# Difference between particle and state

I have recently started studying particle physics, and I was surprised by the description of hadrons. At first I was told that the $\Delta^0$ baryon was composed by one up and two down quarks, and that was fine. But then I saw this description:

$\Delta^0 = \frac{1}{\sqrt{3}} \big( udd + dud + ddu \big)$

which is a symmetrization of $udd$. This fact made me think that $\Delta^0$ was just a system of three identical bosons, which could individually be in the states $u$ and $d$ (just like a system of two electrons which can individually be in the states spin up or spin down). Does this make sense at all? In case it does, what's the name of that boson that can be in the $u$ state or $d$ state in $\Delta^0$ ?

(Note that by "$u$" I always mean "up quark", never "spin up", and similar with $d$).

I will restrict my argument to the two-electron system. However, logically the same argument follows in case of $\Delta^{0}$ Baryons also

Consider a system of two electrons $$|\psi^{-}\rangle=\frac{1}{\sqrt{2}}(|\uparrow\rangle|\downarrow\rangle-|\downarrow\rangle|\uparrow\rangle)$$

If we want to find out the state of individual electrons that belongs to the $|\psi^{-}\rangle$ then we need to trace over the second particle (partial trace operation, https://en.wikipedia.org/wiki/Partial_trace) which would return $$\rho=\frac{1}{2}(|\uparrow\rangle\langle \uparrow|+|\downarrow\rangle\langle \downarrow|)$$ Which is nothing but maximally mixed state. But, now the state of individual electron does not belong to the two-electron state $|\psi^{-}\rangle$. It's just a equal mixture of up and down spins. Moreover, as electrons belonging to $|\psi^{-}\rangle$ are indistinguishable in nature we cannot specify the state of first or second electron that belongs to $|\psi^{-}\rangle$ individually.

Similar argument would follow for $\Delta^{0}$ Boson.

• Sorry but I think you didn't understand my question. The two electron case was just something I understand and which looks similar to the case I don't understand. Two electrons are two identical particles, and your answer explains their state. What are the three identical particles that form a $\Delta^0$ ? – MBolin Nov 2 '17 at 20:29
• How do you define electron as a particle? In quantum Mechanics, there is no definition of particles. Only quantum state is defined. – WInterfell Nov 2 '17 at 20:34
• At best you can say quantum state of quarks that belongs to $\Delta^{0}$ Baryons is maximally mixed. However, as I mentioned in the answer , it does not make sense to talk about state of individual "particles" in a system of identical particles. – WInterfell Nov 2 '17 at 20:40
• That does answer my question (partially). Up and down quarks are states, just like spin up or down. If an electron can be in spin up or spin down states, what particle can be in up quark or down quark states? – MBolin Nov 2 '17 at 20:40
• What are those identical particles in the system $\Delta^0$ @WInterfell ? – MBolin Nov 2 '17 at 20:44

which is a symmetrization of udd.

This fact made me think that Δ0 was just a system of three identical bosons, which could individually be in the states u and d (just like a system of two electrons which can individually be in the states spin up or spin down). Does this make sense at all?

Sorry it cannot be bosons, otherwise at very high energies, where all three interactions are supposed to be one, see this theory, this statement would hold in the sense that just the symmetry of upness and downness would characterize the valence quarks, a type of spin. At our energies quarks have masses and are separable so there cannot be one generic quark in interactions.

Hadrons are very complicated anyway, because there is also the color quantum number so simplified models are just a help, not written in stone. Have a look at this to see the quark antiquark gluon content of a hadron.

In case it does, what's the name of that boson that can be in the u state or d state in Δ0

It should always be a fermion, the generic GUT particle in different representations.