Understanding theorem 6 of postBQP=PP paper

Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Stack Overflow for Teams is now called Stack Internal. Bring the best of human thought and AI automation together at your work. Bring the best of human thought and AI automation together at your work. Learn more TeamsQ&A for workConnect and share knowledge within a single location that is structured and easy to search.In this paper, $\mathsf{BQP}_p$ is defined to be similar to $\mathsf{BQP}$ in all aspects except that the Born's measurement rule is changed. For a normalised state $$|{\psi}\rangle =\sum_x \alpha_x |x\rangle$$ , the probability of measuring the basis state $|x\rangle$ equals$$p(|x\rangle) = \frac{|\alpha_x|^p}{\sum_y |\alpha_y|^p} $$rather than $|\alpha_x|^2$.Theorem 6 from the paper says $\mathsf{PP\subseteq BQP_p}$. The proof goes as follows.To simulate $\mathsf{PP}$ in $\mathsf{BQP}_p$, run the algorithm of theorem 4, having initiazed initialized $O(n^2q(n)/|2-p|)$ ancilla qubits to $|0\rangle$ for some sufficiently large polynomial $q$. Suppose the algorithm's state at some point is $\sum_z\alpha_z|z\rangle$, and we want to postselect on the event $|z\rangle \in S$, where $S$ is a subset of basis states. Here is how:If $p>2$, then apply Hadamard gates to $K=2q(n)/(2-p)$ fresh ancilla qubits consitioned on $|z\rangle \in S$. The result is to increase the probability mass of each $|z\rangle \in S$ from $|\alpha_z|^p$ to$$2^K\cdot |\color{red}{2^{-K/2}}\alpha_z|^p = 2^{q(n)}|\alpha_z|^p, \tag{1}$$while the probability mass of each $|z\rangle \notin S$ remains unchanged.My query: In the above eqn (1), the right-hand side has the factor of $2^{-K/2}$. I am unable to understand it. I was anticipating $2^{-K/p}$.Any hint will be much appreciated.Thanks for contributing an answer to Quantum Computing Stack Exchange!But avoid …Use MathJax to format equations. MathJax reference.To learn more, see our tips on writing great answers.Required, but never shown By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy. Start asking to get answersFind the answer to your question by asking.Explore related questionsSee similar questions with these tags.To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA . rev 2025.11.20.37056