# Narrow Width Approximation for a spin 1 resonance

1. Jan 21, 2013

### jkp

Hi

I'm trying to understand how the sum over spin polarizations is treated when using the Narrow Width Approximation(NWA) with a spin 1 resonance. For a spin zero resonance there is no such problem and the factorization is quite straightforward. I'll go through some details to explain where my problem is.

It'll be easier to talk about a specific example so I'm going to look at the decay of a top quark via a W boson, which is a spin 1 resonance. The decay follows like this $t(k)\to b(p_1) W^*(q) \to b(p_1) f(p_2) \overline{f}'(p_3)$ where $f$ and $f'$ are any fermion. So we'd expect that close to the resonance peak we can factorize the width as
$$\Gamma(t\to b f \overline{f}')=\Gamma(t\to b W) BR(W\to f\overline{f}')$$
that is the production and decay of a real W. My aim is to understand how this factorization occurs here for a spin 1 resonance.

Starting from the decay width formula we have,
$$\Gamma_{t\to b f\overline{f}'}= \frac{1}{2 m_{t}}\frac{1}{2}\sum_{spin} |{\mathcal M}(t\to b f \overline{f}')|^2 dLIPS$$
where
$$dLIPS = \frac{d^3 \vec{p}_{1}}{2 E_{p_1}(2\pi)^3} \frac{d^3 \vec{p}_{2}}{2 E_{p_2}(2\pi)^3} \frac{d^3 \vec{p}_{3}}{2 E_{p_3}(2\pi)^3} (2\pi)^4 \delta^4(p_1+p_2+p_3-k)$$
is the lorentz invariant phase space.

So my first job is to factorize the amplitude into $t\to b W$ and $W\to f\overline{f}'$ parts. We can write the amplitude as,
$${\mathcal M}(t\to b f \overline{f}')= {\mathcal M}_1^\mu \,\frac{(-g_{\mu\nu}+q_\mu q_\nu / M_W^2)}{q^2-M_W^2+iM_W\Gamma_W} \,{\mathcal M}_2^\nu = {\mathcal M}_1^\mu \,\sum_\lambda \frac{\epsilon^*_\mu(q,\lambda) \epsilon_\mu(q,\lambda)} {q^2-M_W^2+iM_W\Gamma_W} \,{\mathcal M}_2^\nu$$
where ${\mathcal M}_{1,2}$ come from the $t-b-W$ vertex and $W-f-f'$ vertex respectively and I've used the polarization sum over the W polarization vector identity, $\sum_\lambda \epsilon^*(q,\lambda)\epsilon(q,\lambda)=(-g_{\mu\nu}+q_\mu q_\nu/M_W^2)$, in the second equality where $\lambda$ is the W polarization. To simplify the notation I'm going to define as follows,
$${\mathcal M}^\lambda_{t\to b W}= {\mathcal M}_1^\mu \epsilon^*_\mu(q,\lambda)$$
and
$${\mathcal M}^\lambda_{W\to f\overline{f}'}= {\mathcal M}_2^\mu \epsilon_\mu(q,\lambda)$$
So the amplitude becomes,
$${\mathcal M}(t\to b f \overline{f}') = \sum_\lambda \frac{{\mathcal M}^\lambda_{t\to b W}{\mathcal M}^\lambda_{W\to f\overline{f}'}} {q^2-M_W^2+iM_W\Gamma_W}$$
Now squaring the amplitude we have,
$$|{\mathcal M}(t\to b f \overline{f}')|^2 = \left|\sum_\lambda {\mathcal M}^\lambda_{t\to b W}{\mathcal M}^\lambda_{W\to f\overline{f}'}\right|^2 \frac{1}{{(q^2-M_W^2)^2+M_W^2\Gamma_W^2}}$$
And this is where I get stuck. The propagator part is easy to deal with in the NWA it will change to a delta function like this,
$$\frac{1}{{(q^2-M_W^2)^2+M_W^2\Gamma_W^2}}\to \frac{2\pi\,\delta(q^2-M_W^2)}{2M_W^2\Gamma_W^2}$$
but I'm not sure how to deal with the square of the polarization summed amplitude,
$$\left|\sum_\lambda {\mathcal M}^\lambda_{t\to b W} {\mathcal M}^\lambda_{W\to f\overline{f}'}\right|^2$$
Ideally I want to show that this can be written as something like,
$$\left(\sum_\gamma\left|{\mathcal M}^\gamma_{t\to b W}\right|^2\right) \left( \frac{1}{3}\sum_\lambda \left|{\mathcal M}^\lambda_{W\to f\overline{f}'}\right|^2 \right)$$
so that the decay width will factorize into, $\Gamma(t\to b f\overline{f}')=\Gamma(t\to b W)BR(W\to f\overline{f}')$.

So my question is how do I show that, $\left|\sum_\lambda {\mathcal M}^\lambda_{t\to b W} {\mathcal M}^\lambda_{W\to f\overline{f}'}\right|^2 = \left(\sum_\gamma\left|{\mathcal M}^\gamma_{t\to b W}\right|^2\right) \left( \frac{1}{3}\sum_\lambda \left|{\mathcal M}^\lambda_{W\to f\overline{f}'}\right|^2 \right)$ ?

