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Define $\mathcal{P}_n$ to be the set of all $n$-qubit Pauli strings with phase $+1$. Then any Hermitian operator $H$ can be decomposed into linear combination of these Pauli strings. That is, $$ H=d^{-1}\sum_{P\in\mathcal{P}_n}{\rm tr}(PH)P $$ where $d=2^n$ is the dimension of the space.

If we only require $H$ to be Hermitian, the coefficients ${\rm tr}(PH)$ can be any real numbers. But if we further require $H$ to be positive, meaning that for any state $|\psi\rangle$, $\langle\psi|H|\psi\rangle\geq0$, then what constraints should we impose on these coefficients?

cos
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1 Answers1

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One characterization is the following: $H \succeq 0$ iff the matrix of coefficients

$$\Gamma_{\alpha \beta} = \operatorname{tr}(P_\alpha^\dagger P_\beta H)$$

satisfies $\Gamma \succeq 0$.

This idea is known as a moment matrix or sum of squares proof of positivity.

It derives from the fact that $H \succeq 0$ iff $\operatorname{tr}(H A) \geq 0$ for all $A\succeq 0$.

Because every $A \succeq 0$ can be factorized as $A = B^\dagger B$, it follows that $H \succeq 0$ iff $$\operatorname{tr}(BB^\dagger H) \geq 0$$ for all $B = \sum_\alpha b_\alpha P_\alpha$.

This can be rewritten as $$b^\dagger \Gamma b \geq 0$$ for all vectors $b$, which is equivalent to the claimed condition $\Gamma \succeq 0$.

Fixing the normalization (e.g. if you want a density matrix $\operatorname{tr}(H) = 1$), is done by the requirement $\Gamma_{00} = \operatorname{tr}(\mathbb{1}^\dagger \mathbb{1} H) = 1$, so that $H$ is normalized to trace one.

Requiring that a submatrix of $\Gamma$ is positive semidefinite is a often used relaxation for the positivity of a density matrix (e.g. for quantum Max Cut or maximal violations of Bell non-locality), when one is only given some subset of Pauli coefficients $\operatorname{tr}(P H)$.

One way to think of this is to expand a state not in terms of a vector basis (leading to a density matrix), but a matrix basis (leading to a moment matrix).

Felix Huber
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