How to superpose two composite qubit states?

4

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 $|a\rangle + |b\rangle$ ?

wang1908

Posted 2019-04-17T09:39:43.750

Reputation: 83

Answers

7

Depending on what precisely your assumptions are about a, b, I think this is essentially impossible, and is something called the "no superposing theorem". Please see this paper.

DaftWullie

Posted 2019-04-17T09:39:43.750

Reputation: 35 722

4

Your question is not quite correctly defined.

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.

Danylo Y

Posted 2019-04-17T09:39:43.750

Reputation: 3 940

Can't you just set $t = 0 \mod 2\pi$? Or is $t$ something out of your control? – wizzwizz4 – 2019-04-17T15:01:44.863

2@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 – 2019-04-17T15:30:38.773