Error correction with orbital angular momentum of multiple photons propagating in a turbulent atmosphere

Orbital angular momentum of photons is an intriguing system for the storage and transmission of quantum information, but it is rapidly degraded by atmospheric turbulence. Understanding the noise processes that affect photons is essential if we desire to protect them. In this paper we use the infinitesimal propagation equation of Roux to derive a discrete Lindblad equation, and numerically study the form of the most relevant Lindblad operators. We find that the dominant Lindblad operators are those that shift the angular momentum by one unit. We explore possible schemes to protect quantum information across multiple photons by concatenating a standard quantum error-correcting code with an error-detecting code for orbital angular momentum.

Figure 1

Example of the operator $\mathcal{C}$ when $\lambda =1.0\times {10}^{-6}$ m, ${\omega }_0=0.01$ m, $C_n^2=1.0\times {10}^{-14}{\text{m}}^{-2/3},t_z=100$, and $L=4$.

Figure 2

Example of the operator $\mathcal{D}$ when $\lambda =1.0\times {10}^{-6}$ m, ${\omega }_0=0.01$ m, $C_n^2=1.0\times {10}^{-14}{\text{m}}^{-2/3},t_z=100$, and $L=4$.

Figure 3

Magnitudes of the eigenvalues of the operator $\tilde{\mathcal{D}}$ when $\lambda =1.0\times {10}^{-6}$ m, ${\omega }_0=0.01$ m, $C_n^2=1.0\times {10}^{-14}{\text{m}}^{-2/3},t_z=100$, and $L=4$.

Figure 4

Example of the Lindblad operator ${\mathbf{L}}_1$ when $\lambda =1.0\times {10}^{-6}$ m, ${\omega }_0=0.01$ m, $C_n^2=1.0\times {10}^{-14}{\text{m}}^{-2/3},t_z=100$, and $L=4$.

Figure 5

Example of the Lindblad operator ${\mathbf{L}}_2$ when $\lambda =1.0\times {10}^{-6}$ m, ${\omega }_0=0.01$ m, $C_n^2=1.0\times {10}^{-14}{\text{m}}^{-2/3},t_z=100$, and $L=4$.

Figure 6

Example of the Lindblad operator ${\tilde{\mathbf{L}}}_1$ when $\lambda =1.0\times {10}^{-6}$ m, ${\omega }_0=0.01$ m, $C_n^2=1.0\times {10}^{-14}{\text{m}}^{-2/3},t_z=100,L=4$, and $n=3$.

Figure 7

Example of the Lindblad operator ${\tilde{\mathbf{L}}}_2$ when $\lambda =1.0\times {10}^{-6}$ m, ${\omega }_0=0.01$ m, $C_n^2=1.0\times {10}^{-14}{\text{m}}^{-2/3},t_z=100,L=4$, and $n=3$.

Figure 8

Trace of the state in Eq. (36) after evolving for $t_z$ when $\lambda =1.0\times {10}^{-6}$ m, ${\omega }_0=0.01$ m, $C_n^2=1.0\times {10}^{-14}{\text{m}}^{-2/3}$.

Figure 9

Probability of a detectable error when using a state of the form given in Eq. (36) after evolving for $t_z$ when $\lambda =1.0\times {10}^{-6}$ m, ${\omega }_0=0.01$ m, $C_n^2=1.0\times {10}^{-14}{\text{m}}^{-2/3}$.

Figure 10

Minimum fidelity between an initial state of the form given in Eq. (36) and the state after evolving for $t_z$ when $\lambda =1.0\times {10}^{-6}$ m, ${\omega }_0=0.01$ m, $C_n^2=1.0\times {10}^{-14}{\text{m}}^{-2/3}$.