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During a standard derivation of the eigenvalues of the angular momentum operators, $J^2$ and $J_z$, where $$J^2|\alpha, \beta\rangle =\hbar^2\alpha|\alpha, \beta\rangle$$ and $$J_z|\alpha, \beta\rangle =\hbar\beta|\alpha, \beta\rangle$$ one can show that $\alpha \geq\beta^2$. Textbooks at this point say that there must exist $\beta_{max}$ for a given $\alpha$. My question is, why cannot we say, based on the last inequality, that $\beta_{max}=\sqrt{\alpha}$? |
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The reason that the equality cannot be attained except for the trivial case $\alpha = 0$ is the noncommutativity of the spin operators. If the equality is true then we would have $J_x^2|\alpha, \beta \rangle = 0$ and the same for $J_y$. This would imply that both $J_x$ and $ J_y$ must be represented by the zero matrix in this Hilbert space, but then the commutation relation: $[J_x, J_y]= i \hbar J_z$ cannot be satisfied. |
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You can't just turn an inequality into an equality. If I tell you $x \ge y$, it's not necessarily true that $y = x$. If you follow the derivations in these textbooks a little further, they should show that $\alpha = \beta_\text{max}(\beta_\text{max} + 1)$, using your notation. So it turns out that $\alpha$ is not even a square number, in general, and thus $\beta_\text{max}\neq\sqrt\alpha$. $\beta_\text{max}$ is simply the largest integer or half-integer (as appropriate) which satisfies the condition. |
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