Oblivious Amplitude Amplification & Eigenstate decomposition

Question

Looking at the oblivious AA (OAA) : https://docs.microsoft.com/en-us/azure/quantum/user-guide/libraries/standard/algorithms

I am trying to figure out the eigendecomposition of Q, which leads to below. Any help or suggestion on how to derive these would be highly appreciated!

$$ |\psi\rangle=\frac{-i}{\sqrt{2}}\left(e^{i \theta}\left|\psi_{+}\right\rangle+e^{-i \theta}\left|\psi_{-}\right\rangle\right) $$ where $\left|\psi_{\pm}\right\rangle$are eigenvectors of $Q$ with eigenvalues $e^{\pm 2 i \theta}$ and only have support on the $+1$ eigenvectors of $P_{0}$ and $P_{1}$.

Here are the key sentences from the website

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Each iteration of Amplitude amplification requires that two reflection operators be specified. Specifically, if $Q$ is the amplitude amplification iterate and $P_{0}$ is a projector operator onto the initial subspace and $P_{1}$ is the projector onto the marked subspace then $Q=-\left(\mathbf{1}-2 P_{0}\right)\left(\mathbf{1}-2 P_{1}\right)$. Recall that a projector is a Hermitian operator that has eigenvalues $+1$ and 0 and as a result $\left(\mathbf{1}-2 P_{0}\right)$ is unitary because it has eigenvalues that are roots of unity (in this case $\pm 1$ ). As an example, consider the case of Grover's search with initial state $H^{\otimes n}|0\rangle$ and marked state $|m\rangle, P_{0}=H^{\otimes n}|0\rangle\langle 0| H^{\otimes n}$ and $P_{1}=|m\rangle\langle m| .$ In most applications of amplitude amplification $P_{0}$ will be a projector onto an initial state meaning that $P_{0}=\mathbf{1}-2|\psi\rangle\langle\psi|$ for some vector $|\psi\rangle ;$ however, for oblivious amplitude amplification $P_{0}$ will typically project onto many quantum states (for example, the multiplicity of the $+1$ eigenvalue of $P_{0}$ is greater than 1 ).

The logic behind amplitude amplification follows directly from the eigen-decomposition of $Q .$ Specifically, the eigenvectors of $Q$ that the initial state has non-zero support over can be shown to be linear combinations of the $+1$ eigenvectors of $P_{0}$ and $P_{1}$. Specifically, the initial state for amplitude amplification (assuming it is a $+1$ eigenvector of $P_{0}$ ) can be written as $$ |\psi\rangle=\frac{-i}{\sqrt{2}}\left(e^{i \theta}\left|\psi_{+}\right\rangle+e^{-i \theta}\left|\psi_{-}\right\rangle\right) $$ where $\left|\psi_{\pm}\right\rangle$are eigenvectors of $Q$ with eigenvalues $e^{\pm 2 i \theta}$ and only have support on the $+1$ eigenvectors of $P_{0}$ and $P_{1}$. The fact that the eigenvalues are $e^{\pm i \theta}$ implies that the operator $Q$ performs a rotation in a two-dimensional subspace specified by the two projectors and the initial state where the rotation angle is $2 \theta$. This is why after $m$ iterations of $Q$ the success probability is $\sin ^{2}([2 m+1] \theta)$

Answer 1

This is all proved in pgs. 9-11 of the paper that first introduced OAA, but I can summarize the important points here.

The starting point is assuming access to a block-encoding of some desired unitary action $V$ which lives inside a larger unitary $U$, and is accessed by successfully projecting into the 'good' subspace $|0^\mu\rangle \otimes I$:

\begin{align} U |0^\mu\rangle|\psi\rangle = \sin(\theta)|0^\mu\rangle V|\psi\rangle + \cos(\theta) |\Phi^\perp\rangle \equiv U |\Psi\rangle = \sqrt{p} |\Phi\rangle + \sqrt{1-p} |\Phi^\perp\rangle \end{align}

Since we only succeed with some probability ($p = \sin^2(\theta)$), we want to boost the weight of this term, but standard amplitude amplification would require knowing $|\psi\rangle$, which could be arbitrary. The trick is to attach that $|0^\mu\rangle$ register so that in fact, we do know the subspace that contains the desired state, because its simply the space defined by the projector $R = 2\Pi - I$, with $\Pi = |0^\mu\rangle\langle 0^\mu| \otimes I$. We only need to figure out how to do the iterated reflections about these subspaces, which together should produce a rotation of $2\theta$ between the subspaces. In that paper, it is shown that the action of $U$ takes the following form \begin{align} U = \begin{bmatrix} \sin{\theta} & \cos{\theta} \\ \cos\theta & -\sin{\theta} \end{bmatrix} \end{align} but in a special basis, where $\{ |\Psi\rangle, |\Psi^\perp\rangle \}$ is mapped to $\{ |\Phi\rangle, |\Phi^\perp\rangle \}$. As a result, $U^\dagger$ has the same form, but maps from the $\Phi$-basis back to the $\Psi$-basis. This matrix has eigenvalues $-1,+1$ and so it is a reflection. If you sandwich it with the other reflection defined earlier to produce the operator $S = -U R U^\dagger R$, you can work out the action of $S$ to be \begin{align} S = \begin{bmatrix} \cos{2\theta} & \sin{2\theta} \\ -\sin2\theta & \cos{2\theta} \end{bmatrix} \end{align} which is a now a rotation of angle $2\theta$ between the input and output subspaces. The action of $S$ can be written in terms of its spectral decomposition \begin{align} S = e^{-i2\theta} |v^{\Phi}_1\rangle\langle v_1^{\Psi}| + e^{i2\theta} |v_2^{\Phi} \rangle\langle v_2^{\Psi}| \end{align} where $|v^{\Psi}_1\rangle = i |\Psi\rangle + |\Psi^\perp\rangle$ and $|v^{\Psi}_2\rangle = -i |\Psi\rangle + |\Psi^\perp\rangle$ (and similarly for the $\Phi$ superscripts). The input state $U|\Psi\rangle$ expressed in the $\Phi$-basis eigenvectors has the form given in the Microsoft link (if you also renormalize the final state, I think).

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The secret going on behind the scenes is that a certain result called Jordan's Lemma says that a product of reflections decomposes as a direct sum over 1 and 2-dimensional subspaces, which in this case are invariant under the action of the projectors $\Pi$ and $U^\dagger \Pi U$. Although these projectors are rank-$2^n$, they act as rank-1 projectors in each of these spaces, and therefore define the space(s) in which $S$ performs its rotations.