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1. In this answer we elaborate on Ballentine's method of finding a canonical transformation (CT) $$(q,p)\quad \longrightarrow\quad (Q,P),\tag{1}$$
   cf. NessunDorma's answer.

2. Recall first of all that instead of position and momentum operators, we may equivalently use annihilation & creation operators $$ \begin{align} a^j~=~& \sqrt{\frac{m\omega}{2\hbar}}q^j+\frac{ip_j}{\sqrt{2m\omega\hbar}} , \cr a^{\dagger}_j~=~& \sqrt{\frac{m\omega}{2\hbar}}q^j-\frac{ip_j}{\sqrt{2m\omega\hbar}}, \cr A^j~=~& \sqrt{\frac{m\omega}{2\hbar}}Q^j+\frac{iP_j}{\sqrt{2m\omega\hbar}}, \cr A^{\dagger}_j~=~& \sqrt{\frac{m\omega}{2\hbar}}Q^j-\frac{iP_j}{\sqrt{2m\omega\hbar}} ,\cr j~\in~&\{1,2,3\}. \end{align}\tag{2}$$ Here $m\omega>0$ is some appropriate dimensionful constant whose numerical value is not important for what follows. The orbital angular momentum (OAM) operators are $$\begin{align} L_j~=~&\sum_{k,\ell=1}^3\epsilon_{jk\ell}q^kp_{\ell}\cr ~\stackrel{(2)}{=}~& -i\hbar\sum_{k,\ell=1}^3\epsilon_{jk\ell}a_k^{\dagger}a^{\ell} ,\cr j~\in~&\{1,2,3\}.\end{align} \tag{3}$$ The number operators are $$\begin{align} n_j~:=~&a_j^{\dagger}a^j,\quad(\longleftarrow\text{no $j$-sum})\cr N_j~:=~&A_j^{\dagger}A^j,\quad(\longleftarrow\text{no $j$-sum})\cr j~\in~&\{1,2,3\}. \end{align}\tag{4}$$

3. Consider a 1-parameter linear CT connected to the identity$^1$ $$\begin{align} Q^j~=~& q^j\cos\theta + \frac{1}{m\omega}\sum_{k=1}^2|\epsilon^{jk}| p_k\sin\theta,\cr P_j~=~& p_j\cos\theta - m\omega\sum_{k=1}^2|\epsilon_{jk}| q^k\sin\theta, \cr A^j~=~& a^j\cos\theta - i\sum_{k=1}^2|\epsilon^{jk}| a^k\sin\theta,\cr A_j^{\dagger}~=~& a_j^{\dagger}\cos\theta + i\sum_{k=1}^2|\epsilon^{jk}| a_k^{\dagger}\sin\theta,\cr j~\in~&\{1,2\}, \cr Q^3~=~& q^3,\cr P_3~=~& p_3,\cr A^3~=~& a^3, \end{align} \tag{5}$$

4. Ballentine's observation becomes $$\begin{align} N_1-N_2~\stackrel{(3)+(4)+(5)}{=}&~(n_1-n_2)\cos 2\theta\cr &+ \frac{L_3}{\hbar} \sin 2\theta.\end{align} \tag{6}$$ Of course we are mainly interested in the angle $\theta=\frac{\pi}{4}$, where the OAM $$\frac{L_3}{\hbar}~\stackrel{(6)}{=}~N_1-N_2\tag{7}$$ becomes the difference of two number operators, cf. OP's question.

5. Let $V_{\ell}$ be a finite-dimensional spin-$\ell$ irrep of the OAM Lie algebra $${\rm span}_{\mathbb{R}}(L_1,L_2,L_3)~\cong~ so(3).\tag{8}$$ We would like to prove that $\ell$ cannot be half-spin, cf. OP's question.

6. Note that $$\sum_{j=1}^3N_j~\stackrel{(4)+(5)}{=}~\sum_{j=1}^3n_j\tag{9}$$ is a Casimir operator in the sense that it commutes with $L_1$, $L_2$ & $L_3$. Therefore there exists a common set of eigenbasis $|\ell m\rangle$ such that $$\begin{align}L_3|\ell m\rangle~=~&\hbar m|\ell m\rangle,\cr \vec{L}^2|\ell m\rangle~=~&\hbar^2\ell(\ell+1)|\ell m\rangle, \cr \ell, m~\in~&\frac{1}{2}\mathbb{Z}, \cr \sum_{j=1}^3N_j|\ell m\rangle~=~&\nu |\ell m\rangle, \cr \nu~\geq~&0.\end{align}\tag{10}$$

7. Consider one such state $|\ell m\rangle$. We can act on $|\ell m\rangle$ (finitely many times) with annihilation operators $A^1$, $A^2$ & $A^3$, to reach a Fock vacuum state $|\Omega\rangle\neq 0$ with$^2$ $$A^j|\Omega\rangle~=~0,\qquad j~\in~\{1,2,3\}.\tag{11}$$ (Else there will be negative norm states by a standard argument, cf. e.g. section 6.1 in Ref. 1 or my Phys.SE answer here.)

