# What is the significance of being equivalent up to local isometry?

Background :

I am reading the paper device independent outlook on quantum mechanics. The author mentions the concept of two pure states being equivalent up local isometry. From what I understood two states $|s\rangle$ and $|t\rangle$ are equivalent up to local isometry if by having the option of appending extra degree of freedom to $|s\rangle$ and performing local unitary operations I can extract out ( the meaning of extract out becomes clear in the example below ) state $|t\rangle$. For example consider the state ( here $|k\rangle$ are orthonormal vectors ) $$|\psi\rangle_{AB}=\sum_k c_k\frac{1}{\sqrt{2}}(|2k,2k\rangle+|2k+1,2k+1\rangle)$$ Let the unitary $U=U_A \otimes U_B$ where $U_A$ is such that it acts in the following way ( $|00\rangle$ is the ancilla ) $$U_A|2k,0\rangle = |2k,0\rangle$$ $$U_A|2k+1,0\rangle = |2k,1\rangle$$ $U_B$ acts in the same way. Now if I append the extra degree of freedom $|00\rangle$ to $|\psi\rangle_{AB}$ and apply $U$ $$U|\psi\rangle_{AB}|00\rangle_{AB} = |\phi\rangle \frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)$$ Here $|\phi\rangle$ is not of interest to me. But by appending the extra degree of freedom and applying the local unitary operations I was able to extract state $|\phi^{+}\rangle=\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)$. Thus $|\psi\rangle_{AB}$ is equivalent to $|\phi^{+}\rangle$ upto local isometry.

Doubts:

1. Is my understanding of local isometry correct ?
2. If yes, then is being equivalent up to local isometry symmetric ( I doubt it,but is answer is no the relations becomes wierd to me ). I mean if $|s\rangle$ is equivalent to $|t\rangle$ , is the reverse also true ?
3. The example above I gave was a specific one from the paper itself. But in general how do I prove that if $|s\rangle$ is equivalent to $|t\rangle$ then for every measurement settings on state $|t\rangle$ there will be some measurement settings for state $|s\rangle$ which gives the same statistics ?
• To be clear, the local unitaries act on the ancilla? What are the dimensionalities of the different hilbert spaces involved? – Emilio Pisanty Jun 27 '15 at 8:31
• @EmilioPisanty Yes the local unitaries act on ancilla. There is no constraint on dimension of $|ψ⟩_{AB}$ . I might miss something important to the question , but I tried to include it similar to the paper, same example is given in the beginning of section 4.2.1 "Equivalence up to a local isometry". More precisely equations (49) to (51). – sashas Jun 27 '15 at 9:32

An isometric quantum channel $T_{A\to A^{\prime}}$ is a channel $T_{A\to A^{\prime}}:\mathcal{B}(\mathcal{H}_A)\to\mathcal{B}(\mathcal{H}^{\prime}_A)$ such that $T_A(\rho)=U\rho U^{\dagger}$ with $U^{\dagger}U=1$ and $UU^{\dagger}=P$ with some projection onto a subspace. The conditions on $U$ mean that the channel is an isometry.
A local isometric channel is a channel that acts locally (as in: in the form of products) on a multipartite system, i.e. for a bipartite setting, a local isometric channel would be of the form $T_{A\to A^{\prime}}\otimes T_{B\to B^{\prime}}:\mathcal{B}(\mathcal{H}_A\otimes \mathcal{H}_B)\to \mathcal{B}(\mathcal{H}_A^{\prime}\otimes \mathcal{H}_B^{\prime})$.
Similarly, I would define that a given state $\rho\in \mathcal{B}(\mathcal{H}_A\otimes \mathcal{H}_B)$ is equivalent to another state $\sigma\in \mathcal{B}(\mathcal{H}_A^{\prime}\otimes \mathcal{H}_B^{\prime})$ via local isometries if there exists a local isometric channel $T_{A\to A^{\prime}}\otimes T_{B\to B^{\prime}}$ that sends $\rho$ to $\sigma$ and there is a local isometric channel $T^{\prime}_{A^{\prime}\to A}\otimes T^{\prime}_{B^{\prime}\to B}$ sending $\sigma$ to $\rho$ (otherwise it wouldn't be an equivalence - which also answers your second question).