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Crossposted on Mathematica SE

Is there a systematic/algorithmic/computational way to "complete the square" or find a sum-of-squares decomposition for a trace of non-commuting operators? For example, I know that

$$\operatorname{Tr} \left( \rho A^{2\dagger} \rho A^2 - \rho A^{\dagger} \rho A^2 - \rho A^{2\dagger} \rho A + \rho A^\dagger \rho A \right) \equiv \operatorname{Tr} \left( \rho (A^2 - A)^\dagger \rho (A^2 - A) \right)$$

Given that $\rho$ is a positive semi-definite operator, this reduction reveals that the expression is non-negative for all operators $A$. While this reduction is easy enough to spot "by inspection", I find myself faced with a more complicated expression that I am almost sure is non-negative, but I haven't been able to reduce "by inspection" into a manifestly non-negative form.

The expression I am currently faced with has 5 terms that each involve 2 powers of $\rho$, 2 powers of $A$, and 2 powers of $A^\dagger$, arranged in different orders. For example, one term looks like $\operatorname{Tr} \left( \rho A^\dagger A^2 \rho A^\dagger \right)$.

For the sake of completeness, I've written the full expression below, although I'd love an answer that provides a systematic approach for tackling this sort of problem, rather than a solution to this specific instance of the problem. Any help would be appreciated.

$$ \operatorname{Tr} \left( \rho^2 A^\dagger A^\dagger A A + 2 \rho A^\dagger A \rho A^\dagger A + \rho A A \rho A^\dagger A^\dagger - 2 (A^\dagger A \rho + \rho A^\dagger A) A \rho A^\dagger \right) $$

Rodrigo de Azevedo
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nlupugla
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  • I did come across this expression in the context of quantum information theory. I tagged the question as quadratic-forms because the expression is a quadratic form in $\rho$. That is, one can write the expression as $\mathrm{vec}(\rho)^\dagger Q \mathrm{vec}(\rho)$ for some suitable matrix $Q$ that is a block matrix involving $A$ and $A^\dagger$. – nlupugla Feb 07 '25 at 20:34
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    My experience in SOS is stuff like this and this but that is not what you need, right? – Rodrigo de Azevedo Feb 07 '25 at 20:49
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    It is related to what I need. In the problems you linked, the variables of the polynomial commute with each other, and the coefficients of the polynomial. The problem, as I understand it, is given a polynomial like $P(x, y, z)$ express it as $P(x, y, z) = f(x, y, z)^2 + g(x, y, z)^2 + h(x, y, z)^2$. In my case, I have a polynomial (or rather a trace of a polynomial) of non-commuting terms. I have a polynomial $\mathrm{Tr}(P(\rho, A))$ that I would like to express as $|| f(\rho, A) ||^2 + || g(\rho, A) ||^2 + || h(\rho, A) ||^2$, where $|| X ||^2 = \mathrm{Tr}(X^\dagger X)$. – nlupugla Feb 07 '25 at 21:14
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    You probably want to look at Sums of hermitian squares, they can be solved with SDP techniques similar to the commutative case. – Rammus Feb 21 '25 at 22:24
  • Thanks! Sums of Hermitian squares seems to be headed in the right direction! The main difference with my problem and what I see from a quick read about sums of Hermitian squares is that variables in the polynomials are assumed to be Hermitian. In my case, $\rho$ is Hermitian, but $A$ is not. Nevertheless, the form is real because all the terms in the sum are Hermitian. Do you know if results about sums of Hermitian squares still work for a case like mine? – nlupugla Feb 25 '25 at 14:56

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