Decoherence-free subspaces

Decoherence-free subspace is a special set of quantum states which is insensitive to some particular noise. As a simple example, suppose we have two localized spins-1/2 particles, which are subject to some random static magnetic field (for simplicity) along the z-axis. The Hamiltonian is assumed to be

H = B(σz1 + σz2),

where σz1 and σz2 are Pauli matrices acting on spin 1 and spin 2 respectively, and the value of  B is unknown and may be time-varying. Suppose we can initiate our quantum state in the subspace {∣ ↑  ↓ ⟩, ∣ ↓  ↑ ⟩}, for example, α∣ ↑  ↓ ⟩ + β∣ ↓  ↑ ⟩, then this symmetric noise would not change the state at all.

The structure of Decoherence-free subspace

For the interest of quantum information processing, the noise induced by the environment (also known as bath) is the major obstacle for maintaining the fidelity of the quantum devices (called system). The total dynamics of the system and the bath is formally governed by the total Hamiltonian

H = HS(t) + HB + HSB(t),

where we assumed that the variables associated with the system may be time-dependent. The unitary transformation operator associated with this Hamiltonian can be decomposed into three terms

U(t) = T{e − i∫0tdτHS(τ)} × e − itHB × T{e − i∫0tdτSB(τ)}, where the first and the third term are defined by the Dyson series and  T is the corresponding time-ordering operator. Here

SB(τ) ≡ ei∫0τH0(s)dsHSB(τ)e − i∫0τH0(s)ds and H0(s) ≡ HS(s) + HB.

Suppose we assume the system and the bath are initially uncorrelated, and represent the initial state as a pure state, ∣Qubits⟩ ⊗ ∣Bath⟩, then a sufficient condition (but not necessary, see the exception below) for the system to be decoupled with the bath is that

SB(τ)∣Qubits⟩ ⊗ ∣Bath⟩ = 0

for (at least) all values of τ ≤ t. If this condition is true, then we must conclude that

U(t)∣Qubits⟩ ⊗ ∣Bath⟩ = e − i∫0tHS(τ)dτQubits⟩ ⊗ e − itHBBath⟩, which suggests that the system and bath are uncorrelated at time  t (in other words, no error is introduced for the system).

Physics of Decoherence-free subspace

It is an interesting question: under what circumstances, the above discussion is physically relevant? First of all, if the degrees of freedom of the environment are weakly perturbed, it is possible to represent the bath as a set of independent oscillators,

$$H_{B}=\sum\limits_{k}{\hbar \omega_{k}a_{k}^{\dagger}a_{k}}$$.

This assumption is not essential but very convenient for performing analysis. Apparently, the application of the decoherence-free encoding depends strongly on the type of system-bath coupling. Let us for the moment, restrict ourselves to the type of interaction which is weak (i.e. linear coupling) and the qubits are strongly polarized along the z-axis. These two assumptions imply we may assume the interaction term as

$$H_{SB}=\sum\limits_{jk}{Z_{j}\left( g_{jk}a_{k}+g_{jk}^{*}a_{k}^{\dagger} \right)}$$,

where the subscript  j labels the qubits, and  gjk are some complex constants. We also have assumed the polarized qubits have the system Hamiltonian of the form


where the exact values of  hj is not relevant here. The important point is that it commutes with other terms [HS, HSB + HB] = 0, and hence not relevant in the following discussion.

According to the previous section, the quantity of interest is

$$\tilde{H}_{SB}\left( \tau \right)=\sum\limits_{jk}{Z_{j}\left( g_{jk}a_{k}e^{-i\omega _{k}\tau}+g_{jk}^{*}a_{k}^{\dagger}e^{i\omega _{k}\tau} \right)}$$.

Collective Noise

Here we see that to have the condition SB(τ)∣Qubits⟩ ⊗ ∣Bath⟩ = 0, one possible way is to have the noise acting collectively for the involved qubits. In other words, we require gjk → gk, and hence

$$\tilde{H}_{SB}\left( \tau \right)=\left( \sum\limits_{j}{Z_{j}} \right)\otimes \sum\limits_{k}{\left( g_{k}a_{k}e^{-i\omega _{k}\tau }+g_{k}^{*}a_{k}^{\dagger}e^{i\omega _{k}\tau } \right)}$$.

The decoherence-free subspace can be constructed within the degenerated eigenstates (with zero as the eigenvalue) of the total magnetization operator ∑jZj.

The physical question about this collective noise assumption is that how can this happen? One possible way is that all of the qubit degrees of freedom are localized at one location, so in principle all qubits "see" the same environment. Alternatively, we may regard the bath as a quantum field (e.g. phonon field), and there exists an intrinsic cut-off frequency  ωc. Suppose the physical qubits are distributed within a distance shorter than the length scale defined by the cut-off frequency, then the bath cannot see the spatial distribution of the qubits and hence act collectively. However these two scenarios are not very likely in the current experimental situations.

Effective Collective Noise

In the above discussion, we have focused on the condition

SB(τ)∣Qubits⟩ ⊗ ∣Bath⟩ = 0

for all τ ≤ t. However this is a sufficient condition but not necessary for achieving the purposes of decoherence-free encoding. It is possible that decoherence-free subspace be developed when time is sufficiently long.

As an example, suppose we consider the excitations of the bath has a linear dispersion relation  ωk = ck, and the coupling contains position information in the phase factor

gjk → gkeikrj, which means that the interaction is point-like.

In the limit t → ∞ (practically, when the time is sufficiently long), then the resonating mode (further details see Yung 2007) is the long wavelength mode k → 0, and the position information will be lost, i.e. eikrj → 1. In this case, there is no noise in the decoherence-free subspace. Physically, the time scale has to be at least longer than the minimum time required for the bath to transfer information, which is roughly given by  d/c, where  d is the qubit-qubit separation and  c is the wave speed.


M.-H. Yung, Independent Noise Approximation for Spin-Boson Decoherence, arXiv:0707.2779.


Last modified: 
Monday, October 26, 2015 - 17:56