# Particular solution of Weyl representation of Dirac equation

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 $$\mathcal{O}'$$.

(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!

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 $$K$$.

• Perfect! I worked out the details and got the answer! Thanks very much! You really helped me a lot! Commented May 7, 2021 at 22:05
• 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? Commented May 7, 2021 at 22:09
• @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. Commented May 8, 2021 at 1:56
• @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.) Commented May 8, 2021 at 2:00
• @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 $\theta \to -\theta$. Commented May 10, 2021 at 22:26