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I edited the question as a result of the discussion in the comments. Originally my quesiton was how to interpret the four discarded solutions. Now I'm making a step back and hope that someone can clarify in what sense its sensible to discard four of the eight original solutions of the Dirac equation.

From making the ansatz ${\mathrm{e}}^{+ipx}$ and ${\mathrm{e}}^{-ipx}$, with $E=\pm \sqrt{ (\vec p)^2 +m^2} $ we get eight solutions of the Dirac equation. $u_1, u_2, u_3 , u_4$ and $v_1,v_2,v_3,v_4$.

Conventionally the four solutions ($u_3 , u_4,v_3,v_4$.) following from $E=- \sqrt{ (\vec p)^2 +m^2}$ are said to be linearly depend and of the remaining four solutions with $E=+\sqrt{ (\vec p)^2 +m^2}$ two ($u_1,u_2$) are commonly interpreted as particle and two ($v_1,v_2$) as antiparticle solutions.

Nevertheless, in order to be able to construct chirality eigenstates we need the other four solutions and I'm unsure in how far we can then say that four of the eight solutions are really linearly dependent. A chiral eigenstate must always be of the form $ \psi_L= \begin{pmatrix} f \\ -f \end{pmatrix} $ for some two component object $f$. In order to construct such an object we need all eight solutions. For example $\psi_L= u_1 - u_3$, as can be seen from the explicit form of the solutions recited below.

In addition, I'm unable to see that the eight solutions are really linearly dependent, because for me this means that we can find numbers $a,b,c,d,e,f,g,h \neq 0$, such that $a u_1 + b u_2 + c u_3 +d u_4 + e v_1 + f v_2 + g v_3 + h v_4=0$. As pointed out in the comments, this can be done, but only for one point in time. Is this really enough? In what sense is then for example the basis used in the Fourier expansion $\sum_n (a_n e^{in x} + b_n e^{-in x}) $ linearly independent? With the same reasoning we could find numbers for one $x$ to show that all these $e^{in x}$ and $e^{-in x}$ are linearly dependent...

The explicit solutions

This is derived for example here

Two solutions follow from the ansatz ${\mathrm{e}}^{-ipx}$ with $E=+ \sqrt{ (\vec p)^2 +m^2}$ and two with $E=- \sqrt{ (\vec p)^2 +m^2}$ .

In the rest frame the solutions are

$$ E=+ \sqrt{ (\vec p)^2 +m^2} \rightarrow u_1 = \begin{pmatrix} 1 \\ 0 \\0 \\ 0 \end{pmatrix} {\mathrm{e}}^{-imt} \qquad u_2 = \begin{pmatrix} 0 \\ 1 \\0 \\ 0 \end{pmatrix} {\mathrm{e}}^{-imt} $$

$$ E=- \sqrt{ (\vec p)^2 +m^2} \rightarrow u_3 = \begin{pmatrix} 0 \\ 0 \\1 \\ 0 \end{pmatrix} {\mathrm{e}}^{-imt} \qquad u_4 = \begin{pmatrix} 0 \\ 0 \\0 \\ 1 \end{pmatrix} {\mathrm{e}}^{-imt} $$

Analogous four solutions from the ansatz ${\mathrm{e}}^{+ipx}$, we get four solutions.

$$ E=+ \sqrt{ (\vec p)^2 +m^2} \rightarrow v_1 = \begin{pmatrix} 0 \\ 0 \\1 \\ 0 \end{pmatrix} {\mathrm{e}}^{imt} \qquad v_2 = \begin{pmatrix} 0 \\0 \\0 \\ 1 \end{pmatrix} {\mathrm{e}}^{imt} $$

$$ E=- \sqrt{ (\vec p)^2 +m^2} \rightarrow v_3 = \begin{pmatrix} 1 \\ 0 \\0 \\ 0 \end{pmatrix} {\mathrm{e}}^{imt} \qquad v_4 = \begin{pmatrix} 0 \\ 1 \\0 \\ 0 \end{pmatrix} {\mathrm{e}}^{imt} $$

