For any two states $|\phi\rangle$ and $|\psi\rangle$

Does there exist a gate $U$ such that $U|\phi\rangle = |\psi\rangle$ ?

I suppose that we know for a vector space $V$ then $\forall \quad a, b \quad \exists M$ such that $Ma = b$. Therefore the original question becomes, does the set of quantum states form a vector space?

The space of quantum states does not form a vector space. Consider a quantum state $|\phi\rangle$. If the space of quantum states is a vector space then $2|\phi\rangle$ is also a quantum state.

But $\langle2\phi,2\phi \rangle = 4\langle\phi,\phi \rangle$ by multilinearity of the inner product. And

$4\langle\phi,\phi \rangle = 4$ as $|\phi\rangle$is a quantum state so $\langle\phi,\phi \rangle$ = 1.

Therefore the space of quantum gates is not a vector space.

Then given that, could there still be a $U$ as posed?

  • $\begingroup$ First thing, vector space does not need the inner product. Secondly, $|\psi\rangle\equiv4|\phi\rangle$ does not satisfy the normalized condition, i.e., $\langle \psi|\psi\rangle \neq 1$, hence it's not a legal quantum state. $\endgroup$
    – narip
    Jul 30 at 3:07
  • $\begingroup$ Right, that's the point I was trying to make regarding $\psi$. I suppose I could have made that clearer. And true I forgot that with the inner product we would have an inner product space. In any case I believe the answer I posed is correct and also does not require the space of quantum states to be a vector space. $\endgroup$ Jul 30 at 3:13
  • 2
    $\begingroup$ Isn't it simply linear system? You know solution and right side and your are looking for matrix of the system which is unitary. $\endgroup$ Jul 30 at 7:09
  • $\begingroup$ see quantumcomputing.stackexchange.com/q/17866/55 $\endgroup$
    – glS
    Jul 30 at 15:19

Supposing that by quantum state you mean a unit vector in $\mathbb{C}^n$ for some fixed $n \in \mathbb{N}$ then the answer is yes. I think the easiest way to see this is by construction.

Let $|\psi\rangle$ be the starting state and let $|\phi\rangle$ be the target state. As they are both normalized we can complete them both into two orthonormal bases of $\mathbb{C}^n$. Say $\{|\psi_1\rangle,\dots, |\psi_n\rangle\}$ and $\{|\phi_1\rangle, \dots, |\phi_n\rangle\}$ are two orthonormal bases of $\mathbb{C}^n$ such that $|\psi_1\rangle = |\psi\rangle$ and $|\phi_1\rangle = |\phi\rangle$. Then define the operator $$ U = \sum_{i=1}^n |\phi_i\rangle\langle \psi_i|. $$ It is straightforward to check that this matrix is both unitary (it maps one orthonormal basis to another) and that $|\phi\rangle = U |\psi\rangle$.



While quantum vectors do not comprise a vector space, there is always such a $U$.

For a vector in $C^n$ to be a quantum state, $\langle \psi|\psi\rangle$ = 1. Vectors satisfying this form a unit $n$-ball in $C^n$.

In the case of $C^2$, AKA states consisting of one qubit, this is the Bloch sphere.

From linear algebra we know that matrix of rotations (which are one in the same as unitary matrices) can take us there. All quantum gates can be considered matrices of rotations.

For any two points on a ball there exists a matrix of rotation that takes you from one two the other. It is therefore that such a $U$ exists for all pairs $|\psi\rangle$, $|\phi\rangle$

  • $\begingroup$ A two dimension vector doesn’t equate two qubits, it equates a two-level quantum system, so one qubit. The Bloch sphere represented one qubit only. $\endgroup$
    – epelaaez
    Jul 30 at 4:34

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