2
$\begingroup$

I have been given and expression for a probability distribution

$$P(x,y,z)= \sum_\lambda P(x|y,\lambda)P(y|\lambda,z)P(z)P(\lambda)$$

and I have been asked to show that the above expression can be written in the form

$$P(x,y,z)= Tr(\rho_{AB}(E_A^{x|y} \otimes E_B^{y|z}))P(z)$$

for $\rho_{AB}$ a density matrix and $E_A^{x|y}, E_B^{y|z}$ POVMs on system AB.

I have no clue of even how the trace comes into picture. I have no clue as to how the first expression can be simplified to get a trace, let alone getting the whole expression correctly.


Cross-posted on math.SE

The answer over there is also interesting to look at......

$\endgroup$
7
  • $\begingroup$ Would the Down voter like to reveal his issue. Is it problematic to ask to ask a question on 2 sites. I urge the Down voter to clarify his issue and no necessarily down vote just because he doesn’t understand $\endgroup$
    – Shashaank
    Feb 18 at 13:21
  • $\begingroup$ not necessarily problematic (it depends on the site, but usually it's not here), but it should be pointed out, and the questions linked together, to avoid effort duplication on the answerers' part $\endgroup$
    – glS
    Feb 18 at 14:01
  • $\begingroup$ @glS ok. It would have been good if the downvoter left a reason. I would then least know what's the issue with the question and could probably add to the question by editing it. $\endgroup$
    – Shashaank
    Feb 18 at 14:15
  • $\begingroup$ I repeat, the Down voter explain your reason. Don’t unnecessarily down vote a thing that went above your head. $\endgroup$
    – Shashaank
    Mar 25 at 19:13
  • $\begingroup$ there isn't any downvote on this question, nor on the one on math $\endgroup$
    – glS
    Mar 25 at 23:03
4
$\begingroup$

First, recall that $\mathrm{tr} A = \sum_i \langle i|A|i \rangle$. Each equation is then a sum where all terms are products of $P(z)$ and three other quantities. Further, the sum in the first equation ranges over a single index suggesting that all matrices under the trace are diagonal. In fact, since we are working with a composite system this also suggests that the basis in which the POVM elements are diagonal is the Schmidt basis of $\rho_{AB}$. At this point we could check which way of mapping the factors between the two equations works, but we don't have to do that since the superscripts on the POVM elements helpfully tell us the mapping.

Taking these observations into account we guess

$$ E_A^{x|y} = \sum_\lambda P(x|y,\lambda) |\lambda\rangle\langle\lambda| \\ E_B^{y|z} = \sum_\lambda P(y|\lambda,z) |\lambda\rangle\langle\lambda| \\ |\psi_{AB}\rangle = \sum_\lambda \sqrt{P(\lambda)} |\lambda\rangle|\lambda\rangle \\ \rho_{AB} = |\psi_{AB}\rangle\langle\psi_{AB}| $$

where $|\lambda\rangle$ is an orthonormal basis. Now, $E_A^{x|y}$ is a valid POVM for fixed $y$ and similarly $E_B^{y|z}$ for fixed $z$. It is not clear from the question that this is the desired POVM structure, but it is what is suggested by the conditional sign in the superscripts.

Let's try our guess

$$ \begin{align} P(x,y,z) &= P(z)\,\mathrm{tr}\left(\rho_{AB}(E_A^{x|y} \otimes E_B^{y|z})\right) \\ &= P(z) \sum_{\lambda_1,\lambda_2}\langle\lambda_1|\langle\lambda_2|\rho_{AB}\left(E_A^{x|y} \otimes E_B^{y|z}\right)|\lambda_1\rangle|\lambda_2\rangle \\ &= P(z) \sum_{\lambda_1,\lambda_2}\langle\lambda_1|\langle\lambda_2|\sum_{\lambda_3, \lambda_4} \sqrt{P(\lambda_3)P(\lambda_4)} |\lambda_3\rangle\langle\lambda_4| \otimes |\lambda_3\rangle\langle\lambda_4| \\ & \left(E_A^{x|y} \otimes E_B^{y|z}\right)|\lambda_1\rangle|\lambda_2\rangle \\ &= P(z) \sum_{\lambda_3,\lambda_4}\sqrt{P(\lambda_3)P(\lambda_4)}\langle\lambda_4|\langle\lambda_4|\left(E_A^{x|y} \otimes E_B^{y|z}\right)|\lambda_3\rangle|\lambda_3\rangle \\ &= P(z) \sum_{\lambda_3,\lambda_4}\sqrt{P(\lambda_3)P(\lambda_4)} \\ & \langle\lambda_4|\langle\lambda_4|\left(\sum_{\lambda_5} P(x|y,\lambda_5) |\lambda_5\rangle\langle\lambda_5| \otimes \sum_{\lambda_6} P(y|\lambda_6,z) |\lambda_6\rangle\langle\lambda_6|\right)|\lambda_3\rangle|\lambda_3\rangle \\ &= P(z) \sum_{\lambda}P(\lambda)P(x|y,\lambda) P(y|\lambda,z). \end{align} $$

$\endgroup$
4
  • $\begingroup$ Can I show the opposite, that is the probability distribution can be written as a trace ( not the other way round). The context of the question requires to do just that. $\endgroup$
    – Shashaank
    Feb 17 at 7:43
  • $\begingroup$ I'm not sure I understand... The above calculation shows that the two are equal. Equality is symmetric. $\endgroup$ Feb 17 at 7:48
  • $\begingroup$ The context requires that the probability distribution under suitable assumption reduces to tho the expression with the trace and that this is possible for all such probability distribution but it is not possible to reduce the expression for trace and show it is equal to the probability distribution. Do you have an idea how to do that $\endgroup$
    – Shashaank
    Feb 17 at 11:09
  • $\begingroup$ Maybe, but I can't say for sure without seeing the details. The matter of impossibility of expressing one type of formula as another type of formula is a very different type of problem than the matter of finding a way to express one formula as another. Please submit another question for this. It sounds interesting. $\endgroup$ Feb 17 at 18:18

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.