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As usual let $|l,m\rangle$ denote an eigenstate of $\vec{L}^2$ and $L_z$. I know that

\begin{align} \vec{L}^2 |l,m\rangle &= \hbar^2 l(l+1) |l,m\rangle, \\ L_z |l,m\rangle &= \hbar m |l,m\rangle. \end{align}

Consider $|1,1\rangle$. Then, I can write $|1,1\rangle$ in terms of $|1,m_x\rangle$ with $m_x = \pm 1,0$, where $|1,m_x\rangle$ denotes an eigenbasis of $L_x$, as

\begin{align} |1,1\rangle_z = c_1\cdot |1,-1\rangle_x + c_2 \cdot |1,0\rangle_x + c_3 \cdot |1,1\rangle_x \end{align} with $|c_1|^2 + |c_2|^2 + |c_3|^2 = 1$.

I want to compute the coefficients $c_1, c_2, c_3$ working with

\begin{align} \Delta^2L_x := \langle (L_x - \langle L_x \rangle)^2\rangle = \frac{1}{2} \cdot \left( \langle L^2\rangle - \langle L_z^2\rangle\right) = \frac{\hbar^2}{2}\cdot\left(l(l+1) - m^2\right) \end{align}

as in this case (for $|1,1\rangle$) we have $\Delta^2L_x = \frac{\hbar^2}{2}$. However, I don't understand how this can be done. Can someone please explain it?

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  • $\begingroup$ What do you find if you insert your state in the last constraint? $\endgroup$ Sep 13, 2021 at 19:45
  • $\begingroup$ @CosmasZachos Tbh, I simply don't know how to evaluate it. I know about $L_z |l,m\rangle_z = \hbar m|l,m\rangle_z$. I don't know what I could say about $L_z|1,-1\rangle_x$ (for example). Similarly, I don't know what, for example, how I could rewrite $L_x|l,m\rangle_x$? From what I see, this would be expressions I would get if I insert $|1,1\rangle_z$ into the last equation ... $\endgroup$
    – offline
    Sep 13, 2021 at 20:41
  • $\begingroup$ You defined $L_x|l,m_x\rangle_x$, above, no? $\endgroup$ Sep 13, 2021 at 21:00
  • $\begingroup$ @CosmasZachos Sorry, I do not see where I say anything about expressions of this form. Can you explain to me what I should find if I insert $|1,1\rangle_z$ in the last equation? $\endgroup$
    – offline
    Sep 13, 2021 at 21:07
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    $\begingroup$ @offline that's the definition! Otherwise the labels in your eigenstates mean nothing. Alternatively, remember that there's nothing special about the $z$ direction, so we expect the same set of possible eigenvalues of angular momentum in the $x$ direction as in the $z$ direction $\endgroup$ Sep 14, 2021 at 13:55

3 Answers 3

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Your requested hint: $$ L_x (c_1 |1,-1\rangle_x + c_2 |1,0\rangle_x + c_3 |1,1\rangle_x)=\hbar(-c_1 |1,-1\rangle_x + c_3 |1,1\rangle_x). $$ Can you take it from here? It is straightforward linear algebra.

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  • $\begingroup$ Why do we have $L_x|l,m\rangle_x = \hbar m |l,x\rangle_x$? I genuinely don't see/understand that. You said in one of your earlier comments that I have defined $L_x|l,m\rangle_x$. Where? I can only repeat myself: I have $L_{\color{red}{z}}|l,m\rangle_{\color{red}{z}} = \hbar m|l,m\rangle_{\color{red}{z}}$. $\endgroup$
    – offline
    Sep 14, 2021 at 5:23
  • $\begingroup$ You defined it as its eigenbasis… $\endgroup$ Sep 14, 2021 at 7:15
  • $\begingroup$ Got it! I did not know that I also have $L_x|l,m\rangle_x = \hbar m|l,m\rangle_x$. Based on that the rest is clear! Thank you! $\endgroup$
    – offline
    Sep 14, 2021 at 10:21
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Let's clarify a number of concepts here to help elucidate how to solve this problem. We know for starters that $$\left[\vec{L}^2,L_i\right]=0,\quad i\in(x,y,z),$$ meaning that we can define simultaneous eigenstates of $\vec{L}^2$ and one of the angular momentum projections $L_i$. Moreover, any other angular momentum projection will also suffice, defined through $$L_{\vec{n}}=\vec{n}\cdot\vec{L}.$$

By spherical symmetry, we do not expect any direction to be privileged, so we expect the set of eigenvalues of $L_{\vec{n}}$ and $\vec{L}^2$ to be independent of the direction $\vec{n}$. That means that we should always be able to define a set of eigenstates $|l,m\rangle_{\vec{n}}$ for any direction such that $$\vec{L}^2|l,m\rangle_{\vec{n}}=l(l+1)|l,m\rangle_{\vec{n}},\quad L_{\vec{n}}|l,m\rangle_{\vec{n}}=m|l,m\rangle_{\vec{n}},\quad l\in1/2,1,3/2,\cdots,\quad m\in(-l,-l+1,\cdots,l).$$

How do we know this can be done? We define rotation operators $$R(\theta,\vec{u})=\exp(-i\theta L_{\vec{u}}),$$ which [must be unitary due to Wigner's theorem and] have the nice property $$R\vec{L} R^\dagger=\mathbf{R}^T\vec{L} \qquad\Rightarrow RL_{\vec{n}} R^\dagger=L_{\mathbf{R}\vec{n}},$$ where $\mathbf{R}^T$ is the transpose of a $3\times 3$ rotation matrix that rotates a vector by angle $\theta$ about axis $\vec{u}$. Then, we can do some manipulations starting from the known properties of the states $|l,m\rangle_z$: \begin{align} m \left(R|l,m\rangle_z\right)&=R m|l,m\rangle_z\\ &=R L_z|l,m\rangle_z\\ &=R L_zR^\dagger R|l,m\rangle_z\\ &=L_{\mathbf{R}z} \left(R|l,m\rangle_z\right). \end{align} This tells us that the rotated angular momentum projection operators $L_{\vec{n}}=L_{\mathbf{R}z}$ have the same eigenvalues $m$ before and after the rotation, with the relationships between the eigenstates being given by $$|l,m\rangle_{\mathbf{R}z}=R|l,m\rangle_z.$$

Given this mathematical machinery, we can always define the eigenstates $|l,m\rangle_{\vec{n}}$ as above, then use the completeness of the eigenstates to write $$|l,m\rangle_{\vec{n}}=\sum_k |l,k\rangle_{z}\langle l,k|l,m\rangle_{\vec{n}}\equiv \sum_k c_k |l,k\rangle_{z}.$$ In your question, you are looking for these coefficients $c_k=\,_{z}\langle l,k|l,m\rangle_{\vec{n}}=\,_{z}\langle l,k|R|l,m\rangle_{z}$. Does our mathematical machinery help with that? Yes! These coefficients are exactly the elements of the Wigner $D$-matrices, which have been tabulated and can be calculated analytically in various ways. Your computation helps calculate them for $l=1$.

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You can just work out the LHS of your final equation in the new basis: $L_x^2|\psi\rangle=c_3|1,1\rangle+c_1|1,-1\rangle$ and then complete the expectation value, and similarly for $\langle L_x\rangle$. And there is the normalisation condition on the $c_i$.

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