2

The depolarizing noise channel uniformly deflates the Bloch sphere to a single point, which is $\mathbf{n}= (0,0,0)$ or in terms of quantum qubit states, we get a maximally mixed state $\rho = \frac{1}{2}I$.

Since the depolarizing noise channel models information loss, recovering the original state is impossible without additional help (error correction, etc.). However, we can still construct a CP map that will simply push all mixed states outwards to the surface of the Bloch sphere. Consider the following construction:

We can represent all states as a set: $$B = \left \{\frac{1}{2} \left (I + \mathbf{n} \cdot \sigma \right) : \mathbf{n} \in \mathbb{R}^3 \right \}$$ Here, $|\mathbf{n}|\leq 1$ and $\sigma$ is a vector of Pauli matrices.

Let $|n\rangle \in \mathbb{C}^2$ be a ket vector that is aligned with $\mathbf{n}$ on a Bloch sphere. Then, for $\rho \in B$, there exists $p \in [0,1]$ such that $$\rho = \frac{1}{2} \left (I + \mathbf{n} \cdot \sigma \right) = (1-p)|n\rangle \langle n| + p |n^{\perp}\rangle \langle n^{\perp}|.$$ Now, define a map $\Phi: \mathcal{D}(\mathcal{H}) \times \mathbb{R}^3 \times \mathbb{R} \rightarrow \mathcal{L}(\mathcal{H})$ as $$\Phi(\bullet,\mathbf{n},p) =\sqrt{(1-p)^{-1}}|n\rangle \langle n| \bullet |n\rangle \langle n|\sqrt{(1-p)^{-1}}.$$ $\Phi$ is not a recovery/reconstruction of the original state. It simply pushes all states outwards on the surface. In other words, this map scales the vector $\mathbf{n}$ of $\rho$ so that it is of unit norm. Alternatively, we could view $\Phi$ as a projective measurement.

Given the correct triplet $(\rho, \mathbf{n}, p)$, the map $\Phi$ produces valid density matrices on the Bloch sphere's surface. Also, it is CP but not TP. Therefore, this map is not a channel.

Questions:

  1. I'm confused. It seems it is possible, at least mathematically, to "inflate" the sphere outwards such that all states are on the sphere's surface. However, Preskill, in his lecture recording, states that:

The map that inverts the deflation is not a channel; it maps polarization contained in the Bloch ball to a polarization outside the Bloch ball (not a physical state).

What do I misunderstand here with this $\Phi$? It is not a channel, ok, but it can map physical states to physical states for inputs $(\rho, \mathbf{n}, p)$ such that $\rho = (1-p)|n\rangle \langle n| + p |n^{\perp}\rangle \langle n^{\perp}|$.

  1. Since $\Phi$ looks like a projective measurement, can this operation be implemented physically?

  2. All this rings the bell about state distinguishability, but I can't quite see and remember how this could be related. Do you have any pointers to the basic/introductory theory that would apply here?

glS
  • 27,510
  • 7
  • 37
  • 125
MonteNero
  • 3,344
  • 7
  • 24

1 Answers1

3

I'll try to answer questions 1. and 2.

  1. Your construction is indeed not a quantum channel, because the projective measurements that are being executed are dependent on the dominant eigenvector (assuming $p \leq \frac{1}{2}$) of the input state. Rather, your construction looks like a family of measurement channels $(n, \Phi_n)$ labelled by $n\in S^2$.

  2. In order to find out which measurement channel to apply, however, you need to know $n$, so you have to know the dominant eigenvector that you wanted to filter in the first place, rendering the operation pointless for some kind of error correction (I presume that was the motivation?).

Refik Mansuroglu
  • 782
  • 2
  • 18