# Set of unitary operators such that $U|0\rangle = |\psi\rangle$

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?

• 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? Apr 21, 2021 at 1:31
• 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. Apr 22, 2021 at 19:09
• Yeah, that was stupid of me. Apr 23, 2021 at 6:34

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 $$U' U_0$$.

• 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. Apr 21, 2021 at 2:54
• 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 Apr 21, 2021 at 2:58
• 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! Apr 22, 2021 at 19:08