Bounding the norm of the Choi representation of a channel in terms of its operator norm

Theorem:

Let $\X$ and $\Y$ be complex Euclidean spaces with $\dim(\X) = n$ and let $\Phi\in\Trans(\X,\Y)$. Let $\norm{\cdot}_1$ denote the usual trace norm of an density operator, and $\triplenorm{\cdot}_1$ the operator norm of a channel.
  \[
  \triplenorm{\Phi}_1 \leq \norm{J(\Phi)}_1 \leq n
  \triplenorm{\Phi}_1.  \]

Proof:

Since the Choi representation of $\Phi$ is given by $J(\Phi)=(\Phi\otimes\I_{\Lin(\X)})(\vec(\I_\X)\vec(\I_\X)^\ast)$, then the trace norm is given by and also satisfies
\[\begin{align*}
\norm{J(\Phi)}_1&=\norm{(\Phi\otimes\I_{\Lin(\X)})(\vec(\I_\X)\vec(\I_\X)^\ast)}_1 \\
&\leq \norm{(\Phi\otimes\I_{\Lin(\X)})}_1\norm{\vec(\I_\X)\vec(\I_\X)^\ast}_1 \\
&=\norm{(\Phi\otimes\I_{\Lin(\X)})}_1n \\
&=n  \triplenorm{\Phi}_1,
\end{align*}\]
where the last two lines from $\norm{\vec(\I_\X)\vec(\I_\X)^\ast}_1=n$ and the definition of the completely bounded trace norm $\triplenorm{\Phi}_1 :=\norm{(\Phi\otimes\I_{\Lin(\X)})}_1$.

Now consider an alternate characterization of the completely bounded trace norm:
\[
\triplenorm{\Phi}_1=max\{\norm{(\I_{\Lin(\Y)}\otimes\sqrt{\rho_0})J(\Phi)(\I_{\Lin(\Y)}\otimes\sqrt{\rho_1})}_1 : \rho_0,\rho_1\in\Density(\X)\}.
\]
Since this norm satisfies the property
\[
\norm{(\I_{\Lin(\Y)}\otimes\sqrt{\rho_0})J(\Phi)(\I_{\Lin(\Y)}\otimes\sqrt{\rho_1})}_\infty\leq\norm{\I_{\Lin(\Y)}\otimes\sqrt{\rho_0}}_\infty\norm{J(\Phi)}_1\norm{\I_{\Lin(\Y)}\otimes\sqrt{\rho_1}}_\infty,
\]
and the spectral norm $\norm{A}_\infty$ of an operator $A$ is given by the largest singular value of $A$, then it follows that $\norm{\I_{\Lin(\Y)}\otimes\sqrt{\rho_a}}_\infty\leq1$ with equality holding in the case where $\rho_a$ is a pure state. Therefore,
\[
\norm{(\I_{\Lin(\Y)}\otimes\sqrt{\rho_0})J(\Phi)(\I_{\Lin(\Y)}\otimes\sqrt{\rho_1})}_\infty\leq\norm{J(\Phi)}_1,
\]
implying that
\[
\triplenorm{\Phi}_1=max\{\norm{(\I_{\Lin(\Y)}\otimes\sqrt{\rho_0})J(\Phi)(\I_{\Lin(\Y)}\otimes\sqrt{\rho_1})}_1 : \rho_0,\rho_1\in\Density(\X)\}\leq \norm{J(\Phi)}_1.
\]
Putting these two bounds together then gives
\[
  \triplenorm{\Phi}_1 \leq \norm{J(\Phi)}_1 \leq n
  \triplenorm{\Phi}_1.
\]

A lower bound on the trace distance of tensor copies of states

Theorem:
  Let $\X$ be a complex Euclidean space, let $\rho_0,\rho_1\in\Density(\X)$ be density operators satisfying
\[
    \bignorm{\rho_0 - \rho_1}_1 \geq \varepsilon
\]
  for $\varepsilon > 0$, and let $n$ be an arbitrary positive integer.
  Then
\[
    \Bignorm{\rho_0^{\otimes n} - \rho_1^{\otimes n}}_1
    \geq 2 - 2 \exp\biggl(-\frac{n\varepsilon^2}{8}\biggr).
\]
  (The notation $\rho^{\otimes n}$ means $\rho$ tensored with itself $n$ times. For example, $\rho^{\otimes 4} = \rho\otimes\rho\otimes\rho\otimes\rho$.)

