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
Q&A (4871)
