# How to check if a mapping is unitary?

In the case of the No-cloning theorem, it is argued that a unitary $$U$$ that is capable of performing coping does not exist. Specifically, for any two unknown states $$|\psi_1\rangle$$ and $$|\psi_2\rangle$$ the calculation below is not possible: \begin{align} U |\psi_1\rangle|0\rangle = |\psi_1\rangle|\psi_1\rangle,\\ U |\psi_2\rangle|0\rangle = |\psi_2\rangle|\psi_2\rangle. \end{align} Therefore, we can't have a map $$|\psi\rangle |0\rangle \rightarrow |\psi\rangle|\psi\rangle$$ that works for all $$|\psi\rangle$$.

To proof that such $$U$$ is not possible, we run the computation below and reveal the contradiction: $$\tag{1} \langle \psi_2|\psi_1\rangle = \langle0|\langle\psi_2|U^{\dagger}U |\psi_1\rangle|0\rangle = \langle \psi_2|\psi_1\rangle^2.$$

My question are:

1. Can computation in (1) be viewed as a universal recipe for checking the existence of mappings?
2. More generally, does there exist a test that would allow checking if a map $$|\psi\rangle|\phi\rangle|M\rangle \rightarrow |\psi', \phi', M'\rangle$$ is possible? Here $$|\psi', \phi', M'\rangle$$ denotes a state which might be all entangled and $$|M\rangle$$ stands for some additional helper register.

Any hints, suggestions or links are appreciated.

• The contradiction in (1) arises from the fact that $\langle \psi_1|\psi_2\rangle \neq \langle \psi_1|\psi_2\rangle^2$ for ANY arbitrary pair of $|\psi_2\rangle, |\psi_2\rangle$. However, for fixed $|\psi_2\rangle, |\psi_2\rangle$, you can always construct a $U$ such that $UJ|\psi_1\rangle |0\rangle =|\psi_1\rangle|\psi_1\rangle$ and $U|\psi_2\rangle|0\rangle=|\psi_2\rangle|\psi_2\rangle$. Does that help answer your first question? And could you clarify your second? Since you're reusing the same symbols on both sides of the equals sign, I'm not sure what you're asking for. Nov 9, 2022 at 1:08
• Hi thanks for the comment. I introduced some clarifications. What you wrote is indeed correct and I appreciate that, but my questions still stand. Nov 9, 2022 at 3:33

The cloning strategy certainly gives one potential way to disprove existence of a solution. If you're really just looking for a unitary transformation, it's extremely effective - essentially, you're saying that unitaries are linear, but you're proving that you require a non-linear transformation. But it's not the only one. (Also note: to disprove no-cloning, is it sufficient to look only at unitaries? I would argue that, at the very least, you need to include an ancilla system.)

The strategy that I often used is a bit more complicated, but I get better returns on it (but that may depend on the set of things you're trying to do!). Imagine you have specified for map as taking states $$|\psi_i\rangle$$ to states $$|\phi_i\rangle$$. I'm going to use discrete notation, but it could be continuous. Let's say I have no knowledge of which of the $$N$$ states I could be given, so I'll assume they all arise with probability $$1/N$$ (probability choice doesn't make any difference for the yes/no question of whether the operation can be achieved, but if it can't be achieved perfectly, this method gives an upper bound on what can be achieved). Let's say that the smallest space that the input states are supported on is dimension $$d$$.

If I construct a matrix $$M=\frac{d}{N}\sum_i|\psi_i\rangle\langle\psi_i|^T\otimes|\phi_i\rangle\langle\phi_i|,$$ then the maximum eigenvalue is an upper bound on the fidelity that can be achieved for a CP map implementing the operation. If that value is less than 1, it cannot be implemented perfectly, even by a CP map (and therefore certainly not be a unitary). While this method does not tell you directly if it can be implemented, if you look at the eigenvector(s) corresponding to the maximum eigenvalue, you can often extract some useful information. For example, if the maximum eigenvalue is unique, that strategy can be realised if and only if (if memory serves) that eigenvector is maximally entangled across the partition corresponding to the input space and the output space. It should always be possible to specify this operation in terms of a unitary (possibly on a larger space).

If you were trying to prove no-cloning with this, you might be tempted to integrate over all possible input states. You certainly can do that and if you know a bit about "twirling", the calculation is not too bad. But to disprove it, it's sufficient to disprove it on a finite subset of states. For example, if you select the set of 6 states corresponding to the eigenvalues of Pauli $$X$$, $$Y$$ and $$Z$$, that's sufficient.

• This is very interesting! I would like to understand this technique in more detail, could you suggest any supplementary reading material that could help me? I feel like it is a bit advanced for my current knowledge. Also, what is CP map? Nov 9, 2022 at 15:57
• One more clarification, is it a "transpose" on the first projector? Nov 9, 2022 at 16:05
• CP = completely positive (the generalisation of unitary operation to include ancillas etc) Nov 10, 2022 at 7:48
• yes, it's a transpose Nov 10, 2022 at 7:48
• In terms of further reading, I'd start by understanding the Choi map and the Choi-Jamiolkowski isomorphism. It's not quite this, but is closely related. Norbert's answer here: physics.stackexchange.com/a/274739 is a good starting point. Nov 10, 2022 at 7:54

Given any two states $$|\psi\rangle$$ and $$|\theta\rangle$$, there always exists a unitary mapping that takes $$\psi$$ to $$\theta$$.

There is in fact a stronger result. Given any two orthonormal bases $$|\psi_1\rangle, ... |\psi_n\rangle$$, and $$|\theta_1\rangle, ... |\theta_n\rangle$$, there always exists a unitary operator that maps each $$|\psi_i\rangle$$ to the corresponding $$|\theta_i\rangle$$.

Back to the simple case. There is some unitary function $$U_1$$ that takes you from the all zero state to $$|\psi\rangle$$. There is another unitary function $$U_2$$ that takes you from the all zero state to $$|\theta\rangle$$. Then $$U_2 U_1^\dagger |\psi\rangle = |\theta\rangle$$

Again, cloning doesn't serve as a good example for what you want. You are trying to map every $$|\psi\rangle$$ to a specific form, not just a single one.