The dynamics of two entangled qubits exposed to classical noise: role of spatial and temporal noise correlations

Abstract

We investigate the decay of two-qubit entanglement caused by the influence of classical noise. We consider the whole spectrum of cases ranging from independent to fully correlated noise affecting each qubit. We take into account different spatial symmetries of noises, and the regimes of noise autocorrelation time. The latter can be either much shorter than the characteristic qubit decoherence time (Markovian decoherence), or much longer (approaching the quasi-static bath limit). We express the entanglement of two-qubit states in terms of expectation values of spherical tensor operators which allows for transparent insight into the role of the symmetry of both the two-qubit state and the noise for entanglement dynamics.

Appendices

Appendix 1: Concurrence of the Bell States

The density matrices of the $X_{\mathrm {corr}}$-states, written in standard two-qubit basis $\{ \left| \uparrow \downarrow \right\rangle ,\left| \downarrow \uparrow \right\rangle ,\left| \uparrow \uparrow \right\rangle ,\left| \downarrow \downarrow \right\rangle \}$, have a form

$$\begin{aligned} \varrho = \left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c} a &{} 0 &{} 0 &{} w\\ 0 &{} b &{} z &{} 0\\ 0 &{} z^*&{}b &{}0\\ w^*&{}0&{}0&{}a\\ \end{array}\right) , \end{aligned}$$

(91)

where the condition $2a+2b=1$ insures that the trace equals unity. Matrices of the form (91) are time reversal invariant, hence the concurrence is easily evaluated because it can be expressed in term of eigenvalues of $\varrho$ itself. These eigenvalues can be determined analytically, so the concurrence is given by [45]

$$\begin{aligned} {\mathcal {C}}(\varrho )= 2\,\text {max}\left\{ 0, |z|-a, |w|-b \right\} . \end{aligned}$$

(92)

Straightforward calculations show that the matrix elements $a,\,b,\,w$ and z can be written in terms of the expectation values of products of the spherical tensor operators,

$$\begin{aligned} z{}&=-2\left\langle T^{(1)}_{11}\otimes T^{(2)}_{1{-1}}\right\rangle _{\varrho } =-2\left\langle T^{(1)}_{1-1}\otimes T^{(2)}_{11}\right\rangle ^*_{\varrho }, \end{aligned}$$

(93)

$$\begin{aligned} w{}&=2\left\langle T^{(1)}_{11}\otimes T^{(2)}_{11}\right\rangle _{\varrho } =2\left\langle T^{(1)}_{1-1}\otimes T^{(2)}_{1-1}\right\rangle ^*_{\varrho }, \end{aligned}$$

(94)

$$\begin{aligned} a{}&= \frac{1}{4} +\left\langle T^{(1)}_{10}\otimes T^{(2)}_{10}\right\rangle _{\varrho }, \end{aligned}$$

(95)

$$\begin{aligned} b{}&= \frac{1}{4} -\left\langle T^{(1)}_{10}\otimes T^{(2)}_{10}\right\rangle _{\varrho }. \end{aligned}$$

(96)

Thus we obtain formula (27).

Appendix 2: Diagonalization of the evolution superoperator for the case of transverse noise

The evolution superoperator for transverse white noise is given by Eq. (71):

$$\begin{aligned} {\mathcal {U}}_{\mathrm {tr}}=\hbox {e}^{i\varOmega t{\mathcal {J}}_z}\hbox {e}^{-\left[ (1-\gamma )\left( \varvec{{\mathcal {J}}}_1^{2}+\varvec{{\mathcal {J}}}_2^{2}\right) -{\mathcal {J}}_z^2\right] \frac{t}{T}}\hbox {e}^{-\frac{t}{T}{\mathcal {L}}}, \end{aligned}$$

(97)

where we introduced the superoperator ${\mathcal {L}}$ defined as

$$\begin{aligned} {\mathcal {L}} = \gamma \varvec{{\mathcal {J}}}^2 +(1-\gamma )\left( {\mathcal {J}}_z^{(1)}{\mathcal {J}}_z^{(2)} +{\mathcal {J}}_z^{(2)}{\mathcal {J}}_z^{(1)}\right) . \end{aligned}$$

