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I have a system of two spin 1/2 particles in the z-direction given by $\phi=\frac{1}{\sqrt{2}} \Big( |->|+> +|->|-> \Big)$ where $+$ means up and $-$ means down. The first ket in each term designates the first particle and the second one is linked to the second particle. All the kets in the z-direction will be written without index from now on.

Let's say I measure $S_{1x}$ and I get the value $+\frac{\hbar}{2}$, which would mean that the state points towards the positive direction in the x axis. Then which would be the state after the measurement?

On the one hand, using the projectors and $|+>_x$ in terms of the eigenvectors of $S_{1z}$ ($|+>_x=\frac{|+>+|->}{\sqrt{2}}$), I obtain

$$ \Big(\frac{|+>+|->}{\sqrt{2}} \otimes I \Big) \Big(\frac{<+|+<-|}{\sqrt{2}} \otimes I \Big) \phi= \Big(\frac{|+>+|->}{\sqrt{2}}\otimes I \Big) \frac{1}{2} \Big(I \otimes |+> + I\otimes |-> \Big) = \frac{1}{2\sqrt{2}} \Big(|+>|+>+|->|+>+|+>|->+|->|-> \Big) $$

which, once normalized becomes

$$\frac{1}{2} \Big(|+>|+>+|->|+>+|+>|->+|->|-> \Big)$$

However, on the other hand, $S_{1x}=\frac{1}{2} \Big(S_{1+}+S_{1-} \Big)$, which applied to the initial state $\phi$ gives $$S_{1x} \phi = \frac{\hbar}{2\sqrt{2}} \Big(|+>|+>+|+>|-> \Big)$$

My question is why I'm not obtaining the same result with both methods, as well as which one is the correct method to find the state after the measurement.

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Write your state as $$ \vert{\phi}\rangle=\frac{1}{\sqrt{2}}\vert{-}\rangle_1\left( \vert + \rangle_2+\vert -\rangle_2\right) $$ Since the eigenstate of spin-up along $\hat x$ is $$ \frac{1}{\sqrt{2}}\left( \vert + \rangle_1+\vert -\rangle_1\right) $$ your measurement ought to be (notwithstanding $\otimes 1$) $$ \hat \Pi_{x,1,+} = \frac{1}{2}\Bigl[ \vert + \rangle_1+\vert -\rangle_1\Bigr] \Bigl[_1\langle + \vert +{_1\langle} -\vert\Bigr] $$ which, when acting on $\vert -\rangle_1$ gives $$ \frac{1}{2}\left( \vert + \rangle_1+\vert -\rangle_1\right) $$ which is indeed the $\vert +\rangle_{x,1}$ state as desired (i.e. the outcome is $+x$). When acting on the full $\vert \phi\rangle$ you will get $$ \hat \Pi_{x,1,+}\vert\phi\rangle= \frac{1}{2\sqrt{2}}\Bigl[ \vert + \rangle_1+\vert -\rangle_1\Bigr]\Bigl[ \vert + \rangle_2+\vert -\rangle_2\Bigr]\, . $$ As expected, this is not a normalized state since the projector does not preserve the norm.

Your initial ket in the $1$ subspace $\vert -\rangle_z$ is not an eigenstate of $S_{x,1}$ and $S_{x,1}\ne \hat\Pi_{x,1,+}$ so I don't see why you want to use the action of $S_{x,1}$ in your problem. In particular, $S_{x,1}$ is hermitian but $\hat\Pi_{x,1,+}$ is not so there is no way their action can be equalled.

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  • $\begingroup$ So what effect does the action of $S_{x,1}=S_{1+}+S{1-}$ have on the initial state if it's not the same as using the projector? I mean, if it doesn't give the state after the measurement, what's its physical meaning? Applying this operator definitely changes the state, after all. $\endgroup$ – Sky Apr 18 '17 at 7:57
  • $\begingroup$ Observables are not used to transform states. You compute their averages etc, and in some cases (like $S_x$) they have an interpretation as generators of infinitesimal transformations. Their eigenvalues and eigenvectors have important applications. $\endgroup$ – ZeroTheHero Apr 18 '17 at 8:04

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