your line $\left&lt;0,j_2;0,m_2 \mid 1,j_2;-1,m_2\right&gt; = {j_2}^2+{m_2}^2$ 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 $\left|j_1,j_2;m_1,m_2\right&gt;$ 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, $(\left|U\right&gt;,\left|V\right&gt;)$, you can define a bra as the object that acts on a vector to get this value, $\left&lt;U \mid V\right&gt;=(\left|U\right&gt;,\left|V\right&gt;)$. There's a theorem that says we can do this if we're properly careful, http://en.wikipedia.org/wiki/Riesz_representation_theorem.

Why are two eigen-state-kets with different eigenvalues orthogonal?
I am kind of studying physics on my own now.
I choose University Physics (13th Edition) for myself,is it fine?
I am also studying Calculus using Thomas' textbook.

http://www.amazon.com/University-Physics-13th-Edition-Young/dp/0321696891/ref=sr_1_1?ie=UTF8&amp;qid=1352260382&amp;sr=8-1&amp;keywords=university+physics+13th+edition
