# A way to derive the Yukawa potential without cheating?

Let's say we have a simple Lagrangian that couples together two real scalar fields with a Yukawa $\phi \psi^2$ coupling.

$$\mathcal{L} = \frac{1}{2}(\partial \phi)^2 - \frac{m^2_1}{2} \phi^2 + i\bar{\psi}\gamma^{\mu}\partial_{\mu} \psi - m^2_2\bar{\psi}\psi + g \bar{\psi}\psi \phi.$$

In the appropriate limit, the $\phi$ particles should experience a Yukawa force that comes from the exchange of virtual $\psi$ messenger particles.

$$V(r) \propto - g^2 \frac{e^{-m_2 r}}{r}.$$

Now, when I look into why this should be, I find explanations that fall into one of two categories.

Explanation 1: Look at the S-matrix elements of two $\phi$ particles scattering off each other to the first order in perturbation theory. The Fourier transform of this element is the Yukawa potential. (This explanation is given on the Wikipedia page).

Explanation 2: If you perturb the $\psi$ field's path integral by two sources at two different positions, then the change in energy is the Yukawa energy between the sources. (This explanation is given in Zee's book, section 1.4 "From Field to Particle to Force.")

Explanation 1 is problematic because it does not explain why the Fourier transform of the S-matrix element should be the potential energy experienced between two localized particles. Furthermore, the Yukawa potential should apply even to particles that are not scattering off of each other, so coming up with an explanation that only applies to scattering particles is unsatisfying. (I know that in non-relativistic quantum mechanics, using the Born approximation, the scattering amplitude is equal to the Fourier transform of the central potential. This is still not very satisfying.)

Explanation 2 is problematic because it does not explain why, given specifically a $\phi \psi^2$ coupling, localized $\psi$ particles can be modeled as sources.

I am trying to come up with a satisfying explanation that explains how this potential arises from the specific coupling in question (that does not have anything to do with scattering). Does anyone know of one?

• "simple Lagrangian that couples together two real scalar fields" nothing about that sentence sound simple xD – Noah Cristino Jul 23 '17 at 14:44
• You says : "Lagrangian that couples together two real scalar fields". Where's the second scalar field in your lagrangian ? I only see one real scalar field $\phi$ and a spinor field $\psi$ in this expression. – Cham Aug 20 '17 at 16:09
• Someone must have edited my original question. My question was even simpler than the Yukawa theory, with two scalar fields. – user1379857 Aug 22 '17 at 1:45

Think back to non-relativistic quantum mechanics and the associated perturbation theory. Recall that in the interaction picture, unitary time evolution operator is given by

$$U(t'-t) = \exp\left[\frac{-i}{\hbar} \int_t^{t'} V_I(t'') dt''\right]$$

where $V_I$ is the perturbation to the free Hamiltonian. The amplitude for transitioning from an initial state $|i\rangle$ at time $t$ to a final state $|f\rangle$ at time $t'$ is then

$$\mathcal{A}=\langle f | U(t'-t)| i \rangle \approx \langle f | i \rangle - \frac{i}{\hbar} \int_t^{t'} dt'' \langle f|V_I|i\rangle$$

Assuming that the initial and final state are different, to first order we have that

$$\mathcal{A} = -\frac{i}{\hbar} \int_t^{t'}dt'' \langle f|V_I |i\rangle$$

Does this look familiar? This is exactly what you're calculating in QFT when you find the amplitude for a particular process, and you should be able to see where the relationship between the element of the S-matrix and the potential energy of the system.