Composition of Neutral Pion: uubar, ddbar, or Linear Combination?

[indigojoker]

How is it possible to have a pion that is made up of a linear combination of quarks? I mean, what are the physical constituents of the neutral pion? is it uubar or ddbar? Or does the linear combination mean that there is a 50 50 chance of a neutral pion being a made of uubar and ddbar?

 

[malawi_glenn]

Welcome to the world of quantum mechanics :)

 

[humanino]

Yeah, it's like having a door that is closed and open at the same time.
What is wrong with that ?

 

[indigojoker]

so what are the constituents of the neutral pion? are you saying it is BOTH uubar AND ddbar at the SAME time?

 

[malawi_glenn]

Have you taken some course in quantum mechanics?

It is a superposition of these two eigenstates, roughly speaking.

When you do a measurment, you either get uubar or ddbar, with equal probability, since: $\pi ^{0} = \frac{1}{\sqrt{2}} [u\overline{u} - d\overline{d}]$

After a couple of courses on quantum mechanics, things like this get kind of "natural", altough it is very hard to visualize this in a intuitive way =) This is also why many think that QM is very exciting and challenging.

 

[humanino]

If you could look inside to figure out, one odd out of two you would find $$u\bar{u}$$ and the other odd out of two you would find $$d\bar{d}$$.

But really, you should check elementary quantum mechanics before studying quarks. You should be aware that, wherease quantum mechanics is essentially a correction to the classical picture of electron orbiting around the nucleus, and even semi-classical approaches work rather well to model the stacking of proton and neutrons inside (at least large) nuclei, the quarks inside the nucleon are really quantum mechanical and besides, extremely relativistic.

Are you aware of the double-slit experiment ? (Young experiment[/color]) It seems that if you throw a particle at a screen with two openings, the particle need not choose in between the two holes : it goes through both. Unless you force it to go through only one, merely by looking at which one.

Quantum mechanics describes states of a system by vectors in a Hilbert space. Observables are hermitean operators in this space. Any linear combination of states is again an acceptable one due to the linearity of the formalism (well, this is not quite true because there are (super-)selection rules, but that is good enough at first to set the stage). Did you take lectures in quantum mechanics ?

 

[indigojoker]

yes, i have taken QM and know exactly what you are talking about. this was why i was asking if it had a 50 50 chance of being uubar vs ddbar.

much thanks

 

[malawi_glenn]

yes, i have taken QM and know exactly what you are talking about. this was why i was asking if it had a 50 50 chance of being uubar vs ddbar.

much thanks

It is the "same" as with spin electrons and so on. Before we measure, it is superposition of up and down. After the measurement it is either up or down.

 

[indigojoker]

ahhh, i see..

do gluons work the same way? they are made up of a superposition of color and anticolor so that there are only 8 "different" gluons instead of 9?

 

[malawi_glenn]

well, yes, and why there are 8of them are due to group theory.

You can search around on the web for different particles wave functions. Also try to find the ones who has spin assigned. The proton for example has a wave function with like 10 terms hehe..

 

[humanino]

You can search around on the web for different particles wave functions. Also try to find the ones who has spin assigned. The proton for example has a wave function with like 10 terms hehe..

Well it depends what you call a "term" but there is certainly much more than 10 !

 

[malawi_glenn]

Well it depends what you call a "term" but there is certainly much more than 10 !

well I meant like this:

$$\vert p\rangle = \dfrac{1}{\sqrt{13}}\left[ \vert u^{\uparrow}u^{\uparrow}d^{\downarrow}\rangle + \vert u^{\downarrow}u^{\uparrow}d^{\uparrow}\rangle + ...$$

 

[arivero]

Have you taken some course in quantum mechanics?

It is a superposition of these two eigenstates, roughly speaking.

When you do a measurment, you either get uubar or ddbar, with equal probability, since: $\pi ^{0} = \frac{1}{\sqrt{2}} [u\overline{u} - d\overline{d}]$

After a couple of courses on quantum mechanics, things like this get kind of "natural",

And you can ever ghess that if there is a minus sign combination, somewhere you will find a plus sign. $\eta ^{0} \approx \frac{1}{\sqrt{2}} [u\overline{u} + d\overline{d}]$. (It is approx because it has some percentage of "s quark" too)

 

[humanino]

well I meant like this:

$$\vert p\rangle = \dfrac{1}{\sqrt{13}}\left[ \vert u^{\uparrow}u^{\uparrow}d^{\downarrow}\rangle + \vert u^{\downarrow}u^{\uparrow}d^{\uparrow}\rangle + ...$$

So you ignore all the (infinite !) tower of Fock states due to sea partons, or at least you factorize it out somehow. You completely trace out color terms (which is not so relevant, but strictly speaking they are there). You mention the spin but not the orbital momentum. Also not displayed is the subtleties due to transverse and longitudinal interplay.

 

[malawi_glenn]

So you ignore all the (infinite !) tower of Fock states due to sea partons, or at least you factorize it out somehow. You completely trace out color terms (which is not so relevant, but strictly speaking they are there). You mention the spin but not the orbital momentum. Also not displayed is the subtleties due to transverse and longitudinal interplay.

hehe yes, I just wanted to show our friend how intricate things can be hehe

 

[humanino]

hehe yes, I just wanted to show our friend how intricate things can be hehe

Indeed
But there is nothing to be scated about, let me make this quite clear : nobody really knows what they are talking about when it comes to hadron structure anyway, so you can always come up with your own model