Hermitian conjugate of plane wave spinors for Dirac equation

[bubblehead]

I need to show that

$u^{+}_{r}(p)u_{s}(p)=\frac{\omega_{p}}{m}\delta_{rs}$

where
$\omega_{p}=\sqrt{\vec{p}^2+m^{2}}$
$u_{r}(p)=\frac{\gamma^{\mu}p_{\mu}+m}{\sqrt{2m(m+\omega_{p})}}u_{r}(m{,}\vec{0})[\itex] is the plane-wave spinor for the positive-energy solution of the Dirac equation.<br /> <br /> I think my problem is twofold: I'm not sure I've computed the Hermitian conjugate of the spinor correctly (just the gamma matrix and p have Hermitian conjugates, is that right?) and I'm not sure how/why the normalization term disappears when squared. Either way, I'm not getting the nice simple answer I should!$

 

[member 553083]

So any help would be much appreciated. The Hermitian conjugate of the spinor is given by u^{+}_{r}(p)=\frac{\gamma^{\mu}p_{\mu}+m}{\sqrt{2m(m+\omega_{p})}}u^{+}_{r}(m{,}\vec{0})Using this, we can compute u^{+}_{r}(p)u_{s}(p):u^{+}_{r}(p)u_{s}(p)=\frac{\gamma^{\mu}p_{\mu}+m}{\sqrt{2m(m+\omega_{p})}}u^{+}_{r}(m{,}\vec{0}) \cdot \frac{\gamma^{\nu}p_{\nu}+m}{\sqrt{2m(m+\omega_{p})}}u_{s}(m{,}\vec{0}) =\frac{\gamma^{\mu}p_{\mu}+m}{\sqrt{2m(m+\omega_{p})}} \cdot \frac{\gamma^{\nu}p_{\nu}+m}{\sqrt{2m(m+\omega_{p})}} \cdot u^{+}_{r}(m{,}\vec{0})u_{s}(m{,}\vec{0})=\frac{(\gamma^{\mu}p_{\mu}+m)(\gamma^{\nu}p_{\nu}+m)}{2m(m+\omega_{p})} \cdot u^{+}_{r}(m{,}\vec{0})u_{s}(m{,}\vec{0})Now, note that (\gamma^{\mu}p_{\mu}+m)(\gamma^{\nu}p_{\nu}+m)=(p^2+m^2)+2m\gamma^{\mu}p_{\mu}Substituting this into our expression for u^{+}_{r}(p)u_{s