# What does it mean that the neutral pion is a mixture of quarks?

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The quark composition of the neutral pion ($\pi^0$) is $\frac{u\bar{u} - d\bar{d}}{\sqrt{2}}$. What does this actually mean?

I think it's bizarre that a particle doesn't have a definite composition. There's a difference of 2 MeV between the quark masses and I don't understand how this can be ignored. If I were to somehow manage to make a bound state of an up and an anti-up quark, what would it be? Would it be a variation on the neutral pion or would it somehow transform into the mixture?

This post imported from StackExchange Physics at 2015-03-02 08:05 (UTC), posted by SE-user Kasper Meerts
Welcome to our wierd-yet-wonderful quantum-mechanical Universe! Joking aside, I'm sure someone will be able to offer you a more expansive answer before long.

This post imported from StackExchange Physics at 2015-03-02 08:05 (UTC), posted by SE-user qftme

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I think it's bizarre that a particle doesn't have a definite composition.

Yeah, it is. As qftme said, that's quantum mechanics for you. It really doesn't make sense until you immerse yourself in the subject for long enough (and even then, only somewhat). But it does appear to be the way the universe works.

Anyway, just so everyone is on the same page, let me start from the basics. If you're familiar with linear algebra, you know that a vector in a 2-dimensional vector space, for example, can be written as a linear combination $\alpha|0\rangle + \beta|1\rangle$ of two basis elements $|0\rangle$ and $|1\rangle$. For example, a direction vector of length 1 that points northeast can be written as

$$\frac{|\text{north}\rangle + |\text{east}\rangle}{\sqrt{2}}$$

or it could be written as

$$|\text{northeast}\rangle$$

or

$$\alpha|\text{north-northeast}\rangle + \beta|\text{east-southeast}\rangle$$

etc. You could figure out what the coefficients $\alpha$ and $\beta$ are in that last case, but it doesn't matter. The point is, there are an infinite number of ways to decompose any vector.

The pion state is an example of such a vector. It's often considered to be a member of a three-dimensional vector space. One possible basis for that vector space is $u\bar{u}$, $d\bar{d}$, and $s\bar{s}$. But another possible basis is

$$\pi^0 = \frac{u\bar{u} - d\bar{d}}{\sqrt{2}}$$

$$\eta = \frac{u\bar{u} + d\bar{d} - 2s\bar{s}}{\sqrt{6}}$$

$$\eta' = \frac{u\bar{u} + d\bar{d} + s\bar{s}}{\sqrt{3}}$$

This basis is useful because these particular combinations happen to be relatively stable; in other words, when a particle consisting of any combination of $u\bar{u}$, $d\bar{d}$, and $s\bar{s}$ is detected in a cloud chamber (if you're old-school) or a calorimeter or something like that, it will behave like one of these three particles. It's possible that what was actually emitted was the quantum state $u\bar{u}$, but in terms of the "stable" states, that is

$$u\bar{u} = \frac{1}{\sqrt{2}}\pi^0 + \frac{1}{\sqrt{6}}\eta + \frac{1}{\sqrt{3}}\eta'$$

(hopefully I did the math right). So you would have a probability of $\frac{1}{2}$ that it acts like (or technically, collapses to) a pion, $\frac{1}{6}$ that it collapses to an eta meson, and $\frac{1}{3}$ that it collapses to an eta prime meson. One of those three possibilities is what you'd actually observe in your detector.

You can do this the other way around, too: suppose that instead of $u\bar{u}$, you started with a pion, and instead of measuring the "stable" meson type, you were able to directly measure the quark content. Since the pion state contains equal components of $u\bar{u}$ and $d\bar{d}$, your hypothetical quark flavor measurement would give you one of those outcomes with 50% probability each: half the time you'd find that you had an up quark and an anti-up quark, and the other half of the time you'd find a down and anti-down quark. That's what the state $\frac{u\bar{u} - d\bar{d}}{\sqrt{2}}$ actually means: it governs the probabilities that the pion will interact with a quark flavor measurement as each particular quark type.

This post imported from StackExchange Physics at 2015-03-02 08:05 (UTC), posted by SE-user David Z
answered Jun 13, 2011 by (660 points)
I understand linear algebra, of course, but I don't understand why you can combine quarks like vectors. How did they come up with these compositions? I'm not convinced there's no problems with mass.

