We’re rewarding the question askers & reputations are being recalculated! Read more.

 2 deleted 8 characters in body edited Sep 29 at 15:33 Sanchayan Dutta 11.3k44 gold badges1919 silver badges6666 bronze badges Here is a way to think about the role of entanglement in quantum codes which I think is complementary to Felix Hubers response. Suppose that we take a maximally entangled state $$|\Psi\rangle_{RQ}$$ and record the $$Q$$ system into some quantum error correcting-correcting code. Suppose the code records $$Q$$ into subsystems $$S_1,S_2,S_3$$ such that erasure of any one subsystem can be corrected for (I've taken a simple example, but generalizations are possible). Then, there is an entropic way of thinking about the error correction conditions (as compared to the more algebraic Knill-Laflamme conditions). Specifically, if $$I(R:S_3) =0$$$$I(R:S_3) = 0$$ Then it follows that $$Q$$ can be recovered from $$S_1S_2$$. See for example https://arxiv.org/pdf/quant-ph/0112106.pdfarXiv:quant-ph/0112106 for a nice presentation of this fact. Using this entropic approach to error correction there are fairly direct routes to understanding entanglement in codes. For instance, we can prove that, $$I(S_1S_2:S_3) \geq 2\log d_R$$$$I(S_1S_2:S_3) \geq 2\log d_R$$ as follows. First we write out this mutual information in terms of its definition, $$I(S_1S_2:S_3) = S(S_1S_2) + S(S_3) - S(S_1S_2S_3)$$$$I(S_1S_2:S_3) = S(S_1S_2) + S(S_3) - S(S_1S_2S_3)$$ We'll introduce a purifying system $$X$$, so that the state on $$RS_1S_2S_3X$$ is pure. Then using purity we can write $$I(S_1S_2:S_3) = S(S_3XR) + S(S_3) - S(XR)$$$$I(S_1S_2:S_3) = S(S_3XR) + S(S_3) - S(XR)$$ Note that since we can recover $$Q$$ from $$S_1S_2$$, $$I(R:S_3X)=I(R:X)=0$$. Using this in the above $$I(S_1S_2:S_3) = S(S_3|X) + S(S_3)$$$$I(S_1S_2:S_3) = S(S_3|X) + S(S_3)$$ Finally, we can bound the right hand side here below by $$2 \log d_R$$. The intuition behind how we can do this is that $$S_3$$ is significant'' in the sense that there is a set of shares (say $$S_1$$) which itself reveals no information about $$Q$$, but together with $$S_3$$ allows $$Q$$ to be recovered. Given this, we expect $$S_3$$ must carry $$2\log d_R$$ of entropy, since transferring it can be used to establish $$2\log d_R$$ worth of entanglement. A similar intuition appears in https://arxiv.org/pdf/quant-ph/0608223.pdfarXiv:quant-ph/0608223. More precisely we consider the quantity $$I(R:S_1S_3) - I(R:S_1)$$, which some basic manipulations reveal $$I(R:S_1S_3) - I(R:S_1) = S(S_3|S_1) + S(S_3|XS_2) \leq S(S_3) + S(S_3|X)$$$$I(R:S_1S_3) - I(R:S_1) = S(S_3|S_1) + S(S_3|XS_2) \leq S(S_3) + S(S_3|X)$$ But then we notice $$I(R:S_1S_3) \geq 2\log d_R$$ since $$S_1S_3$$ allows recovery of $$Q$$, while $$I(R:S_1)=0$$ by the entropic error correction condition. This lower bounds $$S(S_3) + S(S_3|X)$$ and so lower bounds $$I(S_1S_2:S_3)$$. Here is a way to think about the role of entanglement in quantum codes which I think is complementary to Felix Hubers response. Suppose that we take a maximally entangled state $$|\Psi\rangle_{RQ}$$ and record the $$Q$$ system into some quantum error correcting code. Suppose the code records $$Q$$ into subsystems $$S_1,S_2,S_3$$ such that erasure of any one subsystem can be corrected for (I've taken a simple example, but generalizations are