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

  1. Jan 12, 2012 #1

    If I have three light quark flavours with massses [itex]m_u, m_d,m_s [/itex], 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:

    [tex]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] [/tex]

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

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

    Even more confusing for me is that the text writes:

    [tex] 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] [/tex]

    I have absolutely no clue how this [itex]\pi^{0} [/itex], or [itex] \eta [/itex] mass is read off from the above, can anyone shed some light?
  2. jcsd
  3. Jan 12, 2012 #2
    I have not checked the maths, but:
    - for a complex field the mass term is proportional to [itex]\phi^\dagger\phi[/itex], if expanded it in two hermitian fields it is [itex]\phi_1^2+\phi_2^2[/itex]
    - [itex]\eta[/itex] and [itex]\pi_0[/itex] have a mass matrix in the Lagrangian you quoted, physical particles are eigenvalues of mass, try to diagonalize it...
  4. Jan 12, 2012 #3
    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 [itex] \pi^{0},\eta[/itex] etc as being the indices of matrix, so maybe top leftmost is filled by coefficient of [itex] (\pi^{0})^2 [/itex] from L_mass etc? and bottom rightmost is eta^2 coeff.

    Then exactly how does one diagonalize this? sorry for all the questions..
  5. Jan 12, 2012 #4
    For example I'm thinking if we focus on the eta, pi_0 piece we can rewrite this as:

    [tex]\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] [/tex]

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

    But then diagonalize this? what would be meant by the eigenvectors etc?
  6. Jan 12, 2012 #5
    Yes, those are some of the coefficients of the mass matrix, to understand also the diagonal part try to expand this product:
    [tex]\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)[/tex]

    Then you can define new fields
    [tex]\left(\begin{array}{l} \pi' \\ \eta'\end{array}\right)=O\left(\begin{array}{l} \pi^0 \\ \eta\end{array}\right)[/tex]
    to diagonalize that matrix without ruining the kinetic term
    Last edited: Jan 12, 2012
  7. Jan 13, 2012 #6
    I see, so the [itex]\eta[/itex], [itex](\pi^0)[/itex] part of my mass Lagrangian is:

    [tex]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][/tex]

    [tex]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) [/tex]

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

    [tex]\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] [/tex]

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

    [tex]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] [/tex]

    Thanks a lot for your help aesir.
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