# Tag Info

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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 ...

8

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 ...

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: $$\left( \gamma_{5}\right) _{ij}=\left( \psi_{+}\right) _{i}\left( \psi _{+}^{\star}\right) _{j}-\left( ... 6 The Dirac equation for a particle with charge e is$$ \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 $$If you know about gauge transformations$$ ...

5

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 ...

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

For the details of the physics involved in the two ways of interpreting the Dirac wave equation I recommend chapters XI and XII of Dirac's "Principles of Quantum Mechanics" 4th edition, and chapters XX and XXI of Messiah's "Quantum Mechanics", vol. II. For the more historical details I recommend chapters 5 and 6 of Crease and Mann's "The Second Creation", ...

5

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 ...

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

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 ...

5

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 ...

4

To write down Dirac's equation over curved spacetime, first express the geometry in terms of vierbeins and spin connections. Spinor bundles are vector bundles over spacetime transforming locally under the local Lorentz gauge group. Using the spin connection, we can write down covariant derivatives for sections of the spinor bundle. To get the Dirac operator, ...

4

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 ...

4

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 ...

4

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 ...

4

First, to get the equation you want apply $(i\gamma^\nu\partial_\nu + m)$ to both sides, then on the left hand side you'll get \begin{align} (i\gamma^\nu\partial_\nu + m)(i\gamma^\mu\partial_\mu - m)\psi &= (-\gamma^\nu\gamma^\mu\partial_\nu\partial_\mu-m^2)\psi \end{align} which, when set to zero, gives $$... 4 Think it with an example, Einstein's field equations are much more precise than Newton's law of gravity, but it's much more complicated to solve a Classical Mechanics problem with General Relativity. More fundamental and precise doesn't mean that it will give easier calculations. If it did, then then chemistry, medicine, etc... wouldn't exist because they ... 3 The four-component wave function \Psi in the Dirac equation may be viewed as a counterpart of \psi(x) in non-relativistic Schrödinger's equation. The Dirac equation may be written (and, in fact, was originally written by Dirac) in the Schrödinger's form$$ i\hbar \frac{\partial}{\partial t} \Psi = H \Psi $$where H=\vec\alpha\cdot \vec p + m\beta where ... 3 This is a matter of convention. You are totally right: the Q operator you have written implies that b annihilates a positive charge and vice versa. The thing is that in QED one usually defines Q in a slightly different way, namely:$$Q=-\left| e\right| \int {d^3p \over (2\pi)^3(E_p/m)} \sum_s \{b^\dagger(p,s)b(p,s)-d^\dagger(p,s)d(p,s)\}$$with ... 3 A\times P – more precisely, an expression proportional to A\times P + P\times A – wasn't set to zero. It was properly evaluated and the result gave the iq\hbar B/c term. Note that if \pi were a vector of c-numbers rather than operators, \pi\times \pi would be equal to zero. That's how the cross product behaves. So any term in the cross product ... 3 Positrons aren't negative energy solutions in the first place. This is quite evidently shown by the ~1 MeV of energy of the photons emitted upon annihilation. If positrons had negative energy then the remaining energy would be zero. So there is no need at all to explain the "suppression of the decay of electrons into positrons" with whatever argument. But ... 3 Luboš' Answer is just fine, and he's adopted a level of presentation that looks just right for you, but I prefer to put this somewhat differently, using the manifestly Lorentz covariant \gamma^\mu operators instead of using the operators (\beta,\alpha^i). For the massive Dirac equation, you need four objects that satisfy the relationships ... 3 @Cedric: It's true that the notion of "fields" (as we have today in QFT) is something that wasn't completely developed at that time. But i never heard that Dirac's Eq was understood as a "wave equation": he was trying to explicitly describe particles (electrons) — besides, the notion that "particles" and "waves" were connected already existed at that time, ... 3 The short answer to the question, "What happens to the Lagrangian of the Dirac theory under charge conjugation?" is, "Nothing." It is invariant with respect to charge conjugation. Before getting to the longer exposition, I'd like to point out a potential misunderstanding about the nature of invariance of the equations of motion under symmetry ... 2 If you are interested in the origins, I think you must start the history with Sommerfeld. It is Sommerfeld who gets the fine structure levels, via an amazing cancellation of two mistakes (Bohr-Sommerfeld quantisation and lack of spin). So Dirac is sailing with a clear goal: his results must agree with Sommerfeld in some way. And of course Sommerfeld orbits ... 2 A description of electron spin and the Pauli exclusion principle needs to go beyond the Schrödinger-equation to the spinor-valued Dirac-equation. I don't remember my atomic physics course very well but at the level of your analysis I think you just add the rule of the two parallel spin orbitals explicitly. For a discussion on this topic the wikipedia-page ... 2 The most general Lorentz transformation that is connected to the identity is given by the conjugation by \exp(-A) where$$ A = \frac 12 \omega_{\mu\nu} \gamma^\mu \gamma^\nu  and $\omega_{\mu\nu}$ is an antisymmetric tensor containing $D(D-1)/2$ parameters. The group of all such transformations is isomorphic to $Spin(D-1,1)$. If $\omega$ only contains ...

2

There is no "gauge symmetry" in any of these manipulations and you haven't offered any evidence for such a non-existent gauge symmetry. If I understand you well, you have just realized that the exact numerical form of the Dirac matrices depends on conventions. In particular, the matrix entries of the Dirac gamma (and therefore also alpha, beta) matrices ...

2

The problem is that this is not the right way to solve the Dirac equation interacting with an electromagnetic field. The method you are using assumes there is a classical field around the single-particle Dirac electron, and uses this field to find the motion of the electron. This is a meaningless approximation. The field produced by an electron is entangled ...

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