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:

${J_1}^2 \left|j_1,j_2;m_1,m_2\right&gt; = j_1 \left(j_1+1\right) \hbar^2 \left|j_1,j_2;m_1,m_2\right&gt;$
${J_2}^2 \left|j_1,j_2;m_1,m_2\right&gt; = j_2 \left(j_2+1\right) \hbar^2 \left|j_1,j_2;m_1,m_2\right&gt;$
$J_{1z} \left|j_1,j_2;m_1,m_2\right&gt; = m_1 \hbar \left|j_1,j_2;m_1,m_2\right&gt;$
$J_{2z} \left|j_1,j_2;m_1,m_2\right&gt; = m_2 \hbar \left|j_1,j_2;m_1,m_2\right&gt;$

and

${J_1}^2 \left|j_1,j_2;j,m\right&gt; = j_1 \left(j_1+1\right) \hbar^2 \left|j_1,j_2;j,m\right&gt;$
${J_2}^2 \left|j_1,j_2;j,m\right&gt; = j_2 \left(j_2+1\right) \hbar^2 \left|j_1,j_2;j,m\right&gt;$
$J^2 \left|j_1,j_2;j,m\right&gt; = j \left(j+1\right) \hbar^2 \left|j_1,j_2;j,m\right&gt;$
$J_z \left|j_1,j_2;j,m\right&gt; = m \hbar \left|j_1,j_2;j,m\right&gt;$

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: $\left|j_1,j_2;m_1,m_2\right&gt;$.  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:
$\left|0,j_2;0,m_2\right&gt;$

for $j_1 = 1$ I can have any of the following, since $\left|m_1\right| \leq j_1$:
$\left|1,j_2;-1,m_2\right&gt;$
$\left|1,j_2;0,m_2\right&gt;$
$\left|1,j_2;1,m_2\right&gt;$  

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.:

$\left&lt;0,j_2;0,m_2 \mid 1,j_2;-1,m_2\right&gt; = {j_2}^2+{m_2}^2$  

which is non-zero unless $j_2 = 0$.  

What am I misunderstanding here? How do you show that eigenkets with different eigenvalues are orthogonal?


Question: When cellphone providers give "unlimited social networking", how do they identify the traffic?
Answer: Strictly speaking, the phones are not completely restricted because of IP over Facebook. Whatever the medium, there's always someone to think about transporting raw IP packets in it, and then go ahead and actually implement it.

For the actual restriction, this can be done with IP filtering, or by enforcing an unescapable Web proxy which filters on the URL (preferably a transparent proxy rather than something in the phone software, so that jailbroken phones may not evade the restriction).