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Consider calculating the probability of the outcome m alone of some composite system $AB$. \begin{align*} p_A(m) &= \sum_{n=0}^{d_B-1} p_{AB}(m,n)\,,\\ &= \sum_{n=0}^{d_B-1}(⟨α_m|⊗⟨β_n|)\rho_{AB}(|α_m⟩⊗|β_n⟩)\,. \end{align*}

I'm trying to show that this value is independent of the basis $\{|β_n\rangle\}\,,$ where $n = 0, 1, \cdots , d_B-1$ (so we can just replace it with the computational basis for example.

I know that I've to use the no signalling theorem for it, and will have to decompose the density matrix $\rho_{AB}$ into $\sum_i p_i |\psi_i⟩⟨\psi_i|$ but beyond that I haven't made much progress

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Independence of the slightly more general expression $\sum_n\langle x\otimes \beta_n|\rho_{AB}|y\otimes \beta_n\rangle$ from the orthonormal basis $\{\beta_n\}_n$ boils down to the fact that the same is true for the resolution of the identity ${\bf1}=\sum_n|\beta_n\rangle\langle \beta_n|$ (=this identity holds for all orthonormal bases). For all $x,y,\rho_{AB}$ we compute: \begin{align*} \sum_n\langle x\otimes \beta_n|\rho_{AB}|y\otimes \beta_n\rangle&=\sum_n{\rm tr}\big(|y\otimes \beta_n\rangle\langle x\otimes \beta_n|\rho_{AB}\big)\\ &=\sum_n{\rm tr}\big((|y\rangle\langle x|\otimes |\beta_n\rangle\langle \beta_n|)\rho_{AB}\big)\\ &={\rm tr}\Big(\Big(|y\rangle\langle x|\otimes \sum_n|\beta_n\rangle\langle \beta_n|\Big)\rho_{AB}\Big)={\rm tr}\big((|y\rangle\langle x|\otimes {\bf1})\rho_{AB}\big) \end{align*} This shows that the expression $p_A(m)={\rm tr}((|\alpha_m\rangle\langle \alpha_m|\otimes {\bf1})\rho_{AB})$ is indeed independent of the chosen orthonormal basis on $B$, as claimed.

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