TraditionalForm\`\(+\(1\/2\)\) \(V(\((\(b\^†\) cosh\ η - b\ sinh\ η)\)\^2 + \((b\ cosh\ η - \(b\^†\) sinh\ η)\)\^2)\)\)
=(b^†) 
          b (ℏ ω (cosh^2) η - 
              V cosh η sinh η)
\(TraditionalForm\`\(+b\)\ \(\(b\^†\)(ℏ\ ω\ \(sinh\^2\) η - V\ cosh\ η\ sinh\ η)\)\)
\(TraditionalForm\`\(+b\)\ \(b(\(-ℏ\)\ ω\ sinh\ η\ cosh\ η + \(1\/2\) V\ \((\(sinh\^2\) η + \(cosh\^2\) η)\))\)\)
\(TraditionalForm\`\(+\(\(b\^†\) \(\(b\^†\)(\(-ℏ\)\ ω\ sinh\ η\ cosh\ η + \(1\/2\) \(V(\ \(cosh\^2\) η + \(sinh\^2\) η)\))\)\)\)\)
Because η is a free parameter, we can choose it at our convenience.  We choose η such that the coefficients of b b and (b^†) (b^†) vanish,
\(TraditionalForm\`\(-ℏ\)\ ω\ sinh\ η\ cosh\ η + \(1\/2\) \(V(\ \(cosh\^2\) η + \(sinh\^2\) η)\) = 0\).
-ℏ ω sinh 2  η + 
          V cosh 2  η = 0. 
and hence
η = (1/2) 
          arctanh (V/(ℏ ω)), or \(TraditionalForm\`e\^\(4  η\) = \(ℏ\ ω + V\)\/\(ℏ\ ω - V\)\).
With this choice, the Hamiltonian has the familiar form
H = (b^†) 
          b (ℏ ω (cosh^2) η - 
              V cosh η sinh η)
\(TraditionalForm\`\(+b\)\ \(\(b\^†\)(ℏ\ ω\ \(sinh\^2\) η - V\ cosh\ η\ sinh\ η)\)\)
\(TraditionalForm\`\(\(=\)\(\(b\^†\) \(b(ℏ\ \(ω(\(cosh\^2\) η + \(sinh\^2\) η)\) - 
                2  V\ cosh\ η\ sinh\ η)\) + \((ℏ\ ω\ \(sinh\^2\) η - V\ cosh\ η\ sinh\ η)\)\)\)\)
\(TraditionalForm\`\(\(=\)\(\(b\^†\) 
            b \@\(\((ℏ\ ω)\)\^2 - V\^2\) + \(1\/2\) \((\@\(\((ℏ\ ω)\)\^2 - V\^2\) - ℏ\ ω)\)\)\)\)
We find that the ground state is annihilated by b, its energy is \(TraditionalForm\`\(1\/2\) \((\@\(\((ℏ\ ω)\)\^2 - V\^2\) - ℏ\ ω)\)\), and the excitated states are obtained by acting (b^†) each time raising the energy by (Sqrt[(((ℏ ω))^2) - (V^2)]).

The system does not have a stable ground state if V > ℏ ω.

(c)

We use the Hausdorff formula, 
(e^A) B (e^(-A)) = 
        B + [A, B] + (1/(2!))[A, [A, B]] + (1/(3!))[
            A, [A, [A, B]]] + ⋯
In our case, A = (a a - (a^†) 
                (a^†)) η/2, B = a.

Let us first calculate the multiple commutators.
B = a
[A, 
          B] = (η/2)[a a - (a^†) (a^†), 
            a] = η (a^†).
\(TraditionalForm\`\([A, \([A, B]\)]\) = \(\(η\/2\)[
            a\ a - \(a\^†\) 
                a\^†, η\ a\^†] = \(\(η\^2\) \(a\)\(\ \)\)\)\).
[A, [A, [A, B]]] = (η/2)[
            a a - (a^†) 
                (a^†), (η^2) a] = (η^3) 
            (a^†)
etc.  By inspection, it is clear that the 2  n-th commutator is (η^(2  n)) a, while the 2  n + 1-th commutator (η^(2  n + 1)) (a^†).

Now back to the Hausdorff formula,
(e^A) B (e^(-A)) = 
        a + η (a^†) + (1/(2!)) (η^2) 
            a + (1/(3!)) (η^3) 
            (a^†) + ⋯
=a cosh η + (a^†) 
              sinh η = b
Putting them together, we find b = U a (U^(-1))indeed.

We can also see that it is a unitarity operator.  The exponent is anti-Hermitian, 
\(TraditionalForm\`A\^† = \(\((\(η\/2\) \((a\ a - \(a\^†\) a\^†)\))\)\^† = \(\(η\/2\) \((\(a\^†\) a\^† - 
                  a\ a)\) = \(\(-\(η\/2\)\) \((a\ a - \(a\^†\) 
                      a\^†)\) = \(-A\)\)\)\)\).
Therefore, \(TraditionalForm\`U\^† = \(\((e\^A)\)\^† = \(e\^\(A\^†\) = \(e\^\(-A\) = U\^\(-1\)\)\)\)\), and hence it is unitary.

