Question about the helicity operator

  • Thread starter jdstokes
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Main Question or Discussion Point

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

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

[itex][\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[/itex]

[itex][\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[/itex]

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

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

[itex]\epsilon_{ijk}(\gamma^j\gamma^k\gamma^l \partial_l\partial^i\psi - \gamma^l\gamma^j\gamma^k \partial_l\partial^i \psi)[/itex]

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

[itex]\epsilon_{ijk}(\gamma^k \partial^j - \gamma^j\partial^k)\partial^i \psi[/itex]

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

Any help would be appreciated.
 
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Answers and Replies

  • #2
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Woops [itex]\epsilon_{ijk}[/itex] is symmetric in j,k since [itex]\epsilon^{ijk}[/itex] is anti-symmetric.

Edit: Hang on, [itex]\epsilon_{ijk}[/itex] should still be anti-symmetric, just differing in sign from [itex]\epsilon^{ijk}[/itex]. I still don't see why this should vanish?
 
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  • #3
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Presumably [tex]\partial^j \partial k^k[/tex] is symmetric in [tex]j,k[/tex].
 
  • #4
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Sorry I didn't understand what you wrote. Are you suggesting something along these lines?

[itex]\epsilon_{ijk}\gamma^k\partial^j = -\epsilon_{ikj}\gamma^k\partial^j = -\epsilon_{ijk}\gamma^j\partial^k[/itex].

Therefore

[itex]\epsilon_{ijk}(\gamma^k\partial^j - \gamma^j\partial^k)\partial^i\psi = -4\epsilon_{ijk}\gamma^j\partial^k\partial^i\psi[/itex].

But

[itex]\epsilon_{ijk}[/itex] is anti-symmetric in i,k whereas [itex]\partial^i\partial^k[/itex] is symmetric in i,k. Therefore each term in the sum over j vanishes.
 
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  • #5
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Um, yes... each term actually vanishes separately, the gamma matrix is simply along for the ride.
 
  • #6
reilly
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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
 

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