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?


1 Answer 1


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}

  • $\begingroup$ Wait sorry, don't all quantum states have sum of modulus of amplitudes equal to 1? I'm a bit confused. $\endgroup$ Jul 24, 2019 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. $\endgroup$ Jul 24, 2019 at 22:07

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