<p>Assume a qubit quantum channel (trace-preserving CP (==completely positive) map) $\mathcal{N}:C^2\to C^2$ whose quantum capacity is known. Now I construct a new channel $\mathcal{M}:C^2\to C^{2+n}$ where $n&gt;0$:
$$
\mathcal{M}=p\mathcal{N}\oplus(1-p)\mathcal{C},
$$
where $\mathcal{C}$ is an arbitrary channel $C^2\to C^n$, whose form I don't need to know, and $0\leq p\leq1$. Assume that I have access just to the two-dim output subspace of $\mathcal{M}$ given by the image of $\mathcal{N}$. So effectively, I constructed a new CP map $\mathcal{M'}:C^2\to C^{2}$ that is trace-decreasing (since $p\leq1$).</p>

<p>Here is my question:
What si the quantum capacity of $\mathcal{M'}$ ?</p>

<p>Here is my answer:
Irrespective of $n$, I can assume that $\mathcal{M'}$ can be written as
$$
\mathcal{M'}=\mathcal{E}\circ\mathcal{N},
$$
where $\mathcal{E}:C^2\to C^3$ is a qubit erasure channel
$$
\mathcal{E}(\rho)=p{\rho}\oplus(1-p)\rho_{fixed},
$$ 
where  $\rho_{fixed}$ is an input-independent positive operator (density matrix) spanning a one-dimensional subspace. Hence, the quantum capacity of $\mathcal{M'}$ is the minimum of the quantum capacities of $\mathcal{N}$ and $\mathcal{E}$ (known as the bottleneck inequality).</p>

<p><strong>EDIT</strong>
Following Shor's comment I wanted to say that the quantum capacity of $\mathcal{M'}$ will be <strong>bounded</strong> from above by the quantum capacities of $\mathcal{N}$ and $\mathcal{E}$.</p>

<p>Do you agree?</p>

CC BY-SA 3.0
2014-01-06T14:27:27.403Z
Have you looked at what happens with classical channels?
Peter Shor 
CC BY-SA 3.0
2014-01-06T15:31:46.837Z
No, what exactly should I look at?
user119381
CC BY-SA 3.0
2014-01-06T15:32:56.307Z
Suppose that classically, you have an erasure channel composed with another channel. Do you get this bottleneck behavior? [Hint: ***no***, assuming I understand your question correctly; so why should we see it in the quantum version of the question?] Does the classical proof carry over to the quantum case?
Peter Shor 
CC BY-SA 3.0
2014-01-06T15:39:14.860Z
The bottleneck inequality is an ***inequality***, not an equality.
Peter Shor 
CC BY-SA 3.0
2014-01-06T15:43:46.393Z
Oh, so I was being imprecise, I wanted to say that the capacity of $\mathcal{M'}$ will be less than either of $\mathcal{N}$ or $\mathcal{E}$. Is this statement correct? My main question is whether the whole model for this particular TDCP map as  $\mathcal{E}\circ\mathcal{N}$ is correct.
user119381
CC BY-SA 3.0
2014-01-06T20:14:11.337Z
I think by adding the erasure symbol, you've effectively turned a trace-decreasing map into a trace-preserving map. But maybe this is an argument about definitions.
Peter Shor 
CC BY-SA 3.0
2014-01-07T13:06:40.393Z
So I will take it as that you agree. Thx! On a general note, it seems to me that any TDCP map can be written as some channel followed by a qubit erasure channel followed by a restriction to an output subspace. So the quantum (or classical) capacity can be bounded from above
user119381
<quantum-information>
The quantum capacity of a trace-decreasing CP map
user119381
# The quantum capacity of a trace-decreasing CP map

(Asked by: username_0 on 2014-01-06)

Assume a qubit quantum channel (trace-preserving CP (==completely positive) map) $\mathcal{N}:C^2\to C^2$ whose quantum capacity is known. Now I construct a new channel $\mathcal{M}:C^2\to C^{2+n}$ where $n>0$:
$$
\mathcal{M}=p\mathcal{N}\oplus(1-p)\mathcal{C},
$$
where $\mathcal{C}$ is an arbitrary channel $C^2\to C^n$, whose form I don't need to know, and $0\leq p\leq1$. Assume that I have access just to the two-dim output subspace of $\mathcal{M}$ given by the image of $\mathcal{N}$. So effectively, I constructed a new CP map $\mathcal{M'}:C^2\to C^{2}$ that is trace-decreasing (since $p\leq1$).

Here is my question:
What si the quantum capacity of $\mathcal{M'}$ ?

Here is my answer:
Irrespective of $n$, I can assume that $\mathcal{M'}$ can be written as
$$
\mathcal{M'}=\mathcal{E}\circ\mathcal{N},
$$
where $\mathcal{E}:C^2\to C^3$ is a qubit erasure channel
$$
\mathcal{E}(\rho)=p{\rho}\oplus(1-p)\rho\_{fixed},
$$
where $\rho\_{fixed}$ is an input-independent positive operator (density matrix) spanning a one-dimensional subspace. Hence, the quantum capacity of $\mathcal{M'}$ is the minimum of the quantum capacities of $\mathcal{N}$ and $\mathcal{E}$ (known as the bottleneck inequality).

**EDIT**
Following Shor's comment I wanted to say that the quantum capacity of $\mathcal{M'}$ will be **bounded** from above by the quantum capacities of $\mathcal{N}$ and $\mathcal{E}$.

Do you agree?

**Tags:** quantum-information
> **Comments:**
> **username_1**: Have you looked at what happens with classical channels?  
> **username_0**: No, what exactly should I look at?  
> **username_1**: Suppose that classically, you have an erasure channel composed with another channel. Do you get this bottleneck behavior? [Hint: ***no***, assuming I understand your question correctly; so why should we see it in the quantum version of the question?] Does the classical proof carry over to the quantum case?  
> **username_1**: The bottleneck inequality is an ***inequality***, not an equality.  
> **username_0**: Oh, so I was being imprecise, I wanted to say that the capacity of $\mathcal{M'}$ will be less than either of $\mathcal{N}$ or $\mathcal{E}$. Is this statement correct? My main question is whether the whole model for this particular TDCP map as  $\mathcal{E}\circ\mathcal{N}$ is correct.  
> **username_1**: I think by adding the erasure symbol, you've effectively turned a trace-decreasing map into a trace-preserving map. But maybe this is an argument about definitions.  
> **username_0**: So I will take it as that you agree. Thx! On a general note, it seems to me that any TDCP map can be written as some channel followed by a qubit erasure channel followed by a restriction to an output subspace. So the quantum (or classical) capacity can be bounded from above  

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## 0 answers

