# How to sample vectors close to the minimum eigenvector of a unitary matrix?

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 - $$| \langle v_{k}|\phi\rangle|^{2} > | \langle v_{i}|\phi\rangle|^{2},\qquad \forall\ i \neq k.$$

• 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 $|\alpha_k|^2$? Jun 22, 2021 at 8:48
• 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). Jun 22, 2021 at 9:34
• 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? Jun 22, 2021 at 10:37
• Do we have to make use only of $U$, or might we be able to use controlled-$U$? Jun 22, 2021 at 11:38
• 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 $U$. Jun 23, 2021 at 15:19

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.

• 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. Jun 22, 2021 at 17:29
• 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. Jun 23, 2021 at 12:05