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2

Fields, I think that the last one was a really useful advice. I can add that you should recognize the derivative as the momentum operator. $W_{\mu}=\frac{1}{2}\epsilon_{\mu\alpha\beta\sigma}\sum^{\alpha\beta}P^{\sigma}$. Then you can go to the rest frame (since it's a massive particle) and evaluate the scalar. You have $P^\mu=(m,0)$ so the index $\sigma$ ...


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Classically fermionic fields (and hence $b,b^{\dagger},k,k^{\dagger}$) would still be anticommuting, i.e. $$ \{ \psi_i(x,t),\psi_j^*(y,t') \} = 0 \quad \forall x,y,t,t',j $$ they are so-called Grassmann numbers (or better said Grassman-number valued functions). Whereas classical scalar fields are just ordinary commuting numbers (or better said real- or ...


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I think your last paragraph really hits on the key point here; in a formal sense, you absolutely cannot go down from the four component spinor wave function to a two component function. But, as you also correctly pointed out, the negative energy components of the spinor wavefunction are suppressed in the non-relativistic limit $c\rightarrow \infty$. Thus we ...


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The square of the static Dirac Hamiltonian is something like $$ H^2= -\nabla^2 +m^2+e {\boldsymbol \sigma}\cdot {\bf B} $$ where $\nabla$ is the gauge covariant derivative.


5

The product of a term that's anti-symmetric in two indices and a term that's symmetric in two indices is always zero. If $A_{ij} = -A_{ji}$ and $S^{ij} = S^{ji}$, then $$ A_{ij} S^{ij} = - A_{ji}S^{ji},$$ but we can just rename the indices we're summing over by exchanging $i\leftrightarrow j$ and so $A_{ij} S^{ij} = -A_{ij} S^{ij} = 0$. The cross product of ...


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$\gamma_a$ lives in the spin/Clifford bundle denoted by Roman indices $a$, while spacetime $dx^\mu$ is characterized by Greek indices $\mu$. Romans and Greeks shake hands with the help of tetrad (or vierbein) $$ e^a_\mu \gamma_a dx^\mu $$ The flat Minkowsky space (with metric $g_{\mu\nu} = \eta_{ab}e^a_\mu e^a_\nu = \eta_{\mu\nu}$) is characterized by $$ e^...


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As you can see, the covariant derivative has two components: $$\nabla_{\mu} =\partial_{\mu} +\Gamma_{\gamma \alpha}^{\mu}, $$ in which the first describes the change of the vector field and the second is a linear transformation of the vector space, related with the possibility of making different choices for the basis of the vector space in each point of the ...


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There is generally no reason for spin to be conserved (i.e., a good quantum number) if it isn't the only form of angular momentum in a system. Total angular momentum ($\mathbf{J}$), consisting of orbital ($\mathbf{L} = \mathbf{r} \times \mathbf{p}$) plus spin ($\mathbf{S}$), is conserved (corresponding to rotation invariance). It happens that the Schrödinger ...


11

The Dirac Hamiltonian for free particles doesn't commute with the $z$ component of the spin operator $\hat S_z$, but it does commute with the helicity operator $\hat h=S⋅p$. This means one can know simultaneously the energy and helicity of a particle. The Hamiltonian does not commute with $\hat S_z$, so the same is not true. For more information about good ...


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The commutator $[\nabla_\mu,\nabla_\nu]\phi$ is $$ (\partial_\mu+A_\mu)(\partial_\nu+A_\nu)\phi- (\partial_\nu+A_\nu)(\partial_\mu+A_\mu)\phi\\ = \partial_{\mu\nu}\phi + A_\mu \partial_\nu\phi+(\partial_\mu A_\nu) \phi + A_\nu \partial_\mu \phi+ A_\mu A_\nu \phi-(\mu\leftrightarrow \nu)\\ = ((\partial_\mu A_\nu)- (\partial_\nu A_\mu))\phi $$ Notice that we ...


2

Is it possible to do this procedure with different functions...? Yes. ...for what reason do we not (at least not in any standard introductory texts I have ever read) do so? No fundamental reason. It's just convenient in some contexts, that's all. In flat spacetime, using solutions with translation symmetry is often convenient, but it's not necessary. The ...


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This is done so that the $S$-matrix for the fields is manifestly invariant under translations. I will illustrate the case explicitly for a real scalar field $\phi$, with creation and annihilation operators $a^\dagger (\vec{p})$ and $a(\vec{p})$, respectively. Under translations $x \to x + b$, these transform as $$ a'(\vec{p}) = U(b) a(\vec{p}) U^{-1} (b) = e^...


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The charge conjugation matrix $C$ is not an antilinear map. It is ordinary matrix with the property that $$ C\gamma^\mu C^{-1} =-(\gamma^\mu)^T. $$ (or maybe $C^{-1} \gamma^\mu C = -(\gamma^\mu)^T$. Conventions differ.).


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