8. In turn, this implies that $|\ell m\rangle$ is a common eigenvector for $N_1$, $N_2$ & $N_3$, with integer eigenvalues. In particular, $m$ must be an integer, cf. eqs. (7) & (10). $\Box$

9. It's interesting to note that while there are of course no genuine (as opposed to projective) half-spin irreps of the 3D rotation group $SO(3)$, the corresponding Lie algebra $so(3)$ has in principle half-spin irreps, i.e. there are no topological obstructions at the Lie algebra level. Nevertheless, as we saw above, for the OAM Lie algebra, the underlying Fock space representation of the Heisenberg algebra says otherwise! In that sense this proof is very different from a topological proof.

References:

1. L.E. Ballentine, QM modern developments, 1998; p. 170.

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$^1$ Our notation is slightly different from Ref. 1. For starters, we prefer to use small letters before the CT (5) and capital letters after. A type 2 generating function for the CT (5) is $$\begin{align} F_2(q,P)~=~&q^3P_3+\frac{\sum_{j=1}^2q^jP_j}{\cos\theta}\cr ~+~&\frac{1}{2}\sum_{j,k=1}^2\left(m\omega|\epsilon_{jk}| q^jq^k+\frac{1}{m\omega}|\epsilon^{jk}| P_jP_k\right) \tan\theta.\end{align}\tag{12}$$

$^2$ The Fock vacuum $|\Omega\rangle$ happens to be invariant under the CT (5): $$\begin{align} a^j|\Omega\rangle ~=~&0 \cr ~\Updownarrow~& \cr A^j|\Omega\rangle ~=~&0,\cr j~\in~&\{1,2,3\}.\end{align} \tag{13}$$ Note that the Fock vacuum $|\Omega\rangle$ depends on the constant $m\omega$.

1. In this answer we elaborate on Ballentine's method of finding a canonical transformation (CT) $$(q,p)\quad \longrightarrow\quad (Q,P),\tag{1}$$
   cf. NessunDorma's answer. This leads to a proof that (a finite-dimensional irrep of) the orbital angular momentum (OAM) operator has integer spin quantum number, cf. OP's title question.

2. Recall first of all that instead of position and momentum operators, we may equivalently use annihilation & creation operators $$ \begin{align} a^j~=~& \sqrt{\frac{m\omega}{2\hbar}}q^j+\frac{ip_j}{\sqrt{2m\omega\hbar}} , \cr a^{\dagger}_j~=~& \sqrt{\frac{m\omega}{2\hbar}}q^j-\frac{ip_j}{\sqrt{2m\omega\hbar}}, \cr A^j~=~& \sqrt{\frac{m\omega}{2\hbar}}Q^j+\frac{iP_j}{\sqrt{2m\omega\hbar}}, \cr A^{\dagger}_j~=~& \sqrt{\frac{m\omega}{2\hbar}}Q^j-\frac{iP_j}{\sqrt{2m\omega\hbar}} ,\cr j~\in~&\{1,2,3\}. \end{align}\tag{2}$$ Here $m\omega>0$ is some appropriate dimensionful constant whose numerical value is not important for what follows. The OAM operators are $$\begin{align} L_j~=~&\sum_{k,\ell=1}^3\epsilon_{jk\ell}q^kp_{\ell}\cr ~\stackrel{(2)}{=}~& -i\hbar\sum_{k,\ell=1}^3\epsilon_{jk\ell}a_k^{\dagger}a^{\ell} ,\cr j~\in~&\{1,2,3\}.\end{align} \tag{3}$$ The number operators are $$\begin{align} n_j~:=~&a_j^{\dagger}a^j,\quad(\longleftarrow\text{no $j$-sum})\cr N_j~:=~&A_j^{\dagger}A^j,\quad(\longleftarrow\text{no $j$-sum})\cr j~\in~&\{1,2,3\}. \end{align}\tag{4}$$

3. Consider a 1-parameter linear CT connected to the identity$^1$ $$\begin{align} Q^j~=~& q^j\cos\theta + \frac{1}{m\omega}\sum_{k=1}^2|\epsilon^{jk}| p_k\sin\theta,\cr P_j~=~& p_j\cos\theta - m\omega\sum_{k=1}^2|\epsilon_{jk}| q^k\sin\theta, \cr A^j~=~& a^j\cos\theta - i\sum_{k=1}^2|\epsilon^{jk}| a^k\sin\theta,\cr A_j^{\dagger}~=~& a_j^{\dagger}\cos\theta + i\sum_{k=1}^2|\epsilon^{jk}| a_k^{\dagger}\sin\theta,\cr j~\in~&\{1,2\}, \cr Q^3~=~& q^3,\cr P_3~=~& p_3,\cr A^3~=~& a^3, \end{align} \tag{5}$$

4. Ballentine's observation becomes $$\begin{align} N_1-N_2~\stackrel{(3)+(4)+(5)}{=}&~(n_1-n_2)\cos 2\theta\cr &+ \frac{L_3}{\hbar} \sin 2\theta.\end{align} \tag{6}$$ Of course we are mainly interested in the angle $\theta=\frac{\pi}{4}$, where the OAM $$\frac{L_3}{\hbar}~\stackrel{(6)}{=}~N_1-N_2\tag{7}$$ becomes the difference of two number operators, cf. OP's question.

