I am reading the paper "Quantum Amplitude Amplification and Estimation", available here (pages 4 to 7 in particular). I will try to summarize my confusion over a statement on page $7$.

Suppose that $H$ is a finite-dimensional Hilbert space representing the state space of a quantum system spanned by the orthonormal basis states ${|x_i \rangle}_{i=1}^{n} \in H$. Every Boolean function $\chi: \mathbb{Z} \rightarrow \{0,1 \}$ induces a partition of $H$ into a direct sum of two subspaces, a good subspace, and a bad subspace.

Let $A$ be any quantum algorithm that acts on $H$ and uses no measurements, with $|\Psi \rangle = A |0 \rangle$. Decompose $|\Psi \rangle = |\Psi_0 \rangle + |\Psi_1 \rangle $, the projection of $|\Psi \rangle$ into the bad subspace and the good subspace, respectively. Define $Q(A, \chi) = -AS_0A^{-1}S_{\chi}$, where $S_\chi |y \rangle = -1^{\chi(y)} |y \rangle $, and $S_0$ changes the sign of the amplitude iff the state is the zero state $|0 \rangle$. The operator $Q$ is well-defined since we assume that $A$ has no measurements, and therefore has an inverse. Note that equivalently, $S_0 = (I-2|0 \rangle \langle 0|)$. Define $a = \langle \Psi_1 | \Psi_1 \rangle$.

Define $|\Psi_{\pm} \rangle = \frac {1}{\sqrt{2}} (\frac {1}{\sqrt{a}} |\Psi_1 \rangle \pm \frac {i}{\sqrt{1-a}} |\Psi_0 \rangle)$, which is an orthonormal basis of $H_{\Psi}$, the subspace spanned by $ |\Psi_0 \rangle$ and $|\Psi_1 \rangle$, with corresponding eigenvalues $\lambda_{\pm}=e^{-i2\theta_a}$ where $\theta_a$ satisfies $\sin^2{\theta_a}=a$.

The author then states:

The state $|\Psi \rangle = A|0 \rangle$ can be expressed in the eigenvector basis as $|\Psi \rangle = A|0 \rangle = \frac{-i}{\sqrt{2}}(e^{i\theta_a}|\Psi_+ \rangle -e^{-i\theta_a}|\Psi_- \rangle)$, and after $j$ applications of $Q$, we have that $Q^j |\Psi \rangle = \frac {-i} {\sqrt{2}}(e^{i(2j+1)\theta_a} |\Psi_+ \rangle - e^{-i(2j+1)\theta_a}|\Psi_- \rangle)$ $ = \frac {1}{\sqrt{a}}\sin((2j+1)\theta_a) |\Psi_1 \rangle + \frac {1}{\sqrt{1-a}}\cos((2j+1)\theta_a) |\Psi_0 \rangle$. If $0<a<1$, then the probability of producing a good state upon measurement is given by $\sin^2((2j+1)\theta_a)$

My question is what happened to the $\frac{1}{\sqrt{a}}$ in computing the probability of measuring $|\Psi_1 \rangle$? Is this a mistake or is there something simple I am missing?


The quantum state $|\Psi_1\rangle$ is not normalized to $1$. Thus the probability \begin{align} &\left|\frac{1}{\sqrt{a}}\sin((2j+1)\theta_a)|\Psi_1\rangle\right|^2\\ &= \frac{|\langle\Psi_1|\Psi_1\rangle|}{a} \left|\sin^2((2j+1)\theta_a)|\right|^2\\ &= \sin^2((2j+1)\theta_a) \end{align}

| improve this answer | |
  • $\begingroup$ Wait sorry, don't all quantum states have sum of modulus of amplitudes equal to 1? I'm a bit confused. $\endgroup$ – IntegrateThis Jul 24 '19 at 21:58
  • $\begingroup$ Quantum states are usually assumed to be normalized to 1, but as defined in the equation $|\Psi\rangle=|\Psi_0\rangle + |\Psi_1\rangle$, one can check that $\langle \Psi | \Psi \rangle$ can only equal one in general if $|\Psi_0\rangle$ and $|\Psi_1\rangle$ are not normalized to 1. $\endgroup$ – Guang Hao Low Jul 24 '19 at 22:07

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.