I have been given and expression for a probability distribution precisely,

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

and I have been asked to show that the above expression can be written in the forma of the following (with correct changes)

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

where $\rho_{AB}, E_A^{x|y}, E_B^{y|z}$ are the density matrix and POVM’s 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, leave alone getting the whole expression correctly.

Cross-posted on quantumcomputing.SE

  • $\begingroup$ @glS are you the down voter. If so please explain the reason $\endgroup$
    – Shashaank
    Feb 18 at 13:22

It's not exactly clear from the question exactly what you want but here's maybe something to get you started.

If you define some density matrices $\rho_A^\lambda$ such that $\mathrm{Tr}[\rho_A^{\lambda} E_A^{x|y}] = p(x|y,\lambda)$ and similarly define $\rho_B^{\lambda}$ such that $\mathrm{Tr}[\rho_B^{\lambda} E_B^{y|z}] = p(y|z,\lambda)$. Then you can define the joint state $$ \rho_{AB} = \sum_\lambda p(\lambda) \rho_A^\lambda \otimes \rho_B^\lambda. $$

Then you have \begin{align*} \mathrm{Tr}[\rho_{AB} (E_A^{x|y}\otimes E_B^{y|z})]p(z) &= \mathrm{Tr}[(\sum_\lambda p(\lambda) \rho_A^\lambda \otimes \rho_B^\lambda) (E_A^{x|y}\otimes E_B^{y|z})]p(z) \\ &= \sum_\lambda p(\lambda)\mathrm{Tr}[(\rho_A^\lambda \otimes \rho_B^\lambda) (E_A^{x|y}\otimes E_B^{y|z})]p(z) \\ &= \sum_\lambda p(\lambda)\mathrm{Tr}[\rho_A^\lambda E_A^{x|y}] \mathrm{Tr}[ \rho_B^\lambda E_B^{y|z}]p(z) \\ &= \sum_\lambda p(\lambda) p(x|y,\lambda) p(y|z,\lambda) p(z)\\ &= p(x,y,z) \end{align*}

  • $\begingroup$ Can it shown the other way, beginning with the probability distribution and showing that it can be written like a trace? $\endgroup$
    – Shashaank
    Feb 16 at 17:48
  • $\begingroup$ Because the context requires that a probability distribution can be written in terms of a trace rather than the other way round $\endgroup$
    – Shashaank
    Feb 16 at 17:55
  • $\begingroup$ Well then if you take the above construction you need only show that given some set of POVMs $E^{x|y}$ there exist states $\rho^{\lambda}$ such that $\mathrm{Tr}[\rho^{\lambda} E^{x|y}] = p(x|y,\lambda)$. $\endgroup$
    – Rammus
    Feb 16 at 18:02

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