Consider a quantum state

$$ \rho = \begin{pmatrix} \rho_{00} & \rho_{01} \\ \rho_{10} & \rho_{11} \\ \end{pmatrix}. $$

Now, consider the effect of the amplitude damping noise $\mathcal{N}$ of noise strength $q$ on this quantum state. The output state is given by

$$ \sigma =\mathcal{N}(\rho) = \begin{pmatrix} \rho_{00} + q \rho_{11} & \sqrt{1 - q} ~\rho_{01} \\ \sqrt{1- q}~\rho_{10} & (1- q)~\rho_{11} \\ \end{pmatrix}. $$

I am trying to upper bound the trace distance between $\rho$ and $\sigma$ as a function of the strength of the noise (and coefficients of $\rho$.) Note that when the strength of the noise is $0$, the two states are exactly similar, but when the noise strength is $1$, unless $\rho$ is $|0\rangle\langle 0|$, the states should be far apart.


1 Answer 1


No upper bounds needed. Firstly, you have a density matrix so you don't have to have all these free parameters, you can take $$ \rho = \begin{pmatrix} a & \beta \\ \beta^* & 1-a \end{pmatrix} $$ for $0 \leq a \leq 1$ and $\beta \in \mathbb{C}$ such that $|\beta|^2 \leq a(1-a)$. Then the trace distance is defined as $\frac12$ of the 1-norm and for a hermitian operator the 1-norm is just the sum of the absolute value of the eigenvalues. The eigenvalues of a $2 \times 2$ matrix have a nice closed form because they're just roots of a quadratic polynomial and so we find that the eigenvalues of $\rho - \sigma$ are $$ \pm \sqrt{q^2(1-a)^2 - (q + 2\sqrt{1-q} -2)|\beta|^2} $$ Hence the trace distance $\frac12 \|\rho - \sigma\|_1$ is exactly $$ \sqrt{q^2(1-a)^2 - (q + 2\sqrt{1-q} -2)|\beta|^2}\,. $$


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