Stern-Gerlach and density operators

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.

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