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The binding energy of the electrons in a silver atom is far less than the rest energy of an electron, so there is no ambiguity about the number of electrons in a silver atom. That makes adding up the spins a straightforward business. By contrast, the combined mass of the two up and one down quarks in a proton is about 10MeV (it isn't precisely known) but ...


3

Schwartz is simply noting that the $\beta$-function has a generic expansion in QED of the form (29) where $\beta_{0,1,2}$ are some numbers that can be computed by explicitly calculating the various Feynman diagrams. For instance the leading $\epsilon/2$ is the tree-level result in $d=4-\epsilon$ dimensions. This can be easily seen as follows. In ...


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For reference, the fermion propagator is $$ \left\langle 0 \right| T\psi(x)\overline\psi(y) \left|0\right \rangle= S(x-y) = \int \frac{d^4k}{(2\pi)^4} \frac{i}{\not k-m}e^{-ik\cdot(x-y)}$$ Depending on the time ordering, this describes a particle moving from $y$ to $x$, or an antiparticle moving from $x$ to $y$. Now, consider a one-loop diagram in which ...


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There are two things to consider: What does the potential look like? Is the wave function of the qubit narrow in the flux or charge basis? Potential shape The Hamiltonian of the transmon (a junction in parallel with a capacitor) is $$H_{\text{charge qubit}} = - E_J \cos(\phi) + \frac{(-2en)^2}{2C}$$ where $E_J\equiv I_c \Phi_0 / 2\pi$, $I_c$ is the ...


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Short answer Quantum optics is a field/branch of physics, with QED being a specific theory. Those working in quantum optics use a fully quantum theory (QED) when needed, but otherwise resort to using simpler tools such as the semiclassical approximation (classical field and quantum matter) which accurately describe a wide range of experiments (primarily ...


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If q is the momentum for positron, then the propagator for it is still $i\frac{\not{q}+m}{\not{q}^2 +m^2}$.


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He is writing the $\beta$ function as a perturbation series in the small parameter $\alpha$, $$\beta(\alpha) = \sum_{n=1}^\infty c_n \alpha^n.$$ By convention, it's common to write this expansion in the form he has given. The first coefficient is always $c_1 = - 4\pi\epsilon$ (where $\epsilon$ is the dimensional regularization parameter). The remaining terms ...



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