Reading off masses of eight goldstone bosons from chiral Lagrangian mass term

[LAHLH]

Hi,

If I have three light quark flavours with massses $m_u, m_d,m_s$, I want to try and calcuate the masses of the eight pseudogoldstone bosons.

I have found from my mass term in the Chiral L that:

$$L_{mass}=-2v^3 f_{\pi}^{-2}\left[(m_u+m_d)\pi^{+}\pi^{-}+(m_u+m_s)K^{+}K^{-}+(m_d+m_s)\bar{K}^{0}K^{0}+\tfrac{1}{2}m_u\left(\eta /\sqrt{3}+\pi^{0}\right)^2+\tfrac{1}{2}m_d\left(\eta/\sqrt{3}-\pi^{0}\right)^2+\tfrac{2}{3}m_s\eta^2\right]$$

which is all well and good and I was hoping to just read of the masses from this by looking for the form $-1/2 m^2 \phi^2$ and then just identifying m^2 for the various fields $\pi^{+/-},\eta, K^{0},...$ etc

My text says $m_{\pi^{\pm}}^2=2v^3 f_{\pi}^{-2}\left(m_u+m_d\right)$ but what does this mixed term of pi+,pi- mean? I was expecting $\left(\pi^{+}\right)^2$ terms to be present to give the $\pi^{+}$ mass, not a mixture of +/-?

Even more confusing for me is that the text writes:

$$m^{2}_{\pi^{0},\eta}=\frac{4}{3} v^3 f_{\pi}^2\left[m_u+m_d+m_s \mp \left( m_u^2+m_d^2+m_s^2-m_sm_d-m_dm_s-m_sm_u\right)^{1/2}\right]$$

I have absolutely no clue how this $\pi^{0}$, or $\eta$ mass is read off from the above, can anyone shed some light?

 

[aesir]

I have not checked the maths, but:
- for a complex field the mass term is proportional to $\phi^\dagger\phi$, if expanded it in two hermitian fields it is $\phi_1^2+\phi_2^2$
- $\eta$ and $\pi_0$ have a mass matrix in the Lagrangian you quoted, physical particles are eigenvalues of mass, try to diagonalize it...

 

[LAHLH]

Thanks a lot for the reply, I think what you say is definitely the way to get these results, but any chance you could say a fraction more? I'm just struggling to see exactly how and where the "mass matrix" is in my L_mass term. Do I just think of $\pi^{0},\eta$ etc as being the indices of matrix, so maybe top leftmost is filled by coefficient of $(\pi^{0})^2$ from L_mass etc? and bottom rightmost is eta^2 coeff.

Then exactly how does one diagonalize this? sorry for all the questions..

 

[LAHLH]

For example I'm thinking if we focus on the eta, pi_0 piece we can rewrite this as:

$$\eta^2\left[m_u/6+m_d/6+2m_s/3\right]+(\pi^0)^2\left[m_u/2+m_d/2\right]+\eta\pi^0\left[m_u/2\sqrt{3}-m_d/2\sqrt{3}\right]+\pi^0\eta\left[m_u/2\sqrt{3}-m_d/2\sqrt{3}\right]$$

Then thinking of these as representing a matrix, we can write that matrix in the obvious way, i.e. $\left[m_u/6+m_d/6+2m_s/3\right]$ in top left hand corner etc..

But then diagonalize this? what would be meant by the eigenvectors etc?

 

[aesir]

Thanks a lot for the reply, I think what you say is definitely the way to get these results, but any chance you could say a fraction more? I'm just struggling to see exactly how and where the "mass matrix" is in my L_mass term. Do I just think of $\pi^{0},\eta$ etc as being the indices of matrix, so maybe top leftmost is filled by coefficient of $(\pi^{0})^2$ from L_mass etc? and bottom rightmost is eta^2 coeff.

Then exactly how does one diagonalize this? sorry for all the questions..

Yes, those are some of the coefficients of the mass matrix, to understand also the diagonal part try to expand this product:
$$\left(\begin{array}{ll} \pi^0 & \eta\end{array}\right)\left(\begin{array}{ll} m_1^2 & m_{12} \\ m_{12} & m_2^2\end{array}\right)\left(\begin{array}{l} \pi^0 \\ \eta\end{array}\right)$$

Then you can define new fields
$$\left(\begin{array}{l} \pi' \\ \eta'\end{array}\right)=O\left(\begin{array}{l} \pi^0 \\ \eta\end{array}\right)$$
to diagonalize that matrix without ruining the kinetic term

 

[LAHLH]

I see, so the $\eta$, $(\pi^0)$ part of my mass Lagrangian is:


$$L_{mass}= -2v^3 f_{\pi}^{-2}\left[\eta^2\left[\frac{m_u}{6}+\frac{m_d}{6}+\frac{2m_s}{3}\right]+(\pi^0)^2\left[\frac{m_u}{2}+\frac{m_d}{2}\right]+\eta\pi^0\left[\frac{m_u}{2\sqrt{3}}-\frac{m_d}{2\sqrt{3}}\right]+\pi^0\eta\left[\frac{m_u}{2\sqrt{3}}-\frac{m_d}{2\sqrt{3}}\right]\right]$$

$$L_{mass}=\frac{-2v^3 f_{\pi}^{-2}}{6}\left(\begin{array}{ll} \eta & \pi^0\end{array}\right)\left(\begin{array}{ll}m_u+m_d+4m_s & \sqrt{3}(m_u-m_d) \\ \sqrt{3}(m_u-m_d) & 3(m_u+m_d)\end{array}\right)\left(\begin{array}{l} \eta \\ \pi^0\end{array}\right)$$

But now if we make a field redefinition with some linear operator S: $\left(\begin{array}{l} \eta' \\ \pi'\end{array}\right)=S\left(\begin{array}{l} \eta^0 \\ \pi^0 \end{array}\right)$ then we can choose S to be the similarity transformation that diagonalises the above matrix $M'=S^T M S$ and $M'=diag(\lambda_{+},\lambda_{-})$ and in terms of these new fields the mass Lagrangian will look more conventional. $L_{mass}=\lambda_+ (\eta^{'})^2+\lambda_-(\pi^{0 '})^2$ from which we can just read off the masses in the usual way.

So indeed, I found the eigenvalues of the above matrix and they came out to be:

$$\lambda_{+,-}=\frac{4}{6} v^3 f_{\pi}^2\left[m_u+m_d+m_s \mp \left( m_u^2+m_d^2+m_s^2-m_sm_d-m_dm_s-m_sm_u\right)^{1/2}\right]$$

meaning $-\frac{1}{2}m_{\eta'}^2=\lambda_{+}$ which leads to $m_{\eta'}^2=-2\lambda_{+}$ etc. So we might as well revert relabel back to old names $\eta'\to\eta$...

$$m_{\eta,\pi^{0}}^2=-\frac{4}{3} v^3 f_{\pi}^2\left[m_u+m_d+m_s \mp \left( m_u^2+m_d^2+m_s^2-m_sm_d-m_dm_s-m_sm_u\right)^{1/2}\right]$$

Thanks a lot for your help aesir.