In https://arxiv.org/abs/quant-ph/0208112, the authors discuss a scheme to, given a discrete probability distribution $\mathbf p\equiv (p_i)_i$, under some assumptions on $\mathbf p$, prepare the superposition state $$|\psi(\mathbf p)\rangle = \sum_i \sqrt{p_i}|i\rangle.$$ To have a toy example in mind and concretise the discussion, say we want to prepare the two-qubit state $$\sum_{i=0}^3\sqrt{p_i}|i\rangle\equiv \sqrt{p_0}|00\rangle+\sqrt{p_1}|01\rangle +\sqrt{p_2}|10\rangle+ \sqrt{p_3}|11\rangle.$$ As also discussed in How does the induction step in the Grover-Rudolph scheme to prepare superpositions from probabilities work?, the gist of the scheme is to iteratively "refine" the superposition iteratively. In other words, you start from $|0\rangle$, evolve it to $\sqrt{p_0+p_1}|0\rangle+\sqrt{p_2+p_3}|1\rangle$, and then implement another evolution such that $$\sqrt{p_0+p_1}|0\rangle \to \sqrt{p_0}|00\rangle + \sqrt{p_1}|01\rangle, \\ \sqrt{p_2+p_3}|1\rangle \to \sqrt{p_2}|10\rangle + \sqrt{p_3}|11\rangle.$$ To implement these evolutions, the authors propose a two step procedure:

  1. Evolve $|0\rangle$ to $|0\rangle|\theta_0\rangle$ for some suitably defined $\theta_0$. This you can do as long as computing $\theta_0$ is efficient classically.
  2. Use $\theta_0$ as control to perform a rotation on another ancillary qubit. Suitably choosing $\theta_0$, you can thus way implement the evolution $$|0\rangle|\theta_0\rangle\to |0\rangle|\theta_0\rangle\frac{\sqrt{p_0}|0\rangle+\sqrt{p_1}|1\rangle}{\sqrt{p_0+p_1}}.$$

The same procedure is applied separately to $|1\rangle$, resulting in $$|1\rangle|\theta_1\rangle\to |1\rangle|\theta_1\rangle\frac{\sqrt{p_2}|0\rangle+\sqrt{p_3}|1\rangle}{\sqrt{p_2+p_3}}.$$ Overall, this process gives us the evolution $$\sqrt{p_0+p_1}|0\rangle+\sqrt{p_2+p_3}|1\rangle \to |0\rangle|\theta_0\rangle(\sqrt{p_0}|0\rangle+\sqrt{p_1}|1\rangle) + |1\rangle|\theta_1\rangle(\sqrt{p_2}|0\rangle+\sqrt{p_3}|1\rangle).$$ That's almost the target superposition, except for the lingering ancillary qubits $|\theta_i\rangle$, which are however entangled with the rest of the qubits and therefore cannot just be ignored.

The authors just write that "we uncompute the register containing $|\theta_i\rangle$ to leave us with the desired state. How does this computation step work precisely, and what's its cost?

  • 1
    $\begingroup$ If you assume that $\theta$ can be computed efficiently, then you can also efficiently build the oracle $|x,y\rangle\to\left|x,y\oplus\theta_x\right\rangle$, don't you? It seems to me that step 1 happens in superposition, so if step 1 is possible, then you have access to such an oracle and you can uncompute the register. Or am I missing something? $\endgroup$ Aug 9, 2023 at 18:27
  • $\begingroup$ @TristanNemoz I'm not doubting it can be done, I'm just asking what the procedure would be like in this context. But I might have got confused thinking $\theta_i$ as a real number. Here it obviously has to be an integer (or the ket doesn't make sense), and so you just need to do the controlled operation as you point out to take $\theta_i$ out of the picture. Or in other words, you run the inverse of the gate used to create the $\theta_i$ states (which in this case is the gate itself). Thanks $\endgroup$
    – glS
    Aug 10, 2023 at 9:02
  • $\begingroup$ @glS is a [grover-rudolph] tag worth creating? $\endgroup$ Aug 11, 2023 at 20:22
  • $\begingroup$ @MarkSpinelli it might, though the term is not very common. We also already have a state-preparation tag $\endgroup$
    – glS
    Aug 13, 2023 at 18:21

1 Answer 1


The uncomputation is performed by simply running the same unitary used to generate the $|\theta_i\rangle$ states in reverse.

We could describe what's going on more abstractly as follows:

  1. Start with some $\alpha|0\rangle+\beta|1\rangle$
  2. Implement a unitary $U$ such that $U|i\rangle|0\rangle=|i\rangle|\theta_i\rangle$, $i=0,1$. Here $\theta_i\in\mathbb{Z}_2^n$ if the ancilla register used to store the rotation angles $\theta_i$ uses $n$ qubits.
  3. Run the controlled rotations. If the corresponding unitary is $R$, then by definition this is an operation such that $$R|i\rangle|\theta_i\rangle|0\rangle = |i\rangle|\theta_i\rangle |\psi_i\rangle$$ for some states $|\psi_i\rangle\equiv R_i|0\rangle$. The overall state we now have is $$RU(\alpha|0\rangle+\beta|1\rangle) = \alpha|0\rangle|\theta_0\rangle|\psi_0\rangle + \beta|1\rangle|\theta_1\rangle|\psi_1\rangle.$$
  4. For the uncomputation, we just use $U^\dagger$. From the definition of $U$, it immediately follows that $U^\dagger|i\rangle|\theta_i\rangle=|i\rangle|0\rangle$, and thus $$U^\dagger R U (\alpha|0\rangle+\beta|1\rangle)=\alpha|0\rangle|0\rangle|\psi_0\rangle+\beta|1\rangle|0\rangle|\psi_1\rangle \simeq \alpha |0\rangle|\psi_0\rangle+\beta|1\rangle|\psi_1\rangle.$$

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