Feynman rules for Majorana fields (based on Srednicki)

[LAHLH]

Hi,

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

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

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

where,

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

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 $$(\tfrac{1}{i}\tfrac{\delta}{\delta \eta^{T}_{\alpha}(x)})^{T}}$$ in there, but how else could one generate a $$\Psi^{T}$$ given that Z_0 is only a function of $$\eta ^{T}$$. Also on that note why is Z_0 only a function of $$\eta ^{T}$$?

 

[LAHLH]

Sorry if the above looks a bit of a mess and it's hard to dig out the question. All I'm really asking is if the $$\Psi^{T}$$ in the interaction Lagrangian translates into $$(\tfrac{\delta}{\delta \eta^{T}_{\alpha}(x)})^{T}}$$ in the path integral?

 

[Avodyne]

The transpose is just keeping track of the index when the indices are suppressed. If you write out all spin indices, then you don't need to use the transpose symbol at all. Thus in your expressions where you have the indices written out, you can drop the T superscripts. A derivative with respect to eta(x) will then act on both eta(y1) and eta(y2) in Z0; the two terms you get are identical, and that cancels the 1/2 (just like with real scalar fields).

 

[LAHLH]

Thanks Avodyne, this has cleared things up for me. So the transpose is really there simply to make the matrix type multiplication correct, and obviously if we are explicit with indices we don't need it. Since for a vector the first component say of the transpose is equal to the first comp of the orig vec etc

 

[LAHLH]

Now that I've cleared that up, I'm trying to calc the rules as he sets out. Firstly I thought I would consider his diagram 49.1 (one incoming scalar field, two outgoing neutrinos). So from the path integral we have V=1 (one vertex), $$P_{\phi}=1$$, $$P_{\Psi}=2$$

After cancelling out all the coefficient factors like 1/V! etc, for the reasons stated in ch9, and using the path integral I quoted in my OP but without the need for all the transposes.

$$~ { ig \int\,\mathrm{d}^4x (\tfrac{1}{i}\tfrac{\delta}{\delta J(x) })(\tfrac{1}{i}\tfrac{\delta}{\delta \eta_{\alpha}(x)})}C_{\alpha \beta}(\tfrac{1}{i}\tfrac{\delta}{\delta \eta_{\beta}(x)}) \times \left[-i\int\,\mathrm{d}^4 y_1\mathrm{d}^4 y_2 \eta_{\gamma}(y_1) S_{\gamma\sigma}(y_1-y_2)C^{-1}_{\sigma\rho}\eta_\rho(y_2) \right] ^2 \times \left[ i \int \mathrm{d}^4 z_1 \mathrm{d}^4 z_2 J(z_1) \Delta(z_1-z_2) J(z_2) \right]$$

This leads to :

$$~ { ig \int\,\mathrm{d}^4x C_{\alpha \beta} \times \left[-i\int\,\mathrm{d}^4 y_2 \tfrac{1}{i}S_{\alpha\sigma}(x-y_2)C^{-1}_{\sigma\rho}\eta_\rho(y_2) \right] \times \left[-i\int\,\mathrm{d}^4 w_2 \tfrac{1}{i}S_{\beta\tau}(x-w_2)C^{-1}_{\tau\lambda}\eta_\lambda(w_2) \right] \times \left[ i \int \mathrm{d}^4 z_1 J(z_1) \tfrac{1}{i}\Delta(z_1-x)\right]$$

As per usual this isn't exactly what we want for the LSZ formula for this process, we will end up needing $$\langle 0|T\phi (x_0)\Psi_{\mu} (y_0)\Psi_{\nu} (z_0)|0 \rangle_{C}=\tfrac{1}{i}\tfrac{\delta}{\delta J(x_0)}\tfrac{1}{i}\tfrac{\delta}{\delta \eta_\mu(y_0)}\tfrac{1}{i}\tfrac{\delta}{\delta \eta_\nu(z_0)}\times iW(\eta, J) |_{\eta=J=0}$$

