# Does the trace distance specify a unique state

In quantum information, we frequently use the trace distance (see definition) to look at how similar two states are.

If I had a known complete set of states $\{\rho_i\}$ and some unknown state $\sigma$ and knew $D(\rho_i,\sigma)=d_i$ where $D$ denotes the trace distance, can I in principle determine $\sigma$ without additional information?

From a geometric perspective, the answer seems to be yes, however, it is not clear to me how to use the properties of the trace distance to actually prove it (or maybe I'm wrong).

Update: I have found a way around addressing the issue for my research but it is still an interesting question that may be worth pursuing.

• Can you please define state distance? Commented Jul 15, 2018 at 4:10
• Commented Jul 15, 2018 at 4:13
• I don't see why this could possibly be true, you can have many states with the same trace of distance to your given state. Why would the distance alone distinguish them? Commented Jul 15, 2018 at 4:21
• @user157879 I believe you misread my question. The key idea is that we know the distance of $\sigma$ to each point in a basis for the space of states. To understand the intuition consider the space Rn with the Euclidean metric. The distance between the vector $\sigma$ and each basis vector $\rho_i$ confines $\sigma$ to lie on a hypersphere. Together, the knowledge of all the hyperspheres allows us to specify $\sigma$ uniquely by their intersection. Commented Jul 15, 2018 at 10:23
• What do you mean by a "complete" set of states? The trace distance is zero only if two states are the same. If I have information about the uncountable set of trace distances between my unknown state and every possible quantum state, then of course I can infer what my state is by looking for the state which has zero distance. I suspect this is not what you mean though. Commented Jul 15, 2018 at 18:49

For qubits, in the Bloch ball representation: $$\rho=\frac{1}{2}(I+\mathbf{r}\cdot\mathbf{\sigma})$$ $$\rho'=\frac{1}{2}(I+\mathbf{r}'\cdot\mathbf{\sigma})$$ The difference $$\rho-\rho'$$ is $$\rho-\rho'=\frac{1}{2}[(\mathbf{r}-\mathbf{r}')\cdot\mathbf{\sigma}]$$ The eigenvalues of $$\rho-\rho'$$ are $$\lambda_{\pm}=\pm\frac{1}{2}\sqrt{(x-x')^2+(y-y')^2+(z-z')^2}=\pm\frac{||\mathbf{r}-\mathbf{r}'||_2}{2}$$. Thus the trace distance in terms of Bloch vectors is $$D(\rho,\rho')=\frac{||\mathbf{r}-\mathbf{r}'||_2}{2}$$. In this picture, $$D(\rho_i,\sigma)$$ is the the radius of a ball containing $$\sigma$$ as a point and centered in $$\rho_i$$.
I can think two answers to your question: if you give me a table of values $$d_i$$ for the six projectors of the Pauli basis I would say that no, $$\sigma$$ is not specified uniquely, since without further restrictions if I give you $$d_i=\epsilon$$ for all $$i$$ there is $$\epsilon$$ such the intersection between the six balls is empty (or any finite number of projectors). Now if you are asking whether a quantum state $$\sigma$$ can be expressed uniquely via a table of values $$d_i$$ I would believe so, since a point in $$\mathbb{R}^3$$can be expressed uniquely as the intersection between six spheres (maybe even less).