5. Let $V_{\ell}$ be a finite-dimensional spin-$\ell$ irrep of the OAM Lie algebra $${\rm span}_{\mathbb{R}}(L_1,L_2,L_3)~\cong~ so(3).\tag{8}$$ We would like to prove that $\ell$ cannot be half-spin, cf. OP's question.

6. Note that $$\sum_{j=1}^3N_j~\stackrel{(4)+(5)}{=}~\sum_{j=1}^3n_j\tag{9}$$ is a Casimir operator in the sense that it commutes with $L_1$, $L_2$ & $L_3$. Therefore there exists a common set of eigenbasis $|\ell m\rangle$ such that $$\begin{align}L_3|\ell m\rangle~=~&\hbar m|\ell m\rangle,\cr \vec{L}^2|\ell m\rangle~=~&\hbar^2\ell(\ell+1)|\ell m\rangle, \cr \ell, m~\in~&\frac{1}{2}\mathbb{Z}, \cr \sum_{j=1}^3N_j|\ell m\rangle~=~&\nu |\ell m\rangle, \cr \nu~\geq~&0.\end{align}\tag{10}$$

7. Consider one such state $|\ell m\rangle$. We can act on $|\ell m\rangle$ (finitely many times) with annihilation operators $A^1$, $A^2$ & $A^3$, to reach a Fock vacuum state $|\Omega\rangle\neq 0$ with$^2$ $$A^j|\Omega\rangle~=~0,\qquad j~\in~\{1,2,3\}.\tag{11}$$ (Else there will be negative norm states by a standard argument, cf. e.g. section 6.1 in Ref. 1 or my Phys.SE answer here.)

8. In turn, this implies that $|\ell m\rangle$ is a common eigenvector for $N_1$, $N_2$ & $N_3$, with integer eigenvalues. In particular, $m$ must be an integer, cf. eqs. (7) & (10). $\Box$

9. It's interesting to note that while there are of course no genuine (as opposed to projective) half-spin irreps of the 3D rotation group $SO(3)$, the corresponding Lie algebra $so(3)$ has in principle half-spin irreps, i.e. there are no topological obstructions at the Lie algebra level. Nevertheless, as we saw above, for the OAM Lie algebra, the underlying Fock space representation of the Heisenberg algebra says otherwise! In that sense this proof is very different from a topological proof.

References:

1. L.E. Ballentine, QM modern developments, 1998; p. 170.

--

$^1$ Our notation is slightly different from Ref. 1. For starters, we prefer to use small letters before the CT (5) and capital letters after. A type 2 generating function for the CT (5) is $$\begin{align} F_2(q,P)~=~&q^3P_3+\frac{\sum_{j=1}^2q^jP_j}{\cos\theta}\cr ~+~&\frac{1}{2}\sum_{j,k=1}^2\left(m\omega|\epsilon_{jk}| q^jq^k+\frac{1}{m\omega}|\epsilon^{jk}| P_jP_k\right) \tan\theta.\end{align}\tag{12}$$

$^2$ The Fock vacuum $|\Omega\rangle$ happens to be invariant under the CT (5): $$\begin{align} a^j|\Omega\rangle ~=~&0 \cr ~\Updownarrow~& \cr A^j|\Omega\rangle ~=~&0,\cr j~\in~&\{1,2,3\}.\end{align} \tag{13}$$ Note that the Fock vacuum $|\Omega\rangle$ depends on the constant $m\omega$.

1. In this answer we elaborate on Ballentine's method of finding a canonical transformation (CT) $$(q,p)\quad \longrightarrow\quad (Q,P),\tag{1}$$
   cf. NessunDorma's answer.

2. Recall first of all that instead of position and momentum operators, we may equivalently use annihilation & creation operators $$ \begin{align} a^j~=~& \sqrt{\frac{m\omega}{2\hbar}}q^j+\frac{ip_j}{\sqrt{2m\omega\hbar}} ,~~\quad a^{\dagger}_j~=~ \sqrt{\frac{m\omega}{2\hbar}}q^j-\frac{ip_j}{\sqrt{2m\omega\hbar}}, \cr A^j~=~& \sqrt{\frac{m\omega}{2\hbar}}Q^j+\frac{iP_j}{\sqrt{2m\omega\hbar}}, \quad A^{\dagger}_j~=~ \sqrt{\frac{m\omega}{2\hbar}}Q^j-\frac{iP_j}{\sqrt{2m\omega\hbar}} ,\quad j~\in~\{1,2,3\}. \end{align}\tag{2}$$ Here $m\omega>0$ is some appropriate dimensionful constant whose numerical value is not important for what follows. The orbital angular momentum (OAM) operators are $$ L_j~=~\sum_{k,\ell=1}^3\epsilon_{jk\ell}q^kp_{\ell} ~\stackrel{(2)}{=}~ -i\hbar\sum_{k,\ell=1}^3\epsilon_{jk\ell}a_k^{\dagger}a^{\ell} , \qquad j~\in~\{1,2,3\}.\tag{3}$$ The number operators are $$ n_j~:=~a_j^{\dagger}a^j,\quad N_j~:=~A_j^{\dagger}A^j,\quad(\longleftarrow\text{no $j$-sum})\quad j~\in~\{1,2,3\}. \tag{4}$$

