# Why isn't the maximum eigenvalue of $J_z$ squared equal to the maximum eigenvalue of $J^2$?

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 $\beta_{max}=\sqrt{\alpha}$?

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.

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.

• I did not turn an inequality into an equality, I said that y is bounded from above by x, hence the maximum possible value for y must be x. – Revo Jul 21 '12 at 7:18
• That is not true, the maximum possible value for $y$ may be less than $x$, the bound should not necessarily be saturated! In this particular case, besides two equalities, you give, there is also $(J^2-J_z^2-J_z)\left|\alpha,\beta_{max}\right>=J_{-}J_{+}\left|\alpha,\beta_{max}\right>=0$ giving you $\alpha=\beta_{max}(\beta_{max}+1)$. – straups Jul 21 '12 at 8:52
• @Revo you started with $\alpha\ge\beta^2$ and concluded that $\alpha = \beta_\text{max}^2$. That's turning an inequality into an equality. As my answer and straups' comment both say, the only conclusion you can correctly draw is that $\alpha\ge\beta_\text{max}^2$. – David Z Jul 21 '12 at 21:51