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A Quantum measurement operators with Poisson distribution

  1. Dec 20, 2017 #1
    The following is a somewhat mathematical question, but I am interested in using the idea to define a set of quantum measurement operators defined as described in the answer to this post.

    Question:
    The Poisson Distribution ##Pr(M|\lambda)## is given by $$Pr(M|\lambda) = \frac{e^{-\lambda}\lambda^M}{M!}~~~~M = 0,1,2...$$ with mean ##\lambda##. In addition, the distribution has the property that $$\int_{0}^{\infty} Pr(M|\lambda) d \lambda = 1~~~~~~~~~~~~~~(*)$$ as is proved in the post. I am interested in adapting this Poisson Distribution so that it can include an additional parameter (say ##\mu##) which determines the width of the distribution (much like the variance does for the normal distribution. Does anyone have any ideas of how this can be achieved while still maintaining that property ##(*)## holds?

    Note I want to specifically work with the poisson distribution rather than the normal distribution due to how the shape of the poisson distribution at zero.

    Thanks for your time and let me know if you have any queries.
     
  2. jcsd
  3. Dec 20, 2017 #2

    Mentz114

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    Gold Member

    It can't be done. If you add another parameter it is not a Poisson distribution. You need a 2-param function lke a beta function that starts at 0 and can be shaped by the parameters. There are some plots in this document
     
  4. Dec 21, 2017 #3
    Thanks for your response. I have previously defined a set of quantum measurement operators by $$A_C = \sum _M \sqrt{Pr(M|C)} |M \rangle \langle M|~~~~~~~~~~~(*)$$ 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##. In order to be a suitable Kraus measurement operators I need the set of operators to satisfy the *completeness condition* $$\int_{C}A_C^{\dagger}A_C dC = I.$$

    This is satisfied if I use the Poisson Distribution since $$
    \int_0^\infty \mathrm{Pr}(M|C) \: \mathrm dC
    = \int_0^\infty \frac{e^{-C}C^M}{M!} \mathrm dC
    =1,
    $$

    The problem is that I want more control of the shape of the probability ##Pr(M|C)##, which is not allowed when using the standard Poisson Distribution. The Beta function as you proposed does allow freedom to manipulate the shape of the distribution but I am having difficulty seeing how we can define ##A_C## in the way above ##(*)## so that the completeness condition is satisified. Is it at all clear to you how this can be done in some way?
     
  5. Dec 21, 2017 #4

    Mentz114

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    Gold Member

    If you want discrete outcomes then one could divide the x-range into intervals ##i_n## and write ##P(x\in i_n) =B \int_n^{n+1}\beta(x;a,b)dx## from which your *completeness condition* follows if ##B^{-1}=\int_{-\infty}^{\infty}\beta(x;a,b)dx ##
     
  6. Jan 19, 2018 #5
    Thanks for your response but I'm not really following your reasoning as to how you would define the measurement operators ##A_C = \sum_{m}\sqrt{Pr(M|C)}|M \rangle \langle M|## using the Beta functions. In my scheme above using the Poisson distribution, I used the mean ##C## as the index for the measurement operators ##\{A_C\}_C## and then used the fact that ##\int_{0}^{\infty}Pr(M|C)dC = 1## to show that the *completenes relation* ##\int A_{C}A_{C}^{\dagger}dC = I## is satisfied.. Explicitly how are you proposing to define measurement operators with the Beta function?
     
  7. Jan 20, 2018 #6
    @Mentz114
    Is the idea you are referring to something like this:
    Define $$A_{C} := \sum_{x}\sqrt{\text{Pr} (x|n)}|x \rangle \langle x |$$ where $$\text{Pr}(x|n) :=
    \begin{align}
    \begin{cases}\text{Pr}(x \in i_n) = \frac{\int_{n}^{n+1}\beta(x: \alpha, \beta)}{\int_{0}^{\infty}\beta(x:\alpha, \beta)}~~~~~~\text{if }x \in i_n\\ 0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\text{if } x \notin i_n
    \end{cases}
    \end{align}$$

    Hence ##\sum_n A_n^{\dagger}A_n = \sum_{x}\bigg[\sum_n \text{Pr(x|n)} \bigg]|x \rangle \langle x | = I##
     
  8. Jan 20, 2018 #7

    Mentz114

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    Gold Member

    I don't know. I don't understand what you've written. I would write
    [tex]
    \text{Pr}(i=n) = \frac{\int_{x_n}^{x_{n+1}}dx\beta(x: \alpha, \beta)}{\int_{0}^{\infty}dx\beta(x:\alpha, \beta)}
    [/tex]
    where the ##x_n## are the boundaries of your your discrete regions. ##i## is a discrete random variable that depends on ##x## and the ##x_n##. It can betailored to suit.
     
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