In Appendix A, Polchinski does the Euclidean path integral for the Harmonic oscillator. After he Pauli-Villars regularizes the determinant of the kinetic term, he obtains the following expression (A.1.62):

$$ \langle q_f, U | q_i, 0 \rangle \to \left( \frac{\omega}{2 \sinh \omega U} \right)^{1/2} \exp \left[ - S_{cl}(q_i,q_f) + \frac{1}{2}\left( \Omega U - \ln \Omega \right) - S_{ct} \right] \tag{A.1.62}. $$

where $\Omega$ is a frequency scale.

In what follows he says:

"To get a finite answer as $\Omega \to \infty$, we need first to include a term $\frac{1}{2} \Omega$ in the Lagrangian $L_{ct}$, canceling the linear divergence in (A.1.62). That is, there is a linearly divergent bare coupling for the operator 1. It may seem strange that we need to renormalize in a quantum mechanics problem, but power counting is completely uniform with quantum field theory. The logarithmic divergence is a wavefunction renormalization.

What does the italic sentence mean? The operator $1$ should change by some $Z. 1$? Why the logarithm is a wave function renormalization? And is he power counting what?

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  1. We allow counterterms $L_{\rm ct}= \sum_X \delta_X ~X$ of all possible field monomials/"operator" $X$ in the Lagrangian. Here we need $\delta_1 =\frac{\Omega}{2}$ for the constant monomial $X=1$.

  2. Since $\Omega$ appear linearly in $\delta_1$, we speak of a linear divergence. In principle $\Omega$ could have appeared in other powers, cf. power counting.

  3. The wave function renormalization (aka. field strenth renormalization) means that the normalization of the inner product$^1$ $\langle q_f, U | q_i, 0 \rangle$ is changed with a factor $\sqrt{\frac{\pi}{\Omega}}$. Upstairs in an exponential, this becomes a logarithm.


$^1$ Here $U=iT$ denotes Euclidian time.

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  • $\begingroup$ For a usual harmonic oscillator, it looks like we fight one artefact ($\Omega$) with another one (counterterm), does it? $\endgroup$ – Vladimir Kalitvianski Jun 4 '18 at 14:23
  • $\begingroup$ Polchinski identifies them. $\endgroup$ – Qmechanic Jun 4 '18 at 14:27
  • $\begingroup$ No, seriously, in a usual harmonic oscillator we do not have any $\Omega$ at all, so its appearing and disappearing look artificial to me. Am I wrong? $\endgroup$ – Vladimir Kalitvianski Jun 4 '18 at 15:13

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