So I believe this correlation function turns out to be (to tree level at least) sum of all diagrams with one J type source and 2 eta type sources (we no longer need arrows on the solid lines as there is only eta, which is equiv to eta^T)

$$~ {(ig)\left(\tfrac{1}{i}\right)^3 \int\,\mathrm{d}^4x\, C_{\alpha \beta} \times \left[- S_{\alpha\sigma}(x-y_0)C^{-1}_{\sigma\mu} \right] \times \left[- S_{\beta\tau}(x-z_0)C^{-1}_{\tau\nu} \right] \times \left[ \Delta(x_0-x)\right]+\mathcal{O}(g^2)$$

But we also must apply the labels in all possible ways, the eta can be applied in two possible ways, so we also get a term with $$(\mu,y_0) \leftrightarrow (\nu, z_0)$$, i.e.

$$~ {(ig)\left(\tfrac{1}{i}\right)^3 \int\,\mathrm{d}^4x\, C_{\alpha \beta} \times \left[- S_{\alpha\sigma}(x-z_0)C^{-1}_{\sigma\nu} \right] \times \left[- S_{\beta\tau}(x-y_0)C^{-1}_{\tau\mu} \right] \times \left[ \Delta(x_0-x)\right]+\mathcal{O}(g^2)$$

So overall for the correlation function I get:

$$\langle 0|T\phi (x_0)\Psi_{\mu} (y_0)\Psi_{\nu} (z_0)|0 \rangle_{C}= {(ig)\left(\tfrac{1}{i}\right)^3 \int\,\mathrm{d}^4x\, C_{\alpha \beta} \times \left[- S_{\alpha\sigma}(x-y_0)C^{-1}_{\sigma\mu} \right] \times \left[- S_{\beta\tau}(x-z_0)C^{-1}_{\tau\nu} \right] \times \left[ \Delta(x_0-x)\right]$$ $$+{(ig)\left(\tfrac{1}{i}\right)^3 \int\,\mathrm{d}^4x\, C_{\alpha \beta} \times \left[- S_{\alpha\sigma}(x-z_0)C^{-1}_{\sigma\nu} \right] \times \left[- S_{\beta\tau}(x-y_0)C^{-1}_{\tau\mu} \right] \times \left[ \Delta(x_0-x)\right]+\mathcal{O}(g^2)$$

 

[LAHLH]

I'm not sure if this is 100% correct. I worry that if I had acted with eta on the "other side" of the S(x-y)C^{-1}, things would turn out different, unless they wouldn't because of some relations amongst C and C^T perhaps.

But I press on with anyway, as I will need this correlation function to calculate the scattering amp of $$\phi \rightarrow \nu\nu$$:

$$\langle f|i\rangle =\langle 0|T b_{s,out}(p) b_{out s'}(p')a^{\dag}_{in}(k)|0\rangle$$

Srednicki notes the two equivalent forms for the b operator, one with a C and one without, and discusses how we can pick judiciously so that we can get rid of all C factors from our end result. So I need to use a different form for my two b ops:

$$\langle 0| T \Bigg[ \Big(-i\int\,\mathrm{d}^4x e^{-ipx}\Psi_{\alpha}(x) (C(+i{\not}\partial_x+m))_{\alpha\beta}(v_s(p))_{\beta} \Big)\Big(+i\int\,\mathrm{d}^4y\, e^{-ip'y} (\bar{u}_{s'}(p'))_{\sigma} (-i{\not}\partial_y+m)_{\sigma\rho}\Psi_{\rho}(y)\Big)\Big(+i\int\,\mathrm{d}^4\,z e^{-ikz} (-\partial^{2}_z+m^2)\phi(z)\Big)\Bigg]|0\rangle$$

This leads to:

$$i\int\,\mathrm{d}^4x\mathrm{d}^4y\mathrm{d}^4z\, e^{-ikz} \times e^{-ipx}\times e^{-ip'y} \times(-\partial^{2}_z+m^2) \Big[ (C(+i{\not}\partial_x+m))_{\alpha\beta}(v_s(p))_{\beta}\Big] \langle 0|T \Psi_{\alpha}(x)\Psi_{\rho}(y)\phi(z) |0\rangle \Big[(\bar{u}_{s'}(p'))_{\sigma} (-i{\not}\partial_y+m)_{\sigma\rho}\Big]$$

