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

  • $\begingroup$ 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
    Commented Jun 22, 2021 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). $\endgroup$ Commented Jun 22, 2021 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
    Commented Jun 22, 2021 at 10:37
  • $\begingroup$ Do we have to make use only of $U$, or might we be able to use controlled-$U$? $\endgroup$
    – DaftWullie
    Commented Jun 22, 2021 at 11:38
  • 1
    $\begingroup$ 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
    Commented Jun 23, 2021 at 15:19

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$ Commented Jun 22, 2021 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. $\endgroup$ Commented Jun 23, 2021 at 12:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.