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If one measures the projection of spin of a spin half particle along the x axis one will always get plus or minus half $\hbar$

Measuring it along the y axis one will always get plus or minus half $\hbar$

Measuring it along the z axis one will always get plus or minus half $\hbar$

Measuring it along "any" axis one will always get plus or minus half $\hbar$

Why the possible outcomes are the same for all directions?

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Possible duplicate: physics.stackexchange.com/q/20581/2451 –  Qmechanic Dec 21 '12 at 10:20
    
Qmechanic's nice answer to an older copy of the same question refers to the representation theory of groups. Let me add that it would be a contradiction in classical physics is the projection of spin were quantized with respect to any axis - in classical physics, the spin $j_z$ has to be continuous because we may continuously change the axes (rotational symmetry) which has a continuous effect on $j_z$. However, in quantum physics, what is changing continuously are probabilities that the spin is up or down, but the choices are still just two, discrete ones, and the rot. symmetry isn't violated. –  Luboš Motl Dec 21 '12 at 10:49
    
@Qmechanic Yeah I saw that answer but it does not say why the possible results are always the same independent of any direction. May be I am missing something –  Revo Dec 21 '12 at 11:23
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Dear Revo, if a derivation shows that the spectrum of $J_z$ is a particular finite set, it doesn't have to say any extra "why it's independent of the direction". Because no assumption was made about the physical orientation of the $z$-axis and because of the underlying and provable rotational symmetry of the formalism as well as observations, the derivation clearly applies to any axis. So the obstacle you are adding or missing is your own creation. –  Luboš Motl Dec 21 '12 at 11:45
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Are you asking why you can randomly choose a direction, measure $S_z$ in that direction and you always get 1/2 or -1/2? If so, isn't because the act of measurement singles out the axis you're measuring along and makes it special e.g. there's an electric field in that direction. –  John Rennie Dec 21 '12 at 12:03
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Two-valued spin projection is a (quantum) feature of spin-1/2 particles. It is like finding a particle "here" or "there" in a classical physics: it must be somewhere withing only two possibilities, like in two channels. By the way, splitting in two channels have a direct physical meaning, see the experiment of Stern-Gerlach in wikipedia.

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It has to do with the act of picking a preferred axis along which to measure the spin projection. Upon observing it, you force the particle into a specific state with a given projection along that (say, the $z$) axis, but now the projection along the other axes are indeterminate, due to the superposition that the particle is now in. They do have a value, it is just uncertain until you actually measure it - but then you screw up the certainty of your first measurement.

What I'm about to state is a bit abusive/misleading, but it may help you visualize things.

Technically speaking, the length squared of the spin vector for a particle of spin $s$ is $\bar{S}^2=s(s+1)$ (I assume $\hbar=1$), which for spin-half particles is $\frac{3}{4}$. The contribution of the $z$ component is $S_z=(\pm\frac{1}{2})^2=\frac{1}{4}$, so the rest must come from the other components. Specifically, since spin-half particles can only have projections of plus or minus half (just like you stated), while the mean value of the other components may vanish, the mean value of their square has to be $\frac{1}{4}$, so together all three components give the total length.

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