The error is most likely that you are using 

$$\left&lt;j_1,j_2;m_1,m_2 \mid j&#39;_1,j&#39;_2;m&#39;_1,m&#39;_2\right&gt; = j_1j&#39;_1+j_2j&#39;_2+m_1m&#39;_1+m_2m&#39;_2,
\quad \mathrm{(Wrong!)}$$

where you should be using 

$$\left&lt;j_1,j_2;m_1,m_2 \mid j&#39;_1,j&#39;_2;m&#39;_1,m&#39;_2\right&gt; = \delta_{j_1,j&#39;_1}\delta_{j_2,j&#39;_2}\delta_{m_1,m&#39;_1}\delta_{m_2,m&#39;_2},$$

where $\delta_{k,\ell}$ is the Kronecker delta function. 

In other words, the $j$'s and $m$'s are not the coefficients $v^i$ of a vector $\vec{v}=\sum_i v^i \vec{e}_i$ in a Hilbert space, where $\vec{e}_i$ is an orthonormal basis, so that

$$ \left&lt;\vec{v}\mid\vec{v}&#39;\right&gt; =  \sum_i (v^i)^*v&#39;^i, \qquad \left&lt;\vec{e}_i\mid\vec{e}_{i&#39;}\right&gt; = \delta_{i,i&#39;}.
$$

Rather, the $j$'s and $m$'s correspond to the $i$-labels of the basis $\vec{e}_i$. For brevity, we often write $\left&lt; i \mid i&#39;\right&gt;$ instead of $\left&lt;\vec{e}_i\mid\vec{e}_{i&#39;}\right&gt;$.

Finally, to give a complete answer, let me include my comment above that it is a general property of eigenvectors for different eigenvalues of a Hermitian operator, that they are orthogonal to each other, see e.g., Lubos Motl's answer or here.

Why are two eigen-state-kets with different eigenvalues orthogonal?
Below is a passage taken from this article


  I don rubber gloves, bravely grab my pickup stick and swallow the remainder of my coffee. It’s not even midnight yet. I remember the advice of my former roommate who — once and again — occupied the same stool I stood up from.


Does the last sentence I italicized means his former roommate used to work at the shop before, and has again came back to work there for the second time?
