|q⟩ Bad Qubits

advanced · Physics · Open Quantum Systems & Lindblad Dynamics

Checkpoint: Solve a Master Equation

The task

This checkpoint puts the whole module together. A single qubit is coupled to a zero-temperature bath through two channels you have now met:

You will solve the resulting master equation for the Bloch vector and report its transverse component after a fixed evolution time.

Bloch equations from the Lindblad equation

Writing ρ=12(1+x σx+y σy+z σz)\rho = \tfrac12(\mathbb{1} + x\,\sigma_x + y\,\sigma_y + z\,\sigma_z) and substituting the two dissipators (with H=0H=0, so no precession), the components decouple into simple exponential decays:

z˙=−1T1 (z−z∞),x˙=−1T2 x,y˙=−1T2 y,\dot z = -\frac{1}{T_1}\,(z - z_\infty), \qquad \dot x = -\frac{1}{T_2}\,x, \qquad \dot y = -\frac{1}{T_2}\,y,

with the longitudinal fixed point z∞=−1z_\infty = -1 (the ground state at T=0T=0) and the transverse rate set by

1T2=12T1+1Tϕ.\frac{1}{T_2} = \frac{1}{2T_1} + \frac{1}{T_\phi}.

The longitudinal and transverse motions are independent: populations relax toward the ground state on the scale T1T_1, while coherences die on the scale T2T_2.

Solving for the transverse component

The transverse equation integrates immediately to

⟨σx⟩(t)=x0 e−t/T2.\langle\sigma_x\rangle(t) = x_0\,e^{-t/T_2}.

There is no constant offset because the transverse steady value is zero. So the entire problem reduces to forming T2T_2 from T1T_1 and TϕT_\phi, then evaluating one exponential.

Try it

This is a numerical exercise — your code should return a number. With T1=50T_1=50, Tϕ=75T_\phi=75, an initial equatorial state ⟨σx⟩(0)=1\langle\sigma_x\rangle(0)=1, and evolution time t=30t=30 (consistent units), compute T2T_2 via 1/T2=1/(2T1)+1/Tϕ1/T_2 = 1/(2T_1)+1/T_\phi and return ⟨σx⟩(t)=x0 e−t/T2\langle\sigma_x\rangle(t) = x_0\,e^{-t/T_2}.

Run your code to see the quantum state.

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