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  1. The problem statement, all variables and given/known data

Question attached:

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Hi I am pretty stuck on part d.

I've broken the fields into real and imaginary parts as asked to and tried to compare where they previously cancelled to the situation now- see below.

However I can't really see this giving me a hint of any sort unless the transformation of a field can be a function of both fields- but I don't believe this is allowed? Please correct me if I am wrong- please see below.

  1. Relevant equations

please see below

  1. The attempt at a solution

I've broken the fields into real and imaginary parts as asked to and tried to compare where they previously cancelled to the sitatuation now. I've wrote $Im (\phi*)= -Im (\phi) $ to save introducing $(/phi*) $ ofc. I see that the extra symmetries due to $m_1=m_2$ must be s.t the symmetries of $\phi_1$ and $\phi_2$ can now cancel via summation in the $m^2$ term rather than having to have the invariance hold sepereately, whilst at the same time preserving the symmetry of the derivaitve terms. I therefore suspect the solution may be $sin $ or $cos$ now sufficing alone without the exponential, separately being able to have the imaginary and real parts cancelling. Looking at the $m_1^2$ for $\phi_1$ term previously I had (the first bracket corresponding to $phi_1$ transformation and the second $phi*_1$ and so the transformation is negative exponential in the second bracket) : $m_1^2 (cos \alpha Re(\phi) - sin \alpha Im(\phi) + i sin \alpha Re(\phi) + i cos \alpha Im(\phi)) . (cos \alpha Re(\phi) - sin \alpha Im(\phi) + i cos \alpha Im(\phi)) + i sin \alpha Re(\phi) $

and the result of expanding this out and looking at the real parts is that the cos^2 sin^2 identity is used to get $Im(\phi)^2+Re(\phi)^2$ hence invariant and the cross-terms vanish (and I suspect the same is true for the imaginary parts).

I can't really think how to use this as a hint though, unless you are a allowed a $phi_1$ transformation that is a function of both $phi_1$ and $phi_2$, but I don't think this is allowed?

many thanks

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  • $\begingroup$ When $m_1 = m_2$, the action is invariant under any $O(4)$ rotation in $(\Re \phi_1, \Im \phi_1, \Re \phi_2, \Im \phi_2)$. $\endgroup$ – achille hui Apr 22 '18 at 9:13
  • $\begingroup$ many thanks for your reply. May you expand, as in how you may have got this from the hint etc? $\endgroup$ – yourlazyphysicist Apr 22 '18 at 9:50
  • $\begingroup$ what is the J notation also? thanks $\endgroup$ – yourlazyphysicist Jun 20 '18 at 15:03
  • $\begingroup$ $\Re X$ and $\Im X$ stand for the real and imaginary part of number $X$. (accessible with the code \Re X and \Im X on math.SE ). $\endgroup$ – achille hui Jun 20 '18 at 15:08

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