This post imported from StackExchange Physics at 2015-03-02 08:05 (UTC), posted by SE-user Kasper Meerts
@kmm: quarks are vectors. More precisely, the state of any quantum mechanical object is an element of some Hilbert space. So it should be obvious that you can add quarks in this way too. The nontrivial part here is that our theories obey certain symmetries. The symmetry here is approximate and is called flavor. We pretend that $u$, $d$ and $s$ have same mass and then (because of group theory) you can create some irreducible representations of $SU(3)$. These are important because they are stable w.r.t. evolution. In reality, this is only approximate and they are only approximately stable.

This post imported from StackExchange Physics at 2015-03-02 08:05 (UTC), posted by SE-user Marek
@kmm: the more elementary theory (but with same ideas) is that based on just $n$ and $p$ (and it is less approximate too because the masses are quite close). These have a $SU(2)$ symmetry which is incidentally related to rotations and spin. That's why people called it isospin. Because $n, p$ act as a components of a $1/2$-spin particle you can guess what happens when you have two of them together. Like for spin, the total system splits into isospin 0 singlet (this is the actual stable deuterium: $pn - np$ state, like for pion) and isospin 1 triplet (which is unstable for other reasons).

This post imported from StackExchange Physics at 2015-03-02 08:05 (UTC), posted by SE-user Marek
@David: I am not an expert on elementary particles but is it really the case that one should restrict themselves to a 3-dimensional space of $\pi$, $\eta$ and $\eta'$? In my mind, one takes a representation $3 \otimes \bar{3} = 1 \oplus 8$. But here $\eta'$ is the singlet and should have nothing to do with the rest of the particles from octet. I guess your splitting is based on some other approach (you obviously have zero strangeness and zero charge) which is probably more relevant to the elements being stable? Perhaps because the mass of the $s$ is too big and it breaks $SU(3)$ too much?

This post imported from StackExchange Physics at 2015-03-02 08:05 (UTC), posted by SE-user Marek
@Marek: well, actually I "invented" this 3D vector space just to keep the explanation simple. In reality, of course, you would use the full 9D vector space of all light quark combinations, which then breaks down into the singlet and octet. Since I was trying to give an intuitive (rather than technical) explanation, the distinction between the singlet and octet didn't seem that important.

This post imported from StackExchange Physics at 2015-03-02 08:05 (UTC), posted by SE-user David Z
@Marek: It's not really obvious to add states of systems that aren't equal. An up-quark is distinguishable from a down-quark, both by mass and charge. If free quarks were to exist, how would the combination $u + d$ react? What way would it bend in a magnetic field?

This post imported from StackExchange Physics at 2015-03-02 08:05 (UTC), posted by SE-user Kasper Meerts
@kmm: we have equations (QFT + standard model) that tell us how any field evolves, including combinations such as $u + d$. But such a calculation (besides being hard) is completely useless as you can't prepare and observe such a state. In colliders you always have pure particle collisions. But be sure that in the formalism of quantum theory you can always add vectors. Do you at least accept that you can add two states of different energy in quantum harmonic oscillator and that you can compute how it would evolve? Yet classically, such a state has no obvious interpretation...

This post imported from StackExchange Physics at 2015-03-02 08:05 (UTC), posted by SE-user Marek
@Marek: Of course I have no problem with adding states of a QHO. There's a big difference between adding to possible states of one system and adding multiple states together. The up and the down quark aren't the same field, in a way $u + d$ makes as much sense as $u + e$. I find it hard to believe a system can be in a superposition of particles with different quantum numbers, as for example deep inelastic scattering shows that the proton is a structure than can be probed, which would be impossible if we don't agree the proton has a clear composition.

This post imported from StackExchange Physics at 2015-03-02 08:05 (UTC), posted by SE-user Kasper Meerts
@Marek: My problem probably is that I'm supposed to learn about particle physics and that I have to work with Feynman diagrams, without ever having had any explanation about QFT. This approach irritates me, as I don't think it's unreasonable to ask to know what I'm actually calculating.

This post imported from StackExchange Physics at 2015-03-02 08:05 (UTC), posted by SE-user Kasper Meerts
@kmm: are you trying to imply that each field lives in its own Hilbert space? This is simply not so. But without prior exposition to QFT, you might not believe this. In particle approximation, Hilbert space is a Fock space: its base consists of vacuum state and any $n$-particle state you can create on top of that (e.g. state of three electrons and two photons). But because this is still quantum mechanics, you can still add any of those vectors. Quantum mechanics doesn't stop to work just because theory is a little harder and deals with fields.

