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?


2 Answers 2


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 $\hat{L^2}$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.