3. Consider a 1-parameter linear CT connected to the identity$^1$ $$\begin{align} Q^j~=~& q^j\cos\theta + \frac{1}{m\omega}\sum_{k=1}^2|\epsilon^{jk}| p_k\sin\theta,\qquad j~\in~\{1,2\}, \cr P_j~=~& p_j\cos\theta - m\omega\sum_{k=1}^2|\epsilon_{jk}| q^k\sin\theta, \cr A^j~=~& a^j\cos\theta - i\sum_{k=1}^2|\epsilon^{jk}| a^k\sin\theta,\cr A_j^{\dagger}~=~& a_j^{\dagger}\cos\theta + i\sum_{k=1}^2|\epsilon^{jk}| a_k^{\dagger}\sin\theta,\cr Q^3~=~& q^3,\cr P_3~=~& p_3,\cr A^3~=~& a^3, \end{align} \tag{5}$$

4. Ballentine's observation becomes $$ N_1-N_2~\stackrel{(3)+(4)+(5)}{=}~(n_1-n_2)\cos 2\theta+ \frac{L_3}{\hbar} \sin 2\theta. \tag{6}$$ Of course we are mainly interested in the angle $\theta=\frac{\pi}{4}$, where the OAM $$\frac{L_3}{\hbar}~\stackrel{(6)}{=}~N_1-N_2\tag{7}$$ becomes the difference of two number operators, cf. OP's question.

5. Let $V_{\ell}$ be a finite-dimensional spin-$\ell$ irrep of the OAM Lie algebra $${\rm span}_{\mathbb{R}}(L_1,L_2,L_3)~\cong~ so(3).\tag{8}$$ We would like to prove that $\ell$ cannot be half-spin, cf. OP's question.

6. Note that $$\sum_{j=1}^3N_j~\stackrel{(4)+(5)}{=}~\sum_{j=1}^3n_j\tag{9}$$ is a Casimir operator in the sense that it commutes with $L_1$, $L_2$ & $L_3$. Therefore there exists a common set of eigenbasis $|\ell m\rangle$ such that $$L_3|\ell m\rangle~=~\hbar m|\ell m\rangle,\quad \vec{L}^2|\ell m\rangle~=~\hbar^2\ell(\ell+1)|\ell m\rangle, \quad \ell, m~\in~\frac{1}{2}\mathbb{Z}, $$ $$ \sum_{j=1}^3N_j|\ell m\rangle~=~\nu |\ell m\rangle, \quad \nu~\geq~0.\tag{10}$$

7. Consider one such state $|\ell m\rangle$. We can act on $|\ell m\rangle$ (finitely many times) with annihilation operators $A^1$, $A^2$ & $A^3$, to reach a Fock vacuum state $|\Omega\rangle\neq 0$ with$^2$ $$A^j|\Omega\rangle~=~0,\qquad j~\in~\{1,2,3\}.\tag{11}$$ (Else there will be negative norm states by a standard argument, cf. e.g. section 6.1 in Ref. 1 or my Phys.SE answer here.)

8. In turn, this implies that $|\ell m\rangle$ is a common eigenvector for $N_1$, $N_2$ & $N_3$, with integer eigenvalues. In particular, $m$ must be an integer, cf. eqs. (7) & (10). $\Box$

9. It's interesting to note that while there are of course no genuine (as opposed to projective) half-spin irreps of the 3D rotation group $SO(3)$, the corresponding Lie algebra $so(3)$ has in principle half-spin irreps, i.e. there are no topological obstructions at the Lie algebra level. Nevertheless, as we saw above, for the OAM Lie algebra, the underlying Fock space representation of the Heisenberg algebra says otherwise! In that sense this proof is very different from a topological proof.

References:

1. L.E. Ballentine, QM modern developments, 1998; p. 170.

--

$^1$ Our notation is slightly different from Ref. 1. For starters, we prefer to use small letters before the CT (5) and capital letters after. A type 2 generating function for the CT (5) is $$F_2(q,P)~=~q^3P_3+\frac{\sum_{j=1}^2q^jP_j}{\cos\theta} +\frac{1}{2}\sum_{j,k=1}^2\left(m\omega|\epsilon_{jk}| q^jq^k+\frac{1}{m\omega}|\epsilon^{jk}| P_jP_k\right) \tan\theta.\tag{12}$$

$^2$ The Fock vacuum $|\Omega\rangle$ happens to be invariant under the CT (5): $$ a^j|\Omega\rangle ~=~0 \qquad \Leftrightarrow \qquad A^j|\Omega\rangle ~=~0,\qquad j~\in~\{1,2,3\}. \tag{13}$$ Note that the Fock vacuum $|\Omega\rangle$ depends on the constant $m\omega$.

