# Equivalence of angular momenta eigenstates

Consider two angular momentum operators $\hat{J}_{1}$ and $\hat{J}_{2}$ and operator $J := J_{1} \otimes 1 + 1 \otimes J_{2}$ where respectively we have common eigenstates $|j_1j_2;m_1 m_2 \rangle$ of $\hat{J}_{1}$ and $\hat{J}_{2}$ and $|j_1,j_2; jm \rangle$ a common eigenstate of $\hat{J}^2$ and $\hat{J}_{z}$.

Consider an example where we have a single spin $\frac{1}{2}$ particle with say $j_{1} = l$, $m_{1} = m_{l}$, $j_{2} = s = \frac{1}{2}$ and $m_{2} = m_{s} = \pm\frac{1}{2}$. Consider now the angular-momentum configuration in which $m_{l}$ and $m_{s}$ are both maximal-that is, $l$ and $\frac{1}{2}$, respectively. The total $m = m_{l} + m_{s}$ is $l + \frac{1}{2}$, which is possible only for $j = l + \frac{1}{2}$ and not for $j = l - \frac{1}{2}$. Why does it followws that $|j_{1}j_{2}; m_{l}= l, m_{s} = \frac{1}{2} \rangle$ must be equal to $|j_1j_2;j=l+\frac{1}{2}, m = l + \frac{1}{2} \rangle$ up to phase factor?

There are two ingredients needed.

1. For a given set of states with angular momentum $j$, only the state with $m=j$ is killed by $\hat L_+$. This follows from the usual \begin{align} \hat L_+\vert \ell,m\rangle &=\sqrt{(\ell-m)(\ell+m+1)}\vert \ell,m+1 \rangle\, ,\\ \hat L_+\vert \ell,\ell\rangle&=0\, . \end{align}
2. State counting: in your full Hilbert space spanned by states of the form $\vert j_1m_1\rangle \vert j_2m_2\rangle$, there is only one state with eigenvalue $m=\frac{1}{2}+\ell$, and it is the state $$\vert s=\textstyle\frac{1}{2},\ell;m_s=\frac{1}{2},m_\ell=\ell\rangle = \vert \frac{1}{2},\frac{1}{2}\rangle\vert \ell,\ell\rangle\, . \tag{1}$$ All other states of the form $\vert s,m_s\rangle\vert \ell,m_\ell\rangle$ have a lower eigenvalue $m_s+m_\ell$ of $\hat L_x$ since by assumption $m_s+m_\ell<\ell+\textstyle\frac{1}{2}$.

Using $\hat L_+=\hat L_{+,1}\otimes 1+1\otimes \hat L_{+,2}$ one easily verifies that \begin{align} \hat L_+\vert s=\textstyle\frac{1}{2},\ell;m_s=\frac{1}{2},m_\ell=\ell\rangle &= \left[\hat L_{+,1}\vert \textstyle\frac{1}{2}\frac{1}{2}\rangle\right]\vert \ell,\ell\rangle+\vert \textstyle\frac{1}{2}\frac{1}{2}\rangle\left[\hat L_{+,2}\vert \ell,\ell\rangle\right]=0 \end{align} so that $\vert s=\textstyle\frac{1}{2},\ell;m_s=\frac{1}{2},m_\ell=\ell\rangle$ must be a state with $m=j$. Since this $m$ is $\ell+\textstyle\frac{1}{2}$, it follows that $j=\ell+\textstyle\frac{1}{2}$. Of course, we could also include an arbitrary phase on the right hand side of (1) and the result would be the same.

To see how the argument does not work for other states, consider the states with $m=\ell-1/2$. Indeed there are two such states: \begin{align} \vert s=\textstyle\frac{1}{2},\ell;m_s=\frac{1}{2},m_\ell=\ell-1\rangle\, ,\qquad \vert s=\textstyle\frac{1}{2},\ell;m_s=-\frac{1}{2},m_\ell=\ell\rangle \end{align}

