##CP^N## model in Peskin & Schroeder problem 13.3

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SUMMARY

The discussion focuses on the CP^N model in Peskin & Schroeder's problem 13.3, specifically addressing the expansion of the logarithm function. The initial attempt to expand the log using a naive approach did not yield useful results, as the second term vanished. A more effective method proposed is the Taylor series expansion of log(1+x), which provides a clearer path to derive the necessary terms for the problem. The correct expansion includes terms up to A^6, allowing for a more accurate approximation.

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  • Understanding of Taylor series expansion
  • Familiarity with logarithmic functions
  • Knowledge of the CP^N model in quantum field theory
  • Basic concepts of perturbation theory
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  • Study the Taylor series expansion in detail
  • Review logarithmic identities and their applications in physics
  • Explore the CP^N model and its significance in quantum field theory
  • Investigate perturbation theory techniques in quantum mechanics
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Physics students, theoretical physicists, and researchers working on quantum field theory, particularly those focusing on the CP^N model and perturbative methods.

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Homework Statement
Hello,

In Peskin & Schroeder exercise 13.3 question d, it is asked to perform an expansion of the term
$$iS=−N.tr[log(−D2−λ)]+ig2∫d2xλ$$

where ##Dμ=(∂μ+iAμ)##, and ##λ##,##N## and ##g## numbers. The expansion should be made around ##A_μ=0##, and we should use this result to prove the expansion is proportional to the vacuum polarization of massive scalar fields. In momentum space, the log can be written as

$$∫ddx(2π)dlog(k2+A2−λ)$$
Relevant Equations
$$iS=−N.tr[log(−D2−λ)]+ig2∫d2xλ$$


$$∫ddx(2π)dlog(k2+A2−λ)$$
My naive attempt to expand the log was##log(k2+A2−λ)=log[(k2−λ)(1+A2(k2−λ))]=log(k2−λ)+log(1+A2(k2−λ))≈log(k2−λ)+A2(k2−λ)##but it did not help me so far since the second term vanishes. Can someone point me to the right direction?
 
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A better way to expand the log is to use the Taylor series expansion. The Taylor series of log(1+x) is##log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \dots##Therefore, ##log(k^2 - \lambda) + A^2(k^2 - \lambda) \approx log(k^2 - \lambda) + A^2(k^2 - \lambda) - \frac{A^4(k^2 - \lambda)^2}{2} + \frac{A^6(k^2 - \lambda)^3}{3} - \dots##
 

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