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The Dirac equation for a particle with charge $e$ is $$\left[\gamma^\mu (i\partial_\mu - e A_\mu) - m \right] \psi = 0$$ We want to know if we can construct a spinor $\psi^c$ with the opposite charge from $\psi$. This would obey the equation $$\left[\gamma^\mu (i\partial_\mu + e A_\mu) - m \right] \psi^c = 0$$ If you know about gauge transformations $$... 13 The interpretation of the Dirac equation states depend on what representation you choose for your \gamma^\mu-matrices or your \alpha_i and \beta-matrices depending on what you prefer. Both are linked via \gamma^\mu=(\beta,\beta\vec{\alpha}). Choosing your representation will (more or less) fix your basis in which you consider the solutions to your ... 11 We know that we can describe a spin 1/2 massless particle using only a single Weyl field (lets say left-handed \psi_{L}). To introduce a mass term we have to use two spinor fields (one left-handed and one right-handed) and this gives the Dirac mass term. The question is now that if we can describe a massive particle with a single Weyl field. Well yes, ... 11 This is standard theory. Try Birrell, N. D., & Davies, P. C. W. (1982). Quantum Fields in Curved Space. Cambridge: Cambridge University Press. Bog standard Curved space QFT text. Don't remember how much is said specifically about spinors though. Brill, D., & Wheeler, J. (1957). Interaction of Neutrinos and Gravitational Fields. Reviews of Modern ... 11 The Zitterbewegung is more of a relic of the early Dirac equation days. It does not exist in the standard position, velocity and acceleration operators of the single particle field, only in alternatively derived versions. These alternative versions were developed because people thought the standard operators were wrong. In fact they didn't understand the ... 11 Spin is a property of the representation of the rotation group SO(3) that describes how a field transforms under a rotation. This can be worked out for each kind of field or field equation. The Klein-Gordon field gives a spin 0 representation, while the Dirac equation gives two spin 1/2 representations (which merge to a single representation if one also ... 10 The mistake you are making is in "daggering" the object \omega_{\mu\nu}. For each \mu, \nu = 0,\dots 3, the symbol \omega_{\mu\nu} is a real number, so its dagger (which is really just complex conjugation in this case) does nothing; (\omega_{\mu\nu})^\dagger = \omega_{\mu\nu}. When we say that \omega_{\mu\nu} is an antisymmetric real matrix, we ... 10 At the risk of telling you how to "suck eggs" (your level in these things is not altogether clear), here goes. Ingredients: The essential ingredients to this explanation are: A physical "system" which evolves in and whose "events" happen in some space \mathcal{U} (ordinary Euclidean 3-space or Minkowsky spacetime, for example); in physics this space is ... 9 Dear rubenb, yes, what your professor says is surely based on solid maths. The reason is that the 4-component Dirac spinor is actually composed of two separate 2-component pieces. The elementary "spinors" for 3+1 dimensions have two complex components. That results from the isomorphism between groups$$SL(2,C) \sim Spin (3,1).$$Note that both groups have 6 ... 9 Symmetric under charge conjugation (which gives us positrons) and symmetric under the sign of the energy are two different things, which is where I think you are getting confused. Negative energy electrons aren't positrons, they are negative energy electrons. The absence of a negative energy electron in the "sea of charge" can be viewed as a positive ... 9 Let us generalize from four space-time dimensions to a d-dimensional Clifford algebra C. Define$$\tag{1} p~:=~[\frac{d}{2}], $$where [\cdot] denotes the integer part. OP's question then becomes Why must the dimension n of a finite dimensional representation V be a multiple of 2^p? Proof: If C\subseteq {\rm End}(V) and V are both ... 9 What you have is a good start. If we make the usual assignments that {\partial\over{\partial t}} \to -iE and \nabla \to i{\bf p} then we get$$(E - e\Phi)\psi = (\alpha \cdot ({\bf p} - e{\bf A}) + m\beta)\psi.$$Now, pick a particular representation$$\beta = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix},\text{ }\alpha_i = \begin{pmatrix} 0 & ...

7

The question puts the cart before the horse. It is not that you derive that particles described by the Dirac equation have spin $\frac 1 2$. Rather, the Dirac equation is found as the equation for spin $\frac 1 2$ particles. A Dirac spinor $\psi$ is an element of the representation $(0,\frac 1 2) \oplus (\frac 1 2, 0)$ of the Lorentz group.1 In both ...

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There isn't a good definition of chirality in (2+1)D or any other odd dimension. This is because the $\gamma_5$ matrix can't be defined usefully in a Clifford algebra with an odd number of generators. For instance try to define $\gamma_5 = \gamma^0\gamma^1\gamma^2$. This commutes (not anti-commutes) with $\gamma^0,\gamma^1,\gamma^2$ and thus commutes with ...

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The expression $A^{\mu}B_{\mu}$ simply means that $$A^{\mu}B_{\mu}=A^{0}B_{0}+A^{1}B_{1}+A^{2}B_{2}+A^{3}B_{3}$$ Using the Minkowski metric with signature $(+---)$ you write this as $$A^{\mu}B_{\mu}=A^{\mu}\eta_{\mu\nu}B^{\nu}=A^{0}B^{0}-A^{1}B^{1}-A^{2}B^{2}-A^{3}B^{3}$$ The metric simply tells you have how the components of a vector and its dual vector ...

