Let $\rho_1, \rho_2, \rho_3, \rho_4$ be arbitrary single-qubit density matrices.

Let $A$ be an Hermitian operator and its spectral decomposition as $A = \sum_i \lambda_i \lvert i \rangle \langle i \rvert$. Then, we define projection operators $P(A) = \Pi_{i: \lambda_i \geq 0}\lvert i \rangle \langle i \rvert - \Pi_{i: \lambda_i <0}\lvert i \rangle \langle i \rvert$. From this, we define four projectors:

$P_1 = P(\lvert 0 \rangle \langle 0 \rvert - \rho_1)$

$P_2 = P(\lvert 0 \rangle \langle 0 \rvert - \rho_2)$

$P_3 = P(\lvert 0 \rangle \langle 0 \rvert - (\rho_1+\rho_2+\rho_3+\rho_4))$

$P_4 = P(\lvert 0 \rangle \langle 0 \rvert - (\rho_1+\rho_2-i\rho_3+i\rho_4))$

Are $P_i$'s related to each other (can we express one projector in terms of a linear combination of other projectors)?


1 Answer 1


tl;dr: In the single-qubit case the explicit formula $$P(|0\rangle\langle0|-\rho)=\begin{cases} 0&\rho=|0\rangle\langle 0|\\ \frac{|0\rangle\langle0|-\rho}{\sqrt{|\langle 0|\rho|1\rangle|^2+(\langle 1|\rho|1\rangle)^2}}&\text{ else} \end{cases}\tag1 $$ holds for all density matrices $\rho\in\mathbb C^{2\times 2}$ meaning all that $P$ does in your case is scale the input. This shows that expressing some of your matrices via the others is possible iff the matrices themselves are linearly dependent to begin with.

The remainder of this answer is concerned with proving Eq. (1). Diagonalizing arbitrary $A$ Hermitian as $A=UDU^\dagger$ for some $U$ unitary, $D$ diagonal shows $P(A)=U\,{\rm sgn}(D)U^\dagger$ with ${\rm sgn}$ the usual sign function. Now given any qubit density matrix $$ \rho=\begin{pmatrix}1-a&b\\b^*&a\end{pmatrix}$$ (i.e. $a\in[0,1]$, $b\in\mathbb C$ with $|b|^2\leq a(1-a)$) the difference $$|0\rangle\langle 0|-\rho=\begin{pmatrix}a&-b\\-b^*&-a\end{pmatrix} $$ is a traceless Hermitian matrix; hence it has eigenvalues $\pm\sqrt{-\det(|0\rangle\langle 0|-\rho)}=\pm\sqrt{a^2+|b|^2}$. Thus there exists $U\in\mathbb C^{2\times 2}$ unitary such that \begin{align*} P(|0\rangle\langle0|-\rho)&=P(U\,{\rm diag}(\sqrt{a^2+|b|^2},-\sqrt{a^2+|b|^2})U^\dagger)\\ &=U\,{\rm diag}({\rm sgn}(\sqrt{a^2+|b|^2}),{\rm sgn}(-\sqrt{a^2+|b|^2}))U^\dagger) \end{align*} which—unless $a=0$—evaluates to \begin{align*} P(|0\rangle\langle0|-\rho)&=U\sigma_zU^\dagger\\ &=\frac{U\,{\rm diag}(\sqrt{a^2+|b|^2},-\sqrt{a^2+|b|^2})U^\dagger}{\sqrt{a^2+|b|^2}}=\frac{|0\rangle\langle0|-\rho}{\sqrt{a^2+|b|^2}}\,. \end{align*} This implies (1), as desired.


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