Friday, 3 January 2014

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.