How to superpose two composite qubit states?

Assuming we have two sets of $$n$$ qubits. The first set of $$n$$ qubits is in state $$|a\rangle$$ and second set in $$|b\rangle$$. Is there a fixed procedure that generates a superposed state of the two $$|a\rangle + |b\rangle$$ ?

First of all, $$|a\rangle + |b\rangle$$ is not a state. You need to normalize it by considering $$\frac{1}{|||a\rangle + |b\rangle||}(|a\rangle + |b\rangle)$$.
Secondly, in fact, you don't have access to the states $$|a\rangle$$ and $$|b\rangle$$ but to the states up to some global phase, i.e. you can think that the first register is in the vector-state $$e^{i\phi}|a\rangle$$ and the second register is in the vector-state $$e^{i\psi}|b\rangle$$ with inaccessible $$\phi, \psi$$. Since you don't have access to $$\phi, \psi$$, you can't define sum $$|a\rangle + |b\rangle$$. But you can ask to construct normalized state $$|a\rangle + e^{ it}|b\rangle$$ for some $$t$$. This question is correct. Though, as DaftWullie pointed out in his answer, it is impossible to construct such a state.
• Can't you just set $t = 0 \mod 2\pi$? Or is $t$ something out of your control? – wizzwizz4 Apr 17 at 15:01
• @wizzwizz4 Yes, $t$ is unknown. Again, you can't define sum $e^{i\phi}|a\rangle + e^{i\psi}|b\rangle$ to be equal to some exact state (that depends only on $|a\rangle$ and $|b\rangle$). But you can define this sum to be of some type of state. This type of state can be defined as $|a\rangle + e^{ it}|b\rangle$ since $e^{i\phi}|a\rangle + e^{i\psi}|b\rangle = e^{i\phi} (|a\rangle + e^{ i(\psi - \phi)}|b\rangle) \propto |a\rangle + e^{ i(\psi - \phi)}|b\rangle$. – Danylo Y Apr 17 at 15:30