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I was looking for an explanation that doesn't depend on the representation of the gamma matrices that shows that the orthogonality relations are fulfilled. Let me situate my problem:

So the Dirac equation possesses the following plane wave solutions, $$ \psi(x) = u_r(p) e^{\pm ipx/\hbar}, $$ $$ \psi(x) = v_r(p) e^{\pm ipx/\hbar}, $$ where we dropped the constant multiplication factor and $u_r(p)$ and $v_r(p)$ are the constant four spinors. Where we define the adjoints $$ \bar{u}_r(p) = u^{\dagger}_r(p) \gamma^0,\hspace{1.0cm} \bar{v}_r(p) = v^{\dagger}_r(p) \gamma^0.$$ We know that the constant spinors have to satisfy $$(\gamma^{\mu}p_{\mu}-mc)u_r(p)=0,\hspace{1.0cm} (\gamma^{\mu}p_{\mu}+mc)v_r(p)=0,$$ $$\bar{u}_r(p)(\gamma^{\mu}p_{\mu}-mc)=0,\hspace{1.0cm} \bar{v}_r(p)(\gamma^{\mu}p_{\mu}+mc)=0,$$ and we impose the following normalization $$ u^{\dagger}_r(p)u_r(p) = v^{\dagger}_r(p)v_r(p) = E_p/mc^2.$$ Now everywhere I looked this looks to be enough to obtain the orthonormality relations, $$ u^{\dagger}_r(p)u_s(p) = v^{\dagger}_r(p)v_s(p) = E_p/mc^2 \delta_{rs}$$ $$ u^{\dagger}_r(p)v_s(-p)=0$$ and $$ \bar{u}_r(p)u_s(p) = -\bar{v}_r(p) v_s(p)= \delta_{rs}, $$ $$ \bar{u}_r(p)v_s(p) = \bar{v}_r(p) u_s(p)= 0.$$ I really don't see how these relations follow, it seemed straightforward but I couldn't figure it out? Is this indeed enough to show the orthonormality relations? Any help would be more than welcome.

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To prove $\bar{u}_{r}u_{s}=\delta_{rs}$ you can do the following (I'm using $c=1$):

$$ \left(\gamma^{\mu}p_{\mu}-m\right)u_{s}=0\\ \bar{u}_{r}\left(\gamma^{\mu}p_{\mu}-m\right)=0 $$

Multiply the first equation on the left by $\bar{u}_{r}\gamma^{0}$ and the second on the right by $\gamma^{0}{u}_{s}$

$$ \bar{u}_{r}\gamma^{0}\left(\gamma^{\mu}p_{\mu}-m\right)u_{s}=0\\ \bar{u}_{r}\left(\gamma^{\mu}p_{\mu}-m\right)\gamma^{0}{u}_{s}=0 $$

Adding those two equations you get:

$$ \bar{u}_{r}\gamma^{0}\gamma^{\mu}p_{\mu}u_{s}-\bar{u}_{r}\gamma^{0}mu_{s}+\bar{u}_{r}\gamma^{\mu}p_{\mu}\gamma^{0}u_{s}-\bar{u}_{r}m\gamma^{0}u_{s}=0 $$

Now, applying $\bar{u}_{r}=u_{r}^{\dagger}\gamma^{0}$ to the terms with $m$ and $\left(\gamma^{0}\right)^{2}=I$

$$ \bar{u}_{r}\gamma^{0}\gamma^{\mu}p_{\mu}u_{s}+\bar{u}_{r}\gamma^{\mu}\gamma^{0}p_{\mu}u_{s}-2mu_{r}^{\dagger}u_{s}=0 $$

Using the fact that $\gamma^0\gamma^k=-\gamma^0\gamma^k$ being $k=1,2,3$, the sum over $\mu$ cancels for all terms except for $\mu=0$ which gives:

$$ \bar{u}_{r}p_{0}u_{s}+\bar{u}_{r}p_{0}u_{s}-2mu_{r}^{\dagger}u_{s}=0 $$

$$ 2p_{0}\bar{u}_{r}u_{s}=2mu_{r}^{\dagger}u_{s} $$

Knowing that $p_{\mu}=(\frac{E}{c},-p_x,-p_y,-p_z)$ we get

$$ \bar{u}_{r}u_{s}=\frac{m}{E}u_{r}^{\dagger}u_{s} $$

Applying the normalization you gave and adding the corresponding $c$:

$$ \bar{u}_{r}u_{s}=\frac{mc^2}{E}\frac{E}{mc^2}\delta_{rs}=\delta_{rs} $$

To prove $-\bar{v}_{r}v_{s}=\delta_{rs}$ you can proceed the same way adding these two equations:

$$ \bar{v}_{r}\gamma^{0}\left(\gamma^{\mu}p_{\mu}+m\right)v_{s}=0\\ \bar{v}_{r}\left(\gamma^{\mu}p_{\mu}+m\right)\gamma^{0}{v}_{s}=0 $$

Which proceeding in the same way as before will give you:

$$ \bar{v}_{r}v_{s}=-\frac{m}{E}v_{r}^{\dagger}v_{s} $$

Again using your normalization and adding the $c$:

$$ \bar{v}_{r}v_{s}=-\frac{mc^2}{E}\frac{E}{mc^2}\delta_{rs}=-\delta_{rs} $$

To prove $\bar{u}_{r}v_{s}=\bar{v}_{r}u_{s}=0$ you can do the same as before but adding these set of equations:

$$ \bar{u}_{r}\gamma^{0}\left(\gamma^{\mu}p_{\mu}+m\right)v_{s}=0\\ \bar{v}_{r}\left(\gamma^{\mu}p_{\mu}-m\right)\gamma^{0}{u}_{s}=0 $$

Which will give you $\bar{u}_{r}v_{s}=0$. And:

$$ \bar{v}_{r}\gamma^{0}\left(\gamma^{\mu}p_{\mu}-m\right)u_{s}=0\\ \bar{v}_{r}\left(\gamma^{\mu}p_{\mu}+m\right)\gamma^{0}{u}_{s}=0 $$

Which will give you $\bar{v}_{r}u_{s}=0$

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I won't give the full derivation but I'll give you an outline of how this works.

Use the helicity operator $\sigma_\mathbf{p}$ to prove the first relation. i.e. use $$\sigma_\mathbf{p} u_r(\mathbf{p})= (-)^{r+1}u_r(\mathbf{p}),$$ and similar for the other plane wave solutions (you can find these everywhere).

To prove the second relation you also use the previous trick to find that it gives you zero for $r\ne s$. For $r=s$ you use the Dirac equation:you apply it first to the spinor on the right, and then to the spinor on the left to obtain: $$-\bar{u_r}v_r=\bar{u_r}v_r.$$

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  • $\begingroup$ I did expect an orthogonality argument by using the eigenvalues of the spin operator, but how does the norm of spinors equal unity? $\endgroup$ – Dylan_VM Jan 30 '18 at 15:13

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