Why are two eigen-state-kets with different eigenvalues orthogonal? 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?
 A: The error is most likely that you are using 
$$\left<j_1,j_2;m_1,m_2 \mid j'_1,j'_2;m'_1,m'_2\right> = j_1j'_1+j_2j'_2+m_1m'_1+m_2m'_2,
\quad \mathrm{(Wrong!)}$$
where you should be using 
$$\left<j_1,j_2;m_1,m_2 \mid j'_1,j'_2;m'_1,m'_2\right> = \delta_{j_1,j'_1}\delta_{j_2,j'_2}\delta_{m_1,m'_1}\delta_{m_2,m'_2},$$
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
$$ \left<\vec{v}\mid\vec{v}'\right> =  \sum_i (v^i)^*v'^i, \qquad \left<\vec{e}_i\mid\vec{e}_{i'}\right> = \delta_{i,i'}.
$$
Rather, the $j$'s and $m$'s correspond to the $i$-labels of the basis $\vec{e}_i$. For brevity, we often write $\left< i \mid i'\right>$ instead of $\left<\vec{e}_i\mid\vec{e}_{i'}\right>$.
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.
A: Eigenvalues of a Hermitian operator or their set - such as components of $\vec J$ and/or $J^2$ - corresponding to different eigenvalues are always orthogonal to each other because
$$\langle \psi | M | \phi \rangle = m_\psi \langle \psi | \phi \rangle = m_\phi \langle \psi |  \phi \rangle $$
I could get any eigenvalue $m_\phi$ or $m_\psi$ by acting with $M$ on the two sides. Because those two things are equal, we have
$$ (m_\psi-m_\phi) \langle \psi  | \phi \rangle = 0 $$
which implies - because the eigenvalues differ
$$\langle \psi |  \phi \rangle = 0$$
Your derivation of a nonzero inner product is incorrect and you couldn't have derived it from the formulae above your final result because none of them contains any information about the inner product - in fact, they contain no bra vector whatsoever, so your reversal of a ket vector and its interpretation of an inner product was clearly some beginner's misunderstanding of what the symbols mean.
A: your line $\left<0,j_2;0,m_2 \mid 1,j_2;-1,m_2\right> = {j_2}^2+{m_2}^2$ is making an assumption about what an inner product must look like. A technical way to express your problem is to note that the sequence of eigenvalues identify an element in a tensor product of vector spaces, not a direct sum. If it was a direct sum, your sum of squares would be right, but it isn't.
Luboš's Answer is completely right, enough so that I upvoted it because he needs all the rep he can get, but it looks like it needs you to know what you're doing already for you to understand it. When you use objects like $\left|j_1,j_2;m_1,m_2\right>$ to represent a state, you implicitly claim that the operators that have the eigenvalues $j_1,j_2;m_1,m_2$ are self-adjoint and mutually commutative. In elementary terms, we can take this to define the inner product on the Hilbert space. We don't know what the inner product is until we've defined it. If the eigenvalues are different, the inner product is defined to be $0$, if the eigenvalues are the same, the inner product is defined to be $1$.
As Luboš says, all the lines above your first introduction of a bra are only about operators acting on a vector space, you haven't got a Hilbert space until you've defined an inner product (and more than that, closure in the norm). Once you've defined an inner product, $(\left|U\right>,\left|V\right>)$, you can define a bra as the object that acts on a vector to get this value, $\left<U \mid V\right>=(\left|U\right>,\left|V\right>)$. There's a theorem that says we can do this if we're properly careful, http://en.wikipedia.org/wiki/Riesz_representation_theorem.
