I have been studying the path integral approach to QFT and set myself the challenge of starting from a $\phi^3$ interaction Lagrangian and following the method through to completion.
I started with: $$ \mathcal{L} = \frac{1}{2}\partial_\mu\phi\partial^\mu \phi - \frac{m^2}{2}\phi^2 - \frac{\lambda}{3!}\phi^3 $$ By introducing a source term, $-J\phi$, I was then able to express the partition function in the following form: $$ Z[J] = Z_{free} \exp{\biggl\{\frac{-i\lambda}{3!}\int \biggl(\prod_{i=1}^3 \frac{d^D k_i}{(2\pi)^D} ~i\frac{\delta~}{\delta J(k_i)} \biggr)(2\pi)^D \delta(k_1+k_2+k_3)\biggr\}} \\ \exp{\biggl\{\frac{-i}{2}\int \frac{d^D k}{(2\pi)^D} \frac{J(-k) J(k)}{k^2 - m^2} \biggr\}} $$ So at this point I am quite happy, I have both the vertex contributions and the propagator. However, there is a numerical factor $Z_{free}$ given by: $$ Z_{free} = \int[\mathcal{D}\phi] \exp \biggl\{ i \int \frac{d^D k}{(2\pi)^D} ~\phi(-k) \cdot \frac{1}{2}(k^2-m^2)\cdot \phi(k)\biggr\} $$

After performing a Wick rotation this would look like a Gaussian integral, albeit a functional one. However, I am not sure how to proceed with calculating this term, the integral in the exponent is additionally troubling to me.

Any advice would be much appreciated, I am sure this follows from the generalisation of an $n$-dimensional Gaussian integral.


1 Answer 1


Hint: a source-independent prefactor can always be dropped, because it only affects disconnected amplitudes. In other words, your $Z_\mathrm{free}$ does not depend on $J$, so it doesn't affect physical predictions.

  • $\begingroup$ And I couldn't come up for a more confusing name than $Z_{\text{free}}$ for the normalization coefficient :) $\endgroup$ Commented May 9, 2017 at 2:56
  • $\begingroup$ Well it corresponds to the partition function of an action with no interaction term and no source term. Usually I would write it $Z_{free}[J=0]$ but things started to get messy. $\endgroup$
    – Vielbein
    Commented May 9, 2017 at 10:48

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