Consider the scattering $$e^-(p_1)+e^+(p_2)\rightarrow e^-(p_1^\prime)+e^+(p_2^\prime)$$ at the tree-level via a internal photon line of four-momentum $q$. Using energy momentum conservation at the vertex, we get, $$p_1-p_1^\prime=q\Rightarrow q^2=2p_1\cdot q$$ where I used $p_1^2=p_1^{\prime 2}=m_e^2$. Note that $2p_1\cdot q$ is Lorentz invariant and I can evaluate in any frame. In the rest frame of the initial electron $p_1=(m_e,0,0,0)$ and $q=(E,0,0,E)$, we get, $$2m_eE=q^2$$ If we impose energy-momentum relation on the internal photon line too, i.e., $q^2=0$, I get, $$E=0\Rightarrow q=(0,0,0,0).$$ So the photon is not emitted at all! Therefore, we give up energy-momentum relation for the internal photon line, and treat it as a virtual particle.

But how to argue it for a process mediated by massive internal boson? For the weak process $$e^-(p_1)+e^+(p_2)\rightarrow \mu^-(p_1^\prime)+\mu^+(p_2^\prime)$$ mediated by massive $Z-$boson. Using the same line of argument, i.e., taking $p_1=(m_e,\textbf{0})$ and $q=(E_Z,\textbf{q})$, I arrive at $$2m_eE_z=M_Z^2.$$ But here, I don't find any contradiction in assuming the validity of energy-momentum relation for the internal Z-boson. However, the fact that the internal boson line is always virtual imply that there must be a contradiction! Can someone help me with this?

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    $\begingroup$ wait what? why would you impose $q^2=0$? You clearly have $q^2=(p_1-p_1')^2\neq 0$. $\endgroup$ Dec 2, 2016 at 16:58
  • $\begingroup$ @AccidentalFourierTransform You got the point wrong. I know, I cannot use $q^2=0$ for the internal line. But let's assume I can. Then I show that I arrive at a contradiction and my supposition was wrong. I cannot use $q^2=0$ for the internal line. It must be treated as a virtual particle: it does not obey energy-momentum dispersion relation. $\endgroup$
    – SRS
    Dec 2, 2016 at 17:09
  • $\begingroup$ @AccidentalFourierTransform- Actually, I figured it out a similar contradiction right now for massive internal lines too. It works. But I don't know what should I do. I don't understand whether I can answer my own question here or delete this post. $\endgroup$
    – SRS
    Dec 2, 2016 at 17:16
  • $\begingroup$ This answer of mine might help physics.stackexchange.com/questions/286721/… $\endgroup$
    – anna v
    Dec 2, 2016 at 17:59

1 Answer 1


I've worked out the answer to my own question and here I'll attempt to answer it.

Let us assume the massive internal boson line is on-shell i.e., $q^2=M_Z^2$. Now, $$p_1-p_1^\prime=q\Rightarrow p_1^2+p_2^2-2p_1\cdot p_1^\prime=q^2=M^2$$ Using $p_1=(m_e,\textbf{0})$, $p_1^\prime=(E_1^\prime,\textbf{p}_1^\prime)$, and $p_1^2=p_1^{\prime 2}=m_e^2$, we get, $$2(m^2_e-m_eE^\prime_1)=M^2$$ which implies that the energy of the scattered electron $$E_1^\prime=m_e-\frac{M_Z^2}{2m_e}<m_e$$ which is less than its rest mass and therefore, we again arrive at a contradiction. Hence, my starting assumption that $q^2=M_Z^2$, was wrong!

  • $\begingroup$ while the algebra is indeed correct, you should read the last paragraph of my answer: a kinematic contradiction is essentially irrelevant (though reassuring, if you will) to conclude that internal lines are off-shell. $\endgroup$ Dec 2, 2016 at 17:37
  • $\begingroup$ @AccidentalFourierTransform- But don't you think, in case of free theory, it is the momentum of the Fourier modes, which when quantized appear as the momentum of the particles? So Fourier variables can be physical momentum. Right? $\endgroup$
    – SRS
    Dec 2, 2016 at 18:05
  • $\begingroup$ If you carefully follow the derivation of the Feynman rules, you'll see that the momentum of the particles and the momentum of loops are unrelated. In fact, the Feynman rules are most easily derived in position space, at which point there is no reference to any internal momentum: only external momenta. To simplify the rules we Fourier transform and thus introduce internal momenta. But these momenta are Fourier variables, unrelated to any mode of the fields. Therefore, internal momenta have a very different origin than external momenta; the latter are indispensable, the former simply convenient $\endgroup$ Dec 2, 2016 at 18:10
  • $\begingroup$ @AccidentalFourierTransform Then it means that while virtual particles don't obey dispersion relation it is not really surprising. If they are not physical momenta, they need not! Moreover, it is meaningless to invoke uncertainty principle to explain these virtual particles. $\endgroup$
    – SRS
    Dec 2, 2016 at 18:20

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