A Symmetry & Invariance of Pions: π+, π0, π− and Other Mesons Explained

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Pions are spin-0 particles that form an isospin triplet consisting of π+, π0, and π−, characterized by an intrinsic parity of −1. They are classified as pseudoscalar mesons, while ρ mesons, which include ρ+, ρ0, and ρ−, are vector mesons with spin 1 and the same intrinsic parity. The decay of ρ0 into π+π− occurs through the strong interaction, which conserves parity. However, it is impossible for ρ0 to decay into π0π0 via the strong interaction due to conservation of angular momentum and the properties of bosons. This topic is suitable for a homework forum discussion, requiring proper formatting and submission guidelines.
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Could someone please help me with this problem?
Pions are particles with spin 0 and they form an isospin triplet: π+, π0, π (with the superscript indicating the electric charge). Their intrinsic parity is −1 and they are pseudoscalar mesons. In nature we also find other kind of mesons, like the ρ mesons, ρ+, ρ0 and ρ. As pions, they also form an isospin triplet but they are vector mesons, i.e., they have spin 1 and intrinsic parity −1. The ρ0 decays into π+π via the strong interaction, which preserves parity. Prove that a ρ0 cannot decay into π0π0 via the strong interaction. Hints: Remember that pions are bosons and angular momentum is conserved in a reaction.
 
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This looks like a homework problem. It needs to be posted in the appropriate homework forum, with the homework template filled out.

Also, please do not use the same title for multiple threads with different questions.
 
For the quantum state ##|l,m\rangle= |2,0\rangle## the z-component of angular momentum is zero and ##|L^2|=6 \hbar^2##. According to uncertainty it is impossible to determine the values of ##L_x, L_y, L_z## simultaneously. However, we know that ##L_x## and ## L_y##, like ##L_z##, get the values ##(-2,-1,0,1,2) \hbar##. In other words, for the state ##|2,0\rangle## we have ##\vec{L}=(L_x, L_y,0)## with ##L_x## and ## L_y## one of the values ##(-2,-1,0,1,2) \hbar##. But none of these...

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