-2
$\begingroup$

Neutron and the proton have different values of $I_3$, the third component of the Isospin. Yet the strong interaction cannot distinguish between them. Why is this so? I can't understand this.

The ordinary spin couples to a magnetic field but not to the electric field. Therefore, the electric field cannot distinguish between spinless and spinful particles. Is there a similar reason here? Thanks for any help.

$\endgroup$
  • $\begingroup$ Isospin was defined in analogy to spin because experimentally proton and neutron were indistinguishable in nuclear interactions , one entity , similar to an electron with spin up or down, only Clebsch Gordan coefficients distinguish the interaction crossections of the two states . $\endgroup$ – anna v Aug 19 '17 at 3:53
  • $\begingroup$ You can see my notes on isospin here: drive.google.com/open?id=0B18WrUQykMP2NDJZYy16NGpMckE. May be helpful. $\endgroup$ – UKH Aug 19 '17 at 5:17
0
$\begingroup$

Unlike usual spin the isospin is an internal degree of freedom, the particle type that doesn't have any relation to spacetime. The similarity between ordinary spin and isospin originates from the $2\times 2$ unitary matrices group $SU(2)$ being an universal covering of the 3d rotation group $SO(3)$. That means that they have same Lie algebras and their representations are related to each other.

Indeed if you look at the commutation relations of the Pauli matrices $\{\tau_k\}$ they are the same as the commutation relations of the 3d rotation generators $\{J_k\}$. Because of that if you take some 2-dimensional column and $2\times 2$ Hermitian traceless matrix, \begin{equation} \begin{pmatrix}u\\v\end{pmatrix},\quad \hat{T}=T_1\tau_1+T_2\tau_2+T_3\tau_3 \end{equation} and act with some $2\times 2$ unitary matrix on them they will transform as a spinor and a 3d vector $\vec{T}=(T_1,T_2,T_3)$.

The symmetry under 3d rotations correspond to the conserved 3d vector - angular momentum. Similarly because of the aforementioned relation the symmetry under $SU(2)$ tranformations correspond to the conserved 3d vector - isospin. However by construction the isospin has no relation to our space. We could live in the flat world but if we had any $SU(2)$ symmetry we could get conserved isospin 3d vector.

So any time you have two similar particle species, you get $SU(2)$ symmetry and from there you get some conserved isospin-like vector.

In the nuclear physics this appears in the following way. We have proton and neutron, \begin{equation} \begin{pmatrix}p\\n\end{pmatrix},\quad m_p\simeq m_n \end{equation} Because strong interaction doesn't differentiate proton and neutron we can consider act on that column with arbitrary $2\times 2$ unitary matrix and all strong processes will remain almost the same. Thus we have $SU(2)$ symmetry and from that we have a isospin vector conserved by strong interaction. However weak interaction differentiates them and thus doesn't conserve isospin.

Where this structure comes from? The deepest explanation at present comes from the Standard model. There we have three gauge fields. One of them is $B_mu$ with $U(1)$ gauge group that you may think of as just electromagnetic field (indeed it is one of the precursors of the electromagnetic field). Another two are non-abelian Yang-Mills fields that you may think of as generalizations of the electromagnetism - instead of simple $A_mu$ you have some traceless $n\times n$ matrix. Those two are $\hat{W}_\mu=W^{k}_\mu\tau_k$ with $SU(2)$ gauge group and $g_\mu$ with $SU(3)$ gauge group. All the fermions are organized according to those gauge groups. So with lightest generation of quarks we get, \begin{equation} \begin{pmatrix} (u_r,u_g,u_b)_{SU(3)}\\(d_r,d_g,d_b)_{SU(3)} \end{pmatrix}_{SU(2)} \end{equation}

The $SU(2)$ symmetry acts on the columns whereas $SU(3)$ acts on the rows. Each one of them doesn't care about the structure associated with the other. The initial $SU(2)$ symmetry is associated with the so-called weak isospin $\vec{T}_W$ with $u$ quarks having $T_{W,3}=+\frac{1}{2}$ and $d$ quarks $T_{W,3}=-\frac{1}{2}$.

$SU(3)$ gauge field corresponds to the strong interaction. Its associated structure will not be important for us now because of the confinement. What is important is that it doesn't touch the structure associated with $SU(2)$.

At low energies when the Higgs field becomes nonzero the remaining $U(1)\times SU(2)$ symmetry breaks down to $U(1)$ giving rise to the electromagnetic and weak interactions. The quarks (and many other fields) get different masses. But what's important the strong interaction doesn't care about all that and interacts with all quarks in the same way, so strong interaction doesn't differentiate e.g. $uud$ and $udd$.

However $u$ and $d$ quarks for some unknown reason are very light to the point when they often can safely assumed to be massless. The weak interaction is, well, very weak compared to the strong so most of the mass of the hadrons comes from the strong interaction that doesn't differentiate $u$ and $d$. Thanks to that $p=uud$ and $n=udd$ are very close in mass. That's how we get low energy $SU(2)$ symmetry of $(p,n)$ and fundamental weak isospin structure gives rise to the isospin of the nuclear physics.

Now if you ask, why is it so that $SU(2)$ and $SU(3)$ are independent of each other? Why $u$ and $d$ are so light? No one knows that right now. There are various hypotheses about beyond Standard model physics that explain that but unfortunately we haven't got yet evidence for any of them.

$\endgroup$
  • $\begingroup$ "Unlike usual spin the isospin is an internal degree of freedom, the particle type that doesn't have any relation to spacetime" But the spin degree of freedom is also not related to spacetime. It's also an intrinsic property. $\endgroup$ – mithusengupta123 Oct 18 '18 at 7:38
  • $\begingroup$ @mithusengupta123 The spin originates from nontrivial transformation of the quantum field under Lorentz transformations. One of the consequences is that the spin vector rotates under rotations. The isospin vector - truly internal degree of freedom, doesn't. E.g. helicity (projection of spin on momentum) is invariant under rotations whereas "isohelicity" constructed as a projection of isospin on momentum is not as you can't turn all neutrons into protons by standing on your head. $\endgroup$ – OON Oct 18 '18 at 10:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.