# How to get the lowest order ChPT amplitudes?

1. Apr 28, 2015

### Thor Shen

Note from mfb: I fixed the LaTeX formulas

http://image.keyan.cc/data/bcs/2015/0428/w84h1446064_1430226899_628.jpg

I met some trouble by using latex on this version, so I try to add the image from the pdf,
the Latex code is as follow if any help.
The lowest order lagrangian for ChPT is
$\mathcal{L}^{(2)}=\frac{f^2}{12}\langle(\partial\Phi \Phi-\Phi\partial\Phi)^2+M\Phi^4\rangle$

where the isospin limit is assumed and the $\Phi$ is made of the pseudo Goldstone boson fields,
$$\begin{pmatrix} \frac{\pi^0}{\sqrt{2}}+\frac{\eta}{\sqrt{6}} & \pi^+ & K^+\\ \pi^-& \frac{-\pi^0}{\sqrt{2}}+\frac{\eta}{\sqrt{6}} & K^0\\ K^-& \overline{K}^0&\frac{-2\eta}{\sqrt{6}}\\ \end{pmatrix}$$

Let us consider the physical reaction $$I=0~\pi^+(\vec{k})\pi^-(\vec{p})\leftrightarrow\pi^+(\vec{k'})\pi^-(\vec{p'})$$,where the k,p,k' and p' are the momentums of the pion, respectively. The relevant lagrangian can be derived from the above formula,
$$\begin{eqnarray}\nonumber \mathcal{L}_{\pi\pi\pi\pi}&=&2\partial\pi^+\partial\pi^+\pi^-\pi^- -2\partial\pi^+\partial\pi^-\pi^+\pi^--2\partial\pi^-\partial\pi^+\pi^+\pi^- \\ \nonumber &&+2\partial\pi^-\partial\pi^-\pi^+\pi^+ +2m^2_{\pi}\pi^+\pi^+\pi^-\pi^- \end{eqnarray}$$
and the amplitudes of this reaction,
$$\begin{eqnarray}\nonumber V(s,t,u)&=&\langle\pi^+\pi^-|H|\pi^+\pi^-\rangle \\ \nonumber &=&-\langle\pi^+\pi^-|\mathcal{L}|\pi^+\pi^-\rangle \\ =-\frac{1}{6f^2}&\{&\langle\pi^+(k')\pi^-(p')|\partial\pi^+\partial\pi^+\pi^-\pi^-|\pi^+(k)\pi^-(p)\rangle\\ &-&\langle\pi^+(k')\pi^-(p')|\partial\pi^+\partial\pi^-\pi^+\pi^-|\pi^+(k)\pi^-(p)\rangle\\ &-&\langle\pi^+(k')\pi^-(p')|\partial\pi^-\partial\pi^+\pi^+\pi^-|\pi^+(k)\pi^-(p)\rangle\\ &+&\langle\pi^+(k')\pi^-(p')|\partial\pi^-\partial\pi^-\pi^+\pi^+|\pi^+(k)\pi^-(p)\rangle\\ &+&\langle\pi^+(k')\pi^-(p')|m^2_{\pi}\pi^+\pi^+\pi^-\pi^-|\pi^+(k)\pi^-(p)\rangle\}. \end{eqnarray}$$
Finally, the result can be expressed as $$-\frac{1}{3f^2}(s+t-2u+2m^2_{\pi})$$ with the Mandelstam variables $$s=(k+p)^2,t=(k-k')^2$$ and $$u=(k-p')^2$$, that is $$-\frac{1}{6f^2}(2(k-k')^2-2(k-p')^2-2(k-p')^2+2(k+p)^2+4m^2_{\pi})$$. Then how to get the answers by the five amplitudes, respectively?

Last edited by a moderator: Apr 28, 2015
2. Apr 28, 2015

### Hepth

Do you understand how to get the last one, the $m_{\pi}^2$ term?

Theres symmetry involved. Each operator $\pi^{+}$ can be contracted to either an incoming $\pi^{+}$ or an outgoing $\pi^{+}$. So you have 2 connections. Same with the $\pi^{-}$, so theres a symmetry of $2*2= 4$. Thats why theres $4 m_{\pi}^2$. Draw a physical line from each type of operator to their in.out states and make sure all operators are connected.

The pions are spin-0, so when they operate on a in/out state, it just creates or annhiliates that state. it contracts and you get "1". They also commute with eachother at this level, so theyre free to move around to order them properly.

For the derivatives, these give you a factor of the momentum times the pion operator again. The sign of the momentum depends on how you've defined the incoming/outgoing.

You can get this by taking the derivative of the free-pion wavefunction, which is just an exponential.

3. Apr 29, 2015

### Thor Shen

Well,thank you for your apply, but I still have some confusions.
For example,the operator π+and π- act on the state |π+π->.
π+π-+π->=|π+π->? (1)
π+π-|0 0>=|π+π->? (2)
+π-+π-=<π+π-|? (3)
<0 0|π+π-=<π-π+|? (4)
if we consider the crossing symmetry, we can rewritten the <π1+π2-+π-π+π-|π3+π4-> as <0|π+π-π+π-1-π2+π3+π4->. Then the operators act on the vacuum state as <π1-π2+π3+π4-1-π2+π3+π4->=1.
The four is from the commutative operators because the 4 permutation which is π1+π2+π3-π4-,π1+π2+π4-π3-,π2+π1+π3-π4-,and π2+π1+π4-π3-?

4. Apr 29, 2015

### Hepth

ok, maybe the notation is confusing. Think about it this way, in creation/annihilation operators :
$$\hat{\pi}^{+} | \pi^{+} \rangle \\ \hat{\pi}^{+} \pi^{\dagger+}| 0 \rangle \\ (1) | 0 \rangle$$

Where $\hat{\pi}^{+}$ is the full operator and contains both $\pi^{+}$ and $\pi^{\dagger+}$

Where the last line is from the contraction between the two operators.

We're basically doing LSZ reduction, but at a much more algebraic way, so we have to understand all of the shortcuts we're using.

I wish i could draw a connecting line for them like in Peskin&Sch.

You should write out all the creation/annhilation operators so you're left with $<0|0>$ which is 1.

When you have 2 :

$$\langle \pi^{+} |\hat{\pi}^{+} \hat{\pi}^{+} | \pi^{+} \rangle \\ \langle \pi^{+} | \hat{\pi}^{+} \hat{\pi}^{+} \pi^{\dagger+}| 0 \rangle \\ \langle 0 |\pi^{+} \hat{\pi}^{+} \hat{\pi}^{+} \pi^{\dagger+}| 0 \rangle \\$$

then you do the 2 possible contractions between the creation/annhilation operators and the operators with the hat, that contain a sum of both.

5. Apr 29, 2015

### Thor Shen

the operator with hat likes the number operator as $$\hat{\pi}^+=\pi^{\dagger+}\pi^+$$ ,as well as $$\hat{\pi}^-=\pi^{\dagger -}\pi^-$$. And the operators without hat are satisfied with the boson commutation relation?

6. Apr 29, 2015

### Hepth

Technically its more that the hat ones are the plane wave solutions in the LSZ, right?

$\hat{\pi} = (...)(e^{i k \cdot x} \pi^{\dagger} + e^{-i k\cdot x} \pi )$

this also tells you what you get when you take derivatives:

$\partial_{\mu} \hat{\pi} = (...)((i k_{\mu} )e^{i k \cdot x} \pi^{\dagger} + (-i k_{\mu} )e^{-i k\cdot x} \pi )$

but $$k$$ depends on which field it contracted with.