The eigenvalue equation of the $L^2$ operator is given by

$$L^2f_l^m = \hbar ^2l(l+1)f_l^m$$

Side: So a determinate state for some observable $Q$ is a state where every measurement of $Q$ returns the same value (let this value be $q$). Then the standard deviation of $Q$ in a determinate state would be zero, i.e.

$$\sigma ^2 = \big\langle (\hat{Q} - \big\langle Q \big\rangle )^2\big \rangle = 0$$

if every measurement gives $q$, then $\big\langle Q \big\rangle$ = $q$. Then using the rules of Hermitian operators, we can get

$$\hat{Q}f = qf$$

So my question is, given the very first equation, why is the magnitude of $L$, in units of $\hbar$, $\sqrt{l(l+1)}$? The eigenvalue of $L^2$ in units of $\hbar$ is $l(l+1)$.

From what I wrote above, $\big \langle L^2 \big \rangle = l(l+1)$.

In general, $\sqrt{\big \langle L^2 \big \rangle} \neq L$. So why is it that, $L = \sqrt{l(l+1)}$?

I'm trying to understand the magnitude of the $\vec{L}$, which in the common illustration shown to demonstrate that $L_z$ can never be greater than $L$, is said to be $\sqrt{l(l+1)}$. But all they did in Griffiths was take the square root of the eigenvalue of the $L^2$ operator.


Let's take the spin, it's the simplest case, $Q = \mathbf{s}$. The operator $\mathbf{s}$ is a vector,

\begin{equation}s = \mathbf{i}s_x + \mathbf{j}s_y + \mathbf{k}s_z\end{equation}

while the operator s^2 is a scalar

\begin{equation}s^2 = s_x^2 + s_y^2 + s_z^2\end{equation}.

The operators $s_x$, $s_y$, and $s_z$ don't commute two by two, but $s_x^2$, $s_y^2$, $s_z^2$ are numbers, each one is equal to $\frac{1}{4}\hbar^2$. Their sum is therefore $\frac{3}{ 4}\hbar^2$, i.e. $s(s+1)$. But the eigenvalues of each one of $s_x$, $s_y$, and $s_z$, are $+\frac{1}{2}\hbar$ and $-\frac{1}{2}\hbar$ .

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  • $\begingroup$ What's the difference between a scalar and a scalar operator? $\endgroup$ – DWade64 Nov 27 '14 at 21:26

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