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

  • $\begingroup$ 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

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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.