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For spin-1 particle, I don't quite understand how the following relationship is derived: $$\left|+1\right>=-\frac{1}{\sqrt2}(\hat e_x+i\hat e_y)$$ $$\left|0\right>=\hat e_z$$ $$\left|-1\right>=\frac{1}{\sqrt2}(\hat e_x-i\hat e_y),$$ where $\left|-1\right>$, $\left|0\right>$, $\left|+1\right>$ are eigenstate of angular momentum in the z direction. It appears that one need to first make an ad hoc assertion on the middle equation, and then somehow use the raising and lowering operator to obtain the other equations, but I am not sure how the raising and lowering operator will give relationship on the unit vector.

I am also not clear on the interpretation of the above relation: naively I would say the middle equation shows that the $\left|0\right>$ state is linearly polarized in the z direction, whereas the $\left|+1\right>$ and $\left|-1\right>$ states are linearly polarized along the $y=x$ and $y=-x$ line respectively.

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    $\begingroup$ I do not understand what the equations in your relationship are supposed to mean. The l.h.s. are abstract state vectors, the r.h.s. seem to be vectors in some complex 3d space with fixed axes (fixed by what?). What source did you get them from? What is the representation of the angular momentum operators on the r.h.s. (since these are eigenstates, you presumably get these three specific vectors just by finding the eigenvectors of the z-operator in that representation!)? $\endgroup$ – ACuriousMind Jan 6 at 0:25
  • $\begingroup$ @ACuriousMind It is from a handwritten quantum mechanics lecture notes my friend gave me, and this section confuses me. $\endgroup$ – Leo L. Jan 6 at 0:29
  • $\begingroup$ Oh it appears that on the RHS of the equation, the representation of the angular momentum used is $J_i=i\hbar \hat e_i \times$, from the relationship $J_i r=i\hbar \hat e_i \times r$ $\endgroup$ – Leo L. Jan 6 at 0:36
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For spin-1 particle, I don't quite understand how the following relationship is derived

The numbers inside the kets on the left-hand side represent the eigenvalue of $L_z$. Therefore, we can figure out the right-hand side by how they transform under physical rotations $R_z(\theta)$ about the $z$-axis.

The $|0 \rangle$ state corresponds to $\hat{e}_z$, because it is unchanged by a rotation about the $z$-axis, $$R_z(\theta) \hat{e}_z = \hat{e}_z.$$ On the other hand, note that $$R_z(\theta) (\hat{e}_x + i \hat{e}_y) = \cos \theta \, \hat{e}_x + \sin \theta \, \hat{e}_y + i \cos \theta \, \hat{e}_y - i \sin \theta \, \hat{e}_x$$ so collecting terms, $$R_z(\theta) (\hat{e}_x + i \hat{e}_y) = (\cos \theta + i \sin \theta)(\hat{e}_x + i \hat{e}_y) = e^{i \theta} (\hat{e}_x + i \hat{e}_y)$$ which indicates that $\hat{e}_x + i \hat{e}_y$ has angular momentum $1$ and hence corresponds to $|1 \rangle$. A similar argument goes for $|-1 \rangle$. The overall phases and normalization factors are just by convention.

I am also not clear on the interpretation of the above relation

The $|0 \rangle$ state represents polarization about the $z$-axis. The $|\pm 1 \rangle$ states represent circular polarizations, rotating clockwise/counterclockwise about the $z$-axis.

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  • $\begingroup$ If we now go back to the spin-1/2 example, how do we make an analogous argument? We shouldn't be able to choose one state to be along z-direction as both $\left|+1/2\right>$ and $\left|-1/2\right>$ are changed by a rotation $R(\theta)$? $\endgroup$ – Leo L. Jan 6 at 0:48
  • $\begingroup$ @LeoL. Well, spin $1/2$ is quite different from spin $1$. For spin $1/2$, spin up and spin down are orthogonal. For spin $1$, spin up and spin down are the exact same thing, up to a phase. You can't use this argument from spin $1/2$, because those states don't transform by ordinary rotation matrices $R(\theta)$. $\endgroup$ – knzhou Jan 6 at 0:53
  • $\begingroup$ Just to make sure, when you say "spin up and spin down are orthogonal," you mean that the polarization angle between the two states has to be 90 degree right? $\endgroup$ – Leo L. Jan 6 at 0:58

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