# Show that while calculating partial traces the probability is independent of the basis of one of the measurements

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

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.