Proof:
By the Fuchs-van de Graaf inequalities (Theorem 3.34) we have that the following two statements are equivalent:
\[
1-\frac{1}{2}\bignorm{\rho_0-\rho_1}_1\leq\fid(\rho_0,\rho_1)\leq\sqrt{1-\frac{1}{4}\bignorm{\rho_0-\rho_1}_1^2},
\]
\[
2-2\fid(\rho_0,\rho_1)\leq\bignorm{\rho_0-\rho_1}_1\leq 2\sqrt{1-\fid(\rho_0,\rho_1)^2}.
\]
Also, by Proposition 3.16, it follows that
\[
\fid(\rho_0^{\otimes n},\rho_1^{\otimes n})=\fid(\rho_0,\rho_1)^{n}.
\]
Therefore,
\[
\fid(\rho_0^{\otimes n},\rho_1^{\otimes n})^2=\fid(\rho_0,\rho_1)^{2n}\leq\left(1-\frac{1}{4}\bignorm{\rho_0-\rho_1}_1^2\right)^n\leq\left(1-\frac{1}{4}\varepsilon^2\right)^n,
\]
since $ \varepsilon\leq\bignorm{\rho_0 - \rho_1}_1$ by assumption.

Since for $x\in\mathbb{R}$ such that $0\leq x\leq1$, and any positive integer $n$,
\[
(1-x)^n\leq \exp(-nx),
\]
 then for $0\leq\varepsilon\leq2$,
\[
\left(1-\frac{1}{4}\varepsilon^2\right)^n\leq  \exp\biggl(-\frac{n\varepsilon^2}{4}\biggr).
\]
This implies that
\[
\fid(\rho_0^{\otimes n},\rho_1^{\otimes n})\leq\left(1-\frac{1}{4}\varepsilon^2\right)^{n/2}\leq\exp\biggl(-\frac{n\varepsilon^2}{8}\biggr).
\]
However, by the Fuchs-van de Graaf inequalities, since
\[
2-2\fid(\rho_0^{\otimes n},\rho_1^{\otimes n})\leq\bignorm{\rho_0^{\otimes n}-\rho_1^{\otimes n}}_1,
\]
the previous result implies
\[
2-2\exp\biggl(-\frac{n\varepsilon^2}{8}\biggr)\leq2-2\fid(\rho_0^{\otimes n},\rho_1^{\otimes n})\leq\bignorm{\rho_0^{\otimes n}-\rho_1^{\otimes n}}_1,
\]
which completes the proof.

The trace distance between pure states

In this post, the relationship between two distance measures on density operators will be analyzed for a few specific cases. Namely, the trace distance and the Euclidean distance of two pure states will be studied.

The trace norm is defined as $||M||_{Tr}=Tr(\sqrt{M^\dagger M})$, and the trace distance between two density operators $\rho$ and $\sigma$ is given by $||\rho-\sigma||_{Tr}$.

Let's calculate an expression for the trace distance between $\ket{0}$ and $\cos(\theta)\ket{0}+\sin(\theta)\ket{1}$ as a function of $\theta$.

Let
\[
\rho=\ket{0}\bra{0}=\begin{pmatrix} 1&0\\0&0\end{pmatrix}
\]
 and
\[ \begin{align*}
 \sigma&=(\cos(\theta)\ket{0}+\sin(\theta)\ket{1})(\cos(\theta)\bra{0}+\sin(\theta)\bra{1})\\
 \sigma&=\cos^2(\theta)\ket{0}\bra{0}+\cos(\theta)\sin(\theta)\ket{0}\bra{1}+\cos(\theta)\sin(\theta)\ket{1}\bra{0}+\sin^2(\theta)\ket{1}\bra{1} \\
 \sigma&=\begin{pmatrix}\cos^2(\theta)&\cos(\theta)\sin(\theta)\\ \cos(\theta)\sin(\theta)&\sin^2(\theta)
 \end{pmatrix}.
 \end{align*}\]

