Irreducible representation of SU(2)

irreps of $\mathrm{SU}(2)$

I will just give a brief introduction. For more details, you can see GTM267 Quantum Theory for Mathematicians.

$\tilde J_i$ is the basis of $\mathfrak{su}(2)$, we define

\[\tilde{J}_{+} =\tilde{J}_{1} +\mathbf{i}\tilde{J}_{2} =\mathbf{i} \pi ( F_{1}) -\pi ( F_{2})\\ \tilde{J}_{-} =\tilde{J}_{1} -\mathbf{i}\tilde{J}_{2} =\mathbf{i} \pi ( F_{1}) +\pi ( F_{2})\]

Here $\tilde{J_i}:=\frac{J_i}{\hbar}$ is to make it dimensionless. If you use the natural unit system, than $\tilde{J}_i=J_i$.

You can use the relationship $[\tilde{J}i,\tilde{J}_j]=\mathrm{i}\epsilon{jkl}\tilde{J}_k\tilde{J}_l$ to prove the following equations

\[\begin{align*} [\tilde{J}_{3} ,\tilde{J}_{+}] & =[\tilde{J}_{3} ,\tilde{J}_{1} +\mathbf{i}\tilde{J}_{2}]\\ & =\mathbf{i}\tilde{J}_{2} -\mathbf{i} \cdot \mathbf{i}\tilde{J}_{1}\\ & =\tilde{J}_{1} +\mathbf{i}\tilde{J}_{2} =\tilde{J}_{+}\\ [\tilde{J}_{3} ,\tilde{J}_{-}] & =[\tilde{J}_{3} ,\tilde{J}_{1} -\mathbf{i}\tilde{J}_{2}]\\ & =\mathbf{i}\tilde{J}_{2} +\mathbf{i} \cdot \mathbf{i}\tilde{J}_{1}\\ & =-\tilde{J}_{1} +\mathbf{i}\tilde{J}_{2}\\ & =-\tilde{J}_{-}\\ [\tilde{J}_{+} ,\tilde{J}_{-}] & =[\tilde{J}_{1} +\mathbf{i}\tilde{J}_{2} ,\tilde{J}_{1} -\mathbf{i}\tilde{J}_{2}]\\ & =-\mathbf{i}[\tilde{J}_{1} ,J_{2}] +\mathbf{i}[\tilde{J}_{2} ,\tilde{J}_{1}]\\ & =2\tilde{J}_{3} \end{align*}\]

Thus for $\tilde{J}_+$ , consider $\psi$ is the eigenstate of $\tilde{J}_3$ with eigenvalue $\lambda$

\[\tilde{J}_{3}\tilde{J}_{+} \psi=\tilde{J}_{+}\tilde{J}_{3} \psi+[\tilde{J}_{3} ,\tilde{J}_{+}] \psi=\tilde{J}_{+}( \lambda \psi) +\tilde{J}_{+} \psi=( \lambda +1)\tilde{J}_{+} \psi\]

So $\tilde{J}+$ can increase the eigenvalue by one. The similar conclusion can be derived for $\tilde{J}-$.

So consider the finite-dimensional representation of $\mathrm{su}(2)$ . By the Schur’s Theorem in Linear Algebra Done Right(Theo 6.38) , a complex operator in finite-dimensional space has at least one eigenstate.

You can apply $\tilde{J}+$ on it to get the maximal eigenvalue $l$ and the eigenstate $\psi_l$, there must exist maximal eigenvalue because it is a finite-dimension space. And the maximal means that $\tilde{J}+\ket{\psi_l}=0$

Then you can apply $\tilde{J}_-$ on $\psi_l$ and define

\[\ket{ \psi _{l-k}}=(\tilde{J}_-)^k\ket{\psi_l}\]

By this way the eigenvalue of $\ket{\psi_k}$ is $k$.

So by some induction, you can find

\[\tilde{J}_+\ket{\psi_k} = (l-k)(l+1+k)\ket{\psi_{k+1}}\]

Again, there must be finitely many eigenstates. So there must exists $m$ such that $\psi_m\neq0$ and $\psi_{m-1}=0$. Thus $\tilde{J_+}\ket{\psi_{m-1}}=0$

\[\tilde{J}_+{\ket{\psi_{m-1}}} =(l-m+1)(l+m)\ket{\psi_{m}}\]

This means $l=-m$ or $l-m=-1$. For $m\leq l$ the second solution is wrong. The whole space is expanded by $\psi_{-l},\cdots,\psi_l$ because the space is $\mathrm{su}(2)$ irreducible representation space. There are totally $2l+1$ eigenstates thus the dimension of the space is $2l+1$.

So $2l+1$ must be an integer, and $l\geq m=-l$ means $l$ must be zero or positive. That is why $l=0,\frac 12,1,\ldots$

For an infinite representation space, the GTM267 do not give a discussion.