Problem 1. Given the Pauli matrices
and a hermitian matrix that has dimensions 8x8 and written as:
Find the eigenvalues and and eigenvectors.
Solution 1. The interesting thing to note here is that the hermitian matrix can be written as the direct sum:
This means that the eigenspectrum can be calculated from the
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begin2
using SymPy3
endx
16
1
begin2
𝑖 = sympy.I3
⊗(x,y)=kron(x,y)4
⊗(x::Array{Sym,2},y::Array{Sym,2}) = kron(x,y)5
𝐓(x::Array{Sym,2}) = sympy.Matrix(sympy.transpose(x))6
7
"""Direct sum of vector space.8
""" 9
⊕(x,y) = begin10
xdim = size(x);11
ydim = size(y);12
M = [ x zeros(Int,xdim[1],ydim[2]);13
zeros(Int,ydim[1],xdim[2]) y ];14
return M15
end16
end;xxxxxxxxxx1
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begin2
σ₁ = [0 1;3
1 0]4
σ₂ = [0 -𝑖;5
𝑖 0]6
σ₃ = [1 0;7
0 -1] 8
end;The hermitian matrix given by the expression above is then:
x
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H = (σ₁ ⊗ σ₁ + σ₂ ⊗ σ₂ + σ₃ ⊗ σ₃) ⊗ σ₁Now just checking the direct sum comparison.
x
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begin2
D = Sym[ 0 -1 0 2;3
-1 0 2 0;4
0 2 0 -1;5
2 0 -1 0];6
H′ = σ₁ ⊕ D ⊕ σ₁7
endx
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H == H′ "Direct sum doesn't equal hermitian matrix"Now finding the eigenvalues and eigenvectors
x
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begin2
ℍ = H.eigenvects(chop=true)3
#unpack :e - eigenvalue, :m - multiplicty (degeneracy)4
eₕ = [Dict(:e=>e[1],:m=>e[2]) for e in ℍ]5
vₕ = [v[end][1:end] for v in ℍ]6
end;Let see how these eigenvalues compare to the eigenvalues of the matrix
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begin2
𝔻 = D.eigenvects(chop=true)3
eᵨ = [Dict(:e=>e[1],:m=>e[2]) for e in 𝔻]4
vᵨ = [v[end][1:end] for v in 𝔻]5
end;we find that the eigenvalues are the same but the degeneracy is now removed. We can look at the eigenvectors the the
eigenvector index 1