(d)

The ground state is annihilated by b, and hence U a (U^(-1)) | 
            gs⟩ = 0.
Acting (U^(-1)) on both sides, we find a((U^(-1)) | gs⟩) = 
        0, namely that the state (U^(-1)) | gs⟩ is annihilated by the operator a.  Therefore,
\(TraditionalForm\`\(\(\(\(U\^\(-1\) | 
              gs\)\(⟩\)\) = \(\(|\)\(0\)\)\)\(⟩\)\)\),
and hence
|gs⟩ = 
          U | 0⟩.
Because U is a power series in (η/2) (a a - (a^†) 
              (a^†)) which changes numbers by two units, we find that |gs⟩ can be expanded as
 \(TraditionalForm\`\(\(\(\(\(|\)\(gs\)\)\(⟩\)\) = ∑\+\(n = 0\)\%∞ c\_\(2  n\) | 
            2  n\)\(⟩\)\)\)

Using the fact that it is annihilated by b, (a cosh η + (a^†) 
                  sinh η) | gs⟩ = 0, we find
 \(TraditionalForm\`\(\(\(\(\(\(\(\(\(\(b | 
                        gs\)\(⟩\)\) = \((a\ cosh\ η + \(a\^†\) 
                              sinh\ η)\) \(∑\+\(n = 0\)\%∞ c\_\(2  n\)\) | 
                      2  n\)\(⟩\)\)

                \(\(=\)\(∑\+\(n = 0\)\%∞\(\( 
                                c\_\(2  n\)\)(\@\(2  n\) | 
                              2  n - 1⟩\) 
                          cosh\ η + \@\(2  n + 1\) | 
                    2  n + 1\)\)\)\(⟩\)\) 
              sinh\ η\)\()\)\)

          \(\(=\)\(∑\+\(n = 0\)\%∞\((\(c\_\(2  n + 
                            2\)\) \(\@\(2  n + 2\)\) 
                      cosh\ η + \(c\_\(2  n\)\) \(\@\(2  n + 1\)\) 
                      sinh\ η)\) | 
              2  n + 1\)\)\)\(⟩\)\) = 0. \)
It means
(Subscript[c, 2  n + 2]) = (Subscript[c, 2  
                n])(-(Sqrt[((2  n + 1)/(2  n + 2))]) 
            tanh η).
Therefore,
\(TraditionalForm\`c\_\(2  n\) = \(\(\(c\_0\)(\(-1\))\)\^n\) \(\@\(\(\((2  n - 1)\)!!\)\/\(\((2  n)\)!!\)\)\) \(tanh\^n\) η\)
The norm is
1=\(TraditionalForm\`\(c\_0\^2\) \(∑\+\(n = 0\)\%∞\(\(\((2  n - 1)\)!!\)\/\(\((2  n)\)!!\)\) \(tanh\^\(2  n\)\) η\)\)
Rewriting it with the Gamma function,
\(TraditionalForm\`c\_0\^\(-2\) = \(∑\+\(n = 0\)\%∞\(\(\( 2\^n\) \(Γ(
                        n + 1/2)\)/\@π\)\/\(\(2\^n\) \(Γ(
                      n + 1)\)\)\) \(tanh\^\(2  
                    n\)\) η = \(1\/\@π\) \(∑\+\(n = 0\)\%∞\(\( Γ(n + 1/2)\)\/\(Γ(
                      n + 1)\)\) \(tanh\^\(2  n\)\) η\)\)\)
Mathematica gives this infinite series

Sum[Gamma[n + (1/2)]/Gamma[n + 1] (x^n), {n, 
        0, ∞}]

Therefore,
\(TraditionalForm\`c\_0\^\(-2\) = \(\(1\/\@π\) \@π\/\@\(1 - \(tanh\^2\) η\) = \((1 - \(tanh\^2\) η)\)\^\(\(-1\)/2\)\)\)
and hence
(Subscript[c, 0]) = (((1 - (tanh^2) η))^(1/4)).

Putting everything together, the ground state of the Hamiltonian is given in terms of the number states
\(TraditionalForm\`\(\(\(\(\(|\)\(gs\)\)\(⟩\)\) = \(\((1 - \(tanh\^2\) η)\)\^\(1/
                    4\)\) \(∑\+\(n = 0\)\%∞\(\((\(-1\))\)\^n\) \(\@\(\(\((2  n - 1)\)!!\)\/\(\((2  n)\)!!\)\)\) \(tanh\^n\) η\) | 
            2  n\)\(⟩\)\)\).

The Bogoliubov transformation appears not only in the BEC but also in optics.  The laser is a coherent state, and one can realize Bogoliubov transformed ones called the "squeezed states." They have been under active research.

