⚡ Pluto.jl ⚡

Problem 3. Let M be a 2 x 2 matrix and $M^{T}$ the transpose. If $| 0 ⟩$ and $| 1 ⟩$ are standard basis in $C^{2}$ , then show that:

$(M \otimes I_{2}) (| 0 ⟩ \otimes | 0 ⟩ + | 1 ⟩ \otimes | 1 ⟩) = (I_{2} \otimes M^{T}) (| 0 ⟩ \otimes | 0 ⟩ + | 1 ⟩ \otimes | 1 ⟩)$

12.4 μs

Solution 3. Let us first look at the tensor product $M \otimes I_{2}$ and $I_{2} \otimes M$ . Note that $I_{2}$ is a 2x2 identity matrix.

18.5 μs

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using SymPy

10.4 s

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begin
    #⊗(x,y) = sympy.tensorproduct(x,y) # Doesn't seem to work properly.
    ⊗(x::Array{Sym,2},y::Array{Sym,2}) = kron(x,y)
    𝐓(x::Array{Sym,2}) = sympy.Matrix(sympy.transpose(x))
    ⋅ = *
end;

57.0 μs

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begin
    M₁,M₂,M₃,M₄ = sympy.symbols("M₁₁ M₁₂ M₂₁ M₂₂")
    M= sympy.Matrix([M₁ M₂;
                     M₃ M₄])
    I₂ = sympy.eye(2,2)
end;

1.6 s

𝓞₁

$[\begin{array}{rrrr} M ₁ ₁ & 0 & M ₁ ₂ & 0 \\ 0 & M ₁ ₁ & 0 & M ₁ ₂ \\ M ₂ ₁ & 0 & M ₂ ₂ & 0 \\ 0 & M ₂ ₁ & 0 & M ₂ ₂ \end{array}]$

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𝓞₁ = M ⊗ I₂

33.9 ms

𝓞₂

$[\begin{array}{rrrr} M ₁ ₁ & M ₁ ₂ & 0 & 0 \\ M ₂ ₁ & M ₂ ₂ & 0 & 0 \\ 0 & 0 & M ₁ ₁ & M ₁ ₂ \\ 0 & 0 & M ₂ ₁ & M ₂ ₂ \end{array}]$

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𝓞₂ = I₂ ⊗ M

1.6 ms

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begin
    ψ₀ = sympy.Matrix([1;
                       0])
    ψ₁ = sympy.Matrix([0;
                       1])
end;

173 ms

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expr1 = 𝓞₁ ⋅ ((ψ₀⊗ψ₀) + (ψ₁⊗ψ₁));

683 ms

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expr2 = 𝓞₂ ⋅ ((ψ₀⊗ψ₀) + (ψ₁⊗ψ₁));

1.8 ms

Lets see if these two expressions are equivalent with $| ψ_{1}^{'} ⟩ = M \otimes I_{2} | ψ_{1} ⟩$ and $| ψ_{2}^{'} ⟩ = I_{2} \otimes M | ψ_{2} ⟩$ :

$| ψ_{1}^{'} ⟩ = [\begin{matrix} M ₁ ₁ \\ M ₁ ₂ \\ M ₂ ₁ \\ M ₂ ₂ \end{matrix}]$

$| ψ_{2}^{'} ⟩ = [\begin{matrix} M ₁ ₁ \\ M ₂ ₁ \\ M ₁ ₂ \\ M ₂ ₂ \end{matrix}] .$

In fact they are not. Now lets see if $M^{T}$ makes a difference for $| ψ_{2}^{'} ⟩ = I_{2} \otimes M^{T} | ψ_{2} ⟩$ .

2.5 ms

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begin
    𝓞₃ = (I₂ ⊗ 𝐓(M))
    expr3 = 𝓞₃ ⋅ ((ψ₀⊗ψ₀) + (ψ₁⊗ψ₁));
end;

34.5 ms

true

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expr1 == expr3

172 μs

Indeed the are equal:

$| ψ_{1}^{'} ⟩ = [\begin{matrix} M ₁ ₁ \\ M ₁ ₂ \\ M ₂ ₁ \\ M ₂ ₂ \end{matrix}]$

$| ψ_{2}^{'} ⟩ = [\begin{matrix} M ₁ ₁ \\ M ₁ ₂ \\ M ₂ ₁ \\ M ₂ ₂ \end{matrix}] .$

What happens for the orthonormal basis:

$| ψ_{1} ⟩ = (\begin{matrix} \cos θ \\ \sin θ \end{matrix}) | ψ_{2} ⟩ = (\begin{matrix} \sin θ \\ - \cos θ \end{matrix})$

2.6 ms

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begin
    θ = sympy.symbols("θ",real=true)
    ψₐ = sympy.Matrix([cos(θ);
                       sin(θ)])
    ψᵦ = sympy.Matrix([sin(θ);
                       -cos(θ)])
end;

521 ms

expr4

$[\begin{array}{r} M ₁ ₁ (\sin^{2} (θ) + \cos^{2} (θ)) \\ M ₁ ₂ (\sin^{2} (θ) + \cos^{2} (θ)) \\ M ₂ ₁ (\sin^{2} (θ) + \cos^{2} (θ)) \\ M ₂ ₂ (\sin^{2} (θ) + \cos^{2} (θ)) \end{array}]$

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expr4 = 𝓞₁ ⋅ (( ψₐ ⊗ ψₐ) + (ψᵦ ⊗ ψᵦ))

5.7 ms

expr5

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expr5 = 𝓞₃ ⋅ (( ψₐ ⊗ ψₐ) + (ψᵦ ⊗ ψᵦ))

1.4 ms

Indeed they are the same:

19.1 μs

true

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sympy.trigsimp(expr4) == sympy.trigsimp(expr5)

278 ms