Noise, such as photon loss or dephasing, is often described with Kraus operators.

Are there examples of Kraus operators describing the process of control error?

  • $\begingroup$ what do you mean exactly with "the process of control error"? $\endgroup$
    – glS
    Commented Jun 20, 2021 at 9:23
  • $\begingroup$ For example, in interferometer, the photon loss can be described by Kraus operators: $$ K_{0}=\left(\begin{array}{cc} 0 & 0 \\ 0 & 0 \\ 0 & \sqrt{1-\eta} \end{array}\right), K_{1}=\left(\begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \sqrt{1-\eta} & 0 \end{array}\right), K_{2}=\left(\begin{array}{cc} \sqrt{\eta} & 0 \\ 0 & \sqrt{\eta} \\ 0 & 0 \end{array}\right) .$$ But can control error, e.g. the phase in laser is not as expected or the measurement direction is not so precise and so on, also be described by Kraus operators? $\endgroup$
    – narip
    Commented Jun 20, 2021 at 9:28
  • 1
    $\begingroup$ so when you say "control error" you are referring to situations in which the parameters defining a setup are only known up to some error? I don't know specific examples, but just to note: Kraus operators are but a way to describe a quantum channel; if these "control errors" can be modeled with a quantum channel, then yes they can be described via Kraus operators $\endgroup$
    – glS
    Commented Jun 20, 2021 at 9:33

1 Answer 1


Control errors

The term control error is generally used to refer to errors due to imperfections of the qubit control system. Hardware devices that control qubit evolution have a number of knobs that the control system sets to various values. In the process known as calibration we learn the settings of the knobs that correspond to each of the supported gates. Denoting by $U(\theta)$ the evolution resulting from setting the knobs to $\theta$ and with $U_1, U_2, \dots, U_k$ the supported gates, calibration aims to find $\theta_1, \theta_2, \dots, \theta_k$ such that

$$ U_1 = U(\theta_1) \\ U_2 = U(\theta_2) \\ \dots \\ U_k = U(\theta_k). $$

We generally find the values $\theta_i$ by optimization, e.g. by maximizing the entanglement fidelity $F_e$. Optimization generally terminates when it finds a solution $\tilde{\theta_i}$ that is considered sufficiently good, e.g. when $F_e(U(\theta_i), U(\tilde{\theta_i})) \ge 1-\epsilon$ where $\epsilon$ is a tolerance setting. Consequently, we know the settings of the control knobs necessary to execute any particular gate up to some finite accuracy. Moreover, even if we knew $\theta_i$ exactly, the control knobs generally have finite precision, resulting for example from finite word-width of digital-to-analog converters.

The upshot is that whenever the qubit control system is asked to execute the gate $U_i = U(\theta_i)$, it actually executes the gate $V_i=U(\tilde{\theta_i})$. Thus, for the input state $|\psi\rangle$ we obtain $V_i|\psi\rangle$ instead of $U_i|\psi\rangle$. Notice that

$$ V_i|\psi\rangle = V_iU_i^\dagger U_i|\psi\rangle = E_iU_i|\psi\rangle $$

where we defined the unitary control error $E_i = V_iU_i^\dagger$. Thus, we can think of the output state as resulting from the execution of the ideal gate $U_i$ followed by a unitary control error $E_i$.

Kraus representation of control errors

The channel which applies a unitary operator $U$ has a simple Kraus representation

$$ \rho \mapsto \mathcal{U}(\rho) = U\rho U^\dagger $$

where the unitary $U$ is the sole Kraus operator. Applying this to the unitary control error $E_i = V_iU_i^\dagger$ we get

$$ \rho \mapsto \mathcal{E}_i(\rho) = E_i\rho E_i^\dagger = V_iU_i^\dagger\rho U_iV_i^\dagger. $$

Simple example

Suppose we wish to execute $S\equiv R_z \left(\frac{\pi}{2}\right)$, where $\equiv$ denotes equality up to global phase, but due to control error end up actually executing $S':=R_z \left(\frac{\pi}{2}\cdot (1 + 10^{-4})\right)$ instead. The unitary control error is

$$ E = S'S^\dagger = R_z \left(\frac{\pi}{2}\cdot 10^{-4}\right) $$

and its Kraus representation is

$$ \rho \mapsto \mathcal{E}(\rho) = R_z \left(\frac{\pi}{2}\cdot 10^{-4}\right)\rho R_z \left(-\frac{\pi}{2}\cdot 10^{-4}\right). $$


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