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I have been reading "Introduction to Quantum Information Science" by Masahito Hayashi, Satoshi Ishizaka,Akinori Kawachi, Gen Kimura and Tomohiro Ogawa; Springer Publication. I'm currently in the density operator section, page 96. There they are considering two ensemble states as follows $s_1 = \left\{ \frac{1}{2}, \frac{1}{2}; |0\rangle,|1\rangle \right\}$ and $s_2 = \left\{ \frac{1}{2}, \frac{1}{2}; |+\rangle,|-\rangle \right\}$. Let there be a arbitary physical quantity be $A = \sum_a aP_a$ (Spectral Decomposition). Hence they are doing this: $$ Pr(A=a|s_2) = \frac{1}{2} \langle+|P_a|+\rangle + \frac{1}{2} \langle-|P_a|-\rangle = \frac{1}{2} \langle0|P_a|0\rangle + \frac{1}{2} \langle1|P_a|1\rangle = Pr(A=a|s_1)\tag1 $$ They are concluding that that the uniqueness of the states $s_1$ and $s_2$ are getting lost in equation (1), hence they are suggesting below form(2): $$ Pr(A=a|s) = \sum_i p_iPr(A=a|\text{ }|\psi_i\rangle) = \sum_i p_i \langle\psi_i|P_a|\psi_i\rangle \tag2 $$ They are claiming that the above representation[2] is unique

In order to get rid of non-uniqueness defect...

. But i cant understand how? To me both(1) & (2) are same. I'm adding the snapshot too.enter image description here

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  • $\begingroup$ to which "representation" in particular are you referring to? Writing $\mathrm{Pr}(A=a|s)=\sum_i p_i \mathrm{Pr}(A=a| \psi_i)$? $\endgroup$ – glS Jan 12 at 20:23
  • $\begingroup$ @glS to the representation in the line number 5 from the top in the snapshot. $\endgroup$ – Saptarshi Sahoo Jan 12 at 20:31
  • $\begingroup$ I don't really see anything that seems like a "representation" in the fifth line. I tried to answer based on the general idea of why two ensembles would lead to the same observed probabilities $\endgroup$ – glS Jan 12 at 20:37
  • $\begingroup$ @gls please give another look at the question, i've given equation numbers to explain it clearly. $\endgroup$ – Saptarshi Sahoo Jan 12 at 20:54
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Write an ensemble as $\{(p_i,\psi_i)\}_i$, with $p_i$ probabilities and $\psi_i$ pure states. Let $\mathcal I_1\equiv \{(p_i,\psi_i)\}_i$ and $\mathcal I_2\equiv \{(q_i,\phi_i)\}_i$ be two such ensembles. Suppose that $$\sum_i p_i \lvert \psi_i\rangle\!\langle\psi_i\rvert = \sum_i q_i \lvert \phi_i\rangle\!\langle\phi_i\rvert$$ (you can verify that this is the case in your example).

Your statement amounts to observing that, performing some measurement $A$, the probability of getting the outcome $a$ with the ensemble $\mathcal I_1$ is the same as that with the ensemble $\mathcal I_2$.

This probability reads, for $\mathcal I_1$, $$\mathrm{Pr}(A=a|\mathcal I_1) = \sum_j p_j \mathrm{Pr}(A=a|\psi_j) = \sum_j p_j \langle\psi_j|P_a|\psi_j\rangle = \mathrm{Tr}\left(P_a \sum_j p_j |\psi_j\rangle\!\langle\psi_j|\right).\tag A$$ You similarly get for $\mathcal I_2$ $$\mathrm{Pr}(A=a|\mathcal I_2) = \mathrm{Tr}\left(P_a \sum_j q_j |\phi_j\rangle\!\langle\phi_j|\right).\tag B$$ But by assumption the sum in the parentheses of (A) and (B) is the same, hence the conclusion

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  • $\begingroup$ What i was asking is that how equation [2] is "getting rid of the non-uniqueness" from the probability expression given earlier[1], which was, as stated by them, "non-unique" for $s_1$ and $s_2$? $\endgroup$ – Saptarshi Sahoo Jan 12 at 20:40
  • $\begingroup$ because that expression gives identical results when you evaluate it in either ensemble $\endgroup$ – glS Jan 12 at 20:57
  • $\begingroup$ Then the claim "getting rid of the non-uniqueness" is wrong, am i right? Becoz the two ensembles are kinda similar in the probability terms? And (1) and (2) are same, right? $\endgroup$ – Saptarshi Sahoo Jan 12 at 21:01
  • $\begingroup$ I don't understand what you mean. The claim is right, in that the two ensemble provide identical results, thus they are all equivalent, i.e. states can be represented uniquely $\endgroup$ – glS Jan 12 at 21:25
  • $\begingroup$ could you please describe what this phrase , "State can be represented uniquely" mean? Actually I'm not a native English speaker. $\endgroup$ – Saptarshi Sahoo Jan 12 at 21:58

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