How to prove teleportation does not violate non-cloning theorem?
$begingroup$
For a given teleportation process as depicted in the figure, how one can say that teleporting the qubit state $|qrangle$ has not cloned at the end of Bob's measurement?
algorithm entanglement teleportation no-cloning-theorem
edited 3 hours ago
Blue♦
6,29541355
asked 4 hours ago
Student404MusStudent404Mus
163
New contributor
Student404Mus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
For a given teleportation process as depicted in the figure, how one can say that teleporting the qubit state $|qrangle$ has not cloned at the end of Bob's measurement?
algorithm entanglement teleportation no-cloning-theorem
edited 3 hours ago
Blue♦
6,29541355
asked 4 hours ago
Student404MusStudent404Mus
163
New contributor
Student404Mus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
For a given teleportation process as depicted in the figure, how one can say that teleporting the qubit state $|qrangle$ has not cloned at the end of Bob's measurement?
algorithm entanglement teleportation no-cloning-theorem
edited 3 hours ago
Blue♦
6,29541355
asked 4 hours ago
Student404MusStudent404Mus
163
New contributor
Student404Mus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
For a given teleportation process as depicted in the figure, how one can say that teleporting the qubit state $|qrangle$ has not cloned at the end of Bob's measurement?
algorithm entanglement teleportation no-cloning-theorem
algorithm entanglement teleportation no-cloning-theorem
edited 3 hours ago
Blue♦
6,29541355
asked 4 hours ago
Student404MusStudent404Mus
163
New contributor
Student404Mus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 3 hours ago
Blue♦
6,29541355
asked 4 hours ago
Student404MusStudent404Mus
163
New contributor
Student404Mus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 3 hours ago
Blue♦
6,29541355
edited 3 hours ago
Blue♦
6,29541355
edited 3 hours ago
Blue♦
6,29541355
6,29541355
asked 4 hours ago
Student404MusStudent404Mus
163
New contributor
Student404Mus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
Student404MusStudent404Mus
163
asked 4 hours ago
Student404MusStudent404Mus
163
163
New contributor
Student404Mus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Student404Mus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Student404Mus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
There's nothing to prove as such here. It is evident from the teleportation protocol itself.
Yes, at the conclusion of the teleportation protocol, Bob's qubit certainly will take on the same state as Alice's qubit's initial state i.e. $|qrangle$. However, Alice's qubit will now exist as an inseparable part of an entangled Bell state. You haven't really been able to create a "copy" of Alice's qubit, in the sense that Alice's qubit's initial state is "destroyed" in the process and is no longer $|qrangle$.
In (crude) analogical terms, in the quantum teleportation protocol you "cut and paste" the state $|qrangle$ rather than "copying and pasting". So there's no violation of the No cloning theorem!
answered 3 hours ago
Blue♦Blue
6,29541355
$endgroup$
add a comment |
$begingroup$
I.e., if we have the entire initial state is written as follows
$| q rangle otimes | beta_{00} rangle =(alpha | 0rangle+beta | 1 rangle ) otimes frac{1}{sqrt{2}}( |00rangle+| 11 rangle )= frac{1}{sqrt{2}}(alpha | 000rangle+alpha | 011 rangle+beta | 100 rangle+beta | 111 rangle )$ ,
then, after the measurement, we obtain the state
$|psi rangle equiv{ |00rangle frac{alpha | 0 rangle+beta | 1rangle}{2}+| 01 rangle frac{alpha | 1 rangle+beta | 0rangle}{2}+| 10 rangle frac{alpha | 0 rangle-beta | 1rangle}{2}+| 11 rangle frac{alpha | 1 rangle-beta | 0rangle}{2} }$ ,
hence we can no longer write
$| q rangle otimes | something rangle$ ,
namely, we can't factorize $| q rangle$ outside the result which implies that it didn't produce again.
answered 2 hours ago
Student404MusStudent404Mus
163
New contributor
Student404Mus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "694"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Student404Mus is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fquantumcomputing.stackexchange.com%2fquestions%2f5598%2fhow-to-prove-teleportation-does-not-violate-non-cloning-theorem%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There's nothing to prove as such here. It is evident from the teleportation protocol itself.
Yes, at the conclusion of the teleportation protocol, Bob's qubit certainly will take on the same state as Alice's qubit's initial state i.e. $|qrangle$. However, Alice's qubit will now exist as an inseparable part of an entangled Bell state. You haven't really been able to create a "copy" of Alice's qubit, in the sense that Alice's qubit's initial state is "destroyed" in the process and is no longer $|qrangle$.
In (crude) analogical terms, in the quantum teleportation protocol you "cut and paste" the state $|qrangle$ rather than "copying and pasting". So there's no violation of the No cloning theorem!
answered 3 hours ago
Blue♦Blue
6,29541355
$endgroup$
add a comment |
$begingroup$
There's nothing to prove as such here. It is evident from the teleportation protocol itself.
Yes, at the conclusion of the teleportation protocol, Bob's qubit certainly will take on the same state as Alice's qubit's initial state i.e. $|qrangle$. However, Alice's qubit will now exist as an inseparable part of an entangled Bell state. You haven't really been able to create a "copy" of Alice's qubit, in the sense that Alice's qubit's initial state is "destroyed" in the process and is no longer $|qrangle$.
