2
$\begingroup$

Suppose I have a three-qubit entangled state of the following form:

$$ |00\rangle|\psi_1\rangle + |01\rangle|\psi_2\rangle + |10\rangle|\psi_3\rangle + |11\rangle|\psi_4\rangle $$

I refer to the first two qubits as address qubits. The third qubit is the data qubit.

I want to be able to extract some arbitrary $|\psi_i\rangle$, but collapsing the first two qubits and post-selecting has a success probability of $1/4$. And this probability gets even worse with more address qubits: $1/2^n$.

Trying amplitude amplification on the address qubits does not work as the amplitudes of the data qubit also get modified.

Is there any procedure that allows me to increase the success probability of $1/2^n$? Or maybe some proof that it is not possible to do such thing?

$\endgroup$
5
  • 1
    $\begingroup$ I'm confused as to your reason for amplitude amplification not working: assuming you fulfil the conditions of the algorithm (most importantly, having a unitary $U$ such that $U|0\rangle$ gives your initial state), it should work. $\endgroup$
    – DaftWullie
    Nov 24 at 15:46
  • $\begingroup$ I assume the $|\psi_i\rangle$ are unknown, and (in general) not orthogonal? $\endgroup$
    – DaftWullie
    Nov 24 at 15:48
  • $\begingroup$ @DaftWullie I think I wasn’t clear enough. I meant amplitude amplification only on the address qubits, which is why it doesn’t work. So this would be on a scenario were $U$ that prepares the three-qubit state is not efficient and we only want to act on the first two qubits after having the initial state. $\endgroup$
    – epelaaez
    Nov 24 at 15:49
  • 2
    $\begingroup$ But even if the U is inefficient, using it $\sqrt{2^n}$ times is better than using it $O(2^n)$ times in your collapsing qubits protocol. $\endgroup$
    – DaftWullie
    Nov 24 at 15:50
  • $\begingroup$ @DaftWullie that makes sense, I’m thinking more of a situation where we want a database-like structure where we would only use the initialization unitary once and this is the extraction part, but I guess AA on the whole system is still the more efficient way. Then maybe QRAM is the way to go although it is more resource inefficient (what I wanted to avoid). $\endgroup$
    – epelaaez
    Nov 24 at 15:55
3
$\begingroup$

The operation you're asking for is equivalent to postselection. You're trying to force a measurement result of the address register. If that were possible it would make BQP = PostBQP. There's no proof that they're not equal, but it would be very surprising. PostBQP can trivially solve NP complete problems (eg. imagine the state was $\sum_k |\text{IsSolution}(k)\rangle |+\rangle|k\rangle$). Basically, you can infer there must be hard cases where the circuit achieving your task is going to have to be exponentially huge.

You need some additional preconditions that guarantee the circuit construction problem is not equivalent to solving NP complete problems.

I meant amplitude amplification only on the address qubits

You can't change what's in register B by unitary operations on register A. At best, non-unitary operations like measurement on A can tell you information about what's in B. Which is different from forcing it to be something. Any correct circuit for this problem must operate on the address register and the value register.

$\endgroup$
3
  • $\begingroup$ This makes sense. Just one thing, could you elaborate a bit more on how the operation I'm asking for is equivalent to postselection? I'm looking for something to increase the probability of having succesful measurement, which would go before the actual measurement and postselection. Or do this operations also count as part of the postselection process? In that case, would applying AA on the complete system (register A and B) also be part of postselection? $\endgroup$
    – epelaaez
    Nov 24 at 18:12
  • 2
    $\begingroup$ @epelaaez Any process that unconditionally increases the probability of a measurement result is a problem for a bunch of different reasons. It violates linearity which gives you superpowers. For example, it would allow FTL communication. Alice has the value register, Bob has the address register, Bob uses the address register to remotely force the value register to be a chosen message to Alice. $\endgroup$ Nov 24 at 18:34
  • $\begingroup$ That clears up my doubts! Thanks! $\endgroup$
    – epelaaez
    Nov 24 at 18:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.