The state $\vert j=\ell-\textstyle{\frac{1}{2}},m_j=\ell-\textstyle{\frac{1}{2}}\rangle$ must be a linear combination of these two. To see note that the action of $\hat L_+$ on $\vert s=\textstyle\frac{1}{2},\ell;m_s=\frac{1}{2},m_\ell=\ell-1\rangle$ or the $\vert s=\textstyle\frac{1}{2},\ell;m_s=-\frac{1}{2},m_\ell=\ell\rangle$ is not enough to kill the states individually. However, one easily verifies that \begin{align} &\hat L_+ \left(\alpha \vert s=\textstyle\frac{1}{2},\ell;m_s=-\frac{1}{2},m_\ell=\ell\rangle+\beta \vert s=\textstyle\frac{1}{2},\ell;m_s=-\frac{1}{2},m_\ell=\ell\rangle \right)\\ &= \alpha \vert s=\textstyle\frac{1}{2},\ell;m_s=\frac{1}{2},m_\ell=\ell\rangle+\beta \sqrt{2\ell}\vert s=\textstyle\frac{1}{2},\ell;m_s=\frac{1}{2},m_\ell=\ell\rangle=0 \end{align} provided $\alpha+\sqrt{2\ell}\beta=0$.

$$e^{-i\beta L_y}=e^{i\beta L_{y,1}}\otimes e^{i\beta L_{y,2}}$$ since the action of $L_{y,1}$ commutes with the action of $L_{y,2}$. (The same observation holds for all rotations in the different spaces, i.e. $[\hat L_{k,1},\hat L_{i,2}]=0$.)
Thus, as the basis $\vert j_1j_2;m_1m_2\rangle$ is uncoupled" in the sense that $$\vert j_1j_2;m_1m_2\rangle=\vert j_1m_1\rangle\otimes \vert j_2m_2\rangle$$ it follows that \begin{align} \langle j_1j_2;m_1'm_2'\vert e^{-i\beta L_y}\vert j_1j_2;m_1m_2\rangle &=\langle j_1m_1'\vert e^{-i\beta L_{y,1}} \vert j_1m_1\rangle \langle j_2m_2'\vert e^{-i\beta L_{y,2}} \vert j_2m_2\rangle\, ,\\ &=D^{j_1}_{m_1'm_1}(0,\beta,0) D^{j_2}_{m_2'm_2}(0,\beta,0) \end{align} and more generally $$\langle j_1j_2;m_1'm_2'\vert R\vert j_1j_2;m_1m_2\rangle =D^{j_1}_{m_1'm_1}(R) D^{j_2}_{m_2'm_2}(R)$$
• Thanks for this substantial answer. Just one thing that I'm not sure of is your reasoning for the section "how the argument does not work for other states". As I understand, your reasoning is that since $\hat{L}_{+}$ does not kill either $|s = \frac{1}{2}, l; m_{s} = \frac{1}{2}, m_{l} = l - 1\rangle$ or $|s = \frac{1}{2}, l; m_{s} = -\frac{1}{2}, m_{l} = l\rangle$ it follows that $|j= l-1, m_{j} = l-1 \rangle$ must be a linear combintation of these two and hence not equivalent to either one. – user100411 May 15 '17 at 13:14
• This doesn't agree with Sakurai's "Modern Quantum Mechanics" which says that eigenstates of form $|j,m_{j} \rangle$ can only be written as linear combintation of eigenstates of the form $|s, l; m_s, m_{l} \rangle$ given that $m_{j} = m_{s} + m_{l}$. I confirmed this in a comment of my other post. In your case $m_{j} = l-1$ while $m_{s} + m_{j} = l -\frac{1}{2}$, thus your $|j,m_{j}\rangle$ can't be the stated linear combintation. What do you think? – user100411 May 15 '17 at 13:15
• You are right: it was a typo on my part. The values of $j$ go in steps of $1$: the largest is $\ell+1/2$, the next largest is $\ell-1/2$. I have corrected "The state $\vert j=\ell-1,m_j=\ell-1\rangle$" to "The state $\vert j=\ell-\textstyle\frac{1}{2},m_j=\ell-\textstyle\frac{1}{2}\rangle$". – ZeroTheHero May 15 '17 at 20:26
• Okay great thanks I understand now. Do you maybe know why when we consider the respective basis kets $|j_1j_2;m_{1}m_{2} \rangle$ and $|j,m \rangle$ with a rotation operator $\mathcal{D}(R)$ we get the equation $$\langle j_1 j_2; m_1 m_2| \mathcal{D}(R)| j_{1}j_{2}; m'_{1}m'_{2} \rangle = \langle j_{1}m_{2}| \mathcal{D}(R)| j_{1}m'_{1} \rangle \langle j_{2}m_{2}| \mathcal{D}(R)|j_{2} m'_{2} \rangle?$$ – user100411 May 16 '17 at 10:59