Feynman diagrams and Lagrangian

Question

In learning the Lagrangian formalism and now Feyman diagrams, when we look at Feynman diagrams we are told contruct terms at each vertex and propagator terms etc to calculate the overall amplitude and so on. But since the Lagrangian describes the physical processes, is there a way to get all the physics from the Lagrangian instead of using Feyman diagrams? Also, I read that in weak interactions the interactions particle flavour is conserved but for this $e^{-} + e^{+} -> \mu^{-} + \mu^{+}$ it appears not to be the case. Also, I saw another diagram where the electron and positron come in then a vertical line joins them and at each point a photon is emitted. And sometimes I see them join at a single vertex and create one photon, but then I read that is not allowed? can there be an interaction with just the electron and positron annihilating to create a photon?

Answer 1

is there a way to get all the physics from the Lagrangian instead of using Feyman diagrams?

All the Feynman diagrams are literally represented by terms in the Lagrangian. Each term in the Lagrangian has a corresponding Feynman diagram. Usually you will see terms up to quadratic powers in the fields (two vertices in the Feynman representation) or even up to higher powers (eg., $\phi^3$ or $\phi^4$ in a scalar field theory). Lagrangians are usually written down with the highest order contributions to the overall amplitude, but there is no reason why you cannot write a Lagrangian up to whatever power you like.

Also, I read that in weak interactions the interactions particle flavour is conserved but for this $e^{-} + e^{+} -> \mu^{-} + \mu^{+}$ it appears not to be the case.

Flavour in this case refers to leptons. Different types of lepton does not imply different flavour. Remember that flavour refers to elementary particle species and in the standard model there are six flavours of quarks and six for leptons. So muons and electrons are from the same flavour, namely lepton. Are you referring to lepton number? It's still conserved in this particular interaction, since the lepton number before $(-1) +(+1) = 0$ which is the same after $(-1) +(+1) = 0$. You can also have interactions like

$$\langle u \bar{d} \rangle \rightarrow W^+ \rightarrow \mu^{+} +\nu_{\mu}$$

where a meson (quark-antiquark state) forms two leptons. As you can see this is mediated by a $W^{+}$ boson. Here both baryon and lepton number are conserved.

And sometimes I see them join at a single vertex and create one photon, but then I read that is not allowed? can there be an interaction with just the electron and positron annihilating to create a photon?

A single on shell photon would violate conservation of momentum. In the centre of mass frame for the electron and positron, conservation of momentum would require this one photon to have zero momentum, which is not possible since a (real) photon cannot have zero momentum. For a photon, momentum is given by

$$p = \frac{E}{c}$$

This is why you have two photons in your original diagram. That will satisfy conservation of momentum.

Answer 2

In learning the Lagrangian formalism and now Feyman diagrams, when we look at Feynman diagrams we are told contruct terms at each vertex and propagator terms etc to calculate the overall amplitude and so on. But since the Lagrangian describes the physical processes, is there a way to get all the physics from the Lagrangian instead of using Feyman diagrams?

You can derive the Feynman rules from the Lagrangian. One approach is to use the path integral. For instance, for a scalar field, one can define the partition function

\begin{equation} Z[J] = \int D \phi e^{i \int {\rm d}^4x \left[\mathcal{L}(\phi,\partial \phi) + \phi J \right]} \end{equation} and then one can derive a perturbative expansion for any correlation function by functional differentiation of $Z[J]$ with respect to $J$. The Feynman rules drop out of this procedure naturally. I won't derive it, but this approach can be found in most QFT textbooks (Srednicki, Peskin and Schroder,...)

Also, I read that in weak interactions the interactions particle flavour is conserved but for this 𝑒−+𝑒+−>𝜇−+𝜇+ it appears not to be the case.

The initial and final electron number are zero, as are the initial and final muon number. So, no problems.

Also, I saw another diagram where the electron and positron come in then a vertical line joins them and at each point a photon is emitted. And sometimes I see them join at a single vertex and create one photon, but then I read that is not allowed? can there be an interaction with just the electron and positron annihilating to create a photon?

You need to emit at least two photons. Imagine going to the center of mass frame of the electron and positron; there is zero net momentum. One photon cannot have zero net momentum.