Noise and Decoherence

6.1 Relaxation

Both kinds of loss come from the qubit's contact with surroundings we do not measure. How does that contact remove coherence, and what law does energy loss follow? Treating the surroundings as an unmeasured mediator gives the mechanism. Energy loss then gives the relaxation time and a matching decay of coherence. A warm environment finally adds excitation as well as decay.

6.1.1 Energy loss and T1

Where does the information go? Recall the unmeasured mediator. When its final state depended on the data branch, the interference contrast fell by the overlap of the two mediator states. A qubit's surroundings act in the same way: the electromagnetic field, vibrations of the material, nearby defects. The difference is that we neither control nor measure them. We call them the environment.

The environment can keep two kinds of record. If the qubit's energy passes into it, the environment's state records that a decay happened. This is energy relaxation. If the environment only responds differently to \(\ket0\) and \(\ket1\), exchanging no energy, it records which state the qubit was in. This is pure dephasing: loss of phase without any change of population.

Consider relaxation first. Let the environment start in \(\ket{E_0}\), and suppose that over a time \(t\)

\[\ket0\ket{E_0}\mapsto\ket0\ket{E_0},\qquad\ket1\ket{E_0}\mapsto\sqrt\eta\,\ket1\ket{E_0}+\sqrt{1-\eta}\,\ket0\ket{E_1},\]

with \(\braket{E_0}{E_1}=0\). The excited state either survives, with amplitude \(\sqrt\eta\), or gives its energy to the environment, which then holds an excitation \(\ket{E_1}\). The two images are orthonormal, so this extends to a unitary. Under the constant-rate assumption we used to budget gate time, \(\eta=e^{-t/T_1}\), with decay rate \(\Gamma_1=1/T_1\).

What happens to a superposition \(a\ket0+b\ket1\)? It becomes \((a\ket0+b\sqrt\eta\ket1)\ket{E_0}+b\sqrt{1-\eta}\ket0\ket{E_1}\). The excited population is \(|b|^2\eta\). To find the coherence we need a description that allows mixtures, since the qubit alone is no longer in a pure state.

For a known pure state we write the outer product \(\rho=\ket\psi\bra\psi\). For a random preparation we average these outer products with their probabilities. The result is the density matrix. Its diagonal entries are populations, and its off-diagonal entries are the coherences that interference measurements see. When part of a joint state goes unmeasured, we project onto each orthonormal state of that part and add the resulting outer products. This operation is the partial trace.

Projecting onto \(\ket{E_0}\) leaves \(a\ket0+b\sqrt\eta\ket1\), and projecting onto \(\ket{E_1}\) leaves \(b\sqrt{1-\eta}\ket0\). So the qubit's density matrix after the wait is \(K_0\rho K_0^\dagger+K_1\rho K_1^\dagger\), with

\[K_0=\ket0\bra0+\sqrt\eta\ket1\bra1,\qquad K_1=\sqrt{1-\eta}\,\ket0\bra1.\]

Matrices that describe a process in this way are called Kraus operators. Their condition \(K_0^\dagger K_0+K_1^\dagger K_1=I\) keeps the total probability one. Here the sum multiplies the excited population by \(\eta\) and the off-diagonal entry by \(\sqrt\eta\), since only the \(\ket{E_0}\) parts pair up in \(\bra0\rho\ket1\). Population decays as \(e^{-t/T_1}\), but coherence decays as \(e^{-t/(2T_1)}\). So energy loss alone gives a coherence decay time \(2T_1\).

6.1.2 Thermal excitation

A cold environment only absorbs energy. What changes if it is warm enough to give energy back? Allow an upward rate \(\Gamma_\uparrow\) as well as a downward rate \(\Gamma_\downarrow\). The population then obeys \(\dot p_1=\Gamma_\uparrow(1-p_1)-\Gamma_\downarrow p_1\). Setting the derivative to zero gives the equilibrium \(p_{\rm eq}=\Gamma_\uparrow/(\Gamma_\uparrow+\Gamma_\downarrow)\). Subtracting this constant leaves an exponential:

\[p_1(t)=p_{\rm eq}+[p_1(0)-p_{\rm eq}]e^{-(\Gamma_\uparrow+\Gamma_\downarrow)t}.\]

So preparing both the ground and the excited state lets us estimate the common equilibrium and the total rate. A fit that forces the equilibrium to zero biases the inferred lifetime.

How are the two rates related? If the environment is itself in equilibrium at temperature \(T\), then \(p_{\rm eq}\) must equal the Boltzmann value \(1/(1+e^{\hbar\omega/(k_BT)})\). Equating the two expressions gives \(\Gamma_\uparrow/\Gamma_\downarrow=e^{-\hbar\omega/(k_BT)}\), a relation called detailed balance. Excitation by stray radiation from a hotter source need not obey it.

