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Suppose a spin-1 system ($l = 1$) has a particle set in the state of the eigenstate in the $L_{x}$ basis, given as such: $$ |1, m_{x} = +1 \rangle = 1/2 \begin{pmatrix} 1 \\ \sqrt{2} \\ 1 \end{pmatrix}$$. The $L_{z}$ component of its angular momentum is measured and the result is $m_{z} = -1 $ which corresponds to the $L_{z}$ eigenstate of $\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$. Immediately afterwards, the $L_{x}$ component of angular momentum is measured. What are the results obtained and with what probability? If you measured $m_{x} = -1$, and now you decide to measure $L_{z}$ again. What are the possible outcomes and with what probability?

My thought process for this is to first define a Projection Operator $|O\rangle = \sum_{j} |\rho_{j}\rangle \langle \rho_{j}|$ and have it be in the $L_{x}$ basis followed by taking the square inner product with the eigenstates of $L_{x}$ but I don't think I'm quite getting the actual logical flow of the measurements and the math.

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The initial outcome $L_z=-1$ will project you initial state to the appropriate eigenstate of $L_z$. After normalization, the state after this measurement will be $\vert 1,-1\rangle=(0,0,1)^T$. To get the various probabilities of $L_x$ will require you to construct all three $\vert 1,m_x\rangle$ eigenstates and then work out $\vert \langle 1,m_x\vert 1,m_z=-1\rangle\vert^2$.

Measuring $m_x=-1$ again would project your $(0,0,1)^T$ back to your $\vert 1,m_x=-1\rangle$ state, so that measuring $L_z$ would be related to the overlap of eigenstates of $L_z$ on $\vert 1,m_x=-1\rangle$.

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  • $\begingroup$ so I have calculated all the eigenstates of $|1, m_{x} \rangle$ as follow: $$ |1, m_{x} = 1 \rangle = 1/2 (1, \sqrt{2}, 1)^{T} , |1, m_{x} = 0 \rangle = 1/\sqrt{2} (1, 0,- 1)^{T}, |1, m_{x} = -1 \rangle = 1/2 (1, -\sqrt{2}, 1)^{T}$$ $\endgroup$ – Sam Sep 19 '17 at 8:33
  • $\begingroup$ So based after doing $ |\langle 1, m_{x} | 1, m_{z} = -1 \rangle | ^{2}$ , based on your explanation I would do the opposite where $m_{x}$ and $m_{z}$ are interchanged? i.e. $ |\langle 1, m_{z} | 1, m_{x} = -1 \rangle | ^{2}$ where the eigenstates of $m_{z}$ are just $ (1,0,0)^{T}, (0,1,0)^{T}, (0,0,1)^{T}$ $\endgroup$ – Sam Sep 19 '17 at 8:43
  • $\begingroup$ @Sam That is correct. $\endgroup$ – ZeroTheHero Sep 19 '17 at 9:23

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