# How do I plot the decoherence of an open system from its density matrix?

If I have a two qubit state interacting with an environment that will decohere it, how do I model the decoherence from the density matrix? For example, if I start with some state $\Psi(0)=|0>_1|1>_2\otimes|\psi_{e}>$ and act some unitary $$U=e^{-iHt}$$ onto the density matrix of this state where $$H=A(\sigma_x\otimes\sigma_x+\sigma_y\otimes\sigma_y+\sigma_z\otimes\sigma_z)$$ where $\sigma_i$ are the Pauli matrices I will get some output density matrix $\rho(t)$. I then want to trace out everything but qubit 1 and use this output state $\rho_1(t)$ to find the coherence time of that particular spin. Is there a way to plot the decoherence time directly from this output density matrix?

• You need to define some measure to quantify decoherence, e.g. you can consider $\operatorname{Tr}\left(\rho^2\right) -1$. – Count Iblis Sep 8 '17 at 8:40

The most general equation that describes the decoherence of a density matrix is the Lindblad equation. It has the form $$\partial_t \rho = - [H_0, \rho] + \sum_n \gamma_n (L_n \rho L_n^{\dagger} - L_n^{\dagger} L_n \rho - \rho L_n^{\dagger} L_n) .$$
For the qubit case you can express the Lindblad operators $L_n$ in terms of Pauli matrices. So by playing around with different such operators and solving the equation, you may get a feeling of how it behaves and see what the effect is on the density matrix.
You can ignore the system $e$, because it does not interact with system $1$. Up to an irrelevant global phase of $e^{i A t}$ we have $$|\psi(t)\rangle = \cos(2 A t) |01\rangle - i \sin(2 A t) |1 0\rangle$$ and the reduced state is $$\rho_A=\left( \begin{array}{cc} \cos ^2(2 A t) & 0 \\ 0 & \sin ^2(2 A t) \\ \end{array} \right).$$ Now as user Count Iblis suggested, you can look at the purity of this state, which is $\mathrm{tr}(\rho_A^2)=(3+\cos(8 A t))/4$. Note how the purity oscillates. This shows that depending on the interaction time and strength the two systems are entangled or not.
Unfortunately your state $|\psi(0)\rangle$ is incoherent, so I think there are better examples to illustrate the above. Consider instead the state $$|\psi(0)\rangle=\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)\otimes |0\rangle,$$ which is maximally coherent with respect to the computational basis, and the Hamiltonian $$H=\frac{\pi}{4}(\sigma_0-\sigma_z)\otimes (\sigma_0-\sigma_x),$$ where $\sigma_0$ is the identity matrix. The reduced state is $$\rho_A=\frac{1}{4}\left( \begin{array}{cc} 2 & 1+e^{-i \pi t} \\ 1+e^{i \pi t} & 2 \\ \end{array} \right).$$ Here you see how the coherence, i.e. the absolute value of the off-diagonal elements, oscillates with time. In particular, for $t=1$ (where the interaction is the controlled-Not gate) the reduced state has no coherence at all.