![\"\"](\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9781482238129\/files\/bge0.png\")
<\/p>\n\n\n\nClassically, the repetition code is the simplest way of introducing redun-<\/p>\n\n\n\n
dancy to protect information. Assume that noise in the channel is modelled<\/p>\n\n\n\n
as a bit \ufb02ip with probability p (and hence 1 \u2212 p for not \ufb02ipping). This is<\/p>\n\n\n\n
schematised in Figure 10.2. This is known as the binary symmetric channel.<\/p>\n\n\n\n
0<\/p>\n\n\n\n
1\u2212p<\/p>\n\n\n\n
\/\/<\/p>\n\n\n\n
p<\/p>\n\n\n\n
”<\/p>\n\n\n\n
0<\/p>\n\n\n\n
1<\/p>\n\n\n\n
1\u2212p<\/p>\n\n\n\n
\/\/<\/p>\n\n\n\n
p<\/p>\n\n\n\n
77<\/p>\n\n\n\n
1<\/p>\n\n\n\n
FIGURE 10.2: Binary symmetric channel for bit \ufb02ips.<\/p>\n\n\n\n
To protect against errors, each logical bit, indicated by the tilde, is encoded<\/p>\n\n\n\n
using three identical physical bits.<\/p>\n\n\n\n
0 \u2192<\/p>\n\n\n\n
\u02dc<\/p>\n\n\n\n
0 = 000, 1 \u2192<\/p>\n\n\n\n
\u02dc<\/p>\n\n\n\n
1 = 111. (10.1)<\/p>\n\n\n\n
If p is su\ufb03ciently small, then the majority value decides what the original bit<\/p>\n\n\n\n
was. The total probability of error is the sum of probability that 2 bits \ufb02ipped<\/p>\n\n\n\n
and that 3 bits \ufb02ipped which is 3p<\/p>\n\n\n\n
(<\/p>\n\n\n\n
1 \u2212p) + p<\/p>\n\n\n\n
3<\/p>\n\n\n\n
= 3p<\/p>\n\n\n\n
2<\/p>\n\n\n\n
\u22122p<\/p>\n\n\n\n
3<\/p>\n\n\n\n
. If p < 1\/2, this is<\/p>\n\n\n\n
much smaller than the probability of error without encoding, which is p.<\/p>\n\n\n\n
Now suppose we have a quantum channel that was susceptible to only<\/p>\n\n\n\n
qubit \ufb02ips. We can model such a channel by X acting with probability p on<\/p>\n\n\n\n
a state passing through the channel. The quantum equivalent of the 3-bit<\/p>\n\n\n\n
repetition code represents each basis state by 3 identical qubits in the same<\/p>\n\n\n\n
basis state:<\/p>\n\n\n\n
|0i \u2192 |<\/p>\n\n\n\n
\u02dc<\/p>\n\n\n\n
0i = |000i; |1i \u2192 |<\/p>\n\n\n\n
\u02dc<\/p>\n\n\n\n
1i = |111i (10.2)<\/p>\n\n\n\n
<\/p>\n\n\n\n
Quantum Error Correction 199<\/p>\n\n\n\n
so that an arbitrary state is encoded as<\/p>\n\n\n\n
|\u03c8i = \u03b1|0i + \u03b2|1i \u2212\u2192 |<\/p>\n\n\n\n
\u02dc<\/p>\n\n\n\n
\u03c8i = \u03b1|000i + \u03b2|111i. (10.3)<\/p>\n\n\n\n
This encoding process is easily achieved by the circuit of Figure 10.3.<\/p>\n\n\n\n
|\u03c8i<\/p>\n\n\n\n
\u2022 \u2022<\/p>\n\n\n\n
|0i<\/p>\n\n\n\n
|<\/p>\n\n\n\n
\u02dc<\/p>\n\n\n\n
\u03c8i<\/p>\n\n\n\n
|0i<\/p>\n\n\n\n
\uf8fc<\/p>\n\n\n\n
\uf8f4<\/p>\n\n\n\n
\uf8f4<\/p>\n\n\n\n
\uf8f4<\/p>\n\n\n\n
\uf8f4<\/p>\n\n\n\n
\uf8fd<\/p>\n\n\n\n
\uf8f4<\/p>\n\n\n\n
\uf8f4<\/p>\n\n\n\n
\uf8f4<\/p>\n\n\n\n
\uf8f4<\/p>\n\n\n\n
\uf8fe<\/p>\n\n\n\n
FIGURE 10.3: Encoding circuit for the 3-qubit bit-\ufb02ip code.<\/p>\n\n\n\n
This does not make three copies of the original state, however, and neither<\/p>\n\n\n\n
can we try to measure the state after it passes through the noise to check how<\/p>\n\n\n\n
