# Expectation of angular momentum operator $L$

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?

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.

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 $$\hat{L^2}$$.

• So the magnitude and expectation value of angular momentum operator the same? Commented Oct 10, 2022 at 16:53
• Sorry I didn't understand your comment. Again, there is no operator for the magnitude of $|\vec{L}|$, only $L^2$
– John
Commented Oct 10, 2022 at 18:16
• 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 $\sqrt{l(l+1)}\hbar$ Commented Oct 11, 2022 at 14:05
• 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.
– John
Commented Oct 11, 2022 at 14:30