# Upper bounding the trace distance between a noisy and noiseless quantum state

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.

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