In this case, since both $\rho$ and $\sigma$ are density operators, they are Hermitian. Thus, $(\rho-\sigma)^\dagger=(\rho-\sigma)$  implying that $(\rho-\sigma)^\dagger(\rho-\sigma)=(\rho-\sigma)^2$. Then
\[
\rho-\sigma=\begin{pmatrix}1-\cos^2(\theta)&-\cos(\theta)\sin(\theta)\\ -\cos(\theta)\sin(\theta)&-\sin^2(\theta)\end{pmatrix}=\begin{pmatrix}\sin^2(\theta)&-\cos(\theta)\sin(\theta)\\ -\cos(\theta)\sin(\theta)&-\sin^2(\theta)\end{pmatrix},
\]
from which it follows that
\[\begin{align*}
(\rho-\sigma)^2&=\begin{pmatrix}\sin^2(\theta)&-\cos(\theta)\sin(\theta)\\ -\cos(\theta)\sin(\theta)&-\sin^2(\theta)\end{pmatrix}\begin{pmatrix}\sin^2(\theta)&-\cos(\theta)\sin(\theta)\\ -\cos(\theta)\sin(\theta)&-\sin^2(\theta)\end{pmatrix}\\
&=\begin{pmatrix}\sin^4(\theta)+\cos^2(\theta)\sin^2(\theta)&-\sin^2(\theta)\cos(\theta)\sin(\theta)+\sin^2(\theta)cos(\theta)\sin(\theta) \\ -\sin^2(\theta)\cos(\theta)\sin(\theta) +\sin^2(\theta)cos(\theta)\sin(\theta)&\sin^4(\theta)+\cos^2(\theta)\sin^2(\theta)\end{pmatrix} \\
&=\begin{pmatrix}
\sin^2(\theta)&0 \\
0&\sin^2(\theta).
\end{pmatrix}
\end{align*}\]
 Thus,
\[
 \sqrt{(\rho-\sigma)^2}=\begin{pmatrix}
\sqrt{\sin^2(\theta)}&0 \\
0&\sqrt{\sin^2(\theta)}\end{pmatrix}
=\begin{pmatrix}
|\sin(\theta)|&0 \\
0&|\sin(\theta)|\end{pmatrix},
\]
and therefore
\[
||(\rho-\sigma)||_{Tr}=Tr\begin{pmatrix}
|\sin(\theta)|&0 \\
0&|\sin(\theta)|\end{pmatrix}=2|\sin(\theta)|.\]

As another example, lets now calculate an expression for the Euclidean distance between the two points in the Bloch sphere that correspond to the pure states $\ket{0}$ and $\cos(\theta)\ket{0}+\sin(\theta)\ket{1}$.

In the Bloch sphere representation, a general density matrix can be expressed as
\[
\frac{1}{2}(I+c_xX+c_yY+c_zZ),
\]
where $X$, $Y$, $Z$ are the Pauli matrices, and the vector $(c_x,c_y,c_z)$ gives the coordinates of the density matrix on the Bloch sphere. In such a representation, $\rho$ and $\sigma$ can be expressed as follows:
\[\begin{align*}
\rho=&\frac{1}{2}(I+Z),\\
\sigma=&\frac{1}{2}(I+2\cos(\theta)\sin(\theta)X+\cos(2\theta)Z),
\end{align*}\]
so that the
coordinate vectors for the Block sphere representations of $\rho$ and $\sigma$ are given by, respectively,
\[\begin{align*}
v_{\rho}=&(0,0,1) \\
v_{\sigma}=&(2\cos(\theta)\sin(\theta),0,\cos(2\theta)).
\end{align*}\]
 Then the Euclidean distance between $\rho$ and $\sigma$ on the Block sphere is given by
 \[\begin{align*}
 ||v_\rho-v_\sigma||_2&=\sqrt{(-2\cos(\theta)\sin(\theta))^2+(1-\cos(2\theta))^2}\\
 &=\sqrt{4\cos^2(\theta)\sin^2(\theta)+4\sin^4(\theta)} \\
 &=\sqrt{4\sin^2(\theta)(\cos^2(\theta)+\sin^2(\theta))}\\
 &=\sqrt{4\sin^2(\theta)}\\
 &=2|\sin(\theta)|.
 \end{align*}\]

 Hence, $||(\rho-\sigma)||_{Tr}=||v_\rho-v_\sigma||_2=2|\sin(\theta)|$, and the two distance measures agree.

