What is the probability of an electron emitting a photon?

[James1238765]

TL;DR Summary

How to calculate numerical probabilities using Feynman diagram?

I have seen many tutorials that provide steps how to transcribe a Feynman diagram into algebra, for instance [here]:

However, I have never seen the final line of the calculation converted into a real number. What are the steps to get from the algebra equations transcribed using the Feynman diagram, into a final concrete (eg 0.1726341) probability number for this interaction to happen?

 

[Demystifier]

Since you don't seem like someone who would read a whole textbook, see e.g. http://nicadd.niu.edu/~dhiman/courses/phys684_10/lectures/process_rates.pdf
It tells you how to compute decay widths and cross sections, which are probability-related quantities that we actually measure. But you will not find a lot of numbers, even in detailed textbooks on particle physics the computation of actual numbers is usually left to exercises.

 

[Spinnor]

Search out the following text, Quarks and Leptons: An Introductory Course in Modern Particle Physics 1st Edition by Francis Halzen (Author), Alan D. Martin (Author).

In the following you get a number at the very end, 70 pico barns,

https://www.physics.purdue.edu/webapps/index.php/course_document/index/phys564/1744/25/15462

From a Google search,

https://www.google.com/search?q=sim...EEMjAuMZgBAKABAcgBCMABAQ&sclient=gws-wiz-serp

 

[Vanadium 50]

The probability of the diagram you have written down is calculable. The answer is 0.000000.

Momentum and energy are not conserved.

This may sound like a quibble, but it highlights @Demystifier's comment about texts. There is a lot that needs to be considered when making these calculations, and that's why we have textbooks.

 

[haushofer]

TL;DR Summary: How to calculate numerical probabilities using Feynman diagram?

I have seen many tutorials that provide steps how to transcribe a Feynman diagram into algebra, for instance [here]:

View attachment 319296

However, I have never seen the final line of the calculation converted into a real number. What are the steps to get from the algebra equations transcribed using the Feynman diagram, into a final concrete (eg 0.1726341) probability number for this interaction to happen?

Such a diagram is not a process at all. It's the leading term in an infinite series which represents the process. As such the mathematical representation/value of that diagram does not give what you want.

Interpreting Feynman diagrams as processes is something which only happens in popular science. Or at least: that ought to be.

 

[James1238765]

Thank you. I realize many factors must be taken into account to get the final probability. It's just easier usually for me to work backward to understand the answer for the first time. From @Spinnor [here],

where s and alpha have been previously defined as:

and

which are jointly used to derive the final numerical result:

I'm unclear how the 1/137 and 398 values above were obtained?

1. Is g_e some constant value that is readable from a table, giving us the 1/137?

2. How does hc equals to 398/1?

 

[Vanadium 50]

It may be easier for you, but it is much, much harder for us. @Demystifier is right - you are asking us essentially to write a textbook for you, only out of order. Further, I don't think it's even true: thinking you understand is not the same as actually understanding.

To the specific question, the equation in slide 35 is the same as the one on slide 33, only with some numbers plugged in. You can't understand Slide 35 without understanding Slide 33, and this goes all the way back.

 

[Vanadium 50]

PS Please use LaTeX instead of pasting images.

 

[James1238765]

1. $\alpha = 1/137$is the famous fine structure constant.

2. $(\hbar c)^2 = (197.3)^2 = 3.89$ MeV. I presume 398 is a typo and involves decimal places conversion from the $\mu b$ units.

The result states that the probability of muon pair production from electron annihilation depends only on the momentum/energy E of the incoming electrons, proportional to a universally applicable coupling constant.

 

[vanhees71]

ad 1. $\alpha=1/137$ is the usual back-of-the-envelope approximation of the fine structure constant. It's much more precisely known today. See, e.g., the pdg webpage for precise values of the constants:

https://pdg.lbl.gov/

ad 2. It is utmost important in physics to use correct units; $\hbar c \simeq 197.3 \; \text{MeV} \, \text{fm}$. Indeed $\hbar$ as the dimension of an action, i.e., of $\text{energy} \times \text{time}$ and $c$ the dimension of a velocity, i.e., $\text{length}/\text{time}$ and thus $\hbar c$ has dimensions $\text{energy} \times \text{length}$ and convenient units in particle and nuclear physics are $\text{MeV}$ and $\text{fm}$. So a correct statement is $(\hbar c)^2 \simeq 38927.29 \text{MeV}^2 \text{fm}^2$.

 

[Vanadium 50]

The result states that the probability of muon pair production from electron annihilation depends only on the momentum/energy E of the incoming electrons,

That's because you integrated all the other variables out.

 

[James1238765]

@Vanadium 50 Can sigma for other types of interactions depend on other factors than E?

 

[Vanadium 50]

I don't understand the question, "Sigma" usually represents the cross-section in some fiducial region, so it is just a number, not a function. You can talk about its derivative with respect to some variable, but sigma by itself is just a number.

 

[James1238765]

@Vanadium 50 Sigma the "scattering cross section" means the rate at which this particle interaction will occur, is that right? If we increase the momentum of collision, it makes sense that the rate of interaction should go up. I was wondering if anything else can affect this rate of interaction, other than the momentum with which we pump the incoming particles?

 

[Vanadium 50]

First, relying on intuition is a bad idea if anyone is just starting out. Intuition takes time to develop.

Second, it is especially bad if one adopts the "start in the middle and don't bother with the prereqs" strategy. Apart from being inefficient for you, it adds to the burden of people who are trying to teach you - never a good idea, but particularly bad for volunteers.

The road you are trying to take doesn't really go where you want to go.

Now, your intuition is wrong: $e^+e^- \rightarrow \mu^+\mu^-$ goes down with energy. You just calculated it, or at least looked at a calculation and nodded your head. That doesn't mean $e^+e^- \rightarrow \mu^+\mu^- + {\rm anything}$ goes down. It in fact goes up. But that's not what you calculated (well, watched being calculated) is it?