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Suppose we start with $|00...0\rangle$. We want to build an equal superposition over $|0\rangle + ... + |n-1\rangle$.

When $n=2^m$ for some $m$, I know I can do this using $H^{\otimes m}$.

What is the general circuit for this (i.e. in case $n$ is not power of 2)?

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1 Answer 1

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How to prepare a uniform superposition over a range.

Simple: Repeat until success

The simplest approach is to prepare an $n$-qubit quantum integer register $k$, where $n=\lceil \lg_2 N \rceil$, in the state $|+\rangle^{\otimes n}$. That can be done very cheaply using reset gates and Hadamard gates. Then use a comparison operation to measure whether $k<N$ or not. If not, then retry. The main downside of this repeat-until-success approach is that it isn't reversible, and sometimes these sorts of preparation tasks occur inside subroutines where reversibility is needed due to controlling and uncomputing the preparation.

Example repeat-until-success circuit in Quirk, cycling over N=1 to N=127 (technically should only go from 65 to 127 for a 7 qubit system like this):

enter image description here

Flexible: Amplitude Amplification

There is a deterministic reversible circuit with the same cost as the repeat-until-success strategy. You can use a single step of amplitude amplification, with a less-than-N comparison as the oracle, to get to a uniform superposition over the range [0, N). This has a gate count of $O(\lg N+\lg\frac{1}{\epsilon})$, where the $\epsilon$ comes from how closely you approximate the rotation angle within the amplification.

  • Remove any factors of two in $N$ by adding a qubit in the $|+\rangle$ state. $N$ is now odd. Skip the remaining steps if $N=1$.
  • Let $\theta = \arccos(1 - 2^{\lfloor \lg_2 N \rfloor} / N)$.
  • Perform one step of amplification. Use a diffusion angle of $\theta$. Use $f(k) = k < N$ as the oracle.
  • The system is now in the desired state.

Example amplification circuit in Quirk, for N=100

enter image description here

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