possible). Then, there is an entropic way of thinking about the error correction conditions (as compared to the more algebraic Knill-Laflamme conditions). Specifically, if $$I(R:S_3) =0$$ Then it follows that $$Q$$ can be recovered from $$S_1S_2$$. See for example https://arxiv.org/pdf/quant-ph/0112106.pdf for a nice presentation of this fact. Using this entropic approach to error correction there are fairly direct routes to understanding entanglement in codes. For instance, we can prove that, $$I(S_1S_2:S_3) \geq 2\log d_R$$ as follows. First we write out this mutual information in terms of its definition, $$I(S_1S_2:S_3) = S(S_1S_2) + S(S_3) - S(S_1S_2S_3)$$ We'll introduce a purifying system $$X$$, so that the state on $$RS_1S_2S_3X$$ is pure. Then using purity we can write $$I(S_1S_2:S_3) = S(S_3XR) + S(S_3) - S(XR)$$ Note that since we can recover $$Q$$ from $$S_1S_2$$, $$I(R:S_3X)=I(R:X)=0$$. Using this in the above $$I(S_1S_2:S_3) = S(S_3|X) + S(S_3)$$ Finally, we can bound the right hand side here below by $$2 \log d_R$$. The intuition behind how we can do this is that $$S_3$$ is significant'' in the sense that there is a set of shares (say $$S_1$$) which itself reveals no information about $$Q$$, but together with $$S_3$$ allows $$Q$$ to be recovered. Given this, we expect $$S_3$$ must carry $$2\log d_R$$ of entropy, since transferring it can be used to establish $$2\log d_R$$ worth of entanglement. A similar intuition appears in https://arxiv.org/pdf/quant-ph/0608223.pdf. More precisely we consider the quantity $$I(R:S_1S_3) - I(R:S_1)$$, which some basic manipulations reveal $$I(R:S_1S_3) - I(R:S_1) = S(S_3|S_1) + S(S_3|XS_2) \leq S(S_3) + S(S_3|X)$$ But then we notice $$I(R:S_1S_3) \geq 2\log d_R$$ since $$S_1S_3$$ allows recovery of $$Q$$, while $$I(R:S_1)=0$$ by the entropic error correction condition. This lower bounds $$S(S_3) + S(S_3|X)$$ and so lower bounds $$I(S_1S_2:S_3)$$. Here is a way to think about the role of entanglement in quantum codes which I think is complementary to Felix Hubers response. Suppose that we take a maximally entangled state $$|\Psi\rangle_{RQ}$$ and record the $$Q$$ system into some quantum error-correcting code. Suppose the code records $$Q$$ into subsystems $$S_1,S_2,S_3$$ such that erasure of any one subsystem can be corrected for (I've taken a simple example, but generalizations are possible). Then, there is an entropic way of thinking about the error correction conditions (as compared to the more algebraic Knill-Laflamme conditions). Specifically, if $$I(R:S_3) = 0$$ Then it follows that $$Q$$ can be recovered from $$S_1S_2$$. See for example arXiv:quant-ph/0112106 for a nice presentation of this fact. Using this entropic approach to error correction there are fairly direct routes to understanding entanglement in codes. For instance, we can prove that, $$I(S_1S_2:S_3) \geq 2\log d_R$$ as follows. First we write out this mutual information in terms of its definition, $$I(S_1S_2:S_3) = S(S_1S_2) + S(S_3) - S(S_1S_2S_3)$$ We'll introduce a purifying system $$X$$, so that the state on $$RS_1S_2S_3X$$ is pure. Then using purity we can write $$I(S_1S_2:S_3) = S(S_3XR) + S(S_3) - S(XR)$$ Note that since we can recover $$Q$$ from $$S_1S_2$$, $$I(R:S_3X)=I(R:X)=0$$. Using this in the above $$I(S_1S_2:S_3) = S(S_3|X) + S(S_3)$$ Finally, we can