# Question about the helicity operator

In relativistic quantum field theory the Dirac spinors can be chosen to be eigenstates of the helicity operator $\vec{\Sigma}\cdot \vec{p} /|\vec{p}|$.

I want to show that $\vec{\Sigma}\cdot \vec{a}$ commutes with the Dirac Hamiltonian only if $\vec{a}\propto \vec{p}$. As usual I'm using Einstein summation everywhere.

$[\vec{\Sigma}\cdot \vec{a}, H_\mathrm{Dirac}]\psi=[\vec{\Sigma}\cdot \vec{a}, -i\hbar c \gamma^i\partial_i + mc^2]\psi =-i\hbar c[\Sigma^ia_i, \gamma^i \partial_i]\psi$

$[\Sigma^i a_i, \gamma^l \partial_l]\psi=[\frac{i}{2}\epsilon^{ijk}\gamma_j\gamma_k a_i, \gamma^l \partial_l]\psi= \frac{i}{2}\epsilon_{ijk}[\gamma^j\gamma^k a^i, \gamma^l \partial_l]\psi = \frac{i}{2}\epsilon_{ijk}(\gamma^j\gamma^k a^i \gamma^l \partial_l-\gamma^l \partial_l\gamma^j\gamma^k a^i ]\psi$

since $\epsilon^{ijk} = \epsilon_{ijk}(-1)^3$ and $\gamma_i = -\gamma^i,\, a_i = -a^i$ in flat spacetime with (+,-,-,-) signature.

Now suppose $a^i = \partial^i$. Then using equality of mixed partials and dividing out any constants gives

$\epsilon_{ijk}(\gamma^j\gamma^k\gamma^l \partial_l\partial^i\psi - \gamma^l\gamma^j\gamma^k \partial_l\partial^i \psi)$

Using the relation $\{ \gamma^\mu,\gamma^\nu \} = 2\eta^{\mu\nu}$ twice and the fact that $\eta^{ij}\partial_i = \partial^j$ in flat spacetime gives

$\epsilon_{ijk}(\gamma^k \partial^j - \gamma^j\partial^k)\partial^i \psi$

Now, in the sum over k,j the term in brackets is anti-symmetric but so is $\epsilon_{ijk}$ so I don't see why this should vanish.

Any help would be appreciated.

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Woops $\epsilon_{ijk}$ is symmetric in j,k since $\epsilon^{ijk}$ is anti-symmetric.

Edit: Hang on, $\epsilon_{ijk}$ should still be anti-symmetric, just differing in sign from $\epsilon^{ijk}$. I still don't see why this should vanish?

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Presumably $$\partial^j \partial k^k$$ is symmetric in $$j,k$$.

Sorry I didn't understand what you wrote. Are you suggesting something along these lines?

$\epsilon_{ijk}\gamma^k\partial^j = -\epsilon_{ikj}\gamma^k\partial^j = -\epsilon_{ijk}\gamma^j\partial^k$.

Therefore

$\epsilon_{ijk}(\gamma^k\partial^j - \gamma^j\partial^k)\partial^i\psi = -4\epsilon_{ijk}\gamma^j\partial^k\partial^i\psi$.

But

$\epsilon_{ijk}$ is anti-symmetric in i,k whereas $\partial^i\partial^k$ is symmetric in i,k. Therefore each term in the sum over j vanishes.

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Um, yes... each term actually vanishes separately, the gamma matrix is simply along for the ride.

reilly