Chapter 9 - Dynamical Decoupling Controls

Introduction

In the last chapter, it was shown that a symmetry in the system-bath Hamiltonian, if present, could be used to construct states immune to noise. In this chapter we will see that under certain conditions it is possible to reduce errors, create a symmetry, or even remove errors in the evolution of a quantum system. This is done though repeated use of external controls which act on the system. These controls are often called "dynamical decoupling controls" due to their original objective of decoupling the system from the bath. They are quite generally useful controls to consider for the elimination and/or reduction of errors. In this chapter, a simple introduction to dynamical decoupling controls is given and some important concepts discussed.

General Conditions

As stated in Chapter 8 the Hamiltonian describing the evolution of a system and bath which are coupled together can always be written as

	Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H = H_S\otimes I_B + I_S\otimes H_B + H_I, \,\!}	(9.1)

where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H_S \,\!} acts only on the system, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H_B\,\!} acts only on the bath, and

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H_I = \sum_\alpha S_\alpha\otimes B_\alpha, \,\!}

is the interaction Hamiltonian with the Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_\alpha\,\!} acting only on the system and the Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B_\alpha\,\!} acting only on the bath.

The idea is to modify the evolution of the system and bath such that the errors are reduced or eliminated using external control Hamiltonians. These controls are called dynamical decoupling controls since they are used to decouple (at least approximately decouple) the system from the bath. Since can be difficult to change states of a bath, indeed one often does not know details of the bath, the controls which are to be used for reducing errors should act on the system. As discussed previously, the errors arise from the system-bath interaction Hamiltonian $H_{I}\,\!$ and, in particular, the system operators Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_\alpha\,\!} are the operators which describe the affect of the coupling on the system. In general the interaction Hamiltonian is time-dependent since the bath operators will change in time. However, for short times we may assume the interaction Hamiltonian is unchanged, or at least approximately constant. This is sometimes called the short-time assumption in dynamical decoupling.

The Magnus Expansion

A fairly good starting point to see how this is done is the so-called Magnus expansion. (See Blanes, et al. and references therein.) The general problem is that a time-dependent operation is to be applied to the Hamiltonian making the Hamiltonian itself time-dependent and one would like to solve the time-dependent Schrodinger equation:

	Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i\frac{\partial}{\partial t}\left\vert \Psi(t)\right\rangle = H(t) \left\vert \Psi(t) \right\rangle,\,\! }	(9.2)

which is sometimes written as

	Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle U^\prime(t) = -iH(t)U(t).\,\! }	(9.3)

The question is, what $U(t)\,\!$ will solve this equation? If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle U(t) \,\!} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H(t)\,\!} are just numbers, the solution would be

	Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle U^\prime(t) = \exp\left(-i\int_0^t H(t^\prime)dt^\prime\right). \,\!}	(9.4)

However, when the Schrodinger equation is the equation to be solved, $U(t)\,\!$ and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H(t)\,\!} are matrices. To be specific, $U(t)\,\!$ is a unitary matrix and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H(t)\,\!} is a Hermitian matrix. The solution is often written in the form

	Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle U^\prime(t) = {\mathcal{T}}\left[\exp\left(-i\int_0^t H(t^\prime)dt^\prime\right)\right], \,\!}	(9.5)

where ${\mathcal {T}}\,\!$ denotes the time-ordered exponential. In this case, matrices do not commute so that the exponential must be handled with care. Operators must be ordered according to the time where they appear in the operation, and the solution Eq.(9.2) is not the solution to the problem unless Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H(t)\,\!} is a constant matrix.

