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28

There is no full consensus about what relativistic means, but as a good start, we can take the criterion given by Tim Maudlin in his paper "Space-Time in the Quantum World": [A] theory is compatible with Relativity if it can be formulated without ascribing to space-time any more of different intrinsic structure than the [...] relativistic metric. This ...

25

This answer is meant to add to Luke's excellent concise answer, so please read his answer first. In quantum mechanics, only measurements have the statistical distributions, the "uncertainties" and all the things that are (validly) bothering you. As you point out, this makes notions of measured spacetime co-ordinates problematic. But the underlying theory ...

22

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 $$\... 22 The interpretation of the Dirac equation states depend on what representation you choose for your \gamma^\mu-matrices or your \alpha_i and \beta-matrices depending on what you prefer. Both are linked via \gamma^\mu=(\beta,\beta\vec{\alpha}). Choosing your representation will (more or less) fix your basis in which you consider the solutions to your ... 21 At the risk of telling you how to "suck eggs" (your level in these things is not altogether clear), here goes. Ingredients: The essential ingredients to this explanation are: A physical "system" which evolves in and whose "events" happen in some space \mathcal{U} (ordinary Euclidean 3-space or Minkowsky spacetime, for example); in physics this space is ... 20 Let us generalize from four space-time dimensions to a d-dimensional Clifford algebra C. Define$$\tag{1} p~:=~[\frac{d}{2}], $$where [\cdot] denotes the integer part. OP's question then becomes Why must the dimension n of a finite dimensional representation V be a multiple of 2^p? Proof: If C\subseteq {\rm End}(V) and V are both real, ... 20 Here is an algebraic approach to understand the edge state. Let us start from a generic Dirac Hamiltonian for the bulk fermions in the d-dimensional space.$$H=\sum_{i=1:d}\mathrm{i}\partial_i\alpha^i+m(x_i)\beta,$$where \alpha^i and \beta are anti-commuting gamma matrices (\{\alpha^i,\alpha^j\}=2\delta^{ij}, \{\alpha^i,\beta\}=0, \beta\beta=1), ... 18 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, ... 16 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 ... 15 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 ... 14 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 ... 13 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 ... 13 It can be instructive to see the applications of Clifford algebra to areas outside of quantum mechanics to get a more geometric understanding of what spinors really are. I submit to you I can rotate a vector a = a^1 \sigma_1 + a^2 \sigma_2 + a^3 \sigma_3 in the xy plane using an expression of the following form:$$a' = \psi a \psi^{-1}$$where \psi = \... 13 When you are dealing with Grassmann numbers you have a "left derivative" and a "right" derivative. A left derivative removes the variable from the left, a right derivative removes it from the right. Let's say we have the function: $$f(\theta_1, \theta_2)=f_0+f_1\theta_1+f_2\theta_2+f_3\theta_1\theta_2$$ Then the left derivative ... 12 The Lagrangian density for a Dirac field is$$ \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 $$We treat \psi and \bar\psi as independent dynamical ... 12 Non-conservation of charge in Majorana terms The Dirac mass term is m\bar\psi \psi where one field-factor \bar\psi is complex conjugated (aside from other transpositions included in the Dirac conjugation) and the other is not. So one may assign a fermion number 1 to \psi which means that \bar\psi automatically carries -1 and in the product, the ... 12 \require{cancel} First of all, recall that a super-Lie bracket [\cdot,\cdot]_{LB} (such as, e.g., a super-Poisson bracket \{\cdot,\cdot\} & the super-commutator [\cdot,\cdot]), satisfies super-antisymmetry$$ [f,g]_{LB} ~=~ -(-1)^{|f||g|}[g,f]_{LB},\tag{1} $$and the super-Jacobi identity$$\sum_{\text{cycl. }f,g,h} (-1)^{|f||h|}[[f,g]_{LB},h]_{...

11

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 ...

11

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 ...

11

I think I understand what you mean when you say that you're not satisfied with the “nontrivial bulk topology argument” when it comes to thinking about edge states. The Chern number (for time-reversal breaking) and $\mathbb{Z}_{2}$ invariant (for time-reversal symmetric) systems, as DaniH suggested, does indeed give you information about the edge states; the ...

11

Spin-1/2 admits first order equations simply because $(\mathbf{1/2,1/2})\otimes (\mathbf{0,1/2})$ contains the representation $(\mathbf{1/2,0})$ so that a linear equation for free particles can be written (i.e. it contains a derivative acting on one field and returning one field). The first term in the product is the derivative that transforms as a ...

11

The spinors $u^s(p), v^s(p)$ are not operators, they are just (an array of) numbers and therefore commute with the creation/annihilation operators.

10

From the relativistic covariance of the Dirac equation (see Section 2.1.3 in the QFT book of Itzykson and Zuber for a derivation. I also more or less follow their notation.), you know how a Dirac spinor transforms. One has $$\psi'(x')=S(\Lambda)\ \psi(x)$$ under the Lorentz transformation $$x'^\mu= {\Lambda^\mu}_\nu\ x^\nu= {\exp(\lambda)^\mu}_\nu\ x^\nu=(... 10 You have no other choice than to use 4\times 4 matrices. All these "representations" are different realizations (related by similarity transformations) of the only possible irreducible representation of the Clifford algebra that is spanned by the abstract \gamma^\mu. This representation, in a way, is the definition of what a "Dirac spinor" is, and it is ... 10 Chirality is not well-defined for massive fields. A famous consequence of this fact are pion masses, which can be linked to Chiral Symmetry Breaking. In the Lagrangian, you can define left- and right-handed Weyl fermions independently. A mass term will mix these, giving a massive Dirac fermion. Weyl fermions fulfill either$$ P_{L} \psi_L = \psi_L, \quad \...

10

How exactly did Dirac incorporate SR into his wave equation? Relativity treats space and time on equal footing. Lorentz transformations ("boosts") "mix" space and time more or less analogously to the way a 3-dimensional rotations "mixes" the usual $(x,y,z)$ coordinates of space. The Schrödinger equation describes non-relativistic quantum mechanical systems ...

9

If we say: "A field has a spin 0, spin 1/2 or spin 1 representation" then we in fact say something about how the field parameters transform if we go from one reference frame to another. spin 0: The values of the field do not change if we go from one reference frame to another spin 1: We have to apply the Lorentz transform matrix $\Lambda$ on the field ...

9

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 ...

9

As per Rob's suggestion, I decided to make this an answer. (Addendum: I've been meditating on this very topic for some time, and have been directed to some interesting literature referenced on Streater's webpage. As per Rococo's comment, I've updated my answer, but kept the old version for posterity.) The Answer To talk about "spin-1/2 particles", we ...

9

He basically shows that a local symmetry of the matter field described by Dirac equation will directly give as a consequence the existence of the gauge boson field, here photon. This is wrong. There are perfectly well-defined theories where you have matter fields but no gauge fields. And vice-versa: we have perfectly well-defined theories of gauge fields ...

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