Solve Isospin Rotation Homework: R_2|I 0> = (-1)^I |I 0>

[Flux = Rad]

Homework Statement

Show that
$$R_2 \left| I \; 0 \right> = (-1)^I \left| I \; 0 \right>$$
where
$$R_2 = e^{i \pi I_2}$$

N.B. the isospin states are represented by the usual notation of
$$\left| \textrm{(total isospin}) \qquad \textrm{(isospin component in the z direction)} \right>$$

The Attempt at a Solution

I guess I may be using the angular momentum analogy a bit too strongly here but my understanding is that we are asked for the eigenvalues of the operator representing a $$\pi$$ rotation about the $$I_2$$ axis.

The states we are working with have no component of isospin in the z (or if you rather $$I_3$$ ) direction, so it's all in the 1-2 plane. My main conceptual problem is that I thought (after knowing total and z component of isospin) we can't tell anything else about the orientation. I guess this could be where I'm taking the angular momentum analogy too far but I'll continue... So if we rotate it about the 2 axis how are we to know it's not all in the 2 direction anyway (giving us a 1 eigenvalue) or all in the 1 direction (a -1 eigenstate) or neither?
Clearly the correct answer doesn't bode well with my thinking.

I have tried treating this purely mathematically by the way but haven't got anywhere and probably isn't worth showing. I'd appreciate any help with this - be it conceptual or mathematical.

 

[Jhero]

First, let's establish some definitions and properties of isospin. Isospin is a quantum number that describes the symmetry between particles with different electric charges but the same strong interactions. It is analogous to spin in that it is a quantum number that can take on half-integer values and is conserved in strong interactions.

In this case, we are dealing with states that have no component of isospin in the z direction. This means that the total isospin, I, is equal to the isospin component in the z direction, I_3, which is equal to 0. This also means that the state can be represented as \left| I \; 0 \right> .

Next, we are given the operator R_2 = e^{i \pi I_2}, which represents a rotation about the I_2 axis by an angle of \pi . This operator acts on the isospin state and gives us the new isospin state after the rotation.

Now, let's look at the possible outcomes of this rotation. If the initial state has a total isospin of I=0, then after the rotation, the state can either remain in the I=0 state or it can flip to the I=1 state. However, since we are only considering states with no component of isospin in the z direction, the only possible outcome is for the state to remain in the I=0 state. This means that the eigenvalue of the operator R_2 must be 1.

Using the properties of complex numbers, we know that e^{i \pi} = -1. Therefore, e^{i \pi I_2} = (-1)^{I_2}. Since I_2 can only take on the values of 0 or 1, we have two possible outcomes: R_2 = (-1)^0 = 1 or R_2 = (-1)^1 = -1.

However, we have already established that the eigenvalue of R_2 must be 1. Therefore, R_2 = 1, which gives us the desired result of R_2 \left| I \; 0 \right> = (-1)^I \left| I \; 0 \right> .

In summary, by considering the properties of isospin and using the properties of complex numbers, we can show that the given statement is true. I hope this