# Tag Info

8

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

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 $$... 8 This is standard theory. Try Birrell, N. D., & Davies, P. C. W. (1982). Quantum Fields in Curved Space. Cambridge: Cambridge University Press. Bog standard Curved space QFT text. Don't remember how much is said specifically about spinors though. Brill, D., & Wheeler, J. (1957). Interaction of Neutrinos and Gravitational Fields. Reviews of Modern ... 7 Spin is a property of the representation of the rotation group SO(3) that describes how a field transforms under a rotation. This can be worked out for each kind of field or field equation. The Klein-Gordon field gives a spin 0 representation, while the Dirac equation gives two spin 1/2 representations (which merge to a single representation if one also ... 7 The mistake you are making is in "daggering" the object \omega_{\mu\nu}. For each \mu, \nu = 0,\dots 3, the symbol \omega_{\mu\nu} is a real number, so its dagger (which is really just complex conjugation in this case) does nothing; (\omega_{\mu\nu})^\dagger = \omega_{\mu\nu}. When we say that \omega_{\mu\nu} is an antisymmetric real matrix, we ... 6 For massless particles, helicity coincides with chirality thus you ask to find the basis such that$$ \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 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 ...

6

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 ...

6

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

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, ...

5

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 ...

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

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 ...

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

First, C) isn't the Dirac Equation, it's the Klein-Gordon equation Now, to your main question. A) comes from the classical equation for a free massive particle: $\dfrac{p^2}{2m} = E$ by making the operator (operating on $\phi$) substitutions: $p^2 \rightarrow - \hbar^2 \nabla^2$ $E \rightarrow i \hbar \dfrac{\partial}{\partial t}$ C) comes from B) by ...

4

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 ...

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 $$... 4 The Lagrangian density for a Dirac field is$$ \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 $$We treat \psi and \bar\psi as independent dynamical ... 4$$(\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 ...

4

The second equation is actually incorrect. It should be written as follows: $$i\partial^{\mu}\overline{\Psi}\gamma_{\mu}+m\overline{\Psi}=0.$$ Here, $\overline{\Psi}$ is understood as a 4-component row vector (not in the sense of the vector rep. of the Lorentz group). At any rate, $\overline{\Psi}$(or $\Psi^{\dagger}$) is not what you obtain from $\Psi$ ...

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 ...

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