This post imported from StackExchange Physics at 2015-03-02 08:05 (UTC), posted by SE-user Marek
@kmm: of course, what you write about scattering etc. is true but the reason it works the way it works is because of dynamics (encoded in Lagrangian of the theory) which obeys certain symmetries and consequently conservation laws. But it definitely doesn't concern the linear quantum mechanical structure. As for the Feynman diagrams, that is indeed regrettable. I suppose one can learn to work with them without prior exposure to QFT but one loses a lot because physical and mathematical ideas of QFT are in my opinion one of the most beautiful in all of physics.

This post imported from StackExchange Physics at 2015-03-02 08:05 (UTC), posted by SE-user Marek

This post imported from StackExchange Physics at 2015-03-02 08:05 (UTC), posted by SE-user Kasper Meerts
@DavidZ - the answer certainly proves that you are an articulate speaker, but when you insist ''...when a particle ... is detected in a cloud chamber, it will behave like one of these three particles...'' - that's precisely what the OP was asking - why exactly?

This post imported from StackExchange Physics at 2015-03-02 08:05 (UTC), posted by SE-user New_new_newbie
@New_new_newbie no, I think that's a followup to what the OP was asking. It would be a good thing to ask as a separate question, if you want; I actually don't know the answer. (I'm not sure if it is known at all)

This post imported from StackExchange Physics at 2015-03-02 08:05 (UTC), posted by SE-user David Z
@DavidZ - I also didn't know the answer and that's why I posted that comment. ''I'm not sure if it is known at all'' is so comforting!! :)

This post imported from StackExchange Physics at 2015-03-02 08:05 (UTC), posted by SE-user New_new_newbie
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David gives a complete answer on the mechanism. I will tackle:

There's a difference of 2 MeV between the quark masses and I don't understand how this can be ignored. If I were to somehow manage to make a bound state of an up and an anti-up quark, what would it be? Would it be a variation on the neutral pion or would it somehow transform into the mixture?

You seem to imply that assuming the probabilistic interpretation, if the two photons coming from the $\pi^0$ decay are measured, half of the time their effective mass should be smaller because the $u$ quark has a smaller mass than the $d$ quark.

1) Within a bound state particles are virtual. Virtual means that their mass is not constrained to be the invariant mass they would have as a free particle. Think of the nucleons in the nucleus, the proton and the neutron in deuterium.

The mass of a proton is $\ \ \ 938.272013(23) \ \rm MeV/c^2$,

while that of a neutron is $\ \ \ 939.565378(21)\ \rm MeV/c^2$

While a simple sum gives $\sim 1877 \ \rm MeV/c^2$, the mass of deuteron is $1875.612 859 \ \rm MeV/c^2$

The difference is called binding energy, but the point is that neither the proton nor the neutron can have their invariant mass within the bound nucleus, they have a virtual mass.

2) even though mass and energy are connected through $E=mc^2$, mass is not a conserved quantity in special relativity.

Going back to vector addition in three dimensions: when operating a vector addition, the lengths are not conserved. Two vectors may add up to a zero length vector if they are in opposite direction and of the same magnitude.

Mass is the equivalent measure in special relativity four vectors, it is the "length" of the four vector and follows vector algebra. It is not conserved.

The argument: since $d$ quark has a larger invariant mass than a $u$ quark the combination of down anti-down must have a larger invariant mass than the up anti-up one is wrong. The four vector algebra comes out that both have the mass of the $\pi^0$.

3) As the quarks are always bound in hadrons, there are two definitions of a quark's mass, the current mass, the one entering the QCD equations which are the ones you quote, and the constituent mass. That last is the mass with the accompanying gluons within the hadron and is the same for up and down.

This post imported from StackExchange Physics at 2015-03-02 08:05 (UTC), posted by SE-user anna v
answered Jun 13, 2011 by (1,885 points)
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This is not "just quantum mechanics", it is more than that. Quantum mechanics tells you that states u-ubar and d-dbar are allowed to mix, so that you can consider a u-ubar system as a sum of (u-ubar + d-dbar) plus (u-ubar - d-dbar), but it doesn't tell you that they have to mix.

If these states didn't mix, and they had approximately the same energy, then there would be no paradox, you would be free to think of the pion as a u-ubar, or as d-dbar. If the u and d have a different mass, then u-ubar and d-dbar would be the right way to visualize the "u-pion" and the "d-pion", even when there are reasonably strong interactions.

