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I'm trying to implement Majorana's "stellar representation" of a spin-$j$ system as $2j$ points on the $2$-sphere in python. Consulting papers including Extremal quantum states and their Majorana constellations (Bjork et al., 2015), I convert a complex state vector (nominally indexed from -$j$ to $j$) to its corresponding polynomial with:

def vector_to_polynomial(vector):
    components = vector.tolist()
    j = (len(components)-1.)/2.
    coeffs = []
    i = 0
    for m in numpy.arange(-1*j, j+1, 1):
        coeff = math.sqrt(math.factorial(2*j)/(math.factorial(j-m)*math.factorial(j+m)))*components[i]
        coeffs.append(coeff)
        i += 1
    return coeffs[::-1]

I use a polynomial solver to determine the roots, and stereographically project them to the $2$-sphere, taking into account when the degree of the polynomial is less than $2j$ by adjoining some poles (latter code not included).

def root_to_xyz(root):
    if root == float('inf'):
        return [0,0,1]
    x = root.real
    y = root.imag
    return [(2*x)/(1.+(x**2)+(y**2)),\
            (2*y)/(1.+(x**2)+(y**2)),\
            (-1.+(x**2)+(y**2))/(1.+(x**2)+(y**2))]

See Wikipedia. Now QuTiP has an implementation of the Husimi Q function aka qutip.spin_q_function(state, theta, phi), evaluated at points on the sphere. The zeros of Husimi Q coincide with the Majorana stars. Comparing the results of the above with the QuTiP implementation, however, I find that they only match for integer spins, but not half-integer spins, aka for odd-dimensional systems, but not even dimensional systems. I've tried to code up a few other versions of the Majorana polynomial given in other papers, but the same problem seems to recur. Am I missing something more fundamental? Any advice is welcome!

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  • $\begingroup$ Have you considered that Qutip might be wrong? Have you contacted the authors of Qutip? I've created the "Qutip" tag and added it to your question, since a lot of people use Qutip and it would be nice for the Qutip user community to have a tag they can follow if they wish. $\endgroup$ – user1271772 Jul 8 '18 at 7:19
  • $\begingroup$ I am not sure if this is the cause of your trouble, but I think that there is a minus sign error in the third equation of the stereographic projection which should be (1-x2-y2)/(1+x2+y2) $\endgroup$ – David Bar Moshe Aug 12 '18 at 16:56

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