Deriving the Spinor Completeness Relation without using a Representation

Question

Reference: DAMTP problem set 3, question 5 but ignore the spinor solutions given.

To preface, this has taken up 1 entire day and a further 2 afternoons of work so I will just list the most promising attempts rather than clutter half a notebook !

Given the relations for the spinors, $u(\mathbf{p},\pm\tfrac{1}{2})$ and $v(\mathbf{p},\pm\tfrac{1}{2})$ as,

$$ (\not p - m)u(\mathbf{p},s) = 0\qquad (\not p + m)v(\mathbf{p},s) = 0 $$ $$ \bar{u}(\mathbf{p},s)u(\mathbf{p},r) = 2m\delta_{sr}\qquad \bar{v}(\mathbf{p},s)v(\mathbf{p},r) = -2m\delta_{sr}\qquad \bar{u}(\mathbf{p},s)v(\mathbf{p},s) = 0 $$

I am trying to prove the following,

$$ \tag{1} \Lambda_+ = \sum_{s} u(\mathbf{p},s)\bar{u}(\mathbf{p},s) = \not p + m $$ $$ \tag{2} \Lambda_+ = \sum_{s} v(\mathbf{p},s)\bar{v}(\mathbf{p},s) = \not p - m $$

Choosing a representation is too easy and considered cheating. All the information other than trace identities is given above. I believe the same goes for using the spin projection operator.

Attempts

I have shown linear independence in a previous exercise from the same information, so we can assume that $u(\mathbf{p},\pm\tfrac{1}{2})$ and $v(\mathbf{p},\pm\tfrac{1}{2})$ are linearly independent.

1. Adding (1)+(2) ... result: $0$ as $u\bar{u} = (u\bar{u})^{\dagger} = \bar{u}u$ and by the relations we get $4m - 4m = 0$ and no useful information is gained
2. Subtracting (1)-(2) ... result: $8m$ as above
3. Acting on (1), (2) from the right with $(\not p + m)$ from the right ... result $\pm 4m$ by $ u \bar{u} \not p u$
4. Acting on (1), (2) from the right with $u,v,\bar{u},\bar{v}$ ... result same as 3.

In fact every attempt I try just gives me a result with $m$ which gives me reason to believe that there is something fundamentally wrong in my understanding.

Potential Issue

I can 'force' the result by just using the Dirac equation $(\not p + m)u(\mathbf{p},r) = 0$ to equate $\not p$ to $m$ but I don't think this is what I am supposed to be doing, although I cannot see any other way!

The reason I do not like this is because it doesn't force me to limit myself (I believe) to just changing one $2m \to \pm \not p$. I feel if I were to do this I need some condition that prevents me from picking both as $\not p$ to obtain $\Lambda_+ = 2\not p$ for example.

Answer 1

In the rest frame $\not p \rightarrow - m \gamma^0$ and $$ (\not p - m)u(\mathbf{p},s) = 0 \;\; \rightarrow \;\; (I + \gamma^0) u(\mathbf{0},s) = 0 \\ {\bar u}(\mathbf{p},s)(\not p - m) = 0 \;\; \rightarrow \;\; {\bar u}(\mathbf{0},s) (I + \gamma^0) = 0 \\ (\not p + m)v(\mathbf{p},s) = 0 \;\; \rightarrow \;\; (I - \gamma^0) v(\mathbf{0},s) = 0 \\ {\bar v}(\mathbf{p},s)(\not p + m) = 0 \;\; \rightarrow \;\; {\bar v}(\mathbf{p},s)(I - \gamma^0) = 0 $$ So regardless of representation $u(\mathbf{0},s)$ and $v(\mathbf{0},s)$ become right eigenvectors of $\gamma^0$, while ${\bar u}(\mathbf{0},s)$ and ${\bar v}(\mathbf{0},s)$ become left eigenvectors.

This gives closure relationships for both the identity $I$ and $\gamma^0$ in terms of $u{\bar u}$ and $v{\bar v}$ outer products, wherefrom similar closure relations for $I \pm \gamma^0$ follow immediately.

Boost back to an arbitrary frame and you are left with the required relations for $\not p \pm m$.

On a second thought, just notice that $u(\mathbf{p},s)$, $v(\mathbf{p},s)$ are right eigenvectors of $\not p$, and ${\bar u}(\mathbf{p},s)$, ${\bar v}(\mathbf{p},s)$ are left eigenvectors. Build the expression of $\not p$ on the $\mathbf{p}$ eigensubspace in terms of $u{\bar u}$, $v{\bar v}$, and the corresponding closure relation for the eigensubspace projector. Then use those to extract expressions for $\not p \pm m$. No need to boost anywhere!