In P&S 9.3 the path integral

$$ Z[J]=\int {\cal D}\phi \exp[i\int d^4x ({\cal L} + J\phi)]$$

of the (Minkowski) $\phi^4$-theory when subjected to a Wick-rotation (change of the integration path and redefinition of the time coordinate $t\rightarrow -ix^0$ ) leads to the path integral:

$$ Z[J]=\int {\cal D}\phi \exp[-\int d^4x ({\cal L}_E - J\phi)] \tag{9.47}$$

with the Euclidean Lagrangian

$${\cal L}_E=\frac{1}{2}(\partial_{E\mu}\phi)^2 + \frac{1}{2}m^2\phi^2 +\frac{\lambda}{4!}\phi^4$$

which is supposed to have better convergence properties, in particular as it is bounded from below.

On the other hand in the precedent chapter it is demonstrated that the path integral of the same theory (actually only for $\lambda=0$ and $J=0$, but I don't attribute any importance to this detail) upon discretization $$ \phi(x_i) = \frac{1}{V}\sum_n e^{-ik_n x_i} \phi(k_n)$$ yields to be

$$\int {\cal D}\phi \exp(i\int d^4x [\frac{1}{2}(\partial_\mu\phi)^2 -\frac{1}{2}m^2\phi^2]) = \prod_{all k_n}\sqrt{\frac{-i\pi V}{m^2-k_n^2}} \tag{9.23}$$

which implicates that in the continuous limit $n\rightarrow\infty$ the value will be infinite.

This seems to be a contradiction, or would it be mean that the standard and the Wick-rotated path integral would not be identical ? PS: In Convergence Property of Path-Integral a similar question is posed, however, it does not provide any information on this apparent contradiction.


Right, the exponentially damped Euclidean path integral is mathematically better behaved compared to the oscillatory Minkowski path integral, but it still needs to be regularized, e.g. via zeta function regularization, Pauli-Villars regularization, etc.

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  • $\begingroup$ Thank you for the prompt answer. But both path integrals are formally equal, right? $\endgroup$ – Frederic Thomas Nov 1 '19 at 22:48
  • $\begingroup$ The Minkowski path integral with (without) the Feynman $i\epsilon$ prescription is equal (not equal) to the Euclidean path integral, respectively. $\endgroup$ – Qmechanic Nov 2 '19 at 3:40

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