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${P}_m := \{ \Phi_{1m}, \Phi_{2m}, \ldots , \Phi_{nm} \}$ where $$|\Phi_{im} \rangle = \wp_{1m} | \phi_1 \rangle + \wp_{2m} | \phi_{2m} \rangle + \cdots + \wp_{km} | \phi_{km} \rangle$$ such that $\wp_{im}$ is a probability amplitude for $\Phi_{im} \in \mathbb{P}_m$ being in it's $i^{th}$ state, $\phi_{im} $ Probability amplitudes are defined in such a way that $|\wp_{1m}|^2 + |\wp_{2m}|^2 + \cdots + |\wp_{km}|^2 = 1$ and let $\mathbb{P}_m | i \rangle$ be the set $\mathbb{P}_m$ with each of its elements in the $i^{th}$ state. We define the sets to not be pairwise disjoint in the following way:

Let $N$ be an index set such that $\forall m \in N$ let $\mathbb{P}_m$ be a set. Therefore we have $$ \mathbb{P}_m \mid \bigcap_{m \in N} \mathbb{P}_m | i \rangle\neq O$$ in other words $\exists \phi_{i1} , \phi_{i2} , \ldots , \phi_{in} \in \mathbb{P}_1, \mathbb{P}_2, \ldots, \mathbb{P}_n \mid \phi_{i1} \equiv \phi_{i2} \equiv \cdots \equiv \phi_{in} $

The Mathematics of Probabilistic Sets

A Probabilistic Set can be thought of as a set of normalised wave functions $\Phi_{im}$ that have at least one state in common and therefore have a finite probability of intersection.

Union of Probabilistic Sets

The union of probabilistic sets can be defined quite trivially as $$\bigcup_{m \in N} \mathbb{P}_m : = \{ \Phi_{11}, \Phi_{12}, \ldots, \Phi_{1N}, \Phi_{21}, \Phi_{22}, \ldots, \Phi_{2N}, \ldots ,\Phi_{1n}, \Phi_{2n}, \ldots, \Phi_{nN} \}$$ Intersection of Probabilistic Sets It should be fairly clear that two probabilistic sets $\mathbb{P}_m$ and $\mathbb{P}_K$ do not have a well defined intersection. Instead we may discuss the \textit{probability} of two such sets intersecting. \ The probability of intersection of the probabilistic sets shall be denoted $$\mathcal{P} \Big( \bigcap_{m \in N} \mathbb{P}_m | i \rangle \Big)$$ We can define $\Phi_{im} (\phi_{im})$ to be the wave function for $\Phi_{im}$ where $\phi_{im} \equiv [\alpha_i, \beta_i]$. So we define a state $\phi_{im}$ to be the corresponding interval $[\alpha_i,\beta_i]$ on the wave function $\Phi_{im} (\phi_{im})$. It now follows that the probability of finding the element $\Phi_{im}$ in the state $\phi_{im}$ is given by $$ P_{\phi_{im} \in \Phi_{im}} = \int_{\alpha_i}^{\beta_i} d \phi_{im} \Big| \Phi_{im} (\phi_{im}) \Big|^2$$ If only one such $\phi_{im}$ per set exists satisfying $\phi_{i1} , \phi_{i2} , \ldots , \phi_{in} \in \mathbb{P}_1, \mathbb{P}_2, \ldots, \mathbb{P}_n \mid \phi_{i1} \equiv \phi_{i2} \equiv \cdots \equiv \phi_{in} $, then the probability of the intersection of sets is given by $$ \mathcal{P} \Big( \bigcap_{m \in N} \mathbb{P}_m | i \rangle \Big) = \prod_{m \in N} \Big( \int_{\alpha_i}^{\beta_i} d \phi_{im} \Big| \Phi_{im} (\phi_{im}) \Big|^2 \Big)_m$$ However if two such states are equivalent per set, call this second equivalent state $\phi_{jm}$ the probability becomes $$ \mathcal{P} \Big( \bigcap_{m \in N} \mathbb{P}_m | i \rangle \Big) = \prod_{m \in N} \Big( \int_{\alpha_i}^{\beta_i} d \phi_{im} \Big| \Phi_{im} (\phi_{im}) \Big|^2 + \int_{\alpha_j}^{\beta_j} d \phi_{jm} \Big| \Phi_{jm} (\phi_{jm}) \Big|^2 \Big)_m$$ Now we extend this to the obvious case of: what if such elements share $\mathfrak{R}$ equivalent states? Well intuitively the probability of such an intersection occurring approaches $1$ as $\mathfrak{R} \to n$.
Trivially following from the previous two equations, the probability of intersection is given by $$ \mathcal{P} \Big( \bigcap_{m \in N} \mathbb{P}_m | i \rangle \Big) = \prod_{m \in N} \Big( \sum_{\mathfrak{r} \leq \mathfrak{R}} \Big( \int_{\alpha_i}^{\beta_i} d \phi_{im} \Big| \Phi_{im} (\phi_{im}) \Big|^2 \Big)_\mathfrak{r} \Big)_m$$

Is everything I'm doing thus far correct? Is anything I'm doing thus far correct? I was just brainstorming ideas and this is what I came up with so I'd like to know if I'm right. Thanks Any input is appreciated

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I wasn't sure on the tags. If anyone feels they have more appropriate tags, please, feel free to change them –  Anthony Peter Aug 13 '13 at 1:06
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I would avoid the use of $\wp$ for anything except the Weierstrass elliptic function. –  deoxygerbe Aug 13 '13 at 1:44
    
@deoxygerbe Noted. I'll fix that –  Anthony Peter Aug 13 '13 at 1:52
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@ChrisCulter Well it seems to me without too much rigor, that if $k=1$, the set would simply be a normal set as it can only take on one possible value. While if $k=2$, the set has a probability of being in one of two states. Now since it's only a probability that the element will take on one of two states, each element can have a degree of membership to the set according to its probability amplitude. aka a sort of definition of a fuzzy set. Is my intuition right? –  Anthony Peter Aug 13 '13 at 3:21
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@AnthonyPeter That's impressive! Well, if you make any more progress on explanations and/or examples, please update the question; I'd like to take a look! –  Chris Culter Aug 13 '13 at 4:20

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