it has changed, as that would destroy the superposition. If we assume that<\/p>\n\n\n\n
the channel is capable of \ufb02ipping only one qubit, then the codeword state |<\/p>\n\n\n\n
\u02dc<\/p>\n\n\n\n
\u03c8i<\/p>\n\n\n\n
could have changed into four possible output states:<\/p>\n\n\n\n
S<\/p>\n\n\n\n
0<\/p>\n\n\n\n
: |<\/p>\n\n\n\n
\u02dc<\/p>\n\n\n\n
\u03c8i \u2192 |\u03c8<\/p>\n\n\n\n
0<\/p>\n\n\n\n
i = |<\/p>\n\n\n\n
\u02dc<\/p>\n\n\n\n
\u03c8i = \u03b1|000i + \u03b2|111i (10.4a)<\/p>\n\n\n\n
S<\/p>\n\n\n\n
1<\/p>\n\n\n\n
: |<\/p>\n\n\n\n
\u02dc<\/p>\n\n\n\n
\u03c8i \u2192 |\u03c8<\/p>\n\n\n\n
1<\/p>\n\n\n\n
i = X<\/p>\n\n\n\n
1<\/p>\n\n\n\n
|<\/p>\n\n\n\n
\u02dc<\/p>\n\n\n\n
\u03c8i = \u03b1|001i + \u03b2|110i (10.4b)<\/p>\n\n\n\n
S<\/p>\n\n\n\n
2<\/p>\n\n\n\n
: |<\/p>\n\n\n\n
\u02dc<\/p>\n\n\n\n
\u03c8i \u2192 |\u03c8<\/p>\n\n\n\n
2<\/p>\n\n\n\n
i = X<\/p>\n\n\n\n
2<\/p>\n\n\n\n
|<\/p>\n\n\n\n
\u02dc<\/p>\n\n\n\n
\u03c8i = \u03b1|010i + \u03b2|101i (10.4c)<\/p>\n\n\n\n
S<\/p>\n\n\n\n
3<\/p>\n\n\n\n
: |<\/p>\n\n\n\n
\u02dc<\/p>\n\n\n\n
\u03c8i \u2192 |\u03c8<\/p>\n\n\n\n
3<\/p>\n\n\n\n
i = X<\/p>\n\n\n\n
3<\/p>\n\n\n\n
|<\/p>\n\n\n\n
\u02dc<\/p>\n\n\n\n
\u03c8i = \u03b1|100i + \u03b2|011i (10.4d)<\/p>\n\n\n\n
These states are known as syndromes, since we can diagnose the a\ufb04iction due<\/p>\n\n\n\n
to noise by detecting which one occurred! But how do we detect the syndrome<\/p>\n\n\n\n
without measuring or copying? The way out is to use ancillary qubits, with<\/p>\n\n\n\n
controlled gates acting on them and to measure the ancillaries. We have to<\/p>\n\n\n\n
ensure that we do not get any information about the original state by this<\/p>\n\n\n\n
measurement, but still detect the syndrome. Note that the four syndrome<\/p>\n\n\n\n
states are all mutually orthogonal. Therefore it is possible to distinguish them<\/p>\n\n\n\n
by measuring a 2-qubit ancilla. Consider the circuit in Figure 10.4. Each qubit<\/p>\n\n\n\n
of the input state is indicated by its label.<\/p>\n\n\n\n
3<\/p>\n\n\n\n
\u2022<\/p>\n\n\n\n
|<\/p>\n\n\n\n
\u00af<\/p>\n\n\n\n
\u03c8i<\/p>\n\n\n\n
2<\/p>\n\n\n\n
\u2022 \u2022<\/p>\n\n\n\n
1<\/p>\n\n\n\n
\u2022<\/p>\n\n\n\n
|0i<\/p>\n\n\n\n
x<\/p>\n\n\n\n
|0i<\/p>\n\n\n\n
y<\/p>\n\n\n\n
\uf8f1<\/p>\n\n\n\n
\uf8f4<\/p>\n\n\n\n
\uf8f4<\/p>\n\n\n\n
\uf8f2<\/p>\n\n\n\n
\uf8f4<\/p>\n\n\n\n
\uf8f4<\/p>\n\n\n\n
\uf8f3<\/p>\n\n\n\n
FIGURE 10.4: Syndrome measurement for the 3-qubit bit-\ufb02ip code<\/p>\n\n\n\n
![\"\"](\"https:\/\/learning.oreilly.com\/api\/v2\/epubs\/urn:orm:book:9781482238129\/files\/bge1.png\")
<\/p>\n\n\n\n
200 Introduction to Quantum Physics and Information Processing<\/p>\n\n\n\n
TABLE 10.1: Syndrome measurement: outcomes.<\/p>\n\n\n\n
Syndrome xy<\/p>\n\n\n\n
|\u03c8<\/p>\n\n\n\n
0<\/p>\n\n\n\n
i 00<\/p>\n\n\n\n
|\u03c8<\/p>\n\n\n\n
1<\/p>\n\n\n\n
i 01<\/p>\n\n\n\n
|\u03c8<\/p>\n\n\n\n
2<\/p>\n\n\n\n
i 11<\/p>\n\n\n\n
|\u03c8<\/p>\n\n\n\n
3<\/p>\n\n\n\n