The solution to this problem is the following,

	Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle U^\prime(T) = \exp\left(-i\Omega(T)T\right), \,\!}	(9.6)

where

	$\Omega (T)=\sum _{k=1}^{\infty }\Omega _{k},\,\!$	(9.7)

and

	Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \Omega_1 &= \frac{1}{T}\int_0^T H(t_1)dt_1, \\ \Omega_2 &= -\frac{1}{T^2}\frac{i}{2}\int_0^T dt_1\int_0^{t_1}dt_2 [H(t_1),H(t_2)], \\ \Omega_3 &= -\frac{1}{T^3}\frac{1}{6} \int_0^T dt_1\int_0^{t_1} dt_2 \int_0^{t_2} dt_3 ([H(t_1),[H(t_2),H(t_3)]] + [H(t_3),[H(t_2),H(t_1)]]) \\ & \mbox{etc.}, \end{align}\,\!}	(9.8)

where $T\,\!$ is some characteristic time scale.

A First-Order Theory

To show how this theory of dynamical decoupling controls could work in an ideal case, let us consider a simple example. Suppose that the external controls (decoupling controls) are so strong that the Hamiltonian evolution can be neglected during the time the external controls are turned on. Due to their strength, we will also assume that they can be implemented in a very short time and that there are Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle N \,\!} different controls to be used. We will first use a given control Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle U_n \,\!} and then its inverse. Between the controls the system evolves for a short time Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta t \,\!} . After all control pulses have been implemented, the effective evolution of the system will be

	$U_{eff}(T)=\prod _{n=0}^{N-1}U_{n}^{-1}U(\Delta t)U_{n},\,\!$	(9.9)

where

	Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle U(\Delta t) = \exp(-i H \Delta t), \,\!}	(9.10)

$T=N\Delta t,\,\!$ and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H\,\!} is free evolution given by Eq.(9.1) above. Furthermore, suppose that the time Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta t\,\!} is small so that

	$U(\Delta t)\approx I-iH\Delta t.\,\!$	(9.11)

Now suppose that we let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle U_{eff}=\exp(-iH_{eff}T) \,\!} . Inserting Eq.(9.11) into Eq.(9.9) and keeping only first order terms in the product gives

	Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H_{eff} = \frac{1}{N}\sum_{n=0}^{N-1} U_n^{-1} H U_n. \,\!}	(9.12)

This is a simple expression for the effective Hamiltonian evolution of a system undergoing a series of dynamical decoupling controls. Note that the assumptions are that the operations are strong (since the free evolution is neglected during the control pulses) and fast (since we assume that the Hamiltonian Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H\,\!} of Eq.(9.1) is constant during the entire time $T\,\!$ of this cycle of control pulses). Due to these strong and fast assumptions, these are often referred to as "bang-bang" controls.

It is important to note that these controls are rather unrealistic. That is, these criteria are never met completely. However, they are met approximately in some systems, most notably in nuclear magnetic resonance experiments where the so-called average Hamiltonian theory originated. More realistic pulses can be, and have been, explored for use in actual physical systems where they have been shown to reduce noise very effectively. This has been done by generalizing the theory beyond the first-order limit and without the assumption that the pulses are extremely strong.

The Single Qubit Case

The simplest case involves the elimination of an error on a single qubit. There are several types of errors that can degrade a qubit state as discussed in Section 6.4. There is a bit-flip, phase-flip, or both. In this section the first-order approximation is used to show how to eliminate first phase errors and then arbitrary errors on an arbitrary qubit state.

Phase Errors

Let us suppose that the Hamiltonian for the free evolution contains only an interaction part which induces a phase error,

	$H_{i}=\sigma _{z}\otimes B,\,\!$	(9.13)

where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B\,\!} is the bath operator. This interaction Hamiltonian will couple the system to the bath and thus case errors. The factor Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sigma_z\,\!} indicates that it is a phase error. Using the first-order theory, the objective is to find a series of pulses which will effectively decouple the system from the bath. In this case it can be done with only one decoupling pulse, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle U_X(\pi)\,\!} . This will be denoted $U_{1}=U_{X}(\pi )\,\!$ and the identity (doing nothing) will be denoted Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle U_0=I\,\!} . The effective Hamiltonian is

