⚡ Pluto.jl ⚡

Problem 3. Consider a set of vectors ${S}$ in a Hilbert space $C^{n}$ . Let $u \in {S}$ . A function $f (u)$ that maps ${S}$ to $R$ is called a generalized probability measure when the following conditions two conditions are held:

$u \in {S} : 0 \leq f (u) \leq 1$
Then $u_{1}, \dots, u_{n}$ forms an orthonormal basis in $C^{n}$ . Then it follows that $\sum_{i = 1}^{n} f (u_{i}) = 1$ .

If we take $n \leq 3$ , then the probability measure of $f$ over $C^{n}$ has the form:

$f (ρ) = tr (ρ u u^{†})$

a Find the $f (u_{1}), \dots, f (u_{3})$ given the $C^{3}$ with orthonormal basis and density matrix:

$u_{1} = \frac{1}{\sqrt{2}} (\begin{matrix} 1 \\ 0 \\ 1 \end{matrix}) u_{2} = (\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}) u_{3} = \frac{1}{\sqrt{2}} (\begin{matrix} 1 \\ 0 \\ - 1 \end{matrix}),$

$ρ = \frac{1}{3} (\begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{matrix})$

Do the same but consider Hilbert space $C^{4}$ with orthonormal basis and density matrix:

$u_{1} = \frac{1}{\sqrt{2}} (\begin{matrix} e^{i ϕ} \\ 0 \\ 0 \\ e^{i ϕ} \end{matrix}) u_{2} = (\begin{matrix} e^{i ϕ} \\ 0 \\ 0 \\ e^{- i ϕ} \end{matrix}) u_{3} = \frac{1}{\sqrt{2}} (\begin{matrix} 0 \\ e^{i ϕ} \\ e^{i ϕ} \\ 0 \end{matrix}) u_{4} = \frac{1}{\sqrt{2}} (\begin{matrix} 0 \\ e^{i ϕ} \\ e^{- i ϕ} \\ 0 \end{matrix})$

$ρ = \frac{1}{4} (\begin{matrix} 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{matrix})$

18.2 μs

Solution 3.

8.6 μs

xxxxxxxxxx
 
begin
    using SymPy
    using Plots
    pyplot()
end;

314 μs

x
 
begin
    𝑖,𝑒,π,ϕ = sympy.I, sympy.E, sympy.pi,sympy.symbols("ϕ",real=true)
    halfsqrt = 1/sympy.sqrt(2)
    𝚝𝚛(X) = sympy.trace(X)
    dagger(X) = sympy.adjoint(X)
    ⋅(a::T,X) where T<:Number = a*X
    ⋅(X,Y) = sympy.MatMul(X,Y,evaluate=true)
end;

75.9 ms

x
 
begin
    𝐮₁ = halfsqrt ⋅ [1;
                     0;
                     1]
    𝐮₂ = halfsqrt ⋅ [0;
                     1;
                     0]
    𝐮₃ = halfsqrt ⋅ [1;
                     0;
                    -1]
    ρ = 1/3⋅ ones(Sym,3,3)
end;

135 ms

xxxxxxxxxx
 
"""
    evaluate generalized probability measure. Gleason 1957
"""
function f(ρ::Array{Sym},𝐮ᵢ::Array{Sym})
    ρᵢ = 𝐮ᵢ ⋅ dagger(𝐮ᵢ);
    return 𝚝𝚛(ρ⋅ρᵢ)
end;

220 μs

Evaluating the function $f (ρ, u)$ for each orthonormal basis:

$f (ρ, u_{1}) = \frac{2}{3}$

$f (ρ, u_{2}) = \frac{1}{6}$

$f (ρ, u_{3}) = 0$

51.1 ms

Now calculating the same for a Hilbert space of $C^{4}$ as described in the problem statement.

7.8 μs

x
 
begin
    𝐯₁ = halfsqrt ⋅ [𝑒^(𝑖⋅ϕ);
                     0;
                     0;
                     𝑒^(𝑖⋅ϕ)]
    𝐯₂ = halfsqrt ⋅ [𝑒^(𝑖⋅ϕ);
                     0;
                     0;
                     𝑒^(-𝑖⋅ϕ)];
    𝐯₃ = halfsqrt ⋅ [0;
                     𝑒^(𝑖⋅ϕ);
                     𝑒^(𝑖⋅ϕ);
                     0];
    𝐯₄ = halfsqrt ⋅ [0;
                     𝑒^(𝑖⋅ϕ);
                     𝑒^(-𝑖⋅ϕ);
                     0];
​
    𝐩 = 1/4⋅ ones(Sym,4,4)
end;
​

118 ms

Evaluating the function $f (ρ, u)$ for each of these orthonormal basis:

$f (ρ, v_{1}) = \frac{1}{2}$

$f (ρ, v_{2}) = \frac{\cos (2 ϕ) + 1}{4}$

$f (ρ, v_{3}) = \frac{1}{2}$

$f (ρ, v_{4}) = \frac{\cos (2 ϕ) + 1}{4}$

137 ms

Interestingly we see that in this orthonormal basis set there is a dependence on $ϕ$

8.1 μs

x
 
ϕ⃗ = range(0,stop=pi,length=100);

244 ms

x
 
begin
    g = f(𝐩,𝐯₂).simplify();
    plot(ϕ′->ϕ′,ϕ′->g(ϕ=>ϕ′),ϕ⃗,
               xlabel="ϕ",ylabel="v₂(ϕ)",
               size=(300,200),legend=false,
               annotate=(pi/2,0.4,text("f(ρ,v₂)→v₂(ϕ)",10)),
               xticks=([pi*i/4 for i=0:4],[π*i/4 for i=0:4]))
end

270 ms