1. In this answer we elaborate on Ballentine's method of finding a canonical transformation (CT) $$(q,p)\quad \longrightarrow\quad (Q,P),\tag{1}$$
   cf. NessunDorma's answer.

2. Recall first of all that instead of position and momentum operators, we may equivalently use annihilation & creation operators $$ \begin{align} a^j~=~& \sqrt{\frac{m\omega}{2\hbar}}q^j+\frac{ip_j}{\sqrt{2m\omega\hbar}} , \cr a^{\dagger}_j~=~& \sqrt{\frac{m\omega}{2\hbar}}q^j-\frac{ip_j}{\sqrt{2m\omega\hbar}}, \cr A^j~=~& \sqrt{\frac{m\omega}{2\hbar}}Q^j+\frac{iP_j}{\sqrt{2m\omega\hbar}}, \cr A^{\dagger}_j~=~& \sqrt{\frac{m\omega}{2\hbar}}Q^j-\frac{iP_j}{\sqrt{2m\omega\hbar}} ,\cr j~\in~&\{1,2,3\}. \end{align}\tag{2}$$ Here $m\omega>0$ is some appropriate dimensionful constant whose numerical value is not important for what follows. The orbital angular momentum (OAM) operators are $$\begin{align} L_j~=~&\sum_{k,\ell=1}^3\epsilon_{jk\ell}q^kp_{\ell}\cr ~\stackrel{(2)}{=}~& -i\hbar\sum_{k,\ell=1}^3\epsilon_{jk\ell}a_k^{\dagger}a^{\ell} ,\cr j~\in~&\{1,2,3\}.\end{align} \tag{3}$$ The number operators are $$\begin{align} n_j~:=~&a_j^{\dagger}a^j,\quad(\longleftarrow\text{no $j$-sum})\cr N_j~:=~&A_j^{\dagger}A^j,\quad(\longleftarrow\text{no $j$-sum})\cr j~\in~&\{1,2,3\}. \end{align}\tag{4}$$

3. Consider a 1-parameter linear CT connected to the identity$^1$ $$\begin{align} Q^j~=~& q^j\cos\theta + \frac{1}{m\omega}\sum_{k=1}^2|\epsilon^{jk}| p_k\sin\theta,\cr P_j~=~& p_j\cos\theta - m\omega\sum_{k=1}^2|\epsilon_{jk}| q^k\sin\theta, \cr A^j~=~& a^j\cos\theta - i\sum_{k=1}^2|\epsilon^{jk}| a^k\sin\theta,\cr A_j^{\dagger}~=~& a_j^{\dagger}\cos\theta + i\sum_{k=1}^2|\epsilon^{jk}| a_k^{\dagger}\sin\theta,\cr j~\in~&\{1,2\}, \cr Q^3~=~& q^3,\cr P_3~=~& p_3,\cr A^3~=~& a^3, \end{align} \tag{5}$$

4. Ballentine's observation becomes $$\begin{align} N_1-N_2~\stackrel{(3)+(4)+(5)}{=}&~(n_1-n_2)\cos 2\theta\cr &+ \frac{L_3}{\hbar} \sin 2\theta.\end{align} \tag{6}$$ Of course we are mainly interested in the angle $\theta=\frac{\pi}{4}$, where the OAM $$\frac{L_3}{\hbar}~\stackrel{(6)}{=}~N_1-N_2\tag{7}$$ becomes the difference of two number operators, cf. OP's question.

5. Let $V_{\ell}$ be a finite-dimensional spin-$\ell$ irrep of the OAM Lie algebra $${\rm span}_{\mathbb{R}}(L_1,L_2,L_3)~\cong~ so(3).\tag{8}$$ We would like to prove that $\ell$ cannot be half-spin, cf. OP's question.

6. Note that $$\sum_{j=1}^3N_j~\stackrel{(4)+(5)}{=}~\sum_{j=1}^3n_j\tag{9}$$ is a Casimir operator in the sense that it commutes with $L_1$, $L_2$ & $L_3$. Therefore there exists a common set of eigenbasis $|\ell m\rangle$ such that $$\begin{align}L_3|\ell m\rangle~=~&\hbar m|\ell m\rangle,\cr \vec{L}^2|\ell m\rangle~=~&\hbar^2\ell(\ell+1)|\ell m\rangle, \cr \ell, m~\in~&\frac{1}{2}\mathbb{Z}, \cr \sum_{j=1}^3N_j|\ell m\rangle~=~&\nu |\ell m\rangle, \cr \nu~\geq~&0.\end{align}\tag{10}$$

7. Consider one such state $|\ell m\rangle$. We can act on $|\ell m\rangle$ (finitely many times) with annihilation operators $A^1$, $A^2$ & $A^3$, to reach a Fock vacuum state $|\Omega\rangle\neq 0$ with$^2$ $$A^j|\Omega\rangle~=~0,\qquad j~\in~\{1,2,3\}.\tag{11}$$ (Else there will be negative norm states by a standard argument, cf. e.g. section 6.1 in Ref. 1 or my Phys.SE answer here.)

