QFT, Noether and Invariance, Complex fields, Equal mass

[binbagsss]

Homework Statement

Question attached:

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 canceled 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.

Homework Equations

please see below

The Attempt at a Solution

[/B]

I've broken the fields into real and imaginary parts as asked to and tried to compare where they previously canceled 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?

a thousand thanks to you my friend.

 

[MathematicalPhysicist]

You can find this exercise and its solution in several references in the web, I believe it appears in P&S, Srednicki in other words with a good search through google, you can find a solution to this exercise.
In fact this exercise appears in Radovanovic's problem book, problem 5.11.

 

[binbagsss]

You can find this exercise and its solution in several references in the web, I believe it appears in P&S, Srednicki in other words with a good search through google, you can find a solution to this exercise.
In fact this exercise appears in Radovanovic's problem book, problem 5.11.

many thanks for your reply, I had no idea about this book !

 

[binbagsss]

however, not to sure why no one replied to my question can the transformation be a function of both fields, quick question, yes or no answer, would have helped alot, but hey..

 

[MathematicalPhysicist]

however, not to sure why no one replied to my question can the transformation be a function of both fields, quick question, yes or no answer, would have helped alot, but hey..

With these type of QFT questions it's sort of impossible to solve without a reference at hand... :-D

 

[nrqed]

however, not to sure why no one replied to my question can the transformation be a function of both fields, quick question, yes or no answer, would have helped alot, but hey..

Yes, in general it is possible to do a transfer that contains all the fields.

 

[binbagsss]

With these type of QFT questions it's sort of impossible to solve without a reference at hand... :-D[/QUOTE
I don't suppose you know whether a similar sort of solution book may exist for string theory ?

Thanks ( in particular t-duality, massless states ) ?

 

[nrqed]

You might be interested in the solutions to Zwiebach's book on string theory.

 

[MathematicalPhysicist]

There's also a partial solution manual to Polchinski's 2-set volume, just type you know what into google.