# Question about the helicity operator

1. May 9, 2008

### jdstokes

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.

Last edited: May 9, 2008
2. May 9, 2008

### jdstokes

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?

Last edited: May 9, 2008
3. May 9, 2008

### lbrits

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

4. May 9, 2008

### jdstokes

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.

Last edited: May 9, 2008
5. May 9, 2008

### lbrits

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

6. May 10, 2008

### reilly

You'll find it much easier to work directly with standard Dirac solutions in two component form, which are built around the helicity operator. Or look up the angular momentum operator -- see, for example, F. Gross's Relativistic Quantum Mechanics and Field Theory.

Helicity, with a fixed direction of course, commutes with free Hamiltonians for any spin. This is the basis for the elegant Jacob and Wick formalism -- something every particle physicist should master.
Regards,
Reilly Atkinson