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I've been set the task of showing that:

$$ \bar{\psi^{s}}\psi^{s}=2m $$ For s=0,1. Where: $$ \psi^{0,1}=\sqrt{|E|+m}\begin{pmatrix}\chi^{0,1}\\ \frac{\vec{\sigma}\cdot\vec{p}}{E+m}\chi^{0,1}\end{pmatrix} $$ Where $\chi^{0}=\begin{pmatrix}1\\0\end{pmatrix}$ and $\chi^{1}=\begin{pmatrix}0\\1\end{pmatrix}$.

Now, this is what I have so far:

Let $a$ be any spinor; then, by definition $\bar a\equiv a^\dagger \gamma^0$, where $\dagger$ stands for hermitian conjugation: transpose + complex conjugation: $a^\dagger=(a^T)^*$. Therefore: \begin{align*} \bar{\psi^{s}}\psi^{s}=&\,e^{-ip_{\mu}x^{\mu}}e^{ip_{\mu}x^{\mu}}(|E|+m) \begin{pmatrix} 1 & 0 & \frac{\vec{\sigma}\cdot\vec{p}}{E+m} & 0 \end{pmatrix} \begin{pmatrix} 1&0&0&0\\0&1&0&0\\0&0&-1&0\\0&0&0&-1 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \\ \frac{\vec{\sigma}\cdot\vec{p}}{E+m} \\\\ 0 \end{pmatrix}\\ \bar{\psi^{s}}\psi^{s}=&\,(|E|+m) \begin{pmatrix} 1&0&0&0\\0&0&0&0\\0&0&-\frac{\vec{\sigma}\cdot\vec{p}}{E+m}&0\\0&0&0&0 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \\ \frac{\vec{\sigma}\cdot\vec{p}}{E+m} \\\\ 0 \end{pmatrix}\\\\ \therefore\,\bar{\psi^{s}}\psi^{s}=&\left[(|E|+m)-(|E|+m)\left(\frac{\vec{\sigma}\cdot\vec{p}}{E+m}\right)^{2}\right] \end{align*}

Essentially it is at this point where I am not sure where to go to reduce this obtain the value of $\bar{\psi^{s}}\psi^{s}=2m$. I'm assuming I have done everything correctly up to this point! Please advise...

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  • $\begingroup$ That it must be a multiple of the mass is obvious, because it is the only Lorentz invariant scalar that one can construct from the Dirac Lagrangian. I will try to clear it up to get the right coefficient in front. $\endgroup$ – gented Nov 23 '15 at 20:32
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You have done all the hard steps, what remains is to remember that $(\vec{p}\cdot\vec{\sigma})^2=\vec{p}^2=E^2-m^2$ so that \begin{align} (|E|+m) - (|E|+m)\left(\frac{\vec{p}\cdot\vec{\sigma}}{E+m}\right)^2 &= (|E|+m) - (|E|+m)\frac{\vec{p}^2}{(E+m)2}\\ &= (|E|+m) - (|E|+m)\frac{(|E|+m )(|E|- m)}{(E+m)2} \\ &= (|E|+m) - (|E|-m) \\ &=2m \end{align}

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  • $\begingroup$ Yeh I got this as well. Perfect! $\endgroup$ – DarthPlagueis Nov 24 '15 at 9:07

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