Let $u_\lambda(\vec{k})$ and $v_\lambda (\vec{k})$ be solutions of the following equations $$(\not k-m)u_\lambda(\vec{k})=0$$ $$(\not k+m)v_\lambda(\vec{k})=0$$ Suppose that $u_\lambda(\vec{k})^\dagger u_\sigma(\vec{k})=\frac{\omega(\vec{k})}{m}\delta_{\lambda\sigma}$ and $v_\lambda(\vec{k})^\dagger v_\sigma(\vec{k})=\frac{\omega(\vec{k})}{m}\delta_{\lambda\sigma}$, where $\omega(\vec{k})=\sqrt{m^2+\vec{k}^2}$. I am trying to show that it implies that $$\bar{u}_\lambda(\vec{k})u_\sigma(\vec{k})=\delta_{\lambda \sigma} \qquad \quad \bar{v}_\lambda(\vec{k})v_\sigma(\vec{k})=-\delta_{\lambda \sigma}. $$

My question is: How to do it without taking any special representation of Dirac matrices and without doing it in special reference frame (i.e. rest frame)? Moreover, I don't assume transformation law for this bispinors (so I really do not assume they are bispinors in the proper sense). The only thing which I assume is that Dirac equations for $u$ and $v$ are satisfied and we know anticommutation rules for $\gamma$-matrices and we have given above normalization for $u^\dagger u$.


Hint: you want to calculate $$ \bar u_s(\boldsymbol p)\gamma^\mu u_{s'}(\boldsymbol p)=\ ? \tag{1} $$

Being a covariant object, the vector index in the r.h.s. can only be provided by $p^\mu$, and therefore $$ \bar u_s(\boldsymbol p)\gamma^\mu u_{s'}(\boldsymbol p)=a_{ss'} p^\mu \tag{2} $$ (why is $a$ independent of $\boldsymbol p$?)

To find $a_{ss'}$, contract both sides of this expression with $p_\mu$, and use the equation $\not pu=mu$.

Now let $\mu=0$.

Ok, let us add some more details:

If you contract both sides of $(2)$ with $p_\mu$, you get \begin{equation}m\bar u_s(\boldsymbol p)u_{s'}(\boldsymbol p)=a_{ss'}m^2\end{equation} On the other hand, if you let $\mu=0$ in $(2)$, you get \begin{equation}u^\dagger_s(\boldsymbol p)u_{s'}(\boldsymbol p)=\omega_{\boldsymbol p}a_{ss'}\end{equation} From these two equations, you should be able to solve for $\bar u_s(\boldsymbol p)u_{s'}(\boldsymbol p)$.

  • $\begingroup$ Shouldn't be $u^\dagger$ instead of $\bar{u}$ ? $\endgroup$ – mikis Nov 13 '16 at 12:46
  • $\begingroup$ @mikis no, you get $u^\dagger$ when you let $\mu=0$ (because $\bar u\gamma^0=u^\dagger$) $\endgroup$ – AccidentalFourierTransform Nov 13 '16 at 12:47
  • $\begingroup$ Ok. I still have one question : you used fact that $\bar{u}\gamma^\mu u$ is a covariant object. How to see it using only knowledge that $u$ satisfy given Dirac equation ? I think to do it we need a transformation law for bispinors. $\endgroup$ – mikis Nov 13 '16 at 12:56
  • $\begingroup$ @mikis the Dirac equation is covariant, and therefore so are its solutions (for example, the classical wave equation $(\partial_t^2-\nabla^2)y(x)=0$ has solutions $y=\exp(\omega t-\vec k\cdot\vec x)$, which are covariant, as you already know). $\endgroup$ – AccidentalFourierTransform Nov 13 '16 at 13:04
  • 2
    $\begingroup$ "To find ass′" ( ͡° ͜ʖ ͡°) $\endgroup$ – Feathercrown Nov 13 '16 at 18:10

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