Now we'll repeat the calculations done above for these two states: $\rho=\begin{pmatrix} 1/2 &0 \\ 0 & 1/2 \end{pmatrix}$ and the pure state $\cos(\theta)\ket{0}+\sin(\theta)\ket{1}$ as a function of $\theta$.

 As calculatedabove, the density matrix corresponding to the state $\cos(\theta)\ket{0}+\sin(\theta)\ket{1}$ is given by
\[
 \sigma:=\begin{pmatrix}\cos^2(\theta)&\cos(\theta)\sin(\theta)\\ \cos(\theta)\sin(\theta)&\sin^2(\theta)
 \end{pmatrix}
\]

 Calculating the relevant matrices in order to determine the trace distance $||(\rho-\sigma)||_{Tr}=Tr(\sqrt{(\rho-\sigma)^2})$ yields

 \[\begin{align*}
 \rho-\sigma=\begin{pmatrix}\frac{1}{2}-\cos^2(\theta)&\cos(\theta)\sin(\theta)\\ \cos(\theta)\sin(\theta)&\frac{1}{2}-\sin^2(\theta)
 \end{pmatrix},
 \end{align*}\]
 and after some further calculation
\[ \begin{align*}
 (\rho-\sigma)^2&=\begin{pmatrix}\frac{1}{2}-\cos^2(\theta)&\cos(\theta)\sin(\theta)\\ \cos(\theta)\sin(\theta)&\frac{1}{2}-\sin^2(\theta)
 \end{pmatrix}\begin{pmatrix}\frac{1}{2}-\cos^2(\theta)&\cos(\theta)\sin(\theta)\\ \cos(\theta)\sin(\theta)&\frac{1}{2}-\sin^2(\theta)
 \end{pmatrix}\\
 &=\begin{pmatrix}
 \frac{1}{4}&0\\
 0&\frac{1}{4}
 \end{pmatrix},
 \end{align*}\]
 which implies that
\[
 \sqrt{ (\rho-\sigma)^2}=\begin{pmatrix}
\sqrt{ \frac{1}{4}}&0\\
 0&\sqrt{\frac{1}{4}}
 \end{pmatrix}=\begin{pmatrix}
\frac{1}{2}&0\\
 0&\frac{1}{2}
 \end{pmatrix}
\]
 Therefore,
\[
 ||(\rho-\sigma)||_{Tr}=Tr(\sqrt{(\rho-\sigma)^2})=Tr\begin{pmatrix}
\frac{1}{2}&0\\
 0&\frac{1}{2}
 \end{pmatrix}=1
\]
 is the trace distance between the two state $\rho$ and $\sigma$.

 Now, to calculate the Euclidean distance between the two states in the Block sphere representation observe that

\[ \begin{align*}
\rho=&\frac{1}{2}I,\\
\sigma=&\frac{1}{2}(I+2\cos(\theta)\sin(\theta)X+\cos(2\theta)Z),
\end{align*}\]
so that the
coordinate vectors for the Block sphere representations of $\rho$ and $\sigma$ are given by, respectively,
\[\begin{align*}
v_{\rho}=&(0,0,0) \\
v_{\sigma}=&(2\cos(\theta)\sin(\theta),0,\cos(2\theta)).
\end{align*}\]

 The Euclidean distance is then given by
 \[ \begin{align*}
 ||v_\rho-v_\sigma||_2&=\sqrt{(-2\cos(\theta)\sin(\theta))^2+(-\cos(2\theta))^2}\\
 &=\sqrt{4\cos^2(\theta)\sin^2(\theta)+\cos^2(2\theta)} \\
 &=\sqrt{4\cos^2(\theta)\sin^2(\theta)+(\cos^2(\theta)-\sin^2(\theta))^2}\\
 &=\sqrt{2\cos^2(\theta)\sin^2(\theta)+\cos^4(\theta)+\sin^4(\theta))^2}\\
&=\sqrt{2\cos^2(\theta)\sin^2(\theta)+\cos^4(\theta)+\sin^4(\theta))^2}\\
 &=\sqrt{(\cos^2(\theta)+\sin^2(\theta))^2}\\
 &=\sqrt{1^2}\\
 &=1,
 \end{align*}\]
 and so the trace distance of $\rho$ and $\sigma$ agrees with their Euclidean distance on the Block sphere since $||(\rho-\sigma)||_{Tr}=||v_\rho-v_\sigma||_2=1$.