- #1

LAHLH

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So if we have an interaction Lagrangian for a Majorana field: [tex] L_1=\tfrac{1}{2} g\phi\Psi^{T}C\Psi [/tex]

Now looking at the path integral, I believe this must go like:

[tex] Z (\eta^{T},J) ~ \exp{[\tfrac{1}{2} ig \int\,\mathrm{d}^4x (\tfrac{1}{i}\tfrac{\delta}{\delta J(x) })(\tfrac{1}{i}\tfrac{\delta}{\delta \eta^{T}_{\alpha}(x)})^{T}}C_{\alpha \beta}(\tfrac{1}{i}\tfrac{\delta}{\delta \eta^{T}_{\beta}(x)}) \times Z_0(\eta^{T},J) [/tex]

where,

[tex] Z_0(\eta^{T}, J) =exp\left[-\tfrac{i}{2}\int\,\mathrm{d}^4 y_1\mathrm{d}^4 y_2 \eta^{T}_{\gamma}(y_1) S_{\gamma\sigma}(y_1-y_2)C^{-1}_{\sigma\rho}\eta_\rho(y_2) \right] \times exp\left[ \tfrac{i}{2} \int \mathrm{d}^4 z_1 \mathrm{d}^4 z_2 J(z_1) \Delta(z_1-z_2) J(z_2) \right][/tex]

Is this the correct form of the path integral, as Srednicki is not explicit about the Majorana case, he goes into detail for Dirac, so I'm trying to obtain this via analoogy with that, and some earlier Majorana results.

It feels a bit strange having a [tex] (\tfrac{1}{i}\tfrac{\delta}{\delta \eta^{T}_{\alpha}(x)})^{T}} [/tex] in there, but how else could one generate a [tex] \Psi^{T} [/tex] given that Z_0 is only a function of [tex] \eta ^{T} [/tex]. Also on that note why is Z_0 only a function of [tex] \eta ^{T} [/tex]?