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I'm reading through Srednicki's QFT book, and when we introduce external fields, we often multiply terms together to represent the coupling in our Lagrangian densities. My understanding of what constitutes coupling is rather vague: basically having a pair of fields/particles interact, but I'm unsure as to why that interaction is represented by a multiplication?

Likewise, what does the degree of the polynomial in the interaction terms represent? I've seen $\varphi^3$ interaction terms as "higher order interactions", but what physically does this entail/represent?

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    $\begingroup$ Do you know how to extract Feynman rules given a lagrangian? $\endgroup$ – Davide Morgante Feb 18 at 7:58
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Regarding the first question, a simple way to convince yourself that multiplying fields in the lagrangian corresponds to coupling is to think back to particle mechanics. Consider the lagrangian for two masses $m_1$, $m_2$ coupled by an ideal spring (in one dimension for simplicity): \begin{align} L = \frac{1}{2}m_1 \dot{q}_1^2 + \frac{1}{2}m_2 \dot{q}_2^2 - \frac{1}{2}k(q_1 - q_2)^2 \end{align} Expanding out the potential term gives us \begin{align} L = \frac{1}{2}m_1 \dot{q}_1^2 + \frac{1}{2}m_2 \dot{q}_2^2 - \frac{1}{2}k q_1^2 - \frac{1}{2}k q_2^2 + k q_1 q_2. \end{align} We see that this is the same as the sum of the lagrangians for two independent masses on springs, plus an interaction term in which the coordinates of the two masses are multiplied: \begin{align} L &= L_1 + L_2 + L_{\text{int}}\\ L_1 &= \frac{1}{2}m_1 \dot{q}_1^2 - \frac{1}{2}k q_1^2\\ L_2 &= \frac{1}{2}m_2 \dot{q}_2^2 - \frac{1}{2}k q_2^2\\ L_{\text{int}} &= k q_1 q_2. \end{align} Indeed, we're justified in calling this an interaction term, because it does all the work of coupling the Euler-Lagrange equations of the two masses. Without the interaction term, our EL equations would be those of two independent harmonic oscillators: \begin{align} \ddot{q}_1 + \frac{k}{m_1}q_1 = 0\\ \ddot{q}_2 + \frac{k}{m_2}q_2 = 0 \end{align} With the interaction term, we get a coupled system of oscillators: \begin{align} \ddot{q}_1 + \frac{k}{m_1}(q_1 - q_2) = 0\\ \ddot{q}_2 + \frac{k}{m_2}(q_2 - q_1) = 0. \end{align} I did this for point particles in order to put it on more familiar footing, but it's an easy exercise to do the same thing with the lagrangian density for, say, two real scalar fields $\varphi_1$, $\varphi_2$.

At this point, you might say (at least I hope you'd say) "Ok, this makes sense if we're talking about two different fields interacting, but why would we call a term like $\varphi^3$ an interaction if we only have a single field?" One way to justify calling such terms interaction terms is to point out that they lead to non-linearities in the classical equations of motion. Again, go back to a familiar example: the lagrangian density for transverse oscillations of a string. It is \begin{align} \mathscr{L} = \frac{1}{2}(\partial_t \varphi)^2 - \frac{1}{2} (\partial_x \varphi)^2, \end{align} and of course the equation of motion is the wave equation \begin{align} \partial_t^2 \varphi - \partial_x^2 \varphi = 0. \end{align} (I've started leaving out some dimensionful constants.)

Since the wave equation is linear, its solutions obey the superposition principle. One consequence of this is that if we create two wave pulses on a string moving toward one another, when they get to the same spot each will "pass through" the other, then continue on its own unmolested. In other words, wave packets do not interact because they obey a linear equation of motion. If we subject each point on the string to a linear restoring force, we add a $\varphi^2$ term to the lagrangian density and end up with the Klein-Gordon equation as our eom, which is dispersive but still linear. However, if we add a $\phi^3$ term to our lagrangian density, \begin{align} \mathscr{L} = \frac{1}{2}(\partial_t \varphi)^2 - \frac{1}{2} (\partial_x \varphi)^2 + \frac{g}{6} \varphi^3 \end{align} our classical eom becomes \begin{align} \partial_t^2 \varphi - \partial_x^2 \varphi - \frac{g}{2} \varphi^2 = 0. \end{align} Now we have a non-linear equation of motion, and we can no longer superimpose solutions.

Of course, this is all at the classical level, and some additional work is necessary to say what happens when we quantize. But once we figure out how to do perturbation theory, Feynman diagrams give us a simple pictorial representation of the interaction terms in our lagrangians.

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  • $\begingroup$ So when we omit higher order terms like $\varphi^3$ in our theories, we're effectively limiting ourselves to linear solutions? Also, do higher order interactions between different fields (say something like $(q_1q_2)^3$) correspond physically to a similar idea of removing linear solutions? $\endgroup$ – stanfordude Feb 18 at 21:23

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