I'm sorry for all the details but I wanted to give the full context of my question.

Thanks

2. Jan 23, 2013

### andrien

this is most probably the nifty trick of avoiding polarization.You should have written those λ's explicitly.
|Mμε*λμελvMv|2=(1/3)Ʃλ1,λ2|Mμε*λ1μ|2λ2vMv|2
the factor of 1/3 arises because you must sum over the final spin states and take an average over initial spin states after summing it(i.e. divide by 3 for W Boson).

3. Jan 23, 2013

### jkp

Thanks for the reply Andrien but maybe you can provide a little more detail. I don't understand how you get $\lambda_1$ in one bracket and $\lambda_2$ in the other. As far as I see it we have,
$$\left| \sum_\lambda {\mathcal M}_1^\mu \epsilon^{\lambda *}_\mu \epsilon^\lambda_\nu {\mathcal M}_2^\nu \right|^2 = \left(\sum_\lambda{\mathcal M}_1^\mu \epsilon^{\lambda *}_\mu \epsilon^\lambda_\nu {\mathcal M}_2^\nu\right) \left( \sum_\gamma{\mathcal M}_1^\rho \epsilon^{\gamma *}_\rho \epsilon^\gamma_\sigma {\mathcal M}_2^\sigma\right)^* =\sum_\lambda \sum_\gamma ({\mathcal M}_1^\mu {\mathcal M}_1^{\rho *} \epsilon^{\lambda *}_\mu \epsilon^{\gamma}_\rho) ({\mathcal M}_2^\nu{\mathcal M}_2^{\sigma *} \epsilon^\lambda_\nu \epsilon^{\gamma *}_\sigma)$$
here we still have $\lambda$ and $\gamma$ in each of the brackets. This needs to be factorized so that there is $\lambda$ in one of the brackets and $\gamma$ in the other bracket as you wrote in your reply.

I could switch the polarization vectors so that we have,
$$\sum_\lambda \sum_\gamma ({\mathcal M}_1^\mu {\mathcal M}_1^{\rho *} \epsilon^{\lambda *}_\mu \epsilon^{\gamma}_\rho) ({\mathcal M}_2^\nu{\mathcal M}_2^{\sigma *} \epsilon^\lambda_\nu \epsilon^{\gamma *}_\sigma) =\left(\sum_\lambda {\mathcal M}_1^\mu {\mathcal M}_1^{\rho *} \epsilon^{\lambda *}_\mu \epsilon^\lambda_\nu\right) \left(\sum_\gamma{\mathcal M}_2^\nu{\mathcal M}_2^{\sigma *} \epsilon^{\gamma}_\rho \epsilon^{\gamma *}_\sigma\right)$$
but now the lorentz indices don't match up within the brackets!

So how do I take the final step?

4. Jan 23, 2013

### jkp

It seems to me that I need some identity for the polarization vectors such as the following,
$$\varepsilon_\mu^{\lambda_1 *}\varepsilon_\nu^{\lambda_2} =\delta^{\lambda_1\lambda_2}\frac{1}{3} \left(\sum_\lambda \varepsilon_\mu^{\lambda *}\varepsilon_\nu^{\lambda} \right)$$
that is, the product of two polarization vectors is zero if the polarizations are different and equal to 1/3 the sum over polarizations if the two polarizations are equal. Does anybody know if such an identity actually exists?

5. Jan 23, 2013

### andrien

No,the square of the modulus is just extra complication.It has nothing to do with it.It is used in all decay processes where you don't observe the final polarization states.you just sum over the final spin states and take an average over the initial spin states after summing over initial spin states.There is no use of any identity.It is the usual way.since W boson has 3 spin states,you will have to divide by three.In electron scattering you divide by two after summing because of two polarization possible like here in mott scattering on page 203-204
also see here eqn. 3
http://arxiv.org/abs/0807.4112(see the full paper)

6. Jan 29, 2013

### jkp

Thanks again Andrien

From the reference you provide I now see that the Narrow Width Approximation isn't just that the propagator becomes,
$$\frac{1}{{(q^2-M_W^2)^2+M_W^2\Gamma_W^2}}\to \frac{2\pi\,\delta(q^2-M_W^2)}{2M_W^2\Gamma_W^2}$$
but also that we must approximate,
$$\left|\sum_\lambda {\mathcal M}^\lambda_{t\to b W} {\mathcal M}^\lambda_{W\to f\overline{f}'}\right|^2 \to \left(\sum_\gamma\left|{\mathcal M}^\gamma_{t\to b W}\right|^2\right) \left( \frac{1}{3}\sum_\lambda \left|{\mathcal M}^\lambda_{W\to f\overline{f}'}\right|^2 \right)$$
So this second transformation isn't an exact equality as I had assumed in my original question but a further simplification of the NWA proceedure.

7. Jan 29, 2013

### andrien

sure,it is a simplification which is done to avoid polarization.