# A Probability amplitude

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1. May 7, 2017

### redtree

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:
$P((\textbf{r}_{n}| \textbf{r}_{n-1})) = \psi^*_{\textbf{r}_{n}} \cdot \psi_{\textbf{r}_{n}} = |\psi_{\textbf{r}_{n}}|^2$

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

2. May 9, 2017

### Stephen Tashi

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?

3. May 9, 2017

### redtree

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:
$P((\textbf{r}_{n}| \textbf{r}_{n})) = \psi^*_{\textbf{r}_{n}} \otimes \psi_{\textbf{r}_{n}} = |\psi_{\textbf{r}_{n}}|^2$

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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
4. May 10, 2017

### redtree

Upon further reflection, it seems that the equation should read as follows:
$\psi_{\textbf{r}_{n}} \doteq \sqrt{P((\textbf{r}_{n}| \textbf{r}_{n-1})) }$

5. May 10, 2017

### Stephen Tashi

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 $\psi_3$ ?