Flavour-SU(2) for Pion-Multiplett

[blue2script]

Hello!

I hear an elementary particle course and I wonder about one question: There is the multiplett of $$\pi^+, \pi^0, \pi^-$$. It is said that all have Isospin 1 and $$I_z = -1,0,1$$. So far so good. So these particles are to form a two-dimensional irreducible representation of SU(2). They are composed of a quark and an antiquark in the form:

$$\pi^+ : u\bar d$$
$$\pi^0 : u\bar u - d\bar d$$
$$\pi^- : d\bar u$$

So, this should be the I = 1 part of $$\frac{1}{2}\otimes\frac{1}{2} = 0\oplus 1$$. But on p. 26 my Maggiore says that these states are given as above but with a plus sign! The one with the minus sign is said to be the scalar representation.

So where is the mistake if there is any? I also can't see why the state with a minus should be a scalar representation. Hope somebody can help me out of confusion.

Thanks!

Blue2script

PS: By the way, maybe someone can also explain me the deeper meaning and use of the hypercharge. Up to now for me its only a definition used in representing multipletts in Young diagrams...

 

[pam]

That sign can be ambiguous. It depends on how the dbar and ubar are related by isospin.
The ispin lowering operator on dbar can give plus or minus ubar. You have to use one sign consistently, which makes comparing different books tricky.
I recall that the minus sign is used when G parity is introduced. Otherwise, it doesn't matter, as long as you are consistent. I think the minus sign is more common.

 

[samalkhaiat]

Hello!

I hear an elementary particle course and I wonder about one question: There is the multiplett of $$\pi^+, \pi^0, \pi^-$$. It is said that all have Isospin 1 and $$I_z = -1,0,1$$. So far so good. So these particles are to form a two-dimensional irreducible representation of SU(2). They are composed of a quark and an antiquark in the form:

$$\pi^+ : u\bar d$$
$$\pi^0 : u\bar u - d\bar d$$
$$\pi^- : d\bar u$$

It is the 3-dimensinal representation. It is carried either by an isovector

$$[3] = \left( \pi_{1}, \pi_{2}, \pi_{3} \right)$$

or by the traceless hermitian martix

$$\left( \begin{array}{rr} \frac{1}{\sqrt{2}}\pi^{0} & \pi^{+} \\ \pi^{-} & -\frac{1}{\sqrt{2}} \pi^{0} \end{array} \right)$$

$$\frac{1}{2}\otimes\frac{1}{2} = 0\oplus 1$$.

This means

$$| T =1/2 \rangle \ \otimes \ | T = 1/2 \rangle = | T = 1 , T_{3} = 1,0,-1 \rangle \oplus | T = 0 ,T_{3} = 0 \rangle$$

i.e., SU(2)-doublet X SU(2)-doublet = Triplet + singlet

or (in dimension)

$$[2] \otimes [2] = [3] \oplus [1]$$

Let us write this in terms of quarks matrices. We take the fundamental doublets to be a column of quarks

$$[2] = \left( \begin{array}{c} u \\ d \end{array} \right)$$

and a row of antiquarks (conjugate doublet)

$$[2] = \left( \bar{u} \ \bar{d} \right)$$

Thus

$$[2] \otimes [2] = \left( \begin{array}{rr} u \bar{u} & u \bar{d} \\ d \bar{u} & d \bar{d} \end{array} \right)$$

Next, we write the RHS as

$$\left( \begin{array}{cc} ( u \bar{u} - d \bar{d})/2 & d \bar{u} \\ u \bar{d} & (d \bar{d} - u \bar{u})/2 \end{array} \right) + \frac{1}{2} (u \bar{u} + d \bar{d}) \left( \begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array} \right)$$

The 1st matrix is traceless and hermitian, therefore we can identify it with the 3-dimensional representation martix [3] (written above), i.e.,

$$| \pi^{+} \rangle = | T=1,T_{3}=1 \rangle = u \bar{d}$$
$$| \pi^{-} \rangle = | T=1,T_{3}=-1 \rangle = d \bar{u}$$
$$| \pi^{0} \rangle = | T=1,T_{3}=0 \rangle = \frac{1}{\sqrt{2}} (u \bar{u} - d \bar{d})$$

The 2nd matrix is the 1-dimensional invariant representation (i.e., singlet)

$$[1] = \frac{1}{2} (u \bar{u} + d \bar{d}) I = | T=0,T_{3}=0 \rangle$$

So, your Maggiore got his signs wrong.

PS: By the way, maybe someone can also explain me the deeper meaning and use of the hypercharge. Up to now for me its only a definition used in representing multipletts in Young diagrams..

In general, the average charge of an iso-multiplet is given by half of the hypercharge. It is better understood with in flavour SU(3). Tell me first, how much you know about SU(3)?

regards

sam

 

[Vanadium 50]

It's a good rule of thumb that the singlet state is totally symmetric, and in this case that would be $$(u\overline{u} + d\overline{d})$$. The $$\pi^0$$ is part of a triplet, so it has the opposite sign: $$(u\overline{u} - d\overline{d})$$

 

[pam]

It's a good rule of thumb that the singlet state is totally symmetric, and in this case that would be $$(u\overline{u} + d\overline{d})$$. The $$\pi^0$$ is part of a triplet, so it has the opposite sign: $$(u\overline{u} - d\overline{d})$$

That is backwards.
It is generally not the case for Clebsch-Gordon coefficients.
It is not true for spin or I spin in adding u and d quarks.
The state with the minus sign for d-dbar is symmetric, but the
minus sign comes in from sign ambiguities in the raising and lowering operators. The + sign is the usual C-G convention, but the minus sign is often used for pions to get G parity straight.
Either sign is correct, if used consistently.