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13

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 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 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 ... 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 Let us generalize from four space-time dimensions to a d-dimensional Clifford algebra C. Define$$\tag{1} p~:=~[\frac{d}{2}], $$where [\cdot] denotes the integer part. OP's question then becomes Why must the dimension n of a finite dimensional representation V be a multiple of 2^p? Proof: If C\subseteq {\rm End}(V) and V are both ... 9 Dear rubenb, yes, what your professor says is surely based on solid maths. The reason is that the 4-component Dirac spinor is actually composed of two separate 2-component pieces. The elementary "spinors" for 3+1 dimensions have two complex components. That results from the isomorphism between groups$$SL(2,C) \sim Spin (3,1).$$Note that both groups have 6 ... 9 Symmetric under charge conjugation (which gives us positrons) and symmetric under the sign of the energy are two different things, which is where I think you are getting confused. Negative energy electrons aren't positrons, they are negative energy electrons. The absence of a negative energy electron in the "sea of charge" can be viewed as a positive ... 9 What you have is a good start. If we make the usual assignments that {\partial\over{\partial t}} \to -iE and \nabla \to i{\bf p} then we get$$(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 ...

7

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 ...

7

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 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 ...

6

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 ...

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.$$In order for us to recover KG, we had to ... 6$$(\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

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

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 ...

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 ...

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Yes. You are missing the fact that he is using the convention $$\nabla = (\partial_1, \partial_2, \partial_3)$$ as opposed to $$\nabla = (\partial^1, \partial^2, \partial^3)$$ The first convention is by far the most common in my experience.

5

This particular extra term may be removed by a field redefinition $$\psi\to \psi' = \psi - K \cdot \gamma^\mu \partial_\mu \psi$$ for an appropriate value of $K\sim 1/\Lambda$, up to terms that are even higher dimension operators. This also modifies the mass. This field redefinition is an explicit off-shell way to realize Vibert's comment that one is just ...

5

Think it with an example, Einstein's field equations are much more precise than Newton's law of gravity, but it's much more complicated to solve a Classical Mechanics problem with General Relativity. More fundamental and precise doesn't mean that it will give easier calculations. If it did, then then chemistry, medicine, etc... wouldn't exist because they ...

5

The reason is as follows. The scalar Klein-Gordon equation is (and has to be) a second-order equation because the box $$\square = \partial_\mu \partial^\mu$$ is the simplest Lorentz-invariant differential operator that may act on scalar fields. However, Dirac wanted to find a first-order equation and it is indeed possible for spinors because $$\partial ... 5 How about just testing the two different cases? I.e. if \mu\not=0 then the LHS becomes $$(\gamma^\mu)^\dagger= (\gamma^i)^\dagger= -\gamma^i \tag{see below}$$ while the RHS becomes $$(\gamma^\mu)^\dagger=\gamma^0\gamma^i\gamma^0 = -\gamma^0\gamma^0\gamma^i=-\gamma^i~~~~~~~~ (\text{OK}).$$ For ... 5 So the key is to understand that \nabla^\dagger = - \nabla. To see why this should be true, we go back to the definition of the adjoint of an operator, namely$$\left< \phi \right|\left. A \psi \right> = \left< A^\dagger\phi \right|\left. \psi \right> \implies \int d^dx \phi(x)^* {\hat A} \psi(x) = \int d^d x \left( {\hat A}^\dagger \phi(x) ...

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