{"id":4115,"date":"2024-09-21T18:29:34","date_gmt":"2024-09-21T18:29:34","guid":{"rendered":"https:\/\/workhouse.sweetdishy.com\/?p=4115"},"modified":"2024-09-24T11:40:25","modified_gmt":"2024-09-24T11:40:25","slug":"qkd-using-entangled-states","status":"publish","type":"post","link":"https:\/\/workhouse.sweetdishy.com\/index.php\/2024\/09\/21\/qkd-using-entangled-states\/","title":{"rendered":"QKD using entangled states"},"content":{"rendered":"\n<p>Here is another variant of the QKD protocol, due to Ekert [32], which<\/p>\n\n\n\n<p>makes use of correlated quantum pairs as a resource shared between Alice<\/p>\n\n\n\n<p>and Bob. For each key, Bob generates entangled photons and sends one to<\/p>\n\n\n\n<p>Alice.<\/p>\n\n\n\n<p>2<\/p>\n\n\n\n<p>Locally each performs a measurement randomly, according to random<\/p>\n\n\n\n<p>bit strings m<\/p>\n\n\n\n<p>A<\/p>\n\n\n\n<p>and m<\/p>\n\n\n\n<p>B<\/p>\n\n\n\n<p>respectively, in the or the H basis. These bit strings<\/p>\n\n\n\n<p>2<\/p>\n\n\n\n<p>It doesn\u2019t matter who generates the pair. It could also be generated by a third party<\/p>\n\n\n\n<p>and sent to both of them. The quantum channel is used for this purpose.<\/p>\n\n\n\n<p><img loading=\"lazy\" decoding=\"async\" alt=\"\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9781482238129\/files\/bgda.png\" width=\"685\" height=\"405\"><\/p>\n\n\n\n<p>Information and Communication 193<\/p>\n\n\n\n<p>are shared over a public channel, and Alice and Bob compare them to see<\/p>\n\n\n\n<p>which bits match. The measured values of those bit positions are retained as<\/p>\n\n\n\n<p>the shared key.<\/p>\n\n\n\n<p>The point is that when m<\/p>\n\n\n\n<p>A<\/p>\n\n\n\n<p>and m<\/p>\n\n\n\n<p>B<\/p>\n\n\n\n<p>match, the measurement results,<\/p>\n\n\n\n<p>though random, are perfectly correlated, while when they don\u2019t match, Alice<\/p>\n\n\n\n<p>and Bob get the same result only 50% of the time.<\/p>\n\n\n\n<p>The drawback of this scheme is that Alice and Bob would have to verify<\/p>\n\n\n\n<p>that their photons retained their entanglement when the key was being gen-<\/p>\n\n\n\n<p>erated. To do this, they would have to perform an additional exercise of, say<\/p>\n\n\n\n<p>making sure Bell\u2019s inequality was violated. (For instance, each of them could<\/p>\n\n\n\n<p>measure their photons in three di\ufb00erent bases and share the values.)<\/p>\n\n\n\n<p>Example 9.4.3. An example of the entangled QKD scheme: suppose Alice<\/p>\n\n\n\n<p>and Bob share a huge supply of qubits in the state |\u03b2<\/p>\n\n\n\n<p>00<\/p>\n\n\n\n<p>i.<\/p>\n\n\n\n<p>Index 1 2 3 4 5 6 7 8 9 10 11 12<\/p>\n\n\n\n<p>m<\/p>\n\n\n\n<p>A<\/p>\n\n\n\n<p>0 1 0 1 0 0 0 1 1 1 0 0<\/p>\n\n\n\n<p>m<\/p>\n\n\n\n<p>B<\/p>\n\n\n\n<p>1 1 1 0 1 0 1 0 0 0 1 0<\/p>\n\n\n\n<p>K 0 0 1<\/p>\n\n\n\n<p>The presence of an eavesdropper Eve is detected the same way as for the<\/p>\n\n\n\n<p>BB84 scheme. The list of possible secure key-distribution schemes is quite<\/p>\n\n\n\n<p>long, and you are invited to contribute to it!<\/p>\n\n\n\n<p>9.5 Information Reconciliation and Privacy Ampli\ufb01ca-<\/p>\n\n\n\n<p>tion<\/p>\n\n\n\n<p>The sifting procedure in QKD protocols is to ensure the degree of security<\/p>\n\n\n\n<p>of the channel. The presence of an eavesdropper is detected by errors above<\/p>\n\n\n\n<p>a certain tolerance margin, say 20%. However, natural errors in the channel<\/p>\n\n\n\n<p>could also cause discrepancies in the shared key. To remove these, and to<\/p>\n\n\n\n<p>ensure further security on the shared key, two classical procedures known as<\/p>\n\n\n\n<p>information reconciliation (a form of error correction) and privacy ampli\ufb01ca-<\/p>\n\n\n\n<p>tion are carried out.