How to express a probability distribution $P(x,y,z)= \sum_\lambda P(x|y,\lambda)P(y|\lambda,z)P(z)P(\lambda)$ in terms of a trace of a density matrix?



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


Posted 2021-02-16T16:17:57.617

Reputation: 141

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 – Shashaank – 2021-02-18T13:21:42.733

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 – glS – 2021-02-18T14:01:38.447

@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. – Shashaank – 2021-02-18T14:15:37.143



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

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

Adam Zalcman

Posted 2021-02-16T16:17:57.617

Reputation: 4 307

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. – Shashaank – 2021-02-17T07:43:17.277

I'm not sure I understand... The above calculation shows that the two are equal. Equality is symmetric. – Adam Zalcman – 2021-02-17T07:48:16.683

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 – Shashaank – 2021-02-17T11:09:20.210

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. – Adam Zalcman – 2021-02-17T18:18:45.577