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Dirac's derivation of the existence of positrons that you described was a totally legitimate and solid argument and Dirac rightfully received a Nobel prize for this derivation. As you correctly say, the same "sea" argument depending on Pauli's exclusion principle isn't really working for bosons. Modern QFT textbooks want to present fermions and bosons in a ...

7

The Lagrangian density for a Dirac field is $$\mathcal{L} = i\bar\psi\gamma^\mu\partial_\mu\psi -m \bar\psi\psi$$ The Euler-Lagrange equation reads $$\frac{\partial\mathcal{L}}{\partial\psi} - \frac{\partial}{\partial x^\mu}\left[\frac{\partial\mathcal{L}}{\partial(\partial_\mu\psi)}\right] = 0$$ We treat $\psi$ and $\bar\psi$ as independent dynamical ...

6

$$(\psi^\dagger \gamma^0 \psi)^* = \psi^\dagger \gamma^0 \psi$$ because $\gamma^0$ is hermitian. Also, \begin{align} (\psi^\dagger i \gamma^0 \gamma^\mu \partial_\mu \psi)^* &= -i \partial_\mu\psi^\dagger \gamma^{\mu\dagger} \gamma^0 \psi\\ &= -i \partial_\mu\psi^\dagger (\gamma^0 \gamma^\mu \gamma^0)\gamma^0 \psi\\ &= -i ... 6 Non-conservation of charge in Majorana terms The Dirac mass term is m\bar\psi \psi where one field-factor \bar\psi is complex conjugated (aside from other transpositions included in the Dirac conjugation) and the other is not. So one may assign a fermion number 1 to \psi which means that \bar\psi automatically carries -1 and in the product, the ... 6 For massless particles, helicity coincides with chirality thus you ask to find the basis such that \psi_{\pm}=\left( \psi_{\mp}\right) ^{\star},\quad\gamma_{5}\psi_{\pm}% =\pm\psi_{\pm}. $$Using the decomposition of hermitian operator:$$ \left( \gamma_{5}\right) _{ij}=\left( \psi_{+}\right) _{i}\left( \psi _{+}^{\star}\right) _{j}-\left( ...

6

Let's review how the KG equation is recovered from the Dirac: (in natural units where $\hbar=c_0=1)$ $$(i\gamma^\mu \partial_\mu - m)\Psi = 0$$ $$(-i \gamma^\mu \partial_\mu - m)(i \gamma^\mu \partial_\mu - m) = 0$$ $$(\gamma^\nu \gamma^\mu \partial_\nu \partial_\mu + m^2) \Psi = 0$$ $$(\partial^2+m^2)\Psi = 0.$$ In order for us to recover KG, we had to ...

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Dirac's explanation of the emergence of antiparticles such as positrons out of the Dirac sea, and the Dirac sea itself, is completely valid and legitimate, and you have described some non-quantitative aspects of it and differences between it and some condensed-matter situations. Dirac just began with the assumption that the Dirac spinor field $\Psi$ is a ...

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Neutrinos interact in the Standard Model only through their left-handed component, via electroweak interactions. However, the propagating neutrinos, which are mass eigenstates, are described by a field that is a Dirac spinor, i.e. with both chiralities $$\nu=\nu_L+\nu_R.$$ Therefore, when neutrinos are created or measured, the Dirac spinor is projected ...

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The term with the derivatives, is really a directional derivative $$\alpha^k\partial_k=\vec \alpha\cdot \nabla$$ and if applied to a function $f$ it measures its change along that particular direction: $$\alpha^k\partial_kf=\vec \alpha\cdot \nabla f=|\vec\alpha|\frac{\partial f}{\partial\vec n} =|\vec\alpha|\lim_{\delta x\to0}\frac{f(\vec x+\delta x\:\vec ... 5 The Schrödinger equation is only correct in the non-relativistic limit v << c, for particles without spin. The correct equation for spinless (=spin 0) particles is the Klein-Gordon equation, which reduces in the non-relativistic limit to the Schrödinger equation. If we want to talk about spin \frac{1}{2}, the correct, relativistic equation is ... 5 Regular numbers could never fulfill$$ AB + BA = 0, \quad AA = 1 = BB. $$The only way to fulfill the first eq. is to have either A = 0  or B = 0 but this violates the second equation. Matrices on the other hand can fulfill such equations and since Dirac knew about matrices he did not discard his idea after finding something that's impossible with ... 5 What you've written down is the spatial part of the electron wavefunction. The spin state is not included. The full wavefunction of the electron involves both the spatial part and the spin part. Sometimes in quantum mechanics books the full electron wavefunction is written as the tensor product of the spatial and spinor parts, sometimes you'll just see it ... 5 The four-component wave function \Psi in the Dirac equation may be viewed as a counterpart of \psi(x) in non-relativistic Schrödinger's equation. The Dirac equation may be written (and, in fact, was originally written by Dirac) in the Schrödinger's form$$ i\hbar \frac{\partial}{\partial t} \Psi = H \Psi  where $H=\vec\alpha\cdot \vec p + m\beta$ where ...

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I think that the first volume of the series "The Quantum Theory of Fields", by Steven Weinberg, is a good text to understand the origin of Dirac equation, QFT, and all these kind of topics. Maybe Weinberg's books are not the best for a first course in QFT (or in General Relativity, he has also a great book on this topic), but his great coverage and unique ...

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