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Say that we have an unknown $2^{n}\times2^{n}$ unitary matrix $U$ with eigenvectors $|v_{i}\rangle$ and eigenvalues $e^{2\pi j \theta_{i}}$and we want to sample a vector, say $|\phi \rangle$. Since the eigenvectors of $U$ form an orthonormal basis, $|\phi \rangle$ can be written as $$ |\phi \rangle = \sum_{i} \alpha_{i}|v_{i} \rangle,$$ where $|\alpha_{i}|^{2}$ is the probability of $|\phi \rangle$ collapsing to any of the eigenvector states $|v_{i}\rangle$.

Let us assume that there is no degeneracy and there exists some minimum eigenvalue $\theta_{k}$ with the associated eigenvector $|v_{k}\rangle$. What I want to know is whether there is a way to sample $|\phi \rangle$ such that amongst all the eigenvectors $|v_{i}\rangle$, it is closest to $|v_{k}\rangle$. In other words, $|\phi \rangle$ is such that - $$ | \langle v_{k}|\phi\rangle|^{2} > | \langle v_{i}|\phi\rangle|^{2},\qquad \forall\ i \neq k.$$

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  • $\begingroup$ I assume you're taking $0\leq\theta_i<1$? Also, I'm a little confused - is $|\phi\rangle$ given to you (i.e. are the $\alpha_i$ fixed)? Or is your question how to construct the $|\phi\rangle$ in such a way as to maximise $|\alpha_k|^2$? $\endgroup$
    – DaftWullie
    Jun 22 at 8:48
  • $\begingroup$ Yes, that's exactly the question. I want to construct the $|\phi \rangle$ state in such a way so as to maximise $|\alpha_{k}|^{2}$. Also, $\theta_{i}$ does belong to the domain [0,1). $\endgroup$ Jun 22 at 9:34
  • $\begingroup$ I'm not sure I fully understand the rules of the game. Can you give a sketch of what you expect such a sampling procedure to look like? $\endgroup$
    – Rammus
    Jun 22 at 10:37
  • $\begingroup$ Do we have to make use only of $U$, or might we be able to use controlled-$U$? $\endgroup$
    – DaftWullie
    Jun 22 at 11:38
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    $\begingroup$ I suspect that this may be impossible based on your definition of "minimum". Here's a vague argument. Suppose you have a unitary $U$ that has a min eigenvalue $e^{2 \epsilon \pi i}$ for some small $\epsilon > 0$. Then define a unitary $U'$ which is the same as $U$ except you perturb this min eigenvalue to $e^{2 (\epsilon - \delta) \pi i}$ with $\delta>0$. This new unitary should be close to $U$ if $\delta$ or $\epsilon$ are small but if $\delta > \epsilon$ then suddenly this eigenvalue is very large by your definition. Thus your algorithm wont work continuously on the space of unitaries $U$. $\endgroup$
    – Rammus
    Jun 23 at 15:19
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I am creating an instance of your generic problem:

  1. If we consider $n=1$ then $Z$ Gate will become your $2^n \times 2^n$ "unknown" unitary matrix $U$.
  2. Then $\vert 0 \rangle$ and $\vert 1\rangle$ will be its eigenvectors and $1$ and $-1$ the corresponding eigenvalues.
  3. Now $\vert \phi\rangle$ is a vector which can be represented as a linear combination of given eigenvectors $\vert 0 \rangle$ and $\vert 1\rangle$.
  4. In this scenario, $\vert 1\rangle$ is the eigenvector associated with minimum eigenvalue of $-1$.

Let us select one of the eigenvector at random, such as $\vert 1\rangle$. Now you want to say that the system is in state $\vert \phi\rangle$ and it's measurement is always closer to $\vert 1 \rangle$ than to $\vert 0\rangle$ i.e. $\vert \langle 1 \vert \phi \rangle\vert^2 > \vert \langle 0 \vert \phi \rangle\vert^2$

Given an eigenvector $\vert V_k\rangle$, you want a formal way to find a $\vert \phi \rangle$ state which is closer to $\vert V_k \rangle$. i.e. in this case, find $\vert \phi \rangle$ which is closer to $\vert 1 \rangle$.

If I have understood the problem properly then I recommend following approach:

$$ \vert \langle V_k \vert \phi \rangle \vert^2 > \vert \langle V_i \vert \phi \rangle \vert^2, \forall i \neq k $$ $$ \vert \phi \rangle = \sum_{i} a_i \vert V_i \rangle = a_k \vert V_k \rangle + \sum_{i \neq k} a_i \vert V_i \rangle $$ $$ \vert a_k \vert ^2 > \sum_{i \neq k} \vert a_i \vert^2 $$

You need to select a state $\vert \phi \rangle$ such that probability of selected eigenvector is greater than sum total of all other states in its wavefunction.

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  • $\begingroup$ Well Sachin thanks for the answer. You understood the question correctly but I'd like to ask 2 things.First, how can we say that a single qubit matrix boils down to just a Z-gate. Second, since the matrix is 'unknown',as is in the case of QPE,we do not have the knowledge of the eigenvectors of U, so I am not sure if we can "select" eigenvectors. $\endgroup$ Jun 22 at 17:29
  • $\begingroup$ I have considered Z-gates for illustration purpose only. If we provide superposition of eigenstates to QPE then we will obtain corresponding superposition of related eigenphases. $\endgroup$ Jun 23 at 12:05

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