# 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?

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.

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 $$\vert\psi\rangle \mapsto A \left(\begin{array}{c} 1\\ -i\\2\end{array}\right)\, ,$$

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.

• Then why (1+1+4)==1 and not (1+1+0)==1? – NotStudent Jun 11 at 0:00
• So magnitude of A is 1/(squar root 6) – SAKhan Jun 11 at 0:09
• Why -1 on my score? – SAKhan Jun 11 at 0:11
• I didn't downvote, but the answer could use better writing. – nicoguaro Jun 11 at 3:21