bound the right hand side here below by $$2 \log d_R$$. The intuition behind how we can do this is that $$S_3$$ is significant'' in the sense that there is a set of shares (say $$S_1$$) which itself reveals no information about $$Q$$, but together with $$S_3$$ allows $$Q$$ to be recovered. Given this, we expect $$S_3$$ must carry $$2\log d_R$$ of entropy, since transferring it can be used to establish $$2\log d_R$$ worth of entanglement. A similar intuition appears in arXiv:quant-ph/0608223. More precisely we consider the quantity $$I(R:S_1S_3) - I(R:S_1)$$, which some basic manipulations reveal $$I(R:S_1S_3) - I(R:S_1) = S(S_3|S_1) + S(S_3|XS_2) \leq S(S_3) + S(S_3|X)$$ But then we notice $$I(R:S_1S_3) \geq 2\log d_R$$ since $$S_1S_3$$ allows recovery of $$Q$$, while $$I(R:S_1)=0$$ by the entropic error correction condition. This lower bounds $$S(S_3) + S(S_3|X)$$ and so lower bounds $$I(S_1S_2:S_3)$$. 1 answered Sep 11 at 21:59 Alex May 16155 bronze badges Here is a way to think about the role of entanglement in quantum codes which I think is complementary to Felix Hubers response. Suppose that we take a maximally entangled state $$|\Psi\rangle_{RQ}$$ and record the $$Q$$ system into some quantum error correcting code. Suppose the code records $$Q$$ into subsystems $$S_1,S_2,S_3$$ such that erasure of any one subsystem can be corrected for (I've taken a simple example, but generalizations are possible). Then, there is an entropic way of thinking about the error correction conditions (as compared to the more algebraic Knill-Laflamme conditions). Specifically, if $$I(R:S_3) =0$$ Then it follows that $$Q$$ can be recovered from $$S_1S_2$$. See for example https://arxiv.org/pdf/quant-ph/0112106.pdf for a nice presentation of this fact. Using this entropic approach to error correction there are fairly direct routes to understanding entanglement in codes. For instance, we can prove that, $$I(S_1S_2:S_3) \geq 2\log d_R$$ as follows. First we write out this mutual information in terms of its definition, $$I(S_1S_2:S_3) = S(S_1S_2) + S(S_3) - S(S_1S_2S_3)$$ We'll introduce a purifying system $$X$$, so that the state on $$RS_1S_2S_3X$$ is pure. Then using purity we can write $$I(S_1S_2:S_3) = S(S_3XR) + S(S_3) - S(XR)$$ Note that since we can recover $$Q$$ from $$S_1S_2$$, $$I(R:S_3X)=I(R:X)=0$$. Using this in the above $$I(S_1S_2:S_3) = S(S_3|X) + S(S_3)$$ Finally, we can bound the right hand side here below by $$2 \log d_R$$. The intuition behind how we can do this is that $$S_3$$ is significant'' in the sense that there is a set of shares (say $$S_1$$) which itself reveals no information about $$Q$$, but together with $$S_3$$ allows $$Q$$ to be recovered. Given this, we expect $$S_3$$ must carry $$2\log d_R$$ of entropy, since transferring it can be used to establish $$2\log d_R$$ worth of entanglement. A similar intuition appears in https://arxiv.org/pdf/quant-ph/0608223.pdf. More precisely we consider the quantity $$I(R:S_1S_3) - I(R:S_1)$$, which some basic manipulations reveal $$I(R:S_1S_3) - I(R:S_1) = S(S_3|S_1) + S(S_3|XS_2) \leq S(S_3) + S(S_3|X)$$ But then we notice $$I(R:S_1S_3) \geq 2\log d_R$$ since $$S_1S_3$$ allows recovery of $$Q$$, while $$I(R:S_1)=0$$ by the entropic error correction condition. This lower bounds $$S(S_3) + S(S_3|X)$$ and so lower bounds $$I(S_1S_2:S_3)$$.