2. Superfluid Vortex

(a)

The solution is z-independent, and hence is a function of r and θ only.  We use the chain rule 
(({{(Subscript[∇, r])}, {(Subscript[∇, θ])}}))=(({{cos θ, sin θ}, {-r sin θ, r cos θ}}))(({{(Subscript[∇, x])}, {(Subscript[∇, y])}}))
and find
(({{(Subscript[∇, x])}, {(Subscript[∇, y])}}))=(1/r)(({{r cos θ, -sin θ}, {r sin θ,  cos θ}}))(({{(Subscript[∇, r])}, {(Subscript[∇, θ])}})).
Therefore,
(({{(Subscript[∇, x])}, {(Subscript[∇, y])}}))ψ=(({{(Subscript[∇, x])}, {(Subscript[∇, y])}}))f(r)(e^(i n θ))=(1/r)(({{r cos θ, -sin θ}, {r sin θ,  cos θ}}))(({{f' (r)}, {i n f(r)}}))(e^(i n θ)).
The momentum density is
(Overscript[j,  ->]) =(ℏ/(2  m i))f(r)(1/r)(({{r cos θ, -sin θ}, {r sin θ,  cos θ}}))(({{f' (r)}, {i n f(r)}}))+c . c . 
=((n ℏ)/(m r))((f(r))^2)(({{-sin θ}, {cos θ}})).
On the other hand, the number density is
\(TraditionalForm\`ρ = \(\(ψ\^*\) ψ = \(f(r)\)\^2\)\).
The velocity field is their ratio,
(Overscript[v,  ->]) =(((Overscript[j,  ->]) )/ρ)=((n ℏ)/(m r))(({{-sin θ}, {cos θ}})).

For the plotting purpose, we take ℏ = m = 1.

<< Graphics`PlotField`

n = 1

PlotVectorField[(1/(Sqrt[(x^2) + (y^2)])) {-y, x}, {x, -1, 
        1}, {y, -1, 1}, ScaleFactor -> None, 
      ScaleFunction -> ( .05  # &)]

n = -2

PlotVectorField[((-2)/(Sqrt[(x^2) + (y^2)])) {-y, x}, {x, -1, 
        1}, {y, -1, 1}, ScaleFactor -> None, 
      ScaleFunction -> ( .05  # &)]

n = 3

PlotVectorField[(3/(Sqrt[(x^2) + (y^2)])) {-y, x}, {x, -1, 
        1}, {y, -1, 1}, ScaleFactor -> None, 
      ScaleFunction -> ( .05  # &)]

(b)

The circulation is obtained from the loop integral of the velocity field (Overscript[v,  ->]) =(((Overscript[j,  ->]) )/ρ)=((n ℏ)/(m r))(({{-sin θ}, {cos θ}})),
κ=∮((n ℏ)/(m r))(({{-sin θ}, {cos θ}}))·r(({{-sin θ}, {cos θ}}))d θ
=((2  π n ℏ)/m) = ((n h)/m)

(c)

We start with the Euler-Lagrange equation obtained from the Lagrangian, 
\(TraditionalForm\`i\ ℏ\ \(ψ\&·\) + \(\(\(ℏ\^2\) Δ\)\/\(2  
                  m\)\) ψ + μ\ ψ - λ\ \(ψ\^*\) ψ\ ψ = 0\).
We now subsitute our Ansatz into the equation:  ψ(r, θ, 
          t) = (e^(i n θ)) f(r).
Because it is static (Overscript[ψ, ·]) = 0, we find
((ℏ^2)/(2  
                  m)) (((d^2)/(d (r^2))) + (1/r) (d/(d r)) - 
                ((n^2)/(r^2))) f + μ f - λ (f^3) = 0.
We now try to cast it to a form as independent of the parameters as possible.
\(TraditionalForm\`\((d\^2\/\(d\ r\^2\) + \(1\/r\) d\/\(d\ r\) - 
                n\^2\/r\^2)\) 
            f + \(2\ m\ μ\)\/ℏ\^2\ f - \(2\ m\ λ\)\/ℏ\^2\ f\^3 = 0\)
We introduce the variable
(((Sqrt[2 m μ]))/ℏ)r=ρ
Then the equation becomes
(((d^2)/(d (ρ^2))) + (1/ρ) 
                  (d/(d ρ)) - ((n^2)/(ρ^2))) f +  
          f - (λ/μ) (f^3) = 0
Finally, we introduce the function
f = (Sqrt[(μ/λ)]) ϕ
which satisfies
(((d^2)/(d (ρ^2))) + (1/ρ) 
                  (d/(d ρ)) - 
                ((n^2)/(ρ^2))) ϕ + ϕ - (ϕ^3) = 0
The equation is now completely free from physical parameters except for the integer n.

We solve numerically with the boundary conditions ϕ(0) = 0, ϕ(∞) = 1.  Instead of imposing the boundary condition stritcly at the origin, which causes all kinds of problems,we cheat a little bit and impose it slightly off the origin.

vortex = 
      NDSolve[{y''[x] + (1/x) y'[x] - (1/(x^2)) y[x] + y[x] - 
              y[x]^3 == 0, 
          y[(10^(-6))] == 0, y'[(10^(-6))] == 
            1.166380374636}, y, {x, (10^(-6)), 20}]

Plot[y[x] /. vortex[[1]], {x, (10^(-6)), 20}, 
      PlotRange -> {0, 1.1}]

This way, we see that the vortex solution exists for the classical Schrödinger field with a repulsive self-interaction potential, as observed in the atomic BEC as well as liquid He-II.