# Tag Info

11

It's c for constant or celeritas, which means speed in Latin. Everyone uses it because it's convention. You could use $\xi$ or $\zeta$ or $\gamma$ or any other symbol you wanted, but then you'd have to explain what it meant, and people would have to go through the trouble to remember this every time they read your papers. Better to go with convention and ...

2

I assume you're thinking about Minkowski space, i.e. the metric $\eta_{\mu\nu}=\text{diag}(c^2,-1,-1,-1)$. You should be aware that the dot notation is purely a notational shorthand, and has no other information contained in it. In particular, by definition we have $$\dot{A}\equiv\partial_0A=\frac{1}{c}\frac{\partial A}{\partial t}$$ Thus, there is no ...

1

If your desired basis is the set ${|n\rangle}$, then the completeness relation tells you: $\hat{O} = \sum_a \sum_b \langle a|\hat{O}|b \rangle |a \rangle \langle b|$. Ideally, we prefer to do this in the orthonormal basis in which the operator $\hat{O}$ is diagonal, in which case this becomes $\hat{O} = \sum_a \langle a|\hat{O}|a \rangle |a \rangle \langle ... 1 The Delta symbol is often used to describe a "difference". Typically $$\Delta x=x_f-x_i$$ for some parameter$x$. That is, it is the final minus the initial. In this case, for potential energy, the final and initial states are explicitly mentioned as the limits of the integral. The way it is written here allows you to write the following:$$\Delta ... 1 Let the undotted index$a$correspond to the irrep$(\frac12,0)$of the Lorentz group. Then,$F_{ab}$corresponds to the tensor product$(\frac12,0)\otimes (\frac12,0)$which decomposes into the two irreps$(1,0)$and$(0,0)$of the Lorentz group.$(1,0)$is given by the symmetric part of$F_{ab}\$ while the antisymmetric part gives the scalar ...

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