Now plugging in the tree level correlation function I found above:

$$i\int\,\mathrm{d}^4x\mathrm{d}^4y\mathrm{d}^4z\, e^{-ikz} \times e^{-ipx}\times e^{-ip'y} \times(-\partial^{2}_z+m^2) \Big[(\bar{u}_{s'}(p'))_{\sigma} (-i{\not}\overrighttarrow\partial_y+m)_{\sigma\rho}\Big]$$
$$\Bigg[{(ig)\left(\tfrac{1}{i}\right)^3 \int\,\mathrm{d}^4x_1\, C_{\alpha_1 \beta_1} \times \left[- S_{\alpha_1\sigma_1}(x_1-y)C^{-1}_{\sigma_1\rho} \right] \times \left[- S_{\beta_1\tau_1}(x_1-x)C^{-1}_{\tau_1\alpha} \right] \times \left[ \Delta(z-x_1)\right]$$

$$+{(ig)\left(\tfrac{1}{i}\right)^3 \int\,\mathrm{d}^4x_2\, C_{\alpha_2 \beta_2} \times \left[- S_{\alpha_2\sigma_2}(x_2-x)C^{-1}_{\sigma_2\alpha} \right] \times \left[- S_{\beta_2\tau_2}(x_2-y)C^{-1}_{\tau_2\rho} \right] \times \left[ \Delta(z-x_2)\right] \Bigg]$$
$$\Big[ (C(+i{\not}\overleftarrow\partial_x+m))_{\alpha\beta}(v_s(p))_{\beta}\Big]$$

 

[LAHLH]

Finally one can use the fact that these props are Green functions for the Majorna/ KG wave

ops:

for example $$(-\partial^{2}_z+m^2)\Delta(z-w)=\delta^{4}(z-w)$$

Now I suspect the other relation needed is $$C(-i{\not}\partial_x+m)S(x-y)C^{-1}=\delta^{4}(x-y)$$, but I'm having some trouble applying it. So in index notation it is :
$$C_{\alpha\beta}(-i{\not}\partial_x+m)_{\beta\gamma}S_{\gamma\epsilon}(x-y)C^{-1}_{\epsilon\lambda}=\delta^{4}(x-y) \delta{\alpha\lambda}$$

If I collect some potential candidates to use this relation with however:

$$\Big[(-i{\not}\overrighttarrow\partial_y+m)_{\sigma\rho} \Big] C_{\alpha_1 \beta_1} \left[ S_{\alpha_1\sigma_1}(x_1-y)C^{-1}_{\sigma_1\rho} \right]$$

So as it stands it doesn't seem like the indices fit together correctly. But we do have $$C^{T}=C^{-1}=-C$$ and maybe this could help?

 

[Avodyne]

Eq.(42.24) looks like it would be helpful here.

 

[LAHLH]

$$\Big[(-i{\not}\overrighttarrow\partial_y+m)_{\sigma\rho} \Big] C_{\alpha_1 \beta_1} \left[ S_{\alpha_1\sigma_1}(x_1-y)C^{-1}_{\sigma_1\rho} \right]$$

and then I have that $$[S(x-y)C^{-1}]_{\alpha\beta}=-[S(y-x)C^{-1}]_{\beta\alpha}$$

So $$\left[ S_{\alpha_1\sigma_1}(x_1-y)C^{-1}_{\sigma_1\rho} \right]$$ which is $$\left[ S(x_1-y)C^{-1} \right]_{\alpha_1\rho}$$ can be written as $$- \left[ S(y-x_1)C^{-1} \right]_{\rho\alpha_1}$$ by (42.24). This in turn is equal to $$- \left[ S_{\rho\sigma_1}(y-x_1)C^{-1}_{\sigma_1\alpha_1} \right]$$

