Strange bra-ket notation I encountered a question, where I need to find the constant. But the state is given like this: $$|\psi\rangle = A(|1,1\rangle -i|1,-1\rangle+2|1,0\rangle)$$ So normally eg. when the state is given like this: $$|\psi\rangle = A
    \begin{bmatrix}
    1 \\
    0 \\
    1 \\
    \end{bmatrix}
$$ I would just do $$\langle\psi|\psi\rangle = 1$$ and multiply the matrices to get $A$. What exactly does the first notation mean?  
 A: The angular momentum states $|j,m\rangle$ are orthogonal and normalized so that
$$\langle j_1, m_1 | j_2, m_2\rangle = \delta_{j_1,j_2} \delta_{m_1,m_2}$$
In this case,
$$\langle \psi | \psi \rangle = |A|^2 \left[\langle 1,1| +i \langle 1,-1| + 2\langle 1,0|\right] \left[|1,1\rangle - i |1,-1\rangle + 2|1,0\rangle \right] = 1$$
Multiplying those terms out and applying the orthonormality condition above gives you your answer.
A: The states $\vert 1,m\rangle$ are basis states in your 3-dimensional space so one has the correspondence
$$
\vert 1,1\rangle \mapsto 
\left(\begin{array}{c} 1\\ 0\\0\end{array}\right)\, ,
\quad
\vert 1,0\rangle \mapsto 
\left(\begin{array}{c} 0\\1\\0\end{array}\right)\, ,\quad
\vert 1,-1\rangle \mapsto 
\left(\begin{array}{c} 0\\0\\1\end{array}\right)\, .
$$
Thus, your specific ket would be represented as the column vector
$$
\vert\psi\rangle \mapsto 
A
\left(\begin{array}{c} 1\\ -i\\2\end{array}\right)\, ,
$$
A: Spin and its component along the z-axis, s=1, m==1,0,-1.you are to normalize it to one. Normally differnt m scalar product will be zero. Mag A square(1+1+4)==1. can get Magnitude of A.
