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If the expectation value of square of angular momentum operator $L^2$ is $\ell(\ell+1)\hbar^2$ then will the expectation of $L$ be square root of the above eigenvalue? Let me rephrase the question: Is the expectation value and magnitude of the angular momentum operator the same?

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If $A = \int_{\sigma(A)} \lambda dP(\lambda)$ is the spectral decomposition of a selfadjont operator and the spectrum of $A$ satisfies $\sigma(A) \subset [0,+\infty)$, then we can define the square root of $A$ as $$\sqrt{A}:= \int_{\sigma(A)}\sqrt{\lambda} dP(\lambda)\:.$$ In the case of $L^2$ we have that its spectral decomposition is $$L^2 = \sum_{\ell=0,1,2,\ldots} \sum_{-\ell \leq m \leq \ell} \hbar^2 \ell(\ell+1) |\ell, m\rangle \langle \ell, m|\:,$$ with domain $D(L^2)$ made of the linear combinations $\sum_{\ell,m} C_{\ell, m}|\ell, m\rangle$ such that $\sum_{\ell, m} \ell^2(\ell+1)^2 |C_{\ell, m}|^2 <+\infty$. Therefore, using the defintion above, $$L = \sum_{\ell=,0,1,2,\ldots} \sum_{-\ell \leq m \leq \ell} \hbar \sqrt{\ell(\ell+1)} |\ell, m\rangle \langle \ell, m|\:,$$ with domain $D(L)$ made of the linear combinations $\sum_{\ell,m} C_{\ell, m}|\ell, m\rangle$ such that $\sum_{\ell, m} \ell(\ell+1) |C_{\ell, m}|^2 <+\infty$.

This definition also reflects the operational definition of $L$:

I first measure $L^2$ and next I compute the square root of the outcome.

We can now tackle the proposed issue.

In order to avoid problems with domains, let us consider a state which is a finite superposition of definite $L^2$ states. Its state vector reads $$\Psi = \sum_{\ell\in A, m \in B_\ell} C_{\ell,m} |\ell, m\rangle$$ where $A$ is a finite number of naturals and each of the associated set of integers $B_\ell\subset \{m\in \mathbb{Z}\:|\: -\ell \leq m\leq \ell\}$ is finite as well. We also assume that the state vector is normalized $$\sum_{\ell\in A, m \in B_\ell} |C_{\ell,m}|^2 =1\:.$$ With these choices $$\langle L^2 \rangle_{\Psi}= \hbar^2\sum_{\ell\in A} \left(\sum_{m \in B_\ell} |C_{\ell,m}|^2\right) \ell(\ell+1) $$ whereas $$\langle L \rangle_{\Psi}= \hbar \sum_{\ell\in A} \left(\sum_{m \in B_\ell} |C_{\ell,m}|^2\right)\sqrt{\ell(\ell+1)}\:. $$ It is easy, at this point, to choose the coefficients $|C_{\ell,m}|^2$ in order that $$\langle L \rangle_{\Psi} \neq \sqrt{\langle L^2 \rangle_{\Psi}}\:.$$ This answers the question.

Notice however that the two values coincide if $\Psi$ is an eigenvector of $L^2$. In the other cases the inequality generally holds.

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I believe there is no proper operator corresponding to $L$ (linear, hermitian, etc). Instead there are three distinct operators $\hat{L}_i$ out of which you can construct an operator $\hat{L^2}\equiv \hat{L}_x^2+\hat{L}_y^2+\hat{L}_z^2$. $L$ then is just a shorthand for the square root of the expectation value of $\hat{L^2}$.

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  • $\begingroup$ So the magnitude and expectation value of angular momentum operator the same? $\endgroup$
    – Igris
    Commented Oct 10, 2022 at 16:53
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    $\begingroup$ Sorry I didn't understand your comment. Again, there is no operator for the magnitude of $|\vec{L}|$, only $L^2$ $\endgroup$
    – John
    Commented Oct 10, 2022 at 18:16
  • $\begingroup$ The uncertainty in any operator, say L is written as $∆L = \sqrt{<L²>-<L>²}$. So if what you are saying is correct then does that mean always there will not be any uncertainty in L? that's why I am asking whether the expectation is always same as the magnitude. Where the magnitude is $\sqrt{l(l+1)}\hbar$ $\endgroup$
    – Igris
    Commented Oct 11, 2022 at 14:05
  • $\begingroup$ You are right. There is indeed to uncertainty in L. For instance, you know that the spin of electron is S=1/2 for sure. $\endgroup$
    – John
    Commented Oct 11, 2022 at 14:30

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