# Pseudoscalar matrix elements

1. Jun 26, 2013

### Einj

Hi all. I have a question. What is the behaviour of the polarization vector of a pseudoscalar particle under a parity transformation??
Let me explain my problem. I know for sure that the effective matrix element which links a $D^*$ and a $\pi$ can be written as:

$$\langle \pi(p)D^*(q,\lambda) | D^*(k,\eta)\rangle=\frac{g}{M_{D^*}}\epsilon_{\alpha\beta\gamma\delta} \lambda^\alpha \eta^\beta p^\gamma q^\delta,$$
where $g$ is an effective coupling.

What I am trying to prove is that such a matrix element is (as it must be) a scalar. Now if, for example, we put ourselves in the rest frame of the $\pi$ we have just $(\vec{\lambda}\times\vec{\eta})\cdot \vec{q}$. Is that a scalar function?

Thank you very much

2. Jun 26, 2013

### Bill_K

Pseudoscalar particles don't have polarization vectors, Einj.

3. Jun 26, 2013

### Einj

I am sorry that's clearly a typo. I am talking about the $D^*$, a pseudovector. Sorry again.

4. Jun 26, 2013

### andrien

why it should strictly have this form.

5. Jun 26, 2013

### Einj

It is a consequence of Heavy Quark Effective Theory. You can take a look, for example, at arXiv:hep-ph/9605342 [hep-ph]. In this article there should be such a result.

6. Jun 26, 2013

### andrien

7. Jun 26, 2013

### Einj

I am sorry but I wasn't able to find such a lagrangian in the reference, where is it?. However, let me just underline that we are talking of a matrix element between a $\pi$ and two $D^*$, not a $D$ and a $D^*$. Does your point still hold also for such matrix element??

8. Jun 26, 2013

### Hepth

$D^*$ is a vector with parity $P = -1$
$\pi$ is a pseudoscalar $P = -1$

So you have $(-1)^3 = -1$ for your matrix in and out states.

You have two vectors (three momentum, but not independent) and two pseudo-vectors(polarizations) $P = +1$ that you can decompose this matrix element into.

It should be proportional to the two polarizations for sure, and have no free indices.

$\eta_1^{\alpha} \eta_2^{* \beta}$

now you could just multiply this by the metric to get an invariant but this is $P = +1$. You need a negative parity thing somewhere. If you introduce a momentum you'll have a free index if the polarizations are contracted. So you have a $P = -1$ thing but its not invariant.

$\eta_1^{\alpha} \eta_2^{* \beta} p_1^{\delta}$

So you use another momenta (assuming some metric multiplications here, any permutation)

$\eta_1^{\alpha} \eta_2^{* \beta} p_1^{\delta} p_2^{\sigma}$

BUT now its positive parity again. Luckily we have a 4-index negative parity pseudo-tensor, the antisymmetric Levi-Civita tensor giving us an object:

$\epsilon_{\alpha \beta \delta \sigma} \eta_1^{\alpha} \eta_2^{* \beta} p_1^{\delta} p_2^{\sigma}$

That has both the parity and lorentz-invariant properties of the amplitude. Throw in a constant out front for good measure.

For the Lagrangian look into Heavy Meson Chiral Perturbation Theory. the $\frac{1}{M}$ is from the expansion in the heavy quark mass.

In heavy meson chiral pt the vector and pseudoscalar mesons (D,DSTAR) are grouped into a spin-doublet "H" due to transformation properties.

You'll find the lagrangian as Eq 40 in http://arxiv.org/abs/hep-ph/9605342

which is a great overview of HMxPT

9. Jun 27, 2013

### Einj

Thank you very much. You have been very clear!