# Tag Info

11

(I henceforth assume $c= \hbar=1$.) It is forbidden by the four-momentum conservation law. Put yourself in the centre of mass reference frame of the couple of massive particles (electron and positron). There $P_{e\overline{e}} = (2E,\vec{0})$ with $E\geq m_e>0$. Just because four momentum is conserved, this four-momentum must be the same as the one of the ...

7

The advantage of unitary gauge is that it completely removes unphysical fields, while adding additional degrees of freedom to the gauge bosons, which consequently become massive. This gauge works well for tree-level calculations, but complications arise when considering loops: The propagators of gauge fields and ghosts (which are needed to impose the ...

3

We already know that QED is incomplete. That's why we needed to develop the standard model that includes the electroweak force and QCD. We know that the standard model is incomplete and that we may need supersymmetry or other similar extensions to make it better, and even then gravity would be missing on the level of current quantum field theory. How to ...

3

We can write the Fourier transform of $\langle 0|\mathcal{T}A_{\nu}(x)\psi(x_1)\bar\psi(x_2)|0\rangle$ as $$S(p) D_{\nu\alpha}(q) \ e\,\Gamma^{\alpha}(p,q,p+q)S(p+q)$$ where $S(p)$ is the full fermion propagator, $D_{\nu\alpha}(q)$ is the full photon propagator, $\Gamma^{\alpha}(p,q,p+q)$ is the proper vertex function, and an overall momentum conservation ...

2

Indeed, do not take Feynman diagrams as literal representations of what is happening in a particle picture. Only the external lines of a diagram correspond to real particles - the internal lines, though called virtual particles, are little more than artifacts of the perturbative expansion we do to calculate QFT amplitudes, and there is little reason to ...

2

There is the (non-genetal) relation between the free energy of interacting of two currents $J^{a}, J^{b}$ and the propagator: $$U = -\frac{1}{2} \int d^{4}xd^{4}y J^{a}(x) D_{ab}(x - y)J^{b}(y).$$ It's not general, but it realizes the simple example which can help you to understand how to get the expression for force. The structure of field which causes ...

2

The classical Coulomb potential can be recovered in the non-relativistic limit of the tree-level Feynman diagram between two charged particles. Applying the Born approximation to QM scattering, we find that the scattering amplitude for a process with interaction potential $V(x)$ is $$\mathcal{A}(\lvert p \rangle \to \lvert p'\rangle) - 1 = 2\pi \delta(E_p ... 2 Let us start with a simple analogy that helps grasping the concept. Say, we need to calculate a electric field of a charged cylinder. What is the first thing you do? You choose your z-axis along the axis of the cylinder. Why won't you choose your axes differently, like pointing whatever they want? Well, you could, but that will make the problem ... 2 So as you say, the denominator is OK. For the numerator we use$$\tag{1}\not a\gamma_{\mu} = 2a_{\mu}-\gamma_\mu\not a, (\text{follows from } \{{\mu,\nu \} = 2\eta_{\mu\nu}}). $$Then the numerator can be written$$Num = 2p_\mu-k_\mu-\zeta_\mu$$where$$\tag{2}\zeta_\mu:=-k_\mu+\not k\gamma_\mu+\gamma_\mu\underbrace{(\not p-m)}_{=0 \text{ when hitting ...

1

I'll give you a draft of the answer to put you on the right track. You should then be able to fill the hole and complete the details. I think you just need to take the non-relativistic limit of a field theory where instead of the tree-level propagator you use the full 1PI propagator $\frac{1}{p^2-m^2-\Pi(p^2)}$ where $\Pi(\pi^2)$ is the self-energy. If the ...

1

At the classical level, the global gauge invariance leads via Noether's theorem to electric charge conservation, cf. e.g. this Phys.SE post. The Ward-Takahashi identity (WTI) can roughly speaking be thought of as a quantum version of this. In particular, we stress that the WTI is intimately tied to electric charge conservation. OP's observation that ...

1

The kinetic term $$\mathcal{L}_\text{fermion} = i\overline{\psi}\gamma^\mu \partial_\mu \psi$$ changes too, by precisely the right quantity to cancel the change in the interaction term. Thus, the total Lagrangian is invariant, and this is what matters.

1

As @RobinEkman mentioned, the kinetic term changes as well. This can be easily computed \begin{split} {\cal L}_D = i {\bar \psi} \gamma^\mu \partial_\mu \psi &\to i {\bar \psi} e^{- i \alpha} \gamma^\mu \partial_\mu \left( e^{i \alpha} \psi \right) = i {\bar \psi} \gamma^\mu \partial_\mu \psi - \partial_\mu \alpha ( {\bar \psi} ...

Only top voted, non community-wiki answers of a minimum length are eligible