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: Is research a viable business model?
Answer: From the perspective of the US, there are many research institutes that operate as a private business entity.  These research institutes can apply for some federal funding in the same capacity as a research at an educational institution.  I see it being possible to start a private company without a lot of capital, although you will need to have a very strong skill set that is in demand.  Whether this is a sustainable model really seems to depend on the specialized skills that the business would offer and the infrastructure that would allow it to efficiently prepare grants. Research-I educational institutions in the US have an incredible infrastructure to prepare and apply for grants that would be difficult to replicate in the private sector.