In (crude) analogical terms, in the quantum teleportation protocol you "cut and paste" the state $|qrangle$ rather than "copying and pasting". So there's no violation of the No cloning theorem!
answered 3 hours ago
Blue♦Blue
6,29541355
$endgroup$
add a comment |
$begingroup$
There's nothing to prove as such here. It is evident from the teleportation protocol itself.
Yes, at the conclusion of the teleportation protocol, Bob's qubit certainly will take on the same state as Alice's qubit's initial state i.e. $|qrangle$. However, Alice's qubit will now exist as an inseparable part of an entangled Bell state. You haven't really been able to create a "copy" of Alice's qubit, in the sense that Alice's qubit's initial state is "destroyed" in the process and is no longer $|qrangle$.
In (crude) analogical terms, in the quantum teleportation protocol you "cut and paste" the state $|qrangle$ rather than "copying and pasting". So there's no violation of the No cloning theorem!
answered 3 hours ago
Blue♦Blue
6,29541355
$endgroup$
There's nothing to prove as such here. It is evident from the teleportation protocol itself.
Yes, at the conclusion of the teleportation protocol, Bob's qubit certainly will take on the same state as Alice's qubit's initial state i.e. $|qrangle$. However, Alice's qubit will now exist as an inseparable part of an entangled Bell state. You haven't really been able to create a "copy" of Alice's qubit, in the sense that Alice's qubit's initial state is "destroyed" in the process and is no longer $|qrangle$.
In (crude) analogical terms, in the quantum teleportation protocol you "cut and paste" the state $|qrangle$ rather than "copying and pasting". So there's no violation of the No cloning theorem!
answered 3 hours ago
Blue♦Blue
6,29541355
edited 2 hours ago
answered 3 hours ago
Blue♦Blue
6,29541355
answered 3 hours ago
Blue♦Blue
6,29541355
answered 3 hours ago
Blue♦Blue
6,29541355
6,29541355
add a comment |
add a comment |
$begingroup$
I.e., if we have the entire initial state is written as follows
$| q rangle otimes | beta_{00} rangle =(alpha | 0rangle+beta | 1 rangle ) otimes frac{1}{sqrt{2}}( |00rangle+| 11 rangle )= frac{1}{sqrt{2}}(alpha | 000rangle+alpha | 011 rangle+beta | 100 rangle+beta | 111 rangle )$ ,
then, after the measurement, we obtain the state
$|psi rangle equiv{ |00rangle frac{alpha | 0 rangle+beta | 1rangle}{2}+| 01 rangle frac{alpha | 1 rangle+beta | 0rangle}{2}+| 10 rangle frac{alpha | 0 rangle-beta | 1rangle}{2}+| 11 rangle frac{alpha | 1 rangle-beta | 0rangle}{2} }$ ,
hence we can no longer write
$| q rangle otimes | something rangle$ ,
namely, we can't factorize $| q rangle$ outside the result which implies that it didn't produce again.
answered 2 hours ago
Student404MusStudent404Mus
163
New contributor
Student404Mus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I.e., if we have the entire initial state is written as follows
$| q rangle otimes | beta_{00} rangle =(alpha | 0rangle+beta | 1 rangle ) otimes frac{1}{sqrt{2}}( |00rangle+| 11 rangle )= frac{1}{sqrt{2}}(alpha | 000rangle+alpha | 011 rangle+beta | 100 rangle+beta | 111 rangle )$ ,
then, after the measurement, we obtain the state
$|psi rangle equiv{ |00rangle frac{alpha | 0 rangle+beta | 1rangle}{2}+| 01 rangle frac{alpha | 1 rangle+beta | 0rangle}{2}+| 10 rangle frac{alpha | 0 rangle-beta | 1rangle}{2}+| 11 rangle frac{alpha | 1 rangle-beta | 0rangle}{2} }$ ,
hence we can no longer write
$| q rangle otimes | something rangle$ ,
namely, we can't factorize $| q rangle$ outside the result which implies that it didn't produce again.
answered 2 hours ago
Student404MusStudent404Mus
163
New contributor
Student404Mus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I.e., if we have the entire initial state is written as follows
$| q rangle otimes | beta_{00} rangle =(alpha | 0rangle+beta | 1 rangle ) otimes frac{1}{sqrt{2}}( |00rangle+| 11 rangle )= frac{1}{sqrt{2}}(alpha | 000rangle+alpha | 011 rangle+beta | 100 rangle+beta | 111 rangle )$ ,
then, after the measurement, we obtain the state
$|psi rangle equiv{ |00rangle frac{alpha | 0 rangle+beta | 1rangle}{2}+| 01 rangle frac{alpha | 1 rangle+beta | 0rangle}{2}+| 10 rangle frac{alpha | 0 rangle-beta | 1rangle}{2}+| 11 rangle frac{alpha | 1 rangle-beta | 0rangle}{2} }$ ,
hence we can no longer write
$| q rangle otimes | something rangle$ ,
namely, we can't factorize $| q rangle$ outside the