8. In turn, this implies that $|\ell m\rangle$ is a common eigenvector for $N_1$, $N_2$ & $N_3$, with integer eigenvalues. In particular, $m$ must be an integer, cf. eqs. (7) & (10). $\Box$

9. It's interesting to note that while there are of course no genuine (as opposed to projective) half-spin irreps of the 3D rotation group $SO(3)$, the corresponding Lie algebra $so(3)$ has in principle half-spin irreps, i.e. there are no topological obstructions at the Lie algebra level. Nevertheless, as we saw above, for the OAM Lie algebra, the underlying Fock space representation of the Heisenberg algebra says otherwise! In that sense this proof is very different from a topological proof.

References:

1. L.E. Ballentine, QM modern developments, 1998; p. 170.

--

$^1$ Our notation is slightly different from Ref. 1. For starters, we prefer to use small letters before the CT (5) and capital letters after. A type 2 generating function for the CT (5) is $$\begin{align} F_2(q,P)~=~&q^3P_3+\frac{\sum_{j=1}^2q^jP_j}{\cos\theta}\cr ~+~&\frac{1}{2}\sum_{j,k=1}^2\left(m\omega|\epsilon_{jk}| q^jq^k+\frac{1}{m\omega}|\epsilon^{jk}| P_jP_k\right) \tan\theta.\end{align}\tag{12}$$

$^2$ The Fock vacuum $|\Omega\rangle$ happens to be invariant under the CT (5): $$\begin{align} a^j|\Omega\rangle ~=~&0 \cr ~\Updownarrow~& \cr A^j|\Omega\rangle ~=~&0,\cr j~\in~&\{1,2,3\}.\end{align} \tag{13}$$ Note that the Fock vacuum $|\Omega\rangle$ depends on the constant $m\omega$.

1. In this answer we elaborate on Ballentine's method of finding a canonical transformation (CT) $$(q,p)\quad \longrightarrow\quad (Q,P),\tag{1}$$
   cf. NessunDorma's answer.

2. Recall first of all that instead of position and momentum operators, we may equivalently use annihilation operators $$ a^j~=~ \sqrt{\frac{m\omega}{2\hbar}}q^j+\frac{ip_j}{\sqrt{2m\omega\hbar}} ,\qquad A^j~=~ \sqrt{\frac{m\omega}{2\hbar}}Q^j+\frac{iP_j}{\sqrt{2m\omega\hbar}} ,\qquad j~\in~\{1,2,3\},\tag{2}$$ and corresponding creation operators. Here $m\omega>0$ is some appropriate dimensionful constant whose numerical value is not important for what follows. The orbital angular momentum (OAM) operators are $$ L_j~=~\sum_{k,\ell=1}^3\epsilon_{jk\ell}q^kp_{\ell} ~\stackrel{(2)}{=}~ -i\hbar\sum_{k,\ell=1}^3\epsilon_{jk\ell}a_k^{\dagger}a^{\ell} , \qquad j~\in~\{1,2,3\}.\tag{3}$$ The number operators are $$ n_j~:=~a_j^{\dagger}a^j,\qquad N_j~:=~A_j^{\dagger}A^j,\qquad(\longleftarrow\text{no $j$-sum})\qquad j~\in~\{1,2,3\}. \tag{4}$$

3. Consider a 1-parameter linear CT connected to the identity$^1$ $$\begin{align} Q^j~=~& q^j\cos\theta + \frac{1}{m\omega}\sum_{k=1}^2|\epsilon^{jk}| p_k\sin\theta,\qquad j~\in~\{1,2\}, \cr P_j~=~& p_j\cos\theta - m\omega\sum_{k=1}^2|\epsilon_{jk}| q^k\sin\theta, \cr A^j~=~& a^j\cos\theta - i\sum_{k=1}^2|\epsilon^{jk}| a^k\sin\theta,\cr A_j^{\dagger}~=~& a_j^{\dagger}\cos\theta + i\sum_{k=1}^2|\epsilon^{jk}| a_k^{\dagger}\sin\theta,\cr Q^3~=~& q^3,\cr P_3~=~& p_3,\cr A^3~=~& a^3, \end{align} \tag{5}$$

4. Ballentine's observation becomes $$ N_1-N_2~\stackrel{(3)+(4)+(5)}{=}~(n_1-n_2)\cos 2\theta+ \frac{L_3}{\hbar} \sin 2\theta. \tag{6}$$ Of course we are mainly interested in the angle $\theta=\frac{\pi}{4}$, where the OAM $$\frac{L_3}{\hbar}~\stackrel{(6)}{=}~N_1-N_2\tag{7}$$ becomes the difference of two number operators, cf. OP's question.