6.2 Dephasing

Energy loss limits the coherence time to \(2T_1\), but coherence often decays faster. What else removes phase without moving population, and can that phase be recovered? A randomly shifting frequency gives each run a different phase, and averaging over runs lowers the contrast. If the shift changes slowly, reversing its effect halfway through a run recovers the phase.

6.2.1 Frequency noise and T2

Suppose the environment shifts the qubit frequency by a small amount \(\delta\omega(s)\) that differs from run to run. In one run the relative phase then gains \(\phi(t)=\int_0^t\delta\omega(s)\,ds\). Each run stays pure, but averaging runs multiplies the coherence by \(\langle e^{i\phi}\rangle\). We quote a standard probability result: for a zero-mean Gaussian phase of variance \(v\), this average is \(e^{-v/2}\). A spread of phases therefore lowers the interference contrast.

How does the variance grow with time? Suppose the noise is memoryless: its value in each short interval is independent of the past. Such noise is called Markovian. The phase is then a sum of independent increments, so its variance grows in proportion to \(t\). Writing \(v=2t/T_\phi\) gives the factor \(e^{-t/T_\phi}\), exponential pure dephasing with time constant \(T_\phi\). Combining it with the factor \(e^{-t/(2T_1)}\) from independent energy loss gives

\[\frac1{T_2}=\frac1{2T_1}+\frac1{T_\phi}.\]

For \(T_1=100\) μs and \(T_\phi=100\) μs, \(T_2=66.7\) μs.

What if the noise is slow instead? If the offset is nearly constant within a run but varies between runs with spread \(\sigma\), then \(\phi=\delta\omega\,t\) and \(v=\sigma^2t^2\). The envelope is the Gaussian \(e^{-\sigma^2t^2/2}\), not an exponential. The experiment that measures this free decay applies a π/2 pulse, waits, applies a second π/2 pulse, and measures. It is called a Ramsey experiment, and its fitted decay time is written \(T_2^*\). Because slow offsets give a Gaussian envelope, \(T_2^*\) need not be a single microscopic rate.

6.2.2 Echo and noise timescales

Can phase lost to a slow offset be recovered? If \(\delta\omega\) is constant during a run, free evolution gives the phase \(\delta\omega\,t\). Insert an X π pulse halfway through. Since \(XZX=-Z\), the pulse exchanges the roles of \(\ket0\) and \(\ket1\). Relative to the original labels, the second half then accumulates phase with the opposite sign, and the total is \(\delta\omega(t/2)-\delta\omega(t/2)=0\). In coordinates that flip with each π pulse, called a toggling frame, the frequency term simply changes sign at each pulse.

When does the cancellation fail? If the noise changes during the run, the two halves see different offsets. In general the phase is \(\int y(t)\,\delta\omega(t)\,dt\), where \(y\) switches between \(+1\) and \(-1\) at each π pulse. The pulse timing thus selects which noise timescales cancel and which survive. Echo cannot undo energy already passed to the environment, and imperfect π pulses add errors of their own.

6.3 Error Models

Relaxation and dephasing describe one qubit inside its two states. A complete error model must also say what happens outside those states and across several qubits, and it must predict something we can measure. Asking what leakage and shared events add shows what a list of random Pauli errors misses. Following one model all the way to a measured probability then shows how to separate its parts.

6.3.1 Leakage and correlated errors

Is every error a random X, Y, or Z applied by mistake? Such a Pauli error keeps the state inside the qubit's two-dimensional space, and leakage does not. A leaked level may respond differently to later gates and may persist until a dedicated removal step. A model built only from Pauli errors therefore misses this long-lived memory in the error process.

Correlations matter even for errors that stay inside the qubit. Independent flips on two qubits occur together with probability \(p^2\). A shared event that flips both with probability \(p\) has the same single-qubit flip probabilities but a much larger joint probability. Measurements on each qubit separately cannot tell these apart. Only simultaneous operations, with joint outcomes recorded, can.

6.3.2 From a model to a prediction

How does a noise model earn its use? By predicting a measurement. Start with \(\ket+\), wait a time \(t\), and measure X. Under the exponential relaxation and pure-dephasing model, \(\langle X\rangle=e^{-t/T_2}\), so \(P(+)=\tfrac12[1+e^{-t/T_2}]\). At \(t=T_2\) this is about 0.684, not the excited-state survival \(e^{-1}\).

With a detuning, the X expectation also contains a factor \(\cos(\Delta t)\), so the oscillation and its fading contrast have separate causes. The assignment relation then maps these probabilities to reported ones. Keeping the three steps separate lets an experiment distinguish a frequency error, a loss of coherence, and limited detector contrast. Fitting every reduced oscillation to “decoherence” would confuse all three.

Each parameter in these models, from \(T_1\) and \(T_2\) to the detuning and the assignment errors, must be inferred from experiments. Which experiments infer them, and what can they establish about the quality of a gate?

Models and further derivations: Krantz et al.: relaxation, dephasing, and noise models. Numerical examples in this chapter are illustrative, not device specifications.