The operators $J_1^2$, $J_2^2$, $J_{1z}$, and $J_{2z}$ are mutually commuting operators. Likewise, $J_1^2$, $J_2^2$, $J^2$, and $J_z$ are mutually commuting operators. The two groups are incompatible, and the simultaneous eigenkets along with their eigenvalues are given by:
${J_1}^2 \left|j_1,j_2;m_1,m_2\right> = j_1 \left(j_1+1\right) \hbar^2 \left|j_1,j_2;m_1,m_2\right>$
${J_2}^2 \left|j_1,j_2;m_1,m_2\right> = j_2 \left(j_2+1\right) \hbar^2 \left|j_1,j_2;m_1,m_2\right>$
$J_{1z} \left|j_1,j_2;m_1,m_2\right> = m_1 \hbar \left|j_1,j_2;m_1,m_2\right>$
$J_{2z} \left|j_1,j_2;m_1,m_2\right> = m_2 \hbar \left|j_1,j_2;m_1,m_2\right>$
and
${J_1}^2 \left|j_1,j_2;j,m\right> = j_1 \left(j_1+1\right) \hbar^2 \left|j_1,j_2;j,m\right>$
${J_2}^2 \left|j_1,j_2;j,m\right> = j_2 \left(j_2+1\right) \hbar^2 \left|j_1,j_2;j,m\right>$
$J^2 \left|j_1,j_2;j,m\right> = j \left(j+1\right) \hbar^2 \left|j_1,j_2;j,m\right>$
$J_z \left|j_1,j_2;j,m\right> = m \hbar \left|j_1,j_2;j,m\right>$
I read that each set of eigenkets are mutually orthogonal [1] (for eigenkets corresponding to different sets of eigenvalues). This is what I don't understand. In principle it makes sense, but when I plug in numbers I don't get zero for the inner product. For example take the first eigenket: $\left|j_1,j_2;m_1,m_2\right>$. If I choose different eigenvalues for this eigenket (e.g. let $j_1 = 0$ and then let $j_1 = 1$) I get the following:
for $j_1 = 0$ I can have:
$\left|0,j_2;0,m_2\right>$
for $j_1 = 1$ I can have any of the following, since $\left|m_1\right| \leq j_1$:
$\left|1,j_2;-1,m_2\right>$
$\left|1,j_2;0,m_2\right>$
$\left|1,j_2;1,m_2\right>$
If I take the inner-product of the $j_1 = 0$ eigenket with any of the $j_1 = 1$ eigenkets I do not get zero, e.g.:
$\left<0,j_2;0,m_2 \mid 1,j_2;-1,m_2\right> = {j_2}^2+{m_2}^2$
which is non-zero unless $j_2 = 0$.
What am I misunderstanding here? How do you show that eigenkets with different eigenvalues are orthogonal?