Timeline for Are there eight or four independent solutions of the Dirac equation?
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| Jul 4, 2020 at 14:00 | history | edited | Frederic Thomas | CC BY-SA 4.0 |
A couple of typos corrected, in particular in the header.
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| Jun 11, 2020 at 9:33 | history | edited | CommunityBot |
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| Jan 15, 2015 at 8:21 | vote | accept | jak | ||
| Jan 5, 2015 at 21:13 | answer | added | QuantumDot | timeline score: 4 | |
| Jan 4, 2015 at 17:08 | answer | added | Paganini | timeline score: 0 | |
| Jan 4, 2015 at 15:59 | answer | added | Eva | timeline score: 1 | |
| Dec 3, 2014 at 21:39 | history | edited | jak | CC BY-SA 3.0 |
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| Dec 3, 2014 at 17:04 | history | edited | jak | CC BY-SA 3.0 |
added 1353 characters in body; edited title
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| Dec 3, 2014 at 15:56 | history | edited | jak | CC BY-SA 3.0 |
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| Dec 3, 2014 at 15:41 | comment | added | jak | @ACuriousMind For some reasons I find this highly unsatisfactory. My definiton of linearly dependent is that we can find numbers $a,b,c,d,e,f,g,h \neq 0$, such that $a u_1 + b u_2 + c u_3 +d u_4 + e v_1 + f v_2 + g v_3 + h v_4=0$. This is only possible for one $t$ and not for all. With a definition of linear dependence like this, we could for example, argue that the terms in the Fourier basis $ \sum_n a_n e^{in x}$ are linearly dependet, because we can easily find numbers now for one $x$, for example for $x=0$. | |
| Nov 21, 2014 at 14:48 | comment | added | ACuriousMind♦ | Decompose $f = (x,y)$. Choose $a = x$, $b = y$, $c = -x$, $d = -y$. Observe that, for $m \neq 0$, this obviously is not a valid choice for all times, since left- and right-chiral parts do not decouple for massive spinors. | |
| Nov 21, 2014 at 14:34 | comment | added | jak | sorry, I still can't see it. Even with complex numbers I'm not able to choose numbers $a,b,c,d$, such that I get a chirality eigenstate $\psi_L = \begin{pmatrix} f \\ -f \end{pmatrix} = a u_1 + b u_2 + c v_1 + d v_1$. This may be obvious, but a concrete example, i.e. a concrete choice for $a,b,c,d$; would help my understanding a lot. | |
| Nov 21, 2014 at 14:18 | comment | added | ACuriousMind♦ | Observe that $a,b,c,d$ are allowed to be complex numbers, since we are in a Hilbert space. Since they are four independent vectors in a 4D Hilbert space, they span the whole of the space. | |
| Nov 21, 2014 at 14:15 | comment | added | jak | I've read this statement before and I think it is correct, but I failed to construct a left-chiral eigenstate from the commonly choosen basis: $(u_1,u_2,v_1,v_2)$. Do you have an idea how such a state can be constructed in this basis? To be an eigenstate of the chiral operator $\gamma_5$ the upper and lower two-component object inside the Dirac spinor must be related by a minus sign. I'm not able to construct such a state using this basis. More mathematically, how can $\psi_L = \begin{pmatrix} f \\ -f \end{pmatrix} = a u_1 + b u_2 + c v_1 + d v_1$ ? | |
| Nov 21, 2014 at 13:59 | comment | added | ACuriousMind♦ | There aren't "eight solutions" of the Dirac equations, since only four of them are independent. The other solutions aren't "discarded", they are just contained in the four normally chosen as basis. | |
| Nov 21, 2014 at 13:55 | history | asked | jak | CC BY-SA 3.0 |