# How to find Internal Symmetries?

1. Dec 7, 2014

### Breo

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

2. Dec 8, 2014

### 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.

3. Dec 8, 2014

### Breo

Unitary transformation?

4. Dec 8, 2014

### dextercioby

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

5. Dec 8, 2014

### 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?

6. Dec 8, 2014

### dextercioby

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

7. Dec 9, 2014

### 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).

8. Dec 9, 2014

### HomogenousCow

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?

9. Dec 10, 2014

### ChrisVer

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

10. Dec 10, 2014

### HomogenousCow

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

11. Dec 10, 2014

### Breo

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

12. Dec 10, 2014

### 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.

Last edited: Dec 10, 2014
13. Dec 10, 2014

### 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?

14. Dec 10, 2014

### 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 $m \bar{\psi}\psi$

Last edited: Dec 10, 2014