# Are this isospin eigenstate same energy states?

Let's consider a system of two nucleons (protons and neutrons). $$\hat T$$ is the total isospin operator and $$\hat T_3$$ it's projection on the axis. The eigenstate are:

singlet state: $$|0,0\rangle$$

triplet states $$|1,1\rangle$$, $$|1,0\rangle$$, $$|1,-1\rangle$$

Since $$[\hat H, \hat T]=0$$ the states I wrote above can be stationary states. My question is, is it possible that all the four states above have the same energy? namely is it possible that the number of energy degeneracies is four?

Your language is ambiguous. There are three, not one, isospin operators, $$\vec T$$. Let's skip the carrets and use capitals for operators and lower case for their eigenvalues. So the eigenvalues of $$\vec T \cdot \vec T$$ are t(t+1). For the singlet, t=0, and for the triplet t=1.
The hamiltonian being isospin invariant means it commutes with all three $$\vec T$$s, but note it may, or may not depend on the above Casimir invariant $$\vec T \cdot \vec T$$. If it does, the triplet and the singlet have different energies; but if it did not, they 'd be degenerate, all right. What world would you design for the second case?
• NB. Advanced users sometimes use the recondite operators $$(-1+\sqrt{1+4\vec T \cdot \vec T})/2$$ with eigenvalues t, but I'd bet that's not what your ambiguous language is trying to evoke.