UVa Physics Review: probability theory

Thursday, June 17, 2010

Particle Decay is a Random Phenomena: Part III

Now that we understand what it means for a process to be random let's delve into the mathematics. The most important distribution to understand is the Poisson distribution. The probability of exactly n Poisson occurrences during a time interval where This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.

This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.

were expected is:

To satisfy the Poisson conditions events must be occurring with an average rate and be independent of each other, decaying pions certainly qualify. There are some interesting and useful features of this distribution. For instance the mean ( This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.

)and the variance ( This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.

) are equal to This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.

. That means the standard deviation of a Poisson distributed data set is This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.

Say we measure 1,000,000 pions decays. We now want to state our measurement of the This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.

branching ratio ( This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.

). How do we determine the statistical error of our measured value ( This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.

)? Since we know the decays obey Poisson statistics we simply use the features we learned above combined with standard statistical estimation rules:

http://en.wikipedia.org/wiki/Standard_error_(statistics)

The important thing to note is that the "one over square root of N" has nothing to do with Poisson statistics but rather is a feature of sampling error.

Particle Decay is a Random Phenomena: Part I

The standard model tells us that all particles of the same type are indistinguishable. Therefore when describing a system of multiple particles (of the same time) we must take great care to insure our particles are treated as indistinguishable. Accidentally treating particles as identical can cause serious miscalculations. For example (not physics but very illustrative):

For the following questions assume I have two children, child A and child B: the probability of any individual child being a boy is 1/2, the probability of the child being born on a Tuesday is 1/7, etc. For each question I will apply a different set of constraints. Only the constraints mentioned in the question affect that question.

Q1: What is the probability I have two boys?

A1: 1/4.

Q2: Child A is a boy. What is the probability I have two boys?

A2: 1/2.

Q3: At least one of my children is a boy. What is the probability I have two boys?

A3: 1/3.

Q4: At least one of my children is a boy born on Tuesday. What is the probability I have two boys?

A4: 13/27.

The reason for the counterintuitive answers to questions 3 and 4 is that we don't know which child (A or B) satisfies the constraint. Since it could be either the probability of overlap changes the result. The more (less) probable the overlap the closer the answer is to 1/3 (1/2).

But what does this have to do with particle decay?

Quantum Mechanics says there are no hidden variables. If the time a particle had been "alive" affected the probability it would decay in the future then there would have to be some hidden variable to fully describe the particle! If we could look at a particle and tell how long it had been alive then particles would not be indistinguishable!