<\/p>\n\n\n\n<p>The basic idea of information reconciliation is to perform a parity check on<\/p>\n\n\n\n<p>a subset of the key, compare, and correct. At the two-bit level, parity is just<\/p>\n\n\n\n<p>an XOR. So Alice could randomly select two bits out of k<\/p>\n\n\n\n<p>A<\/p>\n\n\n\n<p>, announce their<\/p>\n\n\n\n<p>positions and XOR to Bob. He then compares the parity of the same bits in<\/p>\n\n\n\n<p>k<\/p>\n\n\n\n<p>B<\/p>\n\n\n\n<p>. If they do not match these bits are discarded. If they do then they decide<\/p>\n\n\n\n<p><\/p>\n\n\n\n<p>194 Introduction to Quantum Physics and Information Processing<\/p>\n\n\n\n<p>to discard the second bit. This ensures that Eve does not learn anything more<\/p>\n\n\n\n<p>about their key from their discussion.<\/p>\n\n\n\n<p>The more sophisticated version generalizes this process, as \ufb01rst described<\/p>\n\n\n\n<p>in 1992 by Bennett et al. [8]. They proceed in several iterations of essentially<\/p>\n\n\n\n<p>the same process, but \ufb01rst dividing their keys into predetermined blocks and<\/p>\n\n\n\n<p>checking the parity of the block. If the parity doesn\u2019t match then they re-<\/p>\n\n\n\n<p>cursively bisect their blocks to detect the location of the error and discard<\/p>\n\n\n\n<p>it. To ensure that Eve doesn\u2019t learn anything more from their parity discus-<\/p>\n\n\n\n<p>sions (which happen in public), they discard the last bit of each block whose<\/p>\n\n\n\n<p>parity is disclosed. This process is repeated many times with increasing block<\/p>\n\n\n\n<p>sizes, until eventually the two keys are ensured to be reconciled with a large<\/p>\n\n\n\n<p>probability.<\/p>\n\n\n\n<p>At the end of information reconciliation, Alice and Bob have identical<\/p>\n\n\n\n<p>keys but whose privacy has been compromised by all the public discussions.<\/p>\n\n\n\n<p>To undo this e\ufb00ect, they resort to privacy ampli\ufb01cation. To do this they select<\/p>\n\n\n\n<p>something called a universal hash function to encode their strings. There are<\/p>\n\n\n\n<p>many such functions that provide various bounds for the amount of informa-<\/p>\n\n\n\n<p>tion Eve can gain. One such is to select random subsets of their strings and<\/p>\n\n\n\n<p>to retain their parity bits for a new key.<\/p>\n\n\n\n<p>In any case, both these steps amount to classical error correction and<\/p>\n\n\n\n<p>coding, and will not be dealt with at greater depth in this book.<\/p>\n\n\n\n<p>It is clear from this discussion that depending on the degree of privacy<\/p>\n\n\n\n<p>they choose to have, the initial string length must be fairly large, of the order<\/p>\n\n\n\n<p>of 4 times the length of the desired key.<\/p>\n\n\n\n<p>Problems<\/p>\n\n\n\n<p>9.1. How would the teleportation protocol change if the entangled state shared<\/p>\n\n\n\n<p>by Alice and Bob was any of the other Bell states: |\u03b2<\/p>\n\n\n\n<p>01<\/p>\n\n\n\n<p>i, |\u03b2<\/p>\n\n\n\n<p>10<\/p>\n\n\n\n<p>i, or |\u03b2<\/p>\n\n\n\n<p>11<\/p>\n\n\n\n<p>i?<\/p>\n\n\n\n<p>9.2. Consider the teleportation protocol, and suppose that the unknown qubit<\/p>\n\n\n\n<p>with Alice is entangled with another qubit in the possession of a third party,<\/p>\n\n\n\n<p>Charlie. Show how the protocol teleports the entanglement as well, i.e., at<\/p>\n\n\n\n<p>the end of the protocol, Bob\u2019s qubit is entangled with Charlie\u2019s.