5. Let $V_{\ell}$ be a finite-dimensional spin-$\ell$ irrep of the OAM Lie algebra $${\rm span}_{\mathbb{R}}(L_1,L_2,L_3)~\cong~ so(3).\tag{8}$$ We would like to prove that $\ell$ cannot be half-spin, cf. OP's question.

6. Note that $$\sum_{j=1}^3N_j~\stackrel{(4)+(5)}{=}~\sum_{j=1}^3n_j\tag{9}$$ is a Casimir operator in the sense that it commutes with $L_1$, $L_2$ & $L_3$. Therefore there exists a common set of eigenbasis $|\ell m\rangle$ such that $$L_3|\ell m\rangle~=~\hbar m|\ell m\rangle,\quad \vec{L}^2|\ell m\rangle~=~\hbar^2\ell(\ell+1)|\ell m\rangle, \quad \ell, m~\in~\frac{1}{2}\mathbb{Z}, $$ $$ \sum_{j=1}^3N_j|\ell m\rangle~=~\nu |\ell m\rangle, \quad \nu~\geq~0.\tag{10}$$

7. Consider one such state $|\ell m\rangle$. We can act on $|\ell m\rangle$ (finitely many times) with annihilation operators $A^1$, $A^2$ & $A^3$, to reach a Fock vacuum state $|\Omega\rangle\neq 0$ with$^2$ $$A^j|\Omega\rangle~=~0,\qquad j~\in~\{1,2,3\}.\tag{11}$$ (Else there will be negative norm states by a standard argument, cf. e.g. section 6.1 in Ref. 1 or my Phys.SE answer here.)

8. In turn, this implies that $|\ell m\rangle$ is a common eigenvector for $N_1$, $N_2$ & $N_3$, with integer eigenvalues. In particular, $m$ must be an integer, cf. eqs. (7) & (10). $\Box$

9. It's interesting to note that while there are of course no genuine (as opposed to projective) half-spin irreps of the 3D rotation group $SO(3)$, the corresponding Lie algebra $so(3)$ has in principle half-spin irreps, i.e. there are no topological obstructions at the Lie algebra level. Nevertheless, as we saw above, for the OAM Lie algebra, the underlying Fock space representation of the Heisenberg algebra says otherwise! In that sense this proof is very different from a topological proof.

References:

1. L.E. Ballentine, QM modern developments, 1998; p. 170.

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$^1$ Our notation is slightly different from Ref. 1. For starters, we prefer to use small letters before the CT (5) and capital letters after. A type 2 generating function for the CT (5) is $$F_2(q,P)~=~q^3P_3+\frac{\sum_{j=1}^2q^jP_j}{\cos\theta} +\frac{1}{2}\sum_{j,k=1}^2\left(m\omega|\epsilon_{jk}| q^jq^k+\frac{1}{m\omega}|\epsilon^{jk}| P_jP_k\right) \tan\theta.\tag{12}$$

$^2$ The Fock vacuum $|\Omega\rangle$ happens to be invariant under the CT (5): $$ a^j|\Omega\rangle ~=~0 \qquad \Leftrightarrow \qquad A^j|\Omega\rangle ~=~0,\qquad j~\in~\{1,2,3\}. \tag{13}$$

1. In this answer we elaborate on Ballentine's method of finding a canonical transformation (CT) $$(q,p)\quad \longrightarrow\quad (Q,P),\tag{1}$$
   cf. NessunDorma's answer.

2. Recall first of all that instead of position and momentum operators, we may equivalently use annihilation & creation operators $$ \begin{align} a^j~=~& \sqrt{\frac{m\omega}{2\hbar}}q^j+\frac{ip_j}{\sqrt{2m\omega\hbar}} ,~~\quad a^{\dagger}_j~=~ \sqrt{\frac{m\omega}{2\hbar}}q^j-\frac{ip_j}{\sqrt{2m\omega\hbar}}, \cr A^j~=~& \sqrt{\frac{m\omega}{2\hbar}}Q^j+\frac{iP_j}{\sqrt{2m\omega\hbar}}, \quad A^{\dagger}_j~=~ \sqrt{\frac{m\omega}{2\hbar}}Q^j-\frac{iP_j}{\sqrt{2m\omega\hbar}} ,\quad j~\in~\{1,2,3\}. \end{align}\tag{2}$$ Here $m\omega>0$ is some appropriate dimensionful constant whose numerical value is not important for what follows. The orbital angular momentum (OAM) operators are $$ L_j~=~\sum_{k,\ell=1}^3\epsilon_{jk\ell}q^kp_{\ell} ~\stackrel{(2)}{=}~ -i\hbar\sum_{k,\ell=1}^3\epsilon_{jk\ell}a_k^{\dagger}a^{\ell} , \qquad j~\in~\{1,2,3\}.\tag{3}$$ The number operators are $$ n_j~:=~a_j^{\dagger}a^j,\quad N_j~:=~A_j^{\dagger}A^j,\quad(\longleftarrow\text{no $j$-sum})\quad j~\in~\{1,2,3\}. \tag{4}$$

