Appendix A: Norm of the Amplification Matrix
In this appendix we prove that the norm of the matrix A(𝜃) defined in (28) is less than unity unless at least three of the 𝜃
i
coincide, in which case the norm is unity. The proof is due to Eli Hawkins.
Let |ν〉 be any unit spinor. The squared norm of A|ν〉 is ∥A|ν〉∥2=〈ν|A
‡
A|ν〉, whose maximum is the larger eigenvalue of A
‡
A. This value defines the squared norm ∥A∥2.
Using the definition of the spin projection operators (8) and the inner products of the unit vectors \(\hat {n}_{i}\) (1 if i = j and \(-\frac 13\) if i≠j) we find
$$ \text{tr} (A^{\dagger} A) = 1 + \tfrac16 \sum\limits_{(ij)} \cos(\theta_i-\theta_j) $$
(44)
where the sum is over the 6 choices of {i,j}⊂{1,2,3,4}. This trace is at most 2, and therefore the smaller eigenvalue of A
‡
A is less than 1 unless A
‡
A=1. Hence
$$ {\Phi} := \det \left( A^{\dag} A -1\right) $$
(45)
has the same sign as 1−∥A∥.
The matrix A
‡
A is linear in terms of \(e^{i(\theta _{i}-\theta _{j})}\), therefore Φ is quadratic. Because Φ is invariant under all permutations of the 𝜃’s, it can be written as a quadratic function of the cosines cos(𝜃
i
−𝜃
j
). Because Φ vanishes when the 𝜃’s are all equal, it is convenient to write it in terms of the cosines minus 1. It thus takes the form,
$$\begin{array}{@{}rcl@{}} {\Phi} &=& a \sum\limits_{(ij)} \left( 1 - \cos[\theta_i-\theta_j]\right) + b \sum\limits_{(ij)} \left( 1 - \cos[\theta_i-\theta_j]\right)^2 \\ &&+ c\!\!\sum\limits_{(ij)(kl)} \left( 1 - \cos[\theta_i-\theta_j]\right) \left( 1 - \cos[\theta_k-\theta_l]\right) . \end{array} $$
(46)
The last sum is over the 3 partitions of {1,2,3,4} into pairs. When 𝜃
2 = 𝜃
3 = 𝜃
4, A takes the form (16), which obviously has an eigenvalue of unit modulus. Therefore Φ=0 in this case, which shows that a = b=0. To determine the value of c, consider the case that 𝜃
1 = 𝜃
2=0 and 𝜃
3 = 𝜃
4 = π. Then \(A=\frac 12(\hat {n}_{1}+\hat {n}_{2})\cdot \vec {\sigma }\), so A
‡
A=1/3, hence Φ=4/9. The last sum in (46) is 8, so c=1/18.
The determinant Φ is thus given by
$$ {\Phi} = \tfrac1{18}\sum\limits_{(ij)(kl)} \left( 1 - \cos[\theta_i-\theta_j]\right) \left( 1 - \cos[\theta_k-\theta_l]\right). $$
(47)
This satisfies Φ≥0, therefore ∥A∥≤1. Each term is non-negative, therefore Φ=0 only if every term vanishes. This occurs only if at least three of the 𝜃’s are equal.