Suppose we use the metric $(+,-,-,-)$ thus the momentum squared is

$p^2 = p_0^2-\vec{p}^2 = m^2>0$

Defining $p_E:=\mathrm{i}\cdot p_0$ and $\bar{p}:=(\,p_E,\vec{p})$ with Euclidean norm $\bar{p}^2 = p_E^2+\vec{p}^2$.

Here's my question:

If we plug in


instead of $p_E$ we see that $\bar{p}^2 = -p^2 = -m^2$?

So $\bar{p}^2$ is negative?

Also if $\bar{p}^2 = -p^2 = -m^2$ is true...is it always so? Since $p^2$ is a Lorentz invariant, but how do we interpret $\bar{p}^2$ if it's equal to $-m^2$

What I want to get to is the following:


$\mathcal{L}_{int} = \frac{1}{2}g\phi_1^2\phi_3+\frac{1}{2}h\phi_2^2\phi_1$

with: $p_3 = p_1+p_2$ and


Suppose we have a triangle loop with incoming momentum $p_3$ of mass $M>2m$ and two outgoing identical particles $\phi_2$ and $\phi_2$ momenta $p_1$ and $p_2$ each of mass $m$ (sorry bad notation but $\phi_1$ is not related to $p_1$). The incoming particle $\phi_3$ splits into two light ones (two $\phi_1$'s) of mass $\eta$ each and each of these connects to two $\phi_2$'s.

Thus we have the following momenta flowing inside the loop:

$k$, $ k-p_2$ and $k+p_1$.

After a Wick rotation we get the following integral:

\begin{equation} \int{\,\frac{\mathrm{d}^4\bar{k}}{(2\pi)^4} \frac{1}{\bar{k}^2+m^2}\frac{1}{\left(\bar{k}-\bar{p}_2\right)^2+\eta^2}\frac{1}{\left(\bar{k}+\bar{p}_1\right)^2+\eta^2}} \end{equation} Now to evaluate these let's use Schwingers trick:

\begin{equation} \frac{1}{\bar{k}^2+m^2} = \int_0^\infty{\mathrm{d}s\,\mathrm{e}^{-s(\bar{k}^2+m^2)}}, \end{equation} but for this to be OK we need to have $\mathrm{Re}(\bar{k}^2+m^2)>0$ and similarly for the other propagators, and this is where I get confused.

It seems they don't satisfy this condition depending on how we interpret $\bar{k}^2$ and $\bar{p}^2$ and so on...

On the other hand if all the squares in the denominator of the integrand are taken as positive then the convergence condition is trivially satisfied.

So please help me understand what I'm doing wrong and if you guys can show how to satisfy the positivity condition. Thanks in advance.


1 Answer 1


Yep. You've defined $\bar{p}$ such that $\bar{p}^2$ will have the opposite sign from $p^2$. Notice for instance that the spatial components of momentum, $\vec{p}^2$, have a negative sign in your definition of $p^2$ but a positive sign in your definition of $\bar{p}^2$.

Of course, the sign doesn't change whether something is a Lorentz invariant or not. The way you interpret $\bar{p}^2$ is just what it looks like, the negative invariant mass squared. Or in other words, you interpret $\bar{p}^2$ as a Lorentz invariant, from which you can obtain the physical mass of the particle as $m = \sqrt{-\bar{p}^2} = -i\sqrt{\lVert\bar{p}\rVert}$.

  • $\begingroup$ Thanks for the answer. I'm just confused why every QFT-book call $\bar{p}$ a Euclidean vector when its Euclidean norm obviously is negative. $\endgroup$
    – The Noob
    Commented Jan 31, 2013 at 23:05
  • $\begingroup$ It has complex components. A Euclidean vector is only guaranteed to have a nonnegative norm if the components are real. $\endgroup$
    – David Z
    Commented Jan 31, 2013 at 23:14
  • $\begingroup$ Yes that is true but what I mean by "how" to interpret $\bar{p}^2$ is the following (I'll post it as an answer): $\endgroup$
    – The Noob
    Commented Feb 1, 2013 at 12:54
  • $\begingroup$ @user20458 Ah, you mean as in your edit to the question? I'll take a look at that and either edit or delete my answer accordingly. $\endgroup$
    – David Z
    Commented Feb 1, 2013 at 17:56
  • $\begingroup$ Yes exactly, I edited the question. Hope it makes more sense now. $\endgroup$
    – The Noob
    Commented Feb 1, 2013 at 18:50

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