How Do You Diagonalize the Chiral Symmetry Breaking Term for Pion Masses?

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SUMMARY

The discussion focuses on diagonalizing the chiral symmetry breaking term for pion masses, specifically the quadratic form related to the masses of the η and π0 pions. The quadratic form is derived from the chiral symmetry breaking term in the context of the SUL(3) × SUR(3) symmetry. The user initially attempted orthogonal diagonalization but encountered an incorrect result for the mass of the π0 pion. The solution was found by applying a Taylor expansion for the square root under the condition ms >> mu,d, leading to the correct mass expression.

PREREQUISITES
  • Understanding of chiral symmetry in particle physics
  • Familiarity with quadratic forms and diagonalization techniques
  • Knowledge of Taylor series expansions
  • Basic concepts of pion masses and their interactions
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  • Study the process of orthogonal diagonalization in detail
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This discussion is beneficial for theoretical physicists, graduate students in particle physics, and anyone studying the implications of chiral symmetry breaking in the context of pion masses.

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Homework Statement


I want to diagonalize the quadratic form
$$ m_0((m_u+m_d)\pi^3\pi^3+\frac{2}{\sqrt{3}}(m_u-m_d)\pi^3\pi^8+\frac{1}{3}(m_u+m_d+4m_s)\pi^8\pi^8)$$
which can be found under equation 5.47, in order to get the mass of the η and ##\pi^0## pions. This quadratic form is produced by the term that breaks the Chiral Symmetry ##SU_L(3)\times SU_R(3) ##

Homework Equations


I want to know how to produce the result 5.48 as i am unable till now to do the calculation.

The Attempt at a Solution


I tried to diagonalize the expression using the standard procedure of orthogonal diagonalization. After doing all the calculations i get for the ##\pi^0##: $$m^2_{\pi^0} = m_0\bigg(m_u+m_d-\frac{(m_u-m_d)^2}{2\sqrt{m^2_u+m^2_d+m^2_s-m_um_d-m_um_s-m_dm_s}+2m_s-m_u-m_d}\bigg)$$

which is incorrect. Can anyone help me to figure out what i am doing wrong.
Thank you very much!
 
Last edited:
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Actually i found it. Taking the taylor expansion for the square root at ##m_s>>m_{u,d}## produces the right answer! Thank you very much!
 

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