# Uniqueness of Density Operator

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.

• to which "representation" in particular are you referring to? Writing $\mathrm{Pr}(A=a|s)=\sum_i p_i \mathrm{Pr}(A=a| \psi_i)$? – glS Jan 12 at 20:23
• @glS to the representation in the line number 5 from the top in the snapshot. – Saptarshi Sahoo Jan 12 at 20:31
• 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 – glS Jan 12 at 20:37
• @gls please give another look at the question, i've given equation numbers to explain it clearly. – Saptarshi Sahoo Jan 12 at 20:54

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
• 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 $s_2$? – Saptarshi Sahoo Jan 12 at 20:40