I've been reading about the quantisation of the Dirac field $\psi(x)$ and it is stated that the general solution to the Dirac equation $(i\gamma^{\mu}\partial_{\mu}-m)\psi(x)=0$ is given by the superposition of plane waves $$\psi(x)=\int\frac{d^{3}p}{(2\pi)^{3}}\frac{1}{\sqrt{2E_{p}}}\sum_{s=1,2}\left(c_{s}(\mathbf{p})u_{s}(\mathbf{p})e^{-ip\cdot x}+d^{\dagger}_{s}(\mathbf{p})v_{s}(\mathbf{p})e^{ip\cdot x}\right)$$ where $u_{s}(\mathbf{p})$ and $v_{s}(\mathbf{p})$ are spinors, with the index $s$ labelling the two spin components.

Is this assumed to be in the Weyl representation, where one can decompose the four component spinor field $\psi$ into two, two component spinors (i.e. $u_{s}(\mathbf{p})$ and $v_{s}(\mathbf{p})$)?

Basically, should the notation be interpreted as: $u(\mathbf{p})$ and $v(\mathbf{p})$ are four component spinors that have been decomposed into two sets of two components, labelled by $(s,i)$, with $s$ labelling the two spin components of each spinor and $i$ denoting the two components of each spin component? That is, does $$\psi_{1}(x)=c_{s,1}(\mathbf{p})u_{s,1}(\mathbf{p})+d^{\dagger}_{s,1}(\mathbf{p})v_{s,1}(\mathbf{p})$$ label the first component of the two component spinor (with spin index $s$), and $$\psi_{2}(x)=c_{s,2}(\mathbf{p})u_{s,2}(\mathbf{p})+d^{\dagger}_{s,2}(\mathbf{p})v_{s,2}(\mathbf{p})$$ label the second component of the two component spinor (with spin index $s$)? Is it one index $i$ to label the "handedness", left or right, and one index $s$ to label the "spin", spin-up or spin-down. The we have four components overall, equivalently, two two component spinors, a left-handed one with spin-up and spin-down components, and a right-handed one with spin-up and spin-down components).

The reason I ask is that I've become confused over expressions such as the anti-commutation relations, which are written as $$\lbrace\psi_{i}(x),\psi_{j}(y)\rbrace=\delta^{3}(\mathbf{x}-\mathbf{y})\delta_{ij}$$ and also the completeness relations for $u_{s}(\mathbf{p})$ and $\bar{u}_{s}(\mathbf{p})$, $$\sum_{i=1,2}\bar{u}_{s,i}(\mathbf{p})u_{s,j}(\mathbf{p})=\left(\gamma^{\mu}p_{\mu}+m\right)_{ij}$$ (where $\left(\gamma^{\mu}p_{\mu}+m\right)_{ij}$ are the components of a $4\times 4$ matrix).

And similarly for $v_{s}(\mathbf{p})$ and $\bar{v}_{s}(\mathbf{p})$?

Sorry if this is a trivial question, but I'm finding it difficult to understand as all the texts I've read so far seem to suppress the index notation and don't properly explain it.

  • $\begingroup$ @AccidentalFourierTransform What are the four components then? I thought that the labelling of spin separated the spinor $\psi$ up into two two component spinors, each with spin-up and spin-down components?! $\endgroup$ – Will Apr 6 '16 at 17:07
  • $\begingroup$ @AccidentalFourierTransform Thanks for the tip. So do the components represent anything physical? For example, $u_{1}(\mathbf{p})$ is the spin-up component of $u_{s}(\mathbf{p})$, but what do its four components represent, are they the left and right handed chiral components of the spin-up spinor $u_{1}(\mathbf{p})$? $\endgroup$ – Will Apr 6 '16 at 17:15
  • $\begingroup$ @AccidentalFourierTransform ....Also, in the final equation that I wrote in my post, is it correct that when you sum over the spin components you do so for each individual matrix element $\bar{u}_{s,i}(\mathbf{p})u_{s,j}(\mathbf{p})\equiv (\bar{u}_{s}(\mathbf{p})u_{s}(\mathbf{p}))_{ij}$? Is this why you can commute terms when you have spin sums of the form $\sum_{r,s}(\bar{u}_{s}(\mathbf{p})v_{r}(\mathbf{k}))_{ij}^{\dagger}(\bar{u}_{s}(\mathbf{p})v_{r}(\mathbf{k}))_{ij}$? $\endgroup$ – Will Apr 6 '16 at 17:25

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