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Say, I'm trying to solve the system of equations:

enter image description here

for θ, λ and ɸ , where a, b, c and d are complex numbers and the matrix on the LHS is a unitary matrix.

The SymPy code I have at hand does successfully do the job but there are a few edge cases it misses.

from sympy import *
def get_angles(a, b, c, d):
    theta, phi, lamb = symbols('\\theta \\phi \\lambda', real=True)
    a_eq = Eq(cos(theta / 2), a)
    b_eq = Eq(exp(I * phi) * sin(theta / 2), b)
    c_eq = Eq(-exp(I * lamb) * sin(theta / 2), c)
    d_eq = Eq(exp(I * (phi + lamb)) * cos(theta / 2), d)
    res = solve([a_eq, b_eq, c_eq, d_eq],
                theta,
                phi,
                lamb,
                check=False,
                set=True)
    return res

For instance, it doesn't restrict the range of the solutions. I did notice this answer but it only works for single variable cases. So any idea how to add the domain restrictions for the solution set, when dealing with multiple variables?

1 Answers1

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You can declare assumptions on the symbols and solve should check them e.g.:

In [12]: solve(x*(1-x))                                                                                                           
Out[12]: [0, 1]

In [13]: x = Symbol('x', positive=True)                                                                                           

In [14]: solve(x*(1-x))                                                                                                           
Out[14]: [1]

That works for some restrictions but won't work for a restriction like x<y. You can however post-process the output from solve:

In [6]: sol = [{x:1, y:2, z:3}, {x:1, y:0, z:4}, {x:3, y:2, z:1}]                                                                 

In [7]: sol                                                                                                                       
Out[7]: [{x: 1, y: 2, z: 3}, {x: 1, y: 0, z: 4}, {x: 3, y: 2, z: 1}]

In [8]: cond = And(0 < x, x < z, 0 < y)                                                                                           

In [9]: cond                                                                                                                      
Out[9]: x > 0 ∧ y > 0 ∧ x < z

In [10]: cond.subs(sol[0])                                                                                                        
Out[10]: True

In [11]: [s for s in sol if cond.subs(s)]                                                                                         
Out[11]: [{x: 1, y: 2, z: 3}]
Oscar Benjamin
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