On page 388 in section 11.6 of Peskin and Shroeder.
There appears an equation of the inverse propagator(the second functional derivative of the effective action) for a theory that contains several scalar fields: $$ K^2_{ij} =: \int d^4x e ^{ip\cdot (x-y)}\frac{\delta^2 \Gamma}{\delta \phi^i\delta \phi^j}(x,y)=0 \tag{11.105} $$ When diagonalizing
$$ K^2_{ij} = P_{ik}P_{jl}\tilde{K}^2_{kl} = (P\tilde{K}P^t)_{ij} \, , \quad P \,\text{:an orthogonal matrix} $$ the property that $K^2_{ij}\,$: real is needed. After diagonalizing $$ \tilde{K}^2_{ii}=p^2-m_{i0}^2-M_i^2(p^2) \quad\quad\text{i : no sum} \, ,$$ where $m_{i0}$ is the bare mass of the $i$th scalar field and $M_i^2(p^2)$ is the sum of one-particle-irreducivle two-point diagrams.
Is $M_i^2(p^2)$ always a real number?
That's simply because
$$ \tilde{K}^2_{ii}=0 \quad \Leftrightarrow \quad m_i^2=m_{i0}^2-M_i^2(m_i^2) \quad ? $$ ,where $m_i$ is the physical mass.
Is that whole the story? Is there some other reason?

  • $\begingroup$ Note that $M^2(p^2)$ may acquire a non-zero imaginary part at the threshold of pair production (branch cut) if the model has unstable particles. $\endgroup$ – AccidentalFourierTransform Aug 17 '17 at 10:41
  • $\begingroup$ @AccidentalFourierTransform Would you please also comment on Adam's answer, which points out that the effective potential is always real from the viewpoint of path integral after the Wick rotation. $\endgroup$ – gamebm Sep 17 '17 at 13:28

The matrix $K_{ij}$ define by the OP is just the second derivative of the effective potential (see A question about the proof of Goldstone's theorem ). The effective potential being a real function of the fields. Indeed, it is the Legendre transform of the log of the generating functional, which, after Wick rotation (and using notations of quantum statistical physics), is defined as $$Z=\langle e^{-\beta (\hat H+J_i \int_x\hat \phi_i(x))} \rangle,$$ where the Hamiltonian in presence of (constant) sources is hermitian. This implies that $Z$ is real (by just diagonalizing the Hamiltonian), and therefore so is the effective potential.

Its second derivative, evaluated at its minimum, is therefore a real symmetric matrix, which can be diagonalized, with positive (or zero) eigenvalues.

Without Wick rotation, the effective potential might be purely imaginary depending on its definition, but this does not change the argument.

  • $\begingroup$ Concerning AccidentalFourierTransform's comments above, so Z is not necessarily real? But it is necessary in the sense that mass must be non-negative definite. Can you give more comments? $\endgroup$ – gamebm Sep 13 '17 at 3:13
  • $\begingroup$ @gamebm : AccidentalFourierTransform's comments is about the fact that some propagators might have poles in the complex plane (and not just on the real line), due to decay from a particle into another kind of particle. But this happens after Wick rotation, and does not change the fact that one can diagonalize the matrix $K$. $\endgroup$ – Adam Sep 13 '17 at 5:00
  • $\begingroup$ I still do not understand. The second order derivative of $\Gamma$ is related to the inverse propagator (Eqs.(11.90) & (11.92) on the textbook), namely, $p^2-m^2-M^2(p^2)$. If the mass pole is in the complex plane, the $M^2$ part of the expression is not real. According to your derivation, $Z$, $\Gamma$, and hence the second derivatives of the effective potential are real. So I stuck there, the second derivative should be real or not? I know that I obviously missed something. Would you please be more specific on the "after the Wick rotation" part, many thanks! $\endgroup$ – gamebm Sep 13 '17 at 19:07
  • $\begingroup$ @gamebm: $x^2+1$ is real, yet its zeros are imaginary... although it might be too simple an example, that might help you. $\endgroup$ – Adam Sep 13 '17 at 19:46
  • $\begingroup$ thx for the answer! I see you point. But for this particular question, is it literally the case? For resonance states, according to the textbook (around P.235, when two propagators simultaneously go on-shell), $M^2$ indeed possess non-vanishing real as well as imaginary parts. How does this fit in the context? Thanks in advance for the further explanations. $\endgroup$ – gamebm Sep 14 '17 at 0:18

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