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Okay, so I've been having some trouble studying by myself the solutions of the Dirac equation and I need help. Suppose $\psi$ is a solution of $(i\gamma^{\mu}\partial_\mu-m)\psi = 0$ and let $\psi = \binom{\psi_{L}}{\psi_{R}}$ be defined by: $$\binom{\psi_{L}}{0} = \frac{1}{2}(I-\gamma^{5})\psi \quad \mbox{and} \quad \binom{0}{\psi_{R}} = \frac{1}{2}(I+\gamma^{5})\psi$$ I've already showed that $\psi_{L}$ and $\psi_{R}$ satisfy the coupled equations:

$$i\bar{\sigma}^{\mu}\partial_{\mu}\psi_{L}-m\psi_{R} = 0\quad \mbox{and}\quad i\sigma^{\mu}\partial_{\mu}\psi_{R}-m\psi_{L}=0 \tag{1}$$

where $\sigma^{\mu}$ are the Pauli matrices and $\bar{\sigma}^{\mu}$ is the $2\times 2$ identity if $\mu = 0$ and $-\sigma^{\mu}$ otherwise. If the particle described by $\psi$ is at rest at some reference frame $\mathcal{O}'$, the above equations become: $$i\partial_{0}\psi_{L}' = m\psi_{R}'\quad \mbox{and} \quad i\partial_{0}\psi_{R}' = m\psi_{L}'$$ since the spatial derivatives must be zero.

Now consider an inertial frame $\mathcal{O}$ with respect to which $\mathcal{O}'$ and the particle described above are moving with velocity ${\bf{v}} = (0,0,v)$ in the $x^{3}$ direction and let: $$\Lambda = \begin{pmatrix} \cosh\theta & 0 & 0 & -\sinh\theta \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ -\sinh\theta & 0 & 0 & \cosh\theta \end{pmatrix} $$ which is a Lorentz boost taking $\mathcal{O}$ to $\mathcal{O}'$.

The exercise asks to prove:

(a) To show that $m\cosh\theta = m \gamma = E$ and $m\sinh\theta = mv\gamma = p$.

(b) To use item (a) and the solutions $\psi_{R}'$ and $\psi_{L}'$ above to show that: $$\psi_{L} = e^{i(-Et+px^{3})}\binom{e^{-\frac{\theta}{2}}}{0} \quad \mbox{and} \quad \psi_{R}= e^{i(-Et+px^{3})}\binom{e^{\frac{\theta}{2}}}{0}$$ is the solution of the coupled equations (\ref{1}).

I have very little background with special/general relativity and since I've been studying these topics by myself, I don't have any help. But I don't know even how to start addressing (a) and (b). Any help is welcome!

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Hints:

  1. In the rest frame, the four-momentum is $(m,0,0,0)$. Writing this as a column matrix and multiplying by $\Lambda^{-1}(\theta)=\Lambda(-\theta)$ gives $(E,0,0,p)$ in the boosted frame. This is independent of the Dirac equation.

  2. For any coordinate transformation $x\to \bar x$, we have $d\bar x^a\bar \partial_a=d x^a\partial_a$. Use this to deduce that tf $\bar x^a=\Lambda^a_b x^b$ for any linear transformation $\Lambda$, then $\bar\partial_a=(\Lambda^{-1})_a^c\partial_c$, where $\Lambda^a_b(\Lambda^{-1})_a^c=\delta_b^c$. This is independent of the Dirac equation.

  3. Define $$ K\equiv \exp\left(\gamma^0\gamma^3\theta/2\right) =\cosh(\theta/2)+\gamma^0\gamma^3\sinh(\theta/2), \tag{1} $$ and notice that $$ K^{-1}\gamma^a\partial_a K = \gamma^a(\Lambda^{-1})_a^b\partial_b, \tag{2} $$ with $\Lambda$ defined as in the question. To derive (2), use \begin{align} K^{-1}\left(\gamma^0\partial_0 + \gamma^3\partial_3\right)K &= \left(\gamma^0\partial_0 + \gamma^3\partial_3\right)K^2 \\ K^{-1}\left(\gamma^1\partial_1 + \gamma^2\partial_2\right)K &= \gamma^1\partial_1 + \gamma^2\partial_2 \tag{3} \end{align} with $$ K^2 = \exp\left(\gamma^0\gamma^3\theta\right) =\cosh(\theta)+\gamma^0\gamma^3\sinh(\theta). \tag{4} $$

  4. Use 2 and 3 to see that if we are given one solution of the Dirac equation, then we can construct another solution by multiplying by the matrix $K$ and using $\Lambda$ to transform the spacetime variables as usual.

Don't split it into $\psi_{L/R}$ first, because that will only complicate things. Working with the original Dirac equation is easier. You can split it into $\psi_{L/R}$ after you're done with everything else, because $\gamma^5$ commutes with $K$.

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    $\begingroup$ Perfect! I worked out the details and got the answer! Thanks very much! You really helped me a lot! $\endgroup$
    – MathMath
    Commented May 7, 2021 at 22:05
  • $\begingroup$ Just to be sure of one calculation: to show that given $\psi$ solution of the Dirac equation, $K\psi(\Lambda^{-1}x)$ is another solution, I did the following. Let $\bar{\psi}(x) := \psi(\Lambda^{-1}x)$. Then, since $\psi$ is a solution of the Dirac equation, $0 = (i\gamma^{\mu}\partial_{\mu}-m)\bar{\psi} = (i\gamma^{\mu}(\Lambda^{-1})_{\mu}^{\nu}\partial_{\nu}-m)\psi(\Lambda^{-1}x) = i K^{-1}(i\gamma^{\mu}\partial_{\mu}-m)K\psi(\Lambda^{-1}(x))$, so that $(i\gamma^{\mu}\partial_{\mu}-m)K\psi(\Lambda^{-1}(x))=0$. Is this the correct reasoning? $\endgroup$
    – MathMath
    Commented May 7, 2021 at 22:09
  • $\begingroup$ @MathMath You're on the right track. Write the original solution $\psi(x)$ in terms of a new function $\psi'(x)$ like this: $$ \psi(x)=K\psi'(\bar x) $$ with $\bar x = \Lambda x$. Since $K$ and $\Lambda$ are invertible, this determines the new function $\psi'$ uniquely. Now use the identity $$ (i\gamma^a\partial_a-m)\psi(x)=K(i\gamma^a\bar\partial_a-m)\psi'(\bar x), $$ which follows from equation (2) in the answer. The left-hand side is zero because $\psi(x)$ is assumed to be a solution, and therefore the right-hand side must also be zero, so $\psi'$ is also a solution. $\endgroup$ Commented May 8, 2021 at 1:56
  • $\begingroup$ @MathMath You can also re-arrange the logic, so that $\psi'$ is the original solution and $\psi$ is the new one, but while I was thinking about how to reply to your comment, I realized that the conventions used in the answer are more suited to doing it the other way around. (Oops.) $\endgroup$ Commented May 8, 2021 at 2:00
  • $\begingroup$ @ChiralAnolaly, I believe the conventions become compatible with $\psi'$ being the original solution if one defines $K = \operatorname{cosh}(\theta/2)-\gamma^{0}\gamma^{3}\operatorname{sinh}(\theta/2)$ instead, i.e. make the change $\theta \to -\theta$. $\endgroup$
    – MathMath
    Commented May 10, 2021 at 22:26

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