i 10<\/p>\n\n\n\n
You can verify that the measured 2-bit number xy give you the syndrome<\/p>\n\n\n\n
as in Table 10.1.<\/p>\n\n\n\n
The information contained in the input state |<\/p>\n\n\n\n
\u02dc<\/p>\n\n\n\n
\u03c8i is not revealed by the<\/p>\n\n\n\n
measurements. It is easy to see that for each syndrome |\u03c8<\/p>\n\n\n\n
i<\/p>\n\n\n\n
i, the error can be<\/p>\n\n\n\n
corrected by applying X on the i<\/p>\n\n\n\n
th<\/p>\n\n\n\n
qubit. This action can be linked to the xy<\/p>\n\n\n\n
values, which control the action of X on the corresponding qubit: X<\/p>\n\n\n\n
x\u00afy<\/p>\n\n\n\n
on the<\/p>\n\n\n\n
\ufb01rst qubit, X<\/p>\n\n\n\n
xy<\/p>\n\n\n\n
on the second, and X<\/p>\n\n\n\n
\u00afxy<\/p>\n\n\n\n
on the last qubit of the codeword.<\/p>\n\n\n\n
The nice thing about expressing it as this controlled action is that the<\/p>\n\n\n\n
process can be automated, bypassing the need for measurement, by applying<\/p>\n\n\n\n
suitable controlled gates as in Figure 10.5.<\/p>\n\n\n\n
3<\/p>\n\n\n\n
SM<\/p>\n\n\n\n
|<\/p>\n\n\n\n
\u00af<\/p>\n\n\n\n
\u03c8i<\/p>\n\n\n\n
2<\/p>\n\n\n\n
1<\/p>\n\n\n\n
|0i<\/p>\n\n\n\n
\u2022 \u2022<\/p>\n\n\n\n
|0i<\/p>\n\n\n\n
\u2022 \u2022<\/p>\n\n\n\n
\uf8f1<\/p>\n\n\n\n
\uf8f4<\/p>\n\n\n\n
\uf8f4<\/p>\n\n\n\n
\uf8f4<\/p>\n\n\n\n
\uf8f2<\/p>\n\n\n\n
\uf8f4<\/p>\n\n\n\n
\uf8f4<\/p>\n\n\n\n
\uf8f4<\/p>\n\n\n\n
\uf8f3<\/p>\n\n\n\n
FIGURE 10.5: Error detection and correction for 3-qubit bit-\ufb02ip code. Here<\/p>\n\n\n\n
SM is the syndrome measurement circuit<\/p>\n","protected":false},"excerpt":{"rendered":"
Classically, the repetition code is the simplest way of introducing redun- dancy to protect information. Assume that noise in the channel is modelled as a bit \ufb02ip with probability p (and hence 1 \u2212 p for not \ufb02ipping). This is schematised in Figure 10.2. This is known as the binary symmetric channel. 0 1\u2212p \/\/ […]<\/p>\n","protected":false},"author":1,"featured_media":4044,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[499],"tags":[],"class_list":["post-4120","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-quantum-error-correction"],"jetpack_featured_media_url":"https:\/\/workhouse.sweetdishy.com\/wp-content\/uploads\/2024\/09\/error-state.png","_links":{"self":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/4120","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=4120"}],"version-history":[{"count":1,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/4120\/revisions"}],"predecessor-version":[{"id":4121,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/4120\/revisions\/4121"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/media\/4044"}],"wp:attachment":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/media?parent=4120"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/categories?post=4120"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/tags?post=4120"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}