	$H_{eff}={\frac {1}{2}}\sum _{n=0}^{1}U_{n}^{-1}H_{i}U_{n}={\frac {1}{2}}(U_{0}^{\dagger }H_{i}U_{0}+U_{1}^{\dagger }H_{i}U_{1}).\,\!$	(9.14)

A rotation about the x-axis by an angle Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pi\,\!} will rotate the Pauli matrix Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sigma_z\,\!} to $-\sigma _{z}\,\!$. (See Section C.5, in particular Section C.5.1.) This is because Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle U_1 = U_X(\pi) = \sigma_x\,\!} . After the pulse sequence $U_{0},U_{1}\,\!$, the system is decoupled from the bath because the effective Hamiltonian is zero! There is no more interaction between the system and bath! That is

	$H_{eff}={\frac {1}{2}}(\sigma _{z}\otimes B+\sigma _{x}\sigma _{z}\sigma _{x}\otimes B)={\frac {1}{2}}(\sigma _{z}\otimes B-\sigma _{z}\otimes B)=0.\,\!$	(9.15)

Thus the noise has been removed from the system.

This may be considered an averaging method. (As mentioned before, it is sometimes called average Hamiltonian theory in the NMR literature.) In this case, it is also sometimes called a parity kick since the sign of the interaction Hamiltonian is reversed giving just two terms which cancel.

Arbitrary Single Qubit Errors

Now let us consider arbitrary single qubit errors in an interaction Hamiltonian. The interaction will have the form

	$H_{i}=\sigma _{x}\otimes B_{x}+\sigma _{y}\otimes B_{y}+\sigma _{z}\otimes B_{z}.\,\!$	(9.16)

The objective here is to eliminate all terms in the interaction Hamiltonian. It turns out that this may be accomplished in several different ways. Let us first consider the obvious choice of bang-bang pulses, $\{I,\sigma _{x},\sigma _{y},\sigma _{z}\}\,\!$. First, recall that the Pauli matrices have the property that $\sigma _{i}\sigma _{j}\sigma _{i}=-\sigma _{i}\,\!$ if $i\neq j\,\!$ and $\sigma _{i}\sigma _{j}\sigma _{i}=\sigma _{i}\,\!$ if $i=j\,\!$. Then the effective Hamiltonian is

	${\begin{aligned}H_{eff}&=\;\;{\frac {1}{4}}(\sigma _{x}\otimes B_{x}+\sigma _{y}\otimes B_{y}+\sigma _{z}\otimes B_{z})\\&\;\;\;+{\frac {1}{4}}\sigma _{x}(\sigma _{x}\otimes B_{x}+\sigma _{y}\otimes B_{y}+\sigma _{z}\otimes B_{z})\sigma _{x}\\&\;\;\;+{\frac {1}{4}}\sigma _{y}(\sigma _{x}\otimes B_{x}+\sigma _{y}\otimes B_{y}+\sigma _{z}\otimes B_{z})\sigma _{y}\\&\;\;\;+{\frac {1}{4}}\sigma _{z}(\sigma _{x}\otimes B_{x}+\sigma _{y}\otimes B_{y}+\sigma _{z}\otimes B_{z})\sigma _{z}\\&=\;\;{\frac {1}{4}}(\;\sigma _{x}\otimes B_{x}+\sigma _{y}\otimes B_{y}+\sigma _{z}\otimes B_{z})\\&\;\;\;+{\frac {1}{4}}(\;\sigma _{x}\otimes B_{x}-\sigma _{y}\otimes B_{y}-\sigma _{z}\otimes B_{z})\\&\;\;\;+{\frac {1}{4}}(-\sigma _{x}\otimes B_{x}+\sigma _{y}\otimes B_{y}-\sigma _{z}\otimes B_{z})\\&\;\;\;+{\frac {1}{4}}(-\sigma _{x}\otimes B_{x}-\sigma _{y}\otimes B_{y}+\sigma _{z}\otimes B_{z})\\&=\;\;0.\end{aligned}}\,\!$	(9.17)

So again we see that the interaction Hamiltonian has been eliminated, removing the errors.