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In quantum computing, one of the central areas of study is to determine efficient quantum circuits - described by unitaries - to prepare a state $|\psi\rangle$ from the initial computational basis state $|0\cdots 0\rangle$.

I am currently interested the mathematical abstraction of this problem; to be specific, if we have a $d$ dimensional Hilbert space, an initial state $|0\rangle$ and a final state $|\psi\rangle$, can we say anything about the set of unitaries $U$ such that $U|0\rangle = |\psi\rangle$? In particular, I am wondering about the following two things:

  1. Does this set of unitaries form a submanifold of $U(d)$? (It clearly does not form a Lie subgroup.)
  2. If we know a single unitary $U_0$ such that $U_0|0\rangle = |\psi\rangle$, can we obtain other unitaries with this property from $U_0$? Can we obtain all of the unitaries with this property from $U_0$?

My guess to the first question is that this set of unitaries forms a submanifold of $U(d)$ of dimension $d^2 - d$ isomorphic to $U(d) \ / \ \mathbb{C}^d$, but I am completely unsure about the second. Are there known answers to these questions, or in general is there anything that can be said about this class of unitaries?

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    $\begingroup$ Very naive comment: Can't you get (at least some of) them by conjugating U_0 with set of unitaries that commute with U_0? $\endgroup$
    – Arnab
    Apr 21, 2021 at 1:31
  • $\begingroup$ If $U$ commutes with $U_0$, then $U^\dagger U_0 U = (U^\dagger U) U_0 = U_0$, so unfortunately that does not work. $\endgroup$ Apr 22, 2021 at 19:09
  • $\begingroup$ Yeah, that was stupid of me. $\endgroup$
    – Arnab
    Apr 23, 2021 at 6:34

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Say your state space is $D = 2^d$ dimensions. Without loss of generality, consider an orthonormal basis of states we will label by $| i \rangle$, where $i = 1 \ldots D$, where $|1\rangle = | \psi \rangle$, and $|i \rangle$ for $i > 1$ are other random states.

In this basis, a few things become clear. If we have any block diagonal matrix of the form $$ U' = \begin{pmatrix} 1 & \mathbf{0} \\ \mathbf{0} & U_{D-1}\end{pmatrix} $$ where $U_{D-1}$ is any $(D-1) \times (D-1)$ matrix, then we clearly have $$ U' |\psi\rangle = U' |1\rangle = |1 \rangle = |\psi \rangle. $$

Therefore, we have a whole $U(D-1)$ worth of transformations that preserve $|\psi \rangle$. If we have your matrix $U_0$, composing it with one of these matrices $U'$ will preserve the property that $(U' U_0) |0 \ldots 0 \rangle = |\psi \rangle$. In fact, once we have $U_0$, we can get all the other such matrices by simply multiplying it in this way $U' U_0$.

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    $\begingroup$ I like this! I do not like that it is written in a way of “this form $U'U_0$ is sufficient to get what you want”, rather than necessary to get what you want, but it has the main pointer; suppose $U$ is an arbitrary element that does what OP wishes and choose any particular $U_0$ too, then we have $u= U_0^\dagger U$ which is unitary too but now preserves $u |\psi\rangle=|\psi\rangle,$ and then from there you can argue that $u$ must have a certain Jordan normal form which implies your form for $U'$, which I believe gives necessity on top of sufficiency. $\endgroup$
    – CR Drost
    Apr 21, 2021 at 2:54
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    $\begingroup$ And also I like that you are giving the group structure, the group operation is just the weird $(x,y)\mapsto x U_0^\dagger y$ or so, fixing $U_0$ as the identity $\endgroup$
    – CR Drost
    Apr 21, 2021 at 2:58
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    $\begingroup$ So you've reduced the problem to looking at matrices which fix $|\psi\rangle$ instead of matrices which map $|0\cdots 0\rangle$ to $|\psi\rangle$. That's interesting. Thanks for the answer, and for the helpful comments! $\endgroup$ Apr 22, 2021 at 19:08

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