(98)

Our task is to diagonalize the matrix representation of this superoperator. The matrix elements in the basis of spherical tensor products are

$$\begin{aligned} ({\mathcal {L}})_{m_1m_2,m_1'm_2'}=\frac{\left( T^{(1)}_{1 m_1}\!\!\otimes \! T^{(2)}_{1 m_2}\big | {\mathcal {L}}\left( T^{(1)}_{1 m_1'}\!\!\otimes \! T^{(2)}_{1 m_2'}\right) \right) }{\big |\big |T^{(1)}_{1 m_1} \!\otimes \! T^{(2)}_{1 m_2}\big |\big |\,\big |\big |T^{(1)}_{1 m_1'}\!\!\otimes \! T^{(2)}_{1 m_2'}\big |\big |}, \end{aligned}$$

(99)

where $(A|B)=\mathrm {Tr}( A^\dagger B)$ and $||A||=\sqrt{(A|A)}$. The explicit form of the matrix is given by

(100)

where blank spaces are zero. The matrix has a block-diagonal form and each block can be diagonalized analytically. In particular, spherical tensor products $T^{(1)}_{1 \pm 1}\otimes T^{(2)}_{1 \pm 1} = T_{2\pm 2(11)}$ are already eigenoperators of ${\mathcal {L}}$ and the triple of unpolarized products $\{ T^{(1)}_{1 0}~{\!\!\otimes \!}~ T^{(2)}_{1 0},T^{(1)}_{1 1}~{\!\!\otimes \!}~T^{(2)}_{1 {-1}},T^{(1)}_{1 {-1}}~{\!\!\otimes \!}~ T^{(2)}_{1 1} \}$ form one of the blocks. Because the $X_{\mathrm {corr}}$-states (including the Bell states) are spanned by these five operators, it follows that any state which belongs to this class remains in it for all time when it is evolved with ${\mathcal {U}}_{\mathrm {tr}}$.

Below we list the eigenoperators and the corresponding eigenvalues of ${\mathcal {L}}$:

$$\begin{aligned} 2(1+2\gamma ):{}&\; T^{(1)}_{1 \pm 1}\!\!\otimes \! T^{(2)}_{1 \pm 1} \end{aligned}$$

(101)

$$\begin{aligned} -2(1-2\gamma ):{}&\; -T^{(1)}_{1 1}\!\!\otimes \! T^{(2)}_{1 -1}+T^{(1)}_{1 -1}\!\!\otimes \! T^{(2)}_{1 1}, \end{aligned}$$

(102)

$$\begin{aligned} -1+4\gamma -3\eta :{}&\;T^{(1)}_{1 1}\!\!\otimes \! T^{(2)}_{1 -1}+T^{(1)}_{1 -1}\!\!\otimes \! T^{(2)}_{1 1} +\frac{1-3\eta }{2\gamma }T^{(1)}_{1 0}\!\!\otimes \! T^{(2)}_{1 0}, \end{aligned}$$

(103)

$$\begin{aligned} -1+4\gamma +3\eta :{}&\;T^{(1)}_{1 1}\!\!\otimes \! T^{(2)}_{1 -1}+T^{(1)}_{1 -1}\!\!\otimes \! T^{(2)}_{1 1} +\frac{1+3\eta }{2\gamma }T^{(1)}_{1 0}\!\!\otimes \! T^{(2)}_{1 0}, \end{aligned}$$

(104)

$$\begin{aligned} 6\gamma :{}&\; T^{(1)}_{1 0}\!\!\otimes \! T^{(2)}_{1 \pm 1}+T^{(1)}_{1 \pm 1}\!\!\otimes \! T^{(2)}_{1 0}, \end{aligned}$$

(105)

$$\begin{aligned} 2\gamma :{}&\; T^{(1)}_{1 0}\!\!\otimes \! T^{(2)}_{1 \pm 1}-T^{(1)}_{1 \pm 1}\!\!\otimes \! T^{(2)}_{1 0}, \end{aligned}$$

(106)

where $\eta = \frac{1}{3}\sqrt{1+8\gamma ^2}$.