Plugging this into the original equation would then lead to,

$$-\Big[(-i{\not}\vec{\partial}_y+m)_{\sigma\rho} \Big] C_{\alpha_1 \beta_1} \left[ S_{\rho\sigma_1}(y-x_1)C^{-1}_{\sigma_1 \alpha_1} \right]$$

rewriting more suggestively:

$$-C_{\alpha_1 \beta_1}(-i{\not}\vec{\partial}_y+m)_{\sigma\rho} S_{\rho\sigma_1}(y-x_1)C^{-1}_{\sigma_1\alpha_1}$$

This very nearly matches the desired relation : $$C_{\alpha\beta}(-i{\not}\partial_x+m)_{\beta\gamma}S_{\gamma\epsilo n}(x-y)C^{-1}_{\epsilon\lambda}=\delta^{4}(x-y) \delta{\alpha\lambda}$$

But the first C doesn't connect with the wave operator bizzarely, perhaps a mistake in the indices of my derivation that I can't say, although it seems difficult to see how I would get a wave operator here that could connect to both the leading C and the S(x-y), since the wave operators are all also contracted with the u,v spinors, leaving only one index to play with...hmm

 

[LAHLH]

Ah actually, the C and C^-1 seem to hit each other, $$C^{-1}_{\sigma_1\alpha_1} C_{\alpha_1\beta_1} =\delta_{\sigma_1\beta_1}$$

Which leads to:
$$-\delta_{\sigma_1\beta_1}(-i{\not}\vec{\partial}_y+m)_{\sigma\rho} S_{\rho\sigma_1}(y-x_1)$$
Now I can just use
$$(-i{\not}\vec{\partial}_y+m)_{\sigma\rho} S_{\rho\sigma_1}(y-x_1)=\delta_{\sigma\sigma_1}\delta(y-x_1)$$

hopefully this is correct, thanks so much for the help

 

[LAHLH]

So following all this through I end up with:

$$(ig) \int\,\mathrm{d}^4x\,\mathrm{d}^4y\,\mathrm{d}^4z\, e^{+ikz} e^{-ipx} e^{-ip'y} (\bar{u}_{s'}(p'))_{\sigma} (v_{s}(p))_{\sigma} \times \Bigg[\int\,\mathrm{d}^4w\, \delta^4(x-w)\delta^4(y-w)\delta^4(z-w)+\int\,\mathrm{d}^4w \,\delta^4(x-w)\delta^4(y-w)\delta^4(z-w) \Bigg]$$

This close to what I was expecting, but resolves to:

$$2(ig) (\bar{u}_{s'}(p'))_{\sigma} (v_{s}(p))_{\sigma}(2\pi^4) \delta^4(k-(p+p'))$$

I think perhaps I should have got zero? given that Srednicki finds $$\tau_1=-\tau_2$$

 

[Avodyne]

I believe your answer (with the factor of 2) is correct.

 

[LAHLH]

Really? but how does this match Srednicki saying tau_1=-tau_2 for his two diagrams?

 

[LAHLH]

Does tau_1=-tau_2 not been overall the scattering amp is zero? perhaps I am missing something?

 

[Avodyne]

Does tau_1=-tau_2 not been overall the scattering amp is zero?

No! You don't add these! They are the amplitudes corresponding to two equivalent diagrams, and you only keep one of them. But the overall sign of your amplitude will depend on the convention you choose for which diagram to keep.

The factor of 2 you find is the same as the factor of 2! that is discussed at the top of page 304.

 

[LAHLH]

No! You don't add these! They are the amplitudes corresponding to two equivalent diagrams, and you only keep one of them. But the overall sign of your amplitude will depend on the convention you choose for which diagram to keep.

The factor of 2 you find is the same as the factor of 2! that is discussed at the top of page 304.

Oh, yes you're right he does say they are equivalent and we should keep only one of them, etc.

However, I thought the factor of 2 discussed there was already accounted for in cancelling the 1/2 in the Lagrangian?