Is the superposition of mesons (such as pion) a new particle?

[James1238765]

TL;DR Summary

In what sense can the superposition of meson states be called a new particle?

There are 36 hadron composites composed of 2 quarks selectable from the set $[u, d, c, s, t, b, \bar u, \bar d, \bar c, \bar s, \bar t, \bar b]$ satisfying the condition of having total charge = $[-1, 0, 1]$. However, the superposition states of pure hadrons are sometimes also listed as new independent particles, for example [here]:

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

As superposition only means that whenever this particle is detected it will always assume the identity of either $u \bar u$ or $d \bar d$ meson, how is this $\pi^0$ considered a separate new particle?

The superposition of qubits also have a similar form $\frac{|0> + |1>}{\sqrt 2}$, but we don't usually call this superposition state a new qubit, but rather only a superposition of the two basis states |0> and |1>?

 

[PeterDonis]

There are 36 hadron composites composed of 2 quarks selectable from the set $[u, d, c, s, t, b, \bar u, \bar d, \bar c, \bar s, \bar t, \bar b]$ satisfying the condition of having total charge = $[-1, 0, 1]$.

Yes, but which of these are viewed as "composites" and which are not depends on your choice of basis.

However, the superposition states of pure hadrons are sometimes also listed as new independent particles, for example [here]:

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

No, this isn't a "new particle", it's a different choice of basis from the one you were implicitly assuming.

As superposition only means that whenever this particle is detected it will always assume the identity of either $u \bar u$ or $d \bar d$ meson

No, that's not what it means. There aren't two types of $\pi^0$ mesons. There is only one. If your claim here were true there would be two, distinguishable by experiment.

The superposition of qubits also have a similar form $\frac{|0> + |1>}{\sqrt 2}$, but we don't usually call this superposition state a new qubit, but rather only a superposition of the two basis states |0> and |1>?

There are two quarks in a meson, not one. So to draw an analogy with qubits, you would have to look at two-qubit states, not one-qubit states. And with two-qubit states, there are entangled states that cannot be viewed as simple "superpositions" of one-qubit states, or pairs of them. The $\pi^0$ meson is like those entangled two-qubit states.

 

[James1238765]

@PeterDonis thank you. I had not realized that $\pi^0$ can have different experimental properties than the basis $u \bar u$ and $d \bar d$! Could you list one (or a few) ways $\pi^0$ will behave differently than either $u \bar u$ or $d \bar d$'s properties please?

 

[PeterDonis]

I had not realized that $\pi^0$ can have different experimental properties than the basis $u \bar u$ and $d \bar d$! Could you list one (or a few) ways $\pi^0$ will behave differently than either $u \bar u$ or $d \bar d$'s properties please?

Neither $u \bar{u}$ nor $d \bar{d}$ are detected experimentally at all as far as I know, and are not predicted to be. That's why they're not listed in the Wikipedia article that you referenced.

 

[James1238765]

@PeterDonis thank you. It is fascinating that there could be particles that are entangled as one, and have observably different properties than its respective basis states... Are there other examples perhaps like the $\pi^0$ which are superpositions of two basis states, where the two basis states have also been discovered, and how does this "superposition particle" behave differently than its constituent basis particles?

 

[PeterDonis]

It is fascinating that there could be particles that are entangled as one, and have observably different properties than its respective basis states

You seem to be placing way too much weight on "basis states". There is nothing special about them; they are just convenient mathematical objects that are used in models. They are not any "more real" or "preferred" than any other states.

The $\pi^0$ has a mathematical representation in the chosen basis that happens to look like a two-qubit entangled state, but that does not mean it is literally an entangled state of two "other particles". It isn't.

Are there other examples perhaps like the $pi^0$ which are superpositions of two basis states

Sure, just look in the tables in the Wikipedia article you referenced.

how does this "superposition particle" behave differently than its constituent basis particles?

This is a meaningless question, for the same reason I have already given: there are no such things as "constituent basis particles". The fact that a state happens to exist in the math and happens to be used as a convenient basis state does not mean that state is actually observed in experiments, or that other states must be "made of" the basis states. The whole point of the tables in the Wikipedia article is that those are the only states that are actually observed in experiments (or, in a few cases, that our Standard Model of particle physics predicts will be observed in experiments when we get accelerators of high enough energy). Some of those are "basis states" in the mathematical representation that is being used, and some aren't. That's just how it is.

 

[James1238765]

@PeterDonis I am still a little confused about basis states of quarks. We say the electron is a fundamental particle; the electron is not some combination of some other basis... Now let's move to the proton, which consists of $uud$; aren't u and d fundamental "particles"? I guess I am unclear if all basis for the 6 quarks are equally valid, depending how we like to choose for our basis, then is there no physical meaning to call a u, d, c, s, t, and b the fundamental "quark particles"?

 

[PeterDonis]

We say the electron is a fundamental particle; the electron is not some combination of some other basis...

It's more complicated than that, but yes, the state we call "electron" is directly observed in experiments.

Now let's move to the proton, which consists of $uud$; aren't u and d fundamental "particles"?

It depends on what you mean by "fundamental particles". $u$ and $d$ aren't observed in experiments, and can't be because of quark confinement. All we can observe in experiments are particular baryon and meson states that are particular combinations of quarks. Some combinations, like the proton $uud$, "look simpler" than others, like the $\pi^0$. But that's an artifact of the particular basis we choose to write down the states in. We could change basis so that the $\pi^0$ "looked simpler" than the proton if we wanted to; we just don't for various reasons.

I guess I am unclear if all basis for the 6 quarks are equally valid, depending how we like to choose for our basis, then is there no physical meaning to call a u, d, c, s, t, and b "quark particles"?

The term "quark" does not refer to the particular basis states; it refers to the underlying quantum fields. It's not exactly wrong to say that there are 6 quark fields, but it doesn't capture all of the complications either.

The term "particle" is best limited to the particular states we actually observe in experiments.

 

[PeroK]

@PeterDonis I am still a little confused about basis states of quarks. We say the electron is a fundamental particle; the electron is not some combination of some other basis... Now let's move to the proton, which consists of $uud$; aren't u and d fundamental "particles"? I guess I am unclear if all basis for the 6 quarks are equally valid, depending how we like to choose for our basis, then is there no physical meaning to call a u, d, c, s, t, and b the fundamental "quark particles"?

You could try my Insight:

https://www.physicsforums.com/insights/a-beginners-guide-to-baryons/

 

[vanhees71]

@PeterDonis thank you. I had not realized that $\pi^0$ can have different experimental properties than the basis $u \bar u$ and $d \bar d$! Could you list one (or a few) ways $\pi^0$ will behave differently than either $u \bar u$ or $d \bar d$'s properties please?

"Particles" are asymptotic free "one-particle Fock states", be they composite (as all hadrons are when described in the quark model or QCD) or elementary. They must be mass eigen states as well as eigenstates of electric charge and color singlets. The three pseudoscalar pions are also states of fixed isospin 1.