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The Dirac equation is more restrictive than the Klein-Gordon equation. For every solution to the Dirac equation, its components will be a solution of the Klein-Gordon equation, but the converse isn't true: if you form a spinor whose components are solutions of the Klein-Gordon equation, it might not solve the Dirac equation. If we start with the ...

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There are two factors at play here. The Lorentz force which causes the paths to bend with a radius proportional to the particles velocity and with a sense that dependent on both the particles charge and the direction of the particles velocity. In high energy (compared to $m_e$ events) such as the one pictured, the particles are nearly co-linear at the ...

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The electron has three well known properties, its electric charge, its magnetic dipole moment and its intrinsic spin. All three are constant quantities. And to prevent contradition about the reality of this intrinsic spin, it was shown in the Einstein-de-Haas experiment, that this spin really has to do with a rotation of the electron. It has to be stated, ...

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But what we never seem to see is why the electron and positron move the way that they do. Saying "they move like they do because of the force on them" doesn't explain anything at all. It's a non-answer. The equation of motion for charge particle (electron,positron) in magnetic field is $$m\frac{d}{dt}\left(\frac{\mathbf ... 2 Formally, the meaning you assign is just the usual meaning of the derivative.$$\partial_\mu \psi(x^\nu) = \lim_{h \to 0} \frac{\psi(x^\nu + h\delta^\nu_\mu) - \psi(x^\nu)}{h}$$You can indeed compute it componentwise, because you can subtract two spinors, as in the equation above, just by subtracting their components. The object you get has sixteen ... 0 Basically your confusion is caused by a bias about what a vector is. Everyone agrees that you can add two vector and get another. Everyone agrees you can scale a vector and get another vector. Sometimes we square a vector and get a scalar, but some people say that is "merely" an abuse of notation. But that is just a special case of ... 0 If you define spin angular momentum as \Sigma^{i}= 1/2 e^{ijk} \sigma_{jk} and \sigma_{jk}= e^{ajk} \sigma_{a}  then there is some inconsistency about when you place upper versus lower indices but you get:$$\Sigma^{i}=\frac{ 1}{2 }e^{ijk} \sigma_{jk}= \frac{ 1}{2 }e^{ijk} e^{ajk} \sigma_{a} =\frac{ 2\delta^i_a}{2}\sigma_{a} =\sigma_i$$It really ... 1 Comments to the question (v3): Note that the gamma matrices are covariantly conserved^1$$\nabla_{\mu}\gamma^c~=~\omega_{\mu}{}^c{}_b\gamma^b+\frac{1}{4}\omega_{\mu}{}^{ab}[\gamma_{ab},\gamma^c] ~=~0,\qquad \nabla_{\mu}\gamma^{\nu}~=~0,\tag{1}$$cf. e.g. Ref. 1. Consider the vector current$$J^{\mu}~:=~\bar{\psi}\gamma^{\mu}\psi,\tag{2}$$where \psi ... 1 Classically, spinors are insignificant, and should not be expected to fulfill any physical role, since they are not proper representations of the rotation group \mathrm{SO}(3) or the Lorentz group \mathrm{SO}(1,3). The importance of spinors does not arise from any classically intuitive thinking about orientations. Instead, the reason why spinors appear ... 7 There isn't a good definition of chirality in (2+1)D or any other odd dimension. This is because the \gamma_5 matrix can't be defined usefully in a Clifford algebra with an odd number of generators. For instance try to define \gamma_5 = \gamma^0\gamma^1\gamma^2. This commutes (not anti-commutes) with \gamma^0,\gamma^1,\gamma^2 and thus commutes with ... 1 I edited your question (I moved one parenthesis so that your quote was not a misquote), and maybe that answered your question. In case not ... If you had a Dirac spinor \Psi then there is a transformation like \Psi\mapsto e^{\theta\gamma_{13}/2}\Psi, where \gamma_{13}=\gamma_1\gamma_3 is a product of gamma matrices (or just the unit xz plane if you ... 0 No. Let's look at a spin measurement in detail. You can choose any direction \hat n = n_x\hat x+n_y\hat y+n_z\hat z and take any spin,  \alpha |+\rangle+\beta |-\rangle, where we define |\pm\rangle by$$\left(n_x\sigma_x+n_y\sigma_y+n_z\sigma_z\right) |\pm\rangle=\pm|\pm\rangle). Then you can send a beam of particles into a small region with an ...

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