# Clarification of a portion of the paper Quantum Amplitude Amplification and Estimation

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 $$0, then the probability of producing a good state upon measurement is given by $$\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}
• 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. Jul 24, 2019 at 22:07