2
$\begingroup$

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.

$\endgroup$
0

1 Answer 1

3
$\begingroup$

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 $$ \sqrt{q^2(1-a)^2 - (q + 2\sqrt{1-q} -2)|\beta|^2}\,. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.