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They both seem to have the same quark content:

$$\rho^{+} = u\bar{d} = \pi^{+}$$ and $$\rho^{-} = \bar{u}d = \pi^{-}$$

What is different about the two?

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3 Answers 3

up vote 4 down vote accepted

They have 2 main differences. The first of them is very straightforward: They have different spins: As you pointed out, both are bound states of 2 spin 1/2 particles, therefore you can find the possible spins of such a bound state using the usual rules of angular momentum addition in Quantum Mechanics. 1/2 + 1/2 can give you either 3 spin 1 states (the triplet, with projections m= -1, 0 or 1) or a singlet with total spin 0 and thus projection m = 0.

The rho has spin 1 while the pion has spin 0. So that's one difference.

Now you can ask: "Ok, so I just changed the spin, why are their masses so different?"

And here is where the difference is more interesting. The Pion is actually a (Pseudo)Nambu-Goldstone. I'm not gonna write the whole theory here, but basically if electromagnetism was turned off, quarks would have a symmetry known as isospin. Since the up and down quarks have almost the same mass, one could treat them as the up and down components of a doublet, just like spin up and down in normal spin theory (that's why the term isospin).

After breaking that isospin symmetry, Nambu-Goldstone bosons appear (the pions), which would have been massless (by Goldstone theorem). But since their masses are not exactly the same, and of course, their charges are not the same, so EM effects break that symmetry explicitly. So the pions do acquire a mass. But this mass is very tiny when compared to a typical hadron mass or the QCD scale, 140 MeV for the pion and 770 for the rho.

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Great answer! Thanks. Could you suggest some reading material that might help me understand this better (Assume a 4th year undergraduate understanding of particle physics and QM). – Nikhil Mahajan Mar 5 '14 at 3:06
@NikhilMahajan If you know a bit of group theory or don't mind learning it, look at the book by Cheng and Li. – suresh Mar 5 '14 at 8:27
I agree Cheng and Li is a nice book with this stuff, but sometimes may get a little too technical if you don't have the necessary background. maybe Langacker (Standard model and beyond) also helps! sections 3.2 and 3.3 talk about this – gcsantucci Mar 5 '14 at 15:30
@user41847 Besides group theory, what necessary background is required? – Nikhil Mahajan Mar 5 '14 at 17:18
I guess it helps a lot to know about field theory. It doesn't even have to be quantum field theory, classical should do it for these symmetry breaking discussions. Both books explain what they use of field theory. But I guess Langacker is more self-contained and easier to understand. Quigg book is also a great place to see this (the new edition). – gcsantucci Mar 5 '14 at 21:22

A $\rho$ meson is the spin-1 (angular momentum, not isospin) excitation of the $\pi $ meson. We have,

\begin{equation} \rho = \begin{cases} \rho ^+ \quad u \bar{d} \\ \rho ^0 \quad \frac{ u \bar{u} - d \bar{d} }{ \sqrt{ 2}} \\ \rho ^- \quad d \bar{ u } \end{cases} \end{equation} where each $ \rho $ meson has a different isospin. However they are all spin $ 1 $ particles. The pions are analogous, only they have spin $0$.

Since the color-magnetic force is so strong different spin hadrons have wildly different masses and are referred to as distinct particles.

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These are two really great answers, so I don't feel the need to add much; only to supply that the rho can be thought of as an "excited" pion, like the Delta can be thought of as an "excited" nucleon. As the asker and @gcsantucci point out, the rho has a unit more spin than the pion. This means you could think of it as a light quark-antiquark system which has been given a unit of angular momentum. In the same way, we think of the Delta baryon (spin 3/2) as an excited state of a nucleon (spin 1/2).

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