But for the actual pions, the symmetric part is split in energy from the antisymmetric part by hundreds of MeV, five times the mass of the pions. This splitting is what makes the pions counterintuitive, and to answer the question you need to address the splitting.

Saying that pions are made of quarks is like saying that sound is made of atoms. It's true that if there are no atoms, there is no sound, but that's about it. The QCD vacuum is like a condensed matter system, and it has a quark condensate at the pion scale. The eigenstates of motion of the quark condensate define the low-lying excitations of QCD, and the lightest motion of the condensate is moving it's parts chirally against each other. By this, I mean turning the left-handed u/d and right-handed u/d quarks in the condensate by an opposite phase. This would do nothing to the energy if chiral symmetry were exact, that is, if the quarks were massless. This means that you could "move" the vacuum in the chiral direction without any energy cost, and this gives massless "phonons" (Goldstone bosons) for this process, by moving the vacuum over here a little, and not moving the vacuum over there. These phonons carry the same quantum numbers as the isospin triplet u-dbar/symmetric/d-udbar. These phonons are the pions.

The mass of the pions is not zero, but it is small compared to other strongly interacting particles by a lot. This reflects the fact that the up/down quarks are light compared to the QCD scale. While this picture is only accurate to the extent that the pion mass is small (and the pion is not that light), it is indispensible for understanding pion scattering. Because while the pion mass is visible at scales of 7 to 8 fermis, the interactions with stuff like the proton take place at a scale of 1 fermi, where the pion mass is negligble.

The reason pions are split from their isospin zero partner, the eta-prime, is because the gluons in the vacuum already break part of the chiral symmetry by themselves, through instantons. This splits the two kinds of chiral sound, the pion and the eta, and neither of them is made up of quarks like a molecule is made of atoms. The eta-prime vacuum sound mode is five times stiffer than the pion vacuum sound mode.

When doing quark analysis of light mesons, one must always keep in mind that they only tell you the symmetry numbers, the isospin, strangeness (or SU(3)) quantum numbers. It is only at high energies/high masses that quarks become constituents of the hadrons and mesons in the ordinary sense.

This post imported from StackExchange Physics at 2015-03-02 08:05 (UTC), posted by SE-user Ron Maimon
answered Aug 14, 2011 by (7,580 points)
Isn't the Feynman diagram for the weak decay of say, the pi+ calculated from u+dbar -> W -> muon+neutrino, so doesn't that say the pion is a bound state of q qbar?

This post imported from StackExchange Physics at 2015-03-02 08:05 (UTC), posted by SE-user FrankH
@FrankH: No, it doesn't, but this is a good question. You calculate the weak decay (and the electromagnetic mass splitting--- accurate for pions) from the properties of quark currents. These are bilinear in the quark fields, so it looks like a two-quark state, but it is acting on a condensed vacuum, like a Bogoliubov quasiparticle in a superconductor. Is the quasiparticle made out of two electrons? Sort of, but no, because there's a condensate. If you look in the literature, you will see bogus papers that claim that you can make a pion out of two nonrelativistic constituent quarks.

This post imported from StackExchange Physics at 2015-03-02 08:05 (UTC), posted by SE-user Ron Maimon
Thanks. I always learn a lot from your answers and comments. You say q qbar is bogus, but in physics aren't there often multiple equivalent ways of looking at the same thing that give the same answer - Schrodinger equation / matrix mechanics ( for an ancient example )? Or does the q qbar way give wrong answers somewhere?

This post imported from StackExchange Physics at 2015-03-02 08:05 (UTC), posted by SE-user FrankH
@FrankH: A nonrelativistic q-qbar model gives no explanation for the light mass of the pion, for the gradient interactions (vanishing at low energy) for soft pion emissions, or for the splitting of the eta-prime by hundreds of MeV's from the pions. It is a wrong model in the same way that saying a cooper pair is a bound pair of electrons is a wrong model--- it's seductive and simple, but it isn't the Bogoliubov description.

This post imported from StackExchange Physics at 2015-03-02 08:05 (UTC), posted by SE-user Ron Maimon

Saying that pions are made of quarks is like saying that sound is made of atoms. It's true that if there are no atoms, there is no sound, but that's about it.

Not sure how I should understand this analogy. Isn't every particle viewed as some collective excitation in QFT context? So what makes pions different from genuine bound states? That is, what is the essential difference between the collective excitations labeled "pion" and the collective excitations labeled "bound state"? (see this question)

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