Dephasing (T₂)

When a qubit is not actively driven it still interacts weakly with its environment. Random fluctuations in the local electromagnetic field impart tiny, unpredictable phase kicks. Each kick is a small rotation about the $Z$ -axis; the accumulated effect is that the relative phase between $|0\rangle$ and $|1\rangle$ drifts randomly. This process is called dephasing, and its characteristic timescale is $T_2$ .

What dephasing destroys

Write a general single-qubit density matrix as

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

The diagonal entries $\rho_{00}$ and $\rho_{11}$ are the populations (probabilities of the two basis states). The off-diagonal entries $\rho_{01} = \rho_{10}^*$ are the coherences — they encode the quantum phase relation between $|0\rangle$ and $|1\rangle$ and are the signature of genuine superposition.

Dephasing leaves the populations unchanged but erodes the coherences:

\rho(t) = \begin{pmatrix} \rho_{00} & \rho_{01}\,e^{-t/T_2} \\ \rho_{10}\,e^{-t/T_2} & \rho_{11} \end{pmatrix}.

As $t \to \infty$ the state approaches the diagonal mixed state $\rho_{00}|0\rangle\langle 0| + \rho_{11}|1\rangle\langle 1|$ , which has the same measurement probabilities as before but zero quantum coherence. The information about the superposition phase has leaked irreversibly into the environment.

The discrete phase-flip model

In the discrete (gate-based) picture one step of dephasing is modelled by the phase-flip channel

\mathcal{E}_p(\rho) = (1-p)\,\rho + p\,Z\rho Z,

where $p \in [0, \tfrac{1}{2}]$ is the dephasing probability per gate cycle (see the preceding lesson). Computing the action on the off-diagonal element directly:

[\mathcal{E}_p(\rho)]_{01} = (1-p)\,\rho_{01} + p\,(Z\rho Z)_{01} = (1-p)\,\rho_{01} + p\,(-\rho_{01}) = (1-2p)\,\rho_{01}.

The factor $(1-2p)$ is the dephasing factor for a single channel application.

Relating $p$ to $T_2$

In the continuous limit, dephasing accumulates over time $t$ with rate $1/T_2$ . The coherence decays as $e^{-t/T_2}$ . Applying the discrete channel $n$ times with step size $\Delta t$ gives a coherence of $(1-2p)^n$ . Setting $t = n\,\Delta t$ and matching to the exponential decay yields

(1-2p)^{n} = e^{-n\Delta t / T_2} \implies 1 - 2p \approx e^{-\Delta t / T_2},

valid when $p \ll 1$ . A longer $T_2$ means smaller $p$ per cycle: the qubit retains its coherence for more gate operations before it dephases completely.

The Ramsey experiment

The standard way to measure $T_2$ in the lab is a Ramsey experiment:

Apply $H$ to prepare $|+\rangle$ .
Wait a free-evolution time $t$ , during which dephasing shrinks the coherence by $e^{-t/T_2}$ .
Apply $H$ again and measure in the computational basis.

If there were no dephasing the second $H$ would return the qubit to $|0\rangle$ and every shot would read 0. With dephasing the probability of seeing 0 becomes

P(0) = \frac{1 + e^{-t/T_2}}{2}.

Fitting the observed $P(0)$ versus $t$ to this exponential gives $T_2$ . Modern superconducting qubits achieve $T_2$ in the range of tens to hundreds of microseconds, depending on the qubit design and the quality of the local electromagnetic environment.

Try it

A qubit starts in $|+\rangle$ , so its initial off-diagonal coherence is $\rho_{01} = \tfrac{1}{2}$ . The dephasing channel with parameter $p = 0.15$ is applied once. Compute and return the new coherence after the channel.

Run your code to see the quantum state.