How to find Internal Symmetries?

[Breo]

For instance, how to find what internal symmetries a free relativistic lagrangian density has for complex scalar field?

 

[dextercioby]

Well, this is a circular question, since Lagrangian (densities) are built on purpose to encompass the desired (rigid) symmetries. Think about the $\phi\phi^*$ term. What transformation of the field would leave it invariant? Then you can move to differentiated fields.

 

[Breo]

Unitary transformation?

 

[dextercioby]

Fancy language. Spell it out using 'common knowledge'. :)

 

[Breo]

$\phi \longrightarrow e^{i\alpha}\phi$
$\phi^* \longrightarrow \phi^* e^{-i\alpha}$

$(\phi^* e^{-i\alpha})(e^{i\alpha}\phi) = \phi^* \phi$

Also I could do it with the orthogonal matrix representation:

$$ \phi^* \Omega^+ \Omega \phi = \phi^* \phi $$

And this is all? there are no more kinds of internal symmetries?

 

[dextercioby]

Yes, indeed. For the sake of the Standard Model SO(n) is not needed, a simple phase factor (U(1)) will do.

 

[ChrisVer]

You can find more symmetries. In case for the phi* phi, you can also have the parity transformation:

\phi \rightarrow - \phi

This is one reason someone won't write \phi^3 terms a priori in a Lagrangian. Or instead of a U(1) you can have a broken U(1) in the case that it would correspond to a Z_N...

In fact you can find a lot of symmetries in a Lagrangian density. The reason is already stated, that the given Lagrangians will be built in order to contain your desired symmetries.

Take for example the quarks. The quarks are in a 3-dimensional representation of SU(3)_{color}. How can you make neutral quantities from combinations of 3?
Take for example the 3 \otimes 3 = 6 \oplus \bar{3}. Obviously the combination 3 \otimes 3 cannot be decomposed to a singlet representation, but it contains the complex conjugate of 3, that is the reason sometimes 3 \otimes 3 are called anti-quarks.
On the other hand the combination 3 \otimes \bar{3}= 8 \oplus 1 contains the 1-dimensional object [which transforms trivially under SU(3) transformations] and so can "work". The result is that terms which belong to 3, \bar{3} can be combined in the Lagrangian to give you allowed terms. Similarly for 3 \otimes 3 \otimes 3 (so combinations of 3 fields belonging to 3 representation).

 

[HomogenousCow]

You can find more symmetries. In case for the phi* phi, you can also have the parity transformation:

\phi \rightarrow - \phi

This is one reason someone won't write \phi^3 terms a priori in a Lagrangian. Or instead of a U(1) you can have a broken U(1) in the case that it would correspond to a Z_N...

In fact you can find a lot of symmetries in a Lagrangian density. The reason is already stated, that the given Lagrangians will be built in order to contain your desired symmetries.

Take for example the quarks. The quarks are in a 3-dimensional representation of SU(3)_{color}. How can you make neutral quantities from combinations of 3?
Take for example the 3 \otimes 3 = 6 \oplus \bar{3}. Obviously the combination 3 \otimes 3 cannot be decomposed to a singlet representation, but it contains the complex conjugate of 3, that is the reason sometimes 3 \otimes 3 are called anti-quarks.
On the other hand the combination 3 \otimes \bar{3}= 8 \oplus 1 contains the 1-dimensional object [which transforms trivially under SU(3) transformations] and so can "work". The result is that terms which belong to 3, \bar{3} can be combined in the Lagrangian to give you allowed terms. Similarly for 3 \otimes 3 \otimes 3 (so combinations of 3 fields belonging to 3 representation).

You know I've always wondered why they don't have the real scalar field squared in the yukawa interaction term, is parity not relevant in that situation?

 

[ChrisVer]

What do you mean? a term like : \phi^2 \bar{\psi} \psi?
one reason I see is that this term is non-renormalizable.

 

[HomogenousCow]

What do you mean? a term like : \phi^2 \bar{\psi} \psi?
one reason I see is that this term is non-renormalizable.

Does that mean there's something fundamentally wrong with the term, or just that perturbation theory can't handle it?

 

[Breo]

What are the dimensions of the spinors in terms of powercounting?

 

[ChrisVer]

take the mass term : m \bar{\psi} \psi...
If [m]=1 what should be [\psi] in order to have a dimensionless action?

As for the term, In general you can have such terms into an effective Lagrangian [insert some cut-off]. Why would you do that in the fundamental level? Take eg the case for which you don't write a Fermi-term: \bar{\psi} \psi \bar{\psi} \psi. Both these operators are irrelevant.

 

[Breo]

3/2, but where it comes? It is not a should be, it is a "it is". A spinor is 4 component complex vector. The half integer comes from the idea of bispinor?

 

[ChrisVer]

it simply comes from asking the Dirac Lagrangian to be of dimension 4 (in 4D space).
You are not counting degrees of freedom, but mass dimension... so I don't think the components can help you. A better way to write this is [\psi] = m^{3/2} for not getting confused with a simple shorthanded 3/2 (half-integer) which would make you think of spinors.
Of course the fact that you have spin 1/2 particles leads you to Dirac equation, and so an equation having the mass term as m \bar{\psi}\psi