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I have defined a set of quantum measurement operators of the form $$A_C = \sum _M \sqrt{Pr(M|C)} |M \rangle \langle M|$$ where $Pr(M|C)$ is the Poisson distribution $$Pr(M|C) = \frac{e^{-C}C^M}{M!}~~~~M = 0,1,2...$$ with mean $C$ and $|M \rangle$ are eigenstates of some observable and where there are finite members of $\{ |M \rangle \}_M$. I need the set of operators to satisfy the completeness condition $$\sum_{C}A_C^{\dagger}A_C = I.$$ My first attempt at this was to show that this is already satisfied: $$\sum_{C}A_{C}^{\dagger}A_C = \sum_{C}\sum_{M}Pr(M|C)|M \rangle \langle M| = \sum_{M}\sum_{C}Pr(M|C)|M \rangle \langle M| = \sum_M |M \rangle \langle M | = I$$ but having checked, I don't think the second last equation is necessarily valid for Poisson distributions. Can anyone see how I can adapt this so that the completeness relation is satisfied?

Thanks for any assistance.

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    $\begingroup$ You seem to be claiming that $\sum_C \frac{e^{-C}C^M}{M!}=1$ for all $M$ without specifying what set $C$ ranges over. That simply cannot be right - what if $C$ ranges over $\{1\}$? $\endgroup$ – Emilio Pisanty Nov 9 '17 at 17:08
  • $\begingroup$ @EmilioPisanty Point taken. Please see post if you have time. $\endgroup$ – user165535 Nov 10 '17 at 14:23
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Assuming that by $\sum_C$ you mean $\int_0^\infty \mathrm dC$ (though I have no idea why you'd think that that notation is appropriate), then yes: the Poisson distribution obeys $$ \int_0^\infty \mathrm{Pr}(M|C) \: \mathrm dC = \int_0^\infty \frac{e^{-C}C^M}{M!} \mathrm dC =1, $$ which is simply the gamma-function integral. That means that your $A_C$ are suitable Kraus operators for your space, as you have defined them, under the condition that $\sum_M |M\rangle\langle M|=I$ ─ but it doesn't even require you to say anything about the set of $M$'s, and these could range over a finite or infinite set, have gaps, etc.; so long as they're an orthonormal basis for your space of interest ─ and the weights $\mathrm{Pr}(M|C)$ make sense, which rules out $M=-1,-2,-3,\ldots$, because neither $M!$ nor the gamma integral make sense there ─, you're good to go.

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  • $\begingroup$ Thanks for your answe. My problem is that my $\{| M \rangle \}_M$ runs in integers from $-N$ to $N$. I want to define the operators $A_C$ such that for any $M'$ there exists a $C$ such that $A_C = \sum_M \sqrt{Pr(M|C)} |M \rangle \langle M |$ has the probability peaked at $| M' \rangle \langle M' |$. If I consider $C \in [0, \infty)$ then this is not satisfied. But I'm not sure how to change it without losing the property $$\int_{0}^{\infty} Pr(M|C)dC = \int_{0}^{\infty} \frac{e^{-C}C^M}{M!}dC =1$$ necessary for the $A_C$ to be valid Krauss operators. Can you maybe see any way to resolve this? $\endgroup$ – user165535 Nov 10 '17 at 14:37
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    $\begingroup$ If you want your $M$s to be negative integers then you *do* run into trouble, starting from the fact that $M!$ will cease to make sense for $M=-1,-2,-3,\ldots$. Unless you can put in something sensible there, then you're just barking at the moon. On the other hand, if all you want is some vague peak, then there's an infinite number of possible options and there's not enough information to constrain the problem. $\endgroup$ – Emilio Pisanty Nov 10 '17 at 14:55
  • $\begingroup$ Yes I am just interested in a vague peak. Can I not just reduce this to a notational change and use the same working. If I start with $M \in \{0,...,2N\}$ where $$\int_{0}^{\infty} Pr(M|C)dC = \int_{0}^{\infty} \frac{e^{-C}C^M}{M!}dC =1$$ and calculate as I did in the post, I then obtain $$\tilde{\rho}_C = \frac{A_C \rho A_C^{\dagger}}{P_C} = \sum_{M'} \sum_{M} \sqrt{Pr(M'|C)Pr(M|C)} |M' \rangle \langle M'| \psi \rangle \langle \psi | M \rangle \langle M |$$ $\endgroup$ – user165535 Nov 10 '17 at 15:39
  • $\begingroup$ I then use a change of notation $J := M-N$ thereby obtaining $\tilde{\rho}_C = \sum_{J'} \sum_{J} \sqrt{Pr(J'|C)Pr(J|C)}|J' \rangle \langle J'| \psi \rangle \langle \psi | J \rangle \langle J|$ where $J$ runs from $-N$ to $N$. Can you see anything wrong with this way of extending $M$ to have negative integer values? Thanks. $\endgroup$ – user165535 Nov 10 '17 at 15:39
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    $\begingroup$ @JohnDoe If you're already working with Kraus operators, then it might be a good idea to stick to density matrices, but yes, that is correct. For any more detailed questions, you should either post separately or ask in chat. $\endgroup$ – Emilio Pisanty Nov 13 '17 at 17:09

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