2
$\begingroup$

The parity operator is defined as the bases transformation: $$\{\hat e_x,\hat e_y, \hat e_z\}\rightarrow \{-\hat e_x,-\hat e_y, -\hat e_z\}$$ thus we can perform a parity transformation in the real world by simply changing the direction our scales run on our meter sticks which are pointing along the axes (or turning them around), this can be done without doing anything to our system (i.e. it is a passive transformation). This also means that two observers Alice and Bob can look at the same system but one can use the standard basis and the other the inverted basis.

Given the above I am confused about the concept of parity violation. This would mean that Alice and Bob would see the laws of physics differently simply due to the way the lay their rulers. Consider the example on Wikipeida: enter image description here (From Wikipeida, by SiBr4, CC BY-SA 3.0)

This in my interpretation would mean that when looking at the same situation Alice would say the image on the left and Bob that on the right. Is this analysis right? If so how does it come about? and if not (which is more likely) how should I be interpreting parity violation?

$\endgroup$
3
$\begingroup$

A simplified example of parity violation is the following one where I replace properties of real microscopic particles for imaginary but familiar features of macroscopic objects. (N.B. My colors have nothing to do with colors in QCD).

Suppose that there is a "green" particle which naturally decay into a "red" and a "blue" particle and that the three directions (referred to the motion of the particles) red, green, and blue (with that order) form a right-handed triple of unit vectors $i$, $j$, $k$.

Now observe the phenomenon reflected in a mirror: the three vectors red, green, and blue form a left-handed triple of axes.

Is it possible to prepare the state of the green particle in order that the process viewed in the mirror takes place in the real world?

If not, then the theory responsible for that "colored decay process" phenomenon breaks parity symmetry.

Regarding your analysis, I find it a bit misleading, parity violation concerns physical processes -- whether or not they still define possible physical processes when the active parity symmetry acts on them --and not reference frames used to describe them.

$\endgroup$
  • $\begingroup$ So Lorentz force is a parity breaking? $\endgroup$ – HolgerFiedler Dec 21 '16 at 15:12
  • $\begingroup$ No it is not, essentially because $\vec{B}$ is a pseudovector and $\wedge$ a pseudoopration. $\endgroup$ – Valter Moretti Dec 21 '16 at 15:32
  • $\begingroup$ You say '...when the active parity symmetry acts on them.' are you implying their is a 'passive parity symmetry'? Also by 'active parity symmetry' you mean we pick the system up and 'literally' stick it thru a mirror rather then just reflecting our coordinate systems - which goes against my definition? $\endgroup$ – Quantum spaghettification Dec 21 '16 at 20:11
  • $\begingroup$ Yes there is a passive intepretation which just means to use a left-handed coordinate system instead of a right-handed leaving fixed the studied physical system. Yes also for your second supposition in geometrical sense. $\endgroup$ – Valter Moretti Dec 21 '16 at 20:20
0
$\begingroup$

Parity violation means that the process one can observe in the mirror cannot be realized in reality (say, with all the right helicities replaced with the left ones), which is a common situation in Particle Physics. For example, electroweak interactions break parity, while QCD doesn't.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.