We’re rewarding the question askers & reputations are being recalculated! Read more.
2 deleted 8 characters in body
source | link

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
source | link

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 $I(S_1S_2:S_3)$.