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The setup is the following:

We have a particle beam of spin-up (:= $|+\rangle$) particles coming in a Stern-Gerlach apparatus which measures spin in x-direction.

After passing through, the beam splits up and one will get two beams with states $|+_x\rangle$ , and $|-_x\rangle$, denoting the eigenstates of $\hat{S_x}$.

Afterwards both beams are merged together without a phase difference or an additional measurement happening.

The Question is: What is the resulting spin-state?

My answer would be, that the final spin state is a mixed state described by the density-operator: $\hat{\rho} = \frac{1}{2} |+_x\rangle \langle +_x | + \frac{1}{2} |-_x\rangle \langle -_x |$, but our provided answer says, that the final state is $|+\rangle$, the same state we started with.

Does somebody know why this is true, and where my mistake is?

Thank you so much.

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The density matrix you gave would be the case if you had a 50% probability of your state being $|+_x\rangle$ and a 50% probability of your state being $|-_x\rangle$. But you started with a pure wave function, and no measurement was made, so you will end up with a pure wave function, not a mixed state.

Just use the definitions of $|+_x\rangle$ and $|-_x\rangle$ and then add them together and renormalize, since the two beams were combined:

$$ |+_x\rangle = \frac{1}{\sqrt{2}}(|+\rangle + |-\rangle), \\ |-_x\rangle = \frac{1}{\sqrt{2}}(|+\rangle - |-\rangle), \\ \textrm{Final State} = \frac{|+_x\rangle+|-_x\rangle}{\sqrt{\langle+_x|+_x\rangle + \langle-_x|-_x\rangle}} \\ =\frac{\frac{2}{\sqrt{2}}|+\rangle}{\sqrt{1 + 1}} \\ =|+\rangle. $$

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