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.

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

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