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A Probability amplitude

  1. May 7, 2017 #1
    1. Given a Markov state density function:
    ## P((\textbf{r}_{n}| \textbf{r}_{n-1})) ##

    ##P## describes the probability of transitioning from a state at ## \textbf{r}_{n-1}## to a state at ##\textbf{r}_{n} ##. If ## \textbf{r}_{n-1} = \textbf{r}_{n}##, then ##P## describes the probability of the state at ##\textbf{r}_{n}## transitioning to itself.

    2. A probability density function can be decomposed into probability amplitudes, where the probability density function ##P## is the product of a probability amplitude ##\psi_{\textbf{r}_{n-1}} ## and the complex modulus of second probability amplitude ##\psi^*_{\textbf{r}_{n}} ## where:
    ## P((\textbf{r}_{n}| \textbf{r}_{n-1})) = \psi^*_{\textbf{r}_{n}} \cdot \psi_{\textbf{r}_{n-1}} ##

    Again, if ## \textbf{r}_{n-1} = \textbf{r}_{n}##, then:
    ## P((\textbf{r}_{n}| \textbf{r}_{n-1})) = \psi^*_{\textbf{r}_{n}} \cdot \psi_{\textbf{r}_{n}} = |\psi_{\textbf{r}_{n}}|^2##

    -----------------

    Unfortunately, I have not been able to find a derivation for #2. Does anyone know where I can find such a derivation?
     
  2. jcsd
  3. May 9, 2017 #2

    Stephen Tashi

    User Avatar
    Science Advisor

    According to that notation, to speak ##P((\textbf{r}_{n}| \textbf{r}_{n-1}))## as a specific number, we must specify a value for each of ##\textbf{r}_{n}## and ##\textbf{r}_{n-1}##. So the notation ##P( \textbf{r}_{n} = b | \textbf{r}_{n-1} = a)## represents a specific numerical value.

    It isn't clear what is being claimed. In a Markov process, the values of ##P( \textbf{r}_{n} = b | \textbf{r}_{n-1} = a)## and ##P( \textbf{r}_{n} = a | \textbf{r}_{n-1} = b) ## may be different. What non-commutative arithmetic does the claim use to allow such a situation to happen?
     
  4. May 9, 2017 #3
    1.
    You are correct if ##\textbf{r}_{n}## and ##\textbf{r}_{n-1}## represent scalars (##a## and ##b##), such that ##P( \textbf{r}_{n} = b | \textbf{r}_{n-1} = a) = c##, where ##c## is also a scalar. However, if ##\textbf{r}_{n}## and ##\textbf{r}_{n-1}## are vectors, then ##P((\textbf{r}_{n}| \textbf{r}_{n-1}))## represents a transition probability tensor (transformation matrix). In my understanding, in both cases, ##P## represents a probability density for the transformation.

    2.
    You are correct. I made a notational error. I should have written the following:
    ##P((\textbf{r}_{n}| \textbf{r}_{n-1})) = \psi^*_{\textbf{r}_{n}} \otimes \psi_{\textbf{r}_{n-1}}##

    In this form, assuming ## \psi^*_{\textbf{r}_{n}}## and ## \psi_{\textbf{r}_{n-1}}## are vectors, the expression does not commute.

    Such that, if ##\textbf{r}_{n-1} = \textbf{r}_{n}##, then:
    ##P((\textbf{r}_{n}| \textbf{r}_{n})) = \psi^*_{\textbf{r}_{n}} \otimes \psi_{\textbf{r}_{n}} = |\psi_{\textbf{r}_{n}}|^2##

    -----------------------------

    And my question remains: Does anyone know where I can find such a derivation for the following:
    ##P((\textbf{r}_{n}| \textbf{r}_{n-1})) = \psi^*_{\textbf{r}_{n}} \otimes \psi_{\textbf{r}_{n-1}}##

    Or, am I overthinking the problem, and the answer is simply in the definition of ## \psi##?
     
    Last edited: May 9, 2017
  5. May 10, 2017 #4
    Upon further reflection, it seems that the equation should read as follows:
    ## \psi_{\textbf{r}_{n}} \doteq \sqrt{P((\textbf{r}_{n}| \textbf{r}_{n-1})) }##
     
  6. May 10, 2017 #5

    Stephen Tashi

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    Science Advisor

    Does the claim comes from derivation that involves physics or is it suppose to be purely a mathematical result?

    Assume we have the transition matrix ##M## of a Markov process with a 10 states. What does ##\psi_3## denote? Is it a vector with 10 components, each of which is a complex number?

    Is ##\psi^*_3## a vector whose components are the complex conjugates of the components of ##\psi_3## ?
     
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