I have started teaching myself QFT from the textbook by A. Zee. From reading that book, my idea of a path integral in field theory is the probability amplitude to go from a given field configuration to a different field configuration. At a point, he adds a source function $J(x)$ in the free field theory Lagrangian and evaluates the path integral. It turns out that $Z(J)=Ce^{iW(J)}$ where $W(J)$ is a term containing the integral of source function and the propagator. Then, he says

The overall factor C does not depend on $J$ and, as will become clear in the discussion to follow, is often of no interest to us.

Then, he goes on to consider $W(J)$ for some special localized functions $J$ and that leads to the interpretation of particles in field theory and a derivation of inverse square law.

But, I am quite disturbed by this. QM teaches us that only probabilities are something that are physical and the absolute phase factors in the probability amplitude are of no interest to us; that we should just concentrate on $C$ above. Instead, Zee says that it is of no importance to us and such spectacular stuff comes out just by examining the phases.

So, why is it that phase is so important in QFT? Why did people even try to analyze the phase and interpret the results in terms of particles and forces that can physically be observed?

  • 3
    $\begingroup$ The factor $C$ is just an overall (complex) normalization which gets cancelled in all physical quantities and is independent of $J$, or anything. Its absolute value may be determined from some normalizability condition and its phase (phase of $C$) is the overall phase which is completely irrelevant. On the other hand, everything in $W$ matters, both phase and the absolute value. In QM, relative phases are exactly as important as ratios of absolute values. $\endgroup$ Sep 24, 2012 at 18:19
  • $\begingroup$ It seems to me that the real part of $W$ wouldn't matter because it just adds a phase. Why is this added phase important while the phase of $C$ not? Also, shouldn't the modulus of $C$ matter in calculating transition probabilities? $\endgroup$
    – user4235
    Sep 24, 2012 at 18:30

1 Answer 1


As Lubos Motl mentions in a comment, the physically observable object is not the partition function itself

$$ Z[J]~:=~ \int D\varphi~e^{\frac{i}{\hbar}(S+J\varphi)}, $$

but instead closer related to quantum averages$^1$, e.g. the one-point function

$$ \langle \varphi(x) \rangle_J ~:=~\frac{\int D\varphi~e^{\frac{i}{\hbar}(S+J\varphi)}\varphi(x)}{\int D\varphi~e^{\frac{i}{\hbar}(S+J\varphi)}} ~=~\frac{\hbar}{i}\frac{\delta\ln Z[J]}{\delta J(x)}, $$

where the normalization factor $C:=Z[J\!=\!0]$ in the path integral has no physical consequences.

$^1$ See LSZ reduction formula for more details.


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