I'm making my way through a QFT course and at the moment I'm trying to calculate some amplitudes in a toy-model Yukawa theory. The particular scattering I'm interested in is:

$$\phi \psi \rightarrow \phi \psi $$

and I want to do it 'by hand', i.e. using Dyson's formula and Wick's theorem.

My Lagrangian as given here: Lecutre Notes in eqation 3.7

I was looking at a similar example that is solved in the set of notes above. The example is in the section 3.3.3. My equivalent expression for the normal ordered product in 3.47 is:

$$ :\phi(x_1)\psi(x_1)^\dagger \phi(x_2)\psi(x_2): $$

But how do I know which of the $\phi$'s goes to the right? It could be either

$$\psi(x_1)^\dagger\phi(x_1) \phi(x_2)\psi(x_2)$$ or $$\psi(x_1)^\dagger \phi(x_2)\phi(x_1)\psi(x_2)$$ as I'll be using one of them to destroy and one of them to create a meson, but there doesn't seem to be a 'preferred' choice.

On top of that, there's something else I'd like to get a better intuition about.

If I, say, choose the latter one for now, I run into another problem. When I insert an unitary operator and expand my field operators (for readability only the rightmost ones, as I'm not concerned with the remaining ones) and the final state, $a_{k_1}^\dagger b_{k_2}^\dagger |0\rangle$, I obtain:

$\psi(x_1)^\dagger \phi(x_2)|0\rangle \langle 0|\int \frac{d^3p}{(2\pi)^3\sqrt{2E_p}}\left( a_p e^{-ipx_1} + a_p^\dagger e^{ipx_1} \right)\int \frac{d^3q}{(2\pi)^3\sqrt{2E_q}}\left( b_q e^{-iqx_2} + c_q^\dagger e^{iqx_2} \right)a_{k_1}^\dagger b_{k_2}^\dagger |0\rangle $

I can discard the creation operators in my field operators since they kill the vacuum in the unity operator on the LHS, so I'm left with:

$$\psi(x_1)^\dagger \phi(x_2)|0\rangle \langle 0|\int \frac{d^3pd^3q}{(2\pi)^6\sqrt{2E_p}}\left( a_p e^{-ipx_1} \right)\left( b_q e^{-iqx_2} \right)a_{k_1}^\dagger b_{k_2}^\dagger |0\rangle $$

The difference now, compared to the example from the notes, is that where in third line of 3.48 there are two exponential factors in each pair of round brackets, I'll end up with just one (because after a single non-zero commutation among my four creation/annihilation operators the RHS vacuum is exposed to an annihilation operator). Assuming I'm correct up to this point, is this because in this kind of scattering we have a nucleon in our propagator rather than a meson?

Thanks a lot for your help!

EDIT ----------------------

I just had an idea: is the ambiguity in the ordering of $\phi$'s due to the fact that we have two possible diagrams? And we need to account for both cases and as a result our amplitude is twice the amplitude calculated for one particular choice of ordering?


2 Answers 2


I did a similar thing to you and tried the meson-nucleon scattering. The Hamiltonian is $H_{int}=\int d^3x g\psi^{\dagger}\psi\phi$ and we want to find $<f|S|i>$ to second order in $g$. So we're interested in the time-ordered bit:


Applying Wick's Theorem, the relevant parts are:

$$ :\psi^{\dagger}(x_1)\phi(x_1)\psi(x_2)\phi(x_2):\overbrace{\psi(x_1)\psi^{\dagger}(x_2)}+\\ :\psi^{\dagger}(x_1)\phi(x_1)\psi(x_2)\phi(x_2):\overbrace{\psi^{\dagger}(x_1)\psi(x_2)} $$

These two are equivalent. If you solve the top one say (they are already normal ordered in their form above), you get

$$ <f|S|i>=\frac{(-ig)^2}{2}(2\pi)^4\delta^{(4)}(p_1-p_1'+p_2-p_2')\frac{i}{(p_1+q_1)^2-M^2+i\epsilon} $$

Solving the other one gives the same result so, adding them together gets the thing above without the factor of $1/2$.

Now, if you normal order them the second way as you suggested i.e.

$$ :\psi^{\dagger}(x_1)\phi(x_2)\psi(x_2)\phi(x_1):\overbrace{\psi(x_1)\psi^{\dagger}(x_2)}+\\ :\psi^{\dagger}(x_1)\phi(x_2)\psi(x_2)\phi(x_1):\overbrace{\psi^{\dagger}(x_1)\psi(x_2)} $$

which is equally plausible given $\phi^{\dagger}=\phi$, you would get a final amplitude of

$$ <f|S|i>=(-ig)^2(2\pi)^4\delta^{(4)}(p_1-p_1'+p_2-p_2')\frac{i}{(p_1-p_1')^2-M^2+i\epsilon} $$

This total amplitude is what you'd get as a result of drawing the Feynman diagrams. I suppose the result of this is that Wick's theorem needs to consider $\mathit{all}$ normal orderings in order to get a complete picture!


$:\phi(x_1).\phi(x_2):$ is equal to $:\phi(x_2).\phi(x_1):$ So it doesn't matter how you write it. You can check by expanding it out.

  • $\begingroup$ This answer would benefit from explaining why. Also please use mathjax in future $\endgroup$
    – nox
    Jan 14, 2020 at 6:06

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