3. Consider a 1-parameter linear CT connected to the identity$^1$ $$\begin{align} Q^j~=~& q^j\cos\theta + \frac{1}{m\omega}\sum_{k=1}^2|\epsilon^{jk}| p_k\sin\theta,\qquad j~\in~\{1,2\}, \cr P_j~=~& p_j\cos\theta - m\omega\sum_{k=1}^2|\epsilon_{jk}| q^k\sin\theta, \cr A^j~=~& a^j\cos\theta - i\sum_{k=1}^2|\epsilon^{jk}| a^k\sin\theta,\cr A_j^{\dagger}~=~& a_j^{\dagger}\cos\theta + i\sum_{k=1}^2|\epsilon^{jk}| a_k^{\dagger}\sin\theta,\cr Q^3~=~& q^3,\cr P_3~=~& p_3,\cr A^3~=~& a^3, \end{align} \tag{5}$$

4. Ballentine's observation becomes $$ N_1-N_2~\stackrel{(3)+(4)+(5)}{=}~(n_1-n_2)\cos 2\theta+ \frac{L_3}{\hbar} \sin 2\theta. \tag{6}$$ Of course we are mainly interested in the angle $\theta=\frac{\pi}{4}$, where the OAM $$\frac{L_3}{\hbar}~\stackrel{(6)}{=}~N_1-N_2\tag{7}$$ becomes the difference of two number operators, cf. OP's question.

5. Let $V_{\ell}$ be a finite-dimensional spin-$\ell$ irrep of the OAM Lie algebra $${\rm span}_{\mathbb{R}}(L_1,L_2,L_3)~\cong~ so(3).\tag{8}$$ We would like to prove that $\ell$ cannot be half-spin, cf. OP's question.

6. Note that $$\sum_{j=1}^3N_j~\stackrel{(4)+(5)}{=}~\sum_{j=1}^3n_j\tag{9}$$ is a Casimir operator in the sense that it commutes with $L_1$, $L_2$ & $L_3$. Therefore there exists a common set of eigenbasis $|\ell m\rangle$ such that $$L_3|\ell m\rangle~=~\hbar m|\ell m\rangle,\quad \vec{L}^2|\ell m\rangle~=~\hbar^2\ell(\ell+1)|\ell m\rangle, \quad \ell, m~\in~\frac{1}{2}\mathbb{Z}, $$ $$ \sum_{j=1}^3N_j|\ell m\rangle~=~\nu |\ell m\rangle, \quad \nu~\geq~0.\tag{10}$$

7. Consider one such state $|\ell m\rangle$. We can act on $|\ell m\rangle$ (finitely many times) with annihilation operators $A^1$, $A^2$ & $A^3$, to reach a Fock vacuum state $|\Omega\rangle\neq 0$ with$^2$ $$A^j|\Omega\rangle~=~0,\qquad j~\in~\{1,2,3\}.\tag{11}$$ (Else there will be negative norm states by a standard argument, cf. e.g. section 6.1 in Ref. 1 or my Phys.SE answer here.)

8. In turn, this implies that $|\ell m\rangle$ is a common eigenvector for $N_1$, $N_2$ & $N_3$, with integer eigenvalues. In particular, $m$ must be an integer, cf. eqs. (7) & (10). $\Box$

9. It's interesting to note that while there are of course no genuine (as opposed to projective) half-spin irreps of the 3D rotation group $SO(3)$, the corresponding Lie algebra $so(3)$ has in principle half-spin irreps, i.e. there are no topological obstructions at the Lie algebra level. Nevertheless, as we saw above, for the OAM Lie algebra, the underlying Fock space representation of the Heisenberg algebra says otherwise! In that sense this proof is very different from a topological proof.

References:

1. L.E. Ballentine, QM modern developments, 1998; p. 170.

--

$^1$ Our notation is slightly different from Ref. 1. For starters, we prefer to use small letters before the CT (5) and capital letters after. A type 2 generating function for the CT (5) is $$F_2(q,P)~=~q^3P_3+\frac{\sum_{j=1}^2q^jP_j}{\cos\theta} +\frac{1}{2}\sum_{j,k=1}^2\left(m\omega|\epsilon_{jk}| q^jq^k+\frac{1}{m\omega}|\epsilon^{jk}| P_jP_k\right) \tan\theta.\tag{12}$$

$^2$ The Fock vacuum $|\Omega\rangle$ happens to be invariant under the CT (5): $$ a^j|\Omega\rangle ~=~0 \qquad \Leftrightarrow \qquad A^j|\Omega\rangle ~=~0,\qquad j~\in~\{1,2,3\}. \tag{13}$$ Note that the Fock vacuum $|\Omega\rangle$ depends on the constant $m\omega$.