Values for the spin of a nucleon

[says]

Protons and neutrons are nucleons. The spin of a nucleon is it’s intrinsic angular momentum. Spin has no classic analogue and does not mean the particle is spinning on its axis. It’s an intrinsic property of the nucleon.

Protons and neutrons are fermions and have spin quantum numbers of 1/2.

How can we then say the values of spin for a nucleon are ±1/2?

 

[mfb]

The spin has a direction. If you measure it along an axis you see it is either aligned with it or pointing in the opposite direction.

 

[says]

I don't understand why the wiki for protons: https://en.wikipedia.org/wiki/Proton says it's spin is 1/2. Why not say it's spin is ±1/2

 

[Nugatory]

I don't understand why the wiki for protons: https://en.wikipedia.org/wiki/Proton says it's spin is 1/2. Why not say it's spin is ±1/2

There are two quantum numbers associated with a particle. One, which for a proton or neutron is 1/2, gives us the magnitude of the spin vector. It's denoted by $s$ and is what people including the author of that Wikipedia article are talking about when they say the proton is spin-1/2. The other is $s_z$, the component of the spin vector in a given direction; this is the thing that can be can be either +1/2 or -1/2. When people talk about a particle being spin-up or spin-down, they're talking about $s_z$ not $s$.

 

[says]

The reason I ask is that I was asked a question 'what is the spin of a nucleon and what are it's possible values?'

The 'possible values' bit has confused me.

 

[mfb]

"the spin" probably refers to the particle property - its spin is 1/2.
The possible values are the possible measurement results, +1/2 and -1/2.

In a similar way, you can ask for the speed of an aircraft and its current motion. The speed is 250 m/s, and the current motion could be "250 m/s north" or "-250 m/s north" (aka south).

 

[says]

So if we measured the spin of a nucleon along some specific axis the values it can have are +1/2 or -1/2

 

[mathman]

The sign matters in describing reactions, where there is more than one particle.

 

[says]

If there was only one particle though the sign would still matter

 

[mfb]

If you consider a universe with just a single particle the direction does not matter - you don't even have a reference to use for the measurement.

 

[vanhees71]

Well, the answer to this question is just convention. Usually if you say a particle (and I consider only massive particles here) has a certain spin, what's meant is the representation of the rotation group on zero-momentum states. The representations of the rotation group, or rather its covering group SU(2), are fully determined by the commutation relations of angular-momentum components. Written for the spin operators, they read
$$[\hat{s}_i,\hat{s}_j]=\mathrm{i} \epsilon_{ijk} \hat{s}_k.$$
Here $\epsilon_{ijk}$ is the Levi-Civita symbol (totally antisymmtric in its indices with $\epsilon_{123}=1$), and Einstein summation convention over pair-wise equal indices implied.

From this you can easily see that all $\hat{s}_i$ commute with $\hat{\vec{s}}^2$. Thus each irreducible representation of the angular-momentum algebra is characterized by the eigen-value of $\hat{\vec{s}}^2$. One calls such an operator a Casimir operator. The rotation group has only this one Casimir operator, and that's why you need only one number to characterize each irreducible representation. It turns out that the possible eigenvalues of $\hat{\vec{s}}^2$ are $s(s+1)$ with $s \in \{0,1/2,1,\ldots \}$.

Further each representation is spanned by a complete set of eigenvectors for each $s$, which can be specified by being additionally eigenvectors of one spin component. Usually one chooses $\hat{s}_3$, and it turns out that for given $s$ the possible eigenvalues of $\hat{s}_3$ are $\sigma \in \{-s,-s+1,\ldots,s-1,s \}$.

Now indeed the spin of the nucleons is $s=1/2$, and consequently the 3-component of its spin can take the values $\sigma \in \{-1/2,1/2\}$. The spin space is two-dimensional and provides the "fundamental representation" su(2) of the angular-momentum algebra. By exponentiation you can lift this representation to the fundamental representation of the grou SU(2), which is the covering group of the rotation group SO(3).