[proofplan] Fix an eigenvalue $E$ of $H$, and define the corresponding eigenspace by $\mathcal E_E := \{\psi \in \mathcal D(H) : H\psi = E\psi\}$. The domain-preservation and commutation hypotheses imply that applying a rotation to a vector in $\mathcal E_E$ produces another vector in $\mathcal E_E$, so $\mathcal E_E$ is invariant under the $SU(2)$ action. The dimension estimate then uses the assumed irreducible spin $j$ subrepresentation: such a representation has dimension $2j+1$, and it sits inside $\mathcal E_E$. The self-adjointness of $H$, unitarity of $U$, and strong continuity of $U$ are structural hypotheses for the Hamiltonian symmetry setting; the proof of this conditional degeneracy bound does not require the spectral theorem. [/proofplan] [step:Show that the $E$-eigenspace is invariant under the $SU(2)$ action] Let $g \in SU(2)$ and let $\psi \in \mathcal E_E$. By the definition of $\mathcal E_E$, we have $\psi \in \mathcal D(H)$ and $H\psi = E\psi$. The domain-preservation hypothesis gives $U(g)\psi \in \mathcal D(H)$, so $H U(g)\psi$ is defined. Using the commutation relation on $\psi$, we obtain $H U(g)\psi = U(g)H\psi$. Since $H\psi = E\psi$ and $U(g)$ is linear, $U(g)H\psi = U(g)(E\psi) = E U(g)\psi$. Therefore $H U(g)\psi = E U(g)\psi$. Together with $U(g)\psi \in \mathcal D(H)$, this proves $U(g)\psi \in \mathcal E_E$. Since $g \in SU(2)$ and $\psi \in \mathcal E_E$ were arbitrary, $\mathcal E_E$ is invariant under the representation $U$. [/step] [step:Use the spin $j$ subrepresentation to produce $2j+1$ independent eigenvectors] By hypothesis, there is a subspace $V \subset \mathcal E_E$ such that the restricted representation $U_V: SU(2) \to \mathcal L(V)$, defined by $U_V(g) := U(g)|_V$, is an irreducible spin $j$ representation. In the standard spin convention, this means that $j$ is a nonnegative half-integer and that $V$ is the irreducible $SU(2)$ representation of highest weight $2j$. By the standard classification of irreducible representations of $SU(2)$, the [vector space](/page/Vector%20Space) $V$ has dimension $\dim V = 2j+1$. Equivalently, $V$ admits a basis of $2j+1$ linearly independent vectors, often denoted by $v_m \in V$, where $m \in \{-j, -j+1, \dots, j-1, j\}$. Since $V \subset \mathcal E_E$, every vector $v_m$ lies in $\mathcal D(H)$ and satisfies $Hv_m = Ev_m$. Thus these $2j+1$ linearly independent vectors are all eigenvectors of $H$ with eigenvalue $E$. [guided] The point of the representation-theoretic hypothesis is that the eigenspace does not merely contain one eigenvector: it contains an entire irreducible spin $j$ representation space. Let $U_V: SU(2) \to \mathcal L(V)$ denote the restricted representation on the invariant subspace $V$, defined by $U_V(g) := U(g)|_V$. This is well-defined because the hypothesis states that $U(g)V \subset V$ for every $g \in SU(2)$. The phrase “irreducible spin $j$ representation” carries a specific convention: $j$ is a nonnegative half-integer, and the representation is the irreducible $SU(2)$ representation of highest weight $2j$. Equivalently, the standard classification of irreducible representations of $SU(2)$ says that such a representation has dimension $2j+1$. Therefore $\dim V = 2j+1$. Choose a basis of $V$, indexed in the standard angular-momentum notation by $v_m \in V$, where $m \in \{-j, -j+1, \dots, j-1, j\}$. This index set has exactly $2j+1$ elements, so the vectors $v_m$ form a linearly independent family of size $2j+1$. Now we use the containment $V \subset \mathcal E_E$. By definition of $\mathcal E_E$, every vector in $V$ lies in the operator domain $\mathcal D(H)$ and satisfies the eigenvalue equation for $E$. Hence, for every allowed value of $m$, $Hv_m = Ev_m$. Thus the spin multiplet supplies $2j+1$ linearly independent eigenvectors of $H$ with the same energy eigenvalue $E$. [/guided] [/step] [step:Compare the spin subspace with the whole eigenspace] Since $V \subset \mathcal E_E$, monotonicity of vector-space dimension gives $\dim \mathcal E_E \ge \dim V$. From the previous step, $\dim V = 2j+1$, and therefore $\dim \mathcal E_E \ge 2j+1$. By definition, the multiplicity of the eigenvalue $E$ is the dimension of its eigenspace $\mathcal E_E$. Hence $E$ has multiplicity at least $2j+1$. [/step]