<\/p>\n\n\n\n<p>9.3. Formulate the matrix equivalent of the dense coding protocol and show that<\/p>\n\n\n\n<p>it is unitary.<\/p>\n\n\n\n<p>9.4. Analyzing the BB84 more thoroughly, consider that Eve measures every<\/p>\n\n\n\n<p>photon sent by Alice, in the or H basis according to a random string m<\/p>\n\n\n\n<p>e<\/p>\n\n\n\n<p>.<\/p>\n\n\n\n<p><img loading=\"lazy\" decoding=\"async\" alt=\"\" src=\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9781482238129\/files\/bgdc.png\" width=\"210\" height=\"252\"><\/p>\n\n\n\n<p>Problems 195<\/p>\n\n\n\n<p>Suppose Alice and Bob now announce m bits out of their shared set. What<\/p>\n\n\n\n<p>is the probability that no error will be found? What fraction of Eve\u2019s bits<\/p>\n\n\n\n<p>would match with Alice\u2019s and Bob\u2019s? Would the public discussion between<\/p>\n\n\n\n<p>Alice and Bob help Eve at all?<\/p>\n\n\n\n<p>9.5. At one point in history, it was suggested that Eve might bene\ufb01t by measuring<\/p>\n\n\n\n<p>in a basis intermediate between the and H:<\/p>\n\n\n\n<p>|0<\/p>\n\n\n\n<p>e<\/p>\n\n\n\n<p>i = cos<\/p>\n\n\n\n<p>\u03c0<\/p>\n\n\n\n<p>8<\/p>\n\n\n\n<p>|0i + sin<\/p>\n\n\n\n<p>\u03c0<\/p>\n\n\n\n<p>8<\/p>\n\n\n\n<p>|1i (9.10)<\/p>\n\n\n\n<p>|1<\/p>\n\n\n\n<p>e<\/p>\n\n\n\n<p>i = sin<\/p>\n\n\n\n<p>\u03c0<\/p>\n\n\n\n<p>8<\/p>\n\n\n\n<p>|0i \u2212 cos<\/p>\n\n\n\n<p>\u03c0<\/p>\n\n\n\n<p>8<\/p>\n\n\n\n<p>|1i (9.11)<\/p>\n\n\n\n<p>What is the probability that any one measurement by Eve gives the correct<\/p>\n\n\n\n<p>result? If she prepares and then transmits photons in this basis to Bob, what<\/p>\n\n\n\n<p>is the probability that Bob\u2019s string has an error?<\/p>\n\n\n\n<p>9.6. Suppose Alice prepares two qubits in the entangled state<\/p>\n\n\n\n<p>1<\/p>\n\n\n\n<p>\u221a<\/p>\n\n\n\n<p>2<\/p>\n\n\n\n<p>[|01i \u2212 |10i]<\/p>\n\n\n\n<p>and sends one qubit to Bob. Suppose that Eve intercepts and measures<\/p>\n\n\n\n<p>that qubit, and then based on the outcome, prepares and sends a photon<\/p>\n\n\n\n<p>to Bob. What can you say about the correlation between the qubits with<\/p>\n\n\n\n<p>Alice and Bob<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Here is another variant of the QKD protocol, due to Ekert [32], which makes use of correlated quantum pairs as a resource shared between Alice and Bob. For each key, Bob generates entangled photons and sends one to Alice. 2 Locally each performs a measurement randomly, according to random bit strings m A and m [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":4043,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[498],"tags":[],"class_list":["post-4115","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-information-and-communication"],"jetpack_featured_media_url":"https:\/\/workhouse.sweetdishy.com\/wp-content\/uploads\/2024\/09\/informative.png","_links":{"self":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/4115","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/comments?post=4115"}],"version-history":[{"count":2,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/4115\/revisions"}],"predecessor-version":[{"id":4583,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/4115\/revisions\/4583"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/media\/4043"}],"wp:attachment":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/media?parent=4115"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/categories?post=4115"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/tags?post=4115"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}