Examples for chiral eigenstate are, with some two component object $f$

$$\psi_L = \begin{pmatrix} f \\ -f \end{pmatrix} \hat = u_1 - u_3 = \begin{pmatrix} 1 \\ 0 \\0 \\ 0 \end{pmatrix} {\mathrm{e}}^{-imt} - \begin{pmatrix} 0 \\ 0 \\1 \\ 0 \end{pmatrix} {\mathrm{e}}^{-imt} \qquad \text{ or } \qquad \psi_L = \begin{pmatrix} f \\ -f \end{pmatrix} \hat = u_2 -u_4 $$

$$\psi_L = \begin{pmatrix} f \\ -f \end{pmatrix} \hat = v_1 - v_3 \qquad \text{ or } \qquad \psi_L = \begin{pmatrix} f \\ -f \end{pmatrix} \hat = u_2 - u_4 $$

And similar for $\Psi_R = \begin{pmatrix} f \\ f \end{pmatrix}$.

Are four of the eight solutions really dependent? If yes, how can this be shown explicitly ? Any source, book, pdf would be awesome. Is it possible to interpret the soltutions $(u_3,u_4,v_3,v_4)$ that can be discarded for many applications, but that are needed in order to create chirality eigenstates?

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  • $\begingroup$ 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. $\endgroup$ – ACuriousMind Nov 21 '14 at 13:59
  • $\begingroup$ 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 $\psi_L = \begin{pmatrix} f \\ -f \end{pmatrix} = a u_1 + b u_2 + c v_1 + d v_1$ ? $\endgroup$ – jak Nov 21 '14 at 14:15
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    $\begingroup$ 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. $\endgroup$ – ACuriousMind Nov 21 '14 at 14:18
  • $\begingroup$ 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. $\endgroup$ – jak Nov 21 '14 at 14:34
  • $\begingroup$ 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. $\endgroup$ – ACuriousMind Nov 21 '14 at 14:48
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OP's solutions $u_3$, $u_4$, $v_3$, $v_4$ don't solve the Dirac equation. For example, let's try $u_3$:

$$ \begin{align} 0&=(i \gamma^\mu\partial_\mu -m) \psi, \qquad\text{with}\enspace \psi= u_3 = (0,0,1,0)^T e^{-imt}\\ &=\big[i\big(\gamma^0\frac{\partial}{\partial t}+\vec\gamma\cdot\nabla \big)-m\big]u_3\\ &=[i\gamma^0(-i m)-m]u_3\\ &=\textstyle\Big[\pmatrix{m\\&m\\&&-m\\&&&-m}-\pmatrix{m\\&m\\&&m\\&&&m}\Big]\pmatrix{0\\0\\1\\0}e^{-imt}\\ &=\pmatrix{0\\&0\\&&-2m\\&&&-2m}\pmatrix{0\\0\\1\\0}e^{-imt}\qquad\text{(false)} \end{align} $$

It is because these solutions don't solve the Dirac equation that there are only four degrees of freedom. It is said that the Dirac equation projects out four physical solutions out of a possible total of eight degrees of freedom.

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this is not an answer to your question per se but it might be relevant. The solutions you've listed are not the general solutions of the Dirac equation. These are only the solutions for a particle at rest. The general solutions have terms dependent on the momentum, energy and mass. The chirality eigenstates are not derived from the spinors, but rather from the gamma-5 matrix and with the help of the helicity eigenstates in the ultrarelativistic limit ($E >>m$). Check Mark Thompson's book on Modern Particle Physics.

The solutions of the Dirac equation are made of a spinor times a plane wave. The spinors themselves are the 4-component vectors. They are the linearly independent ones. If you check the forms of the general (not yours, which are at rest) forms of the spinors (without the plane wave part) you'll see that indeed only four of them are linearly independent. All of the above can be seen derived from first principles in Thompson's book.

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There are only 4 independent solutions because:

$$v_{1,2}(+E, +\vec{p}) e^{+i(Et-\vec{p}.\vec{x})} = u_{3,4}(-E, -\vec{p}) e^{-i((-E)t-(-\vec{p}).\vec{x})}$$

(where $u$ and $v$ here don't include the propagating part contrary to your notation). You could choose to work with $u_1,u_2, u_3, u_4$ but it is more convenient to use $u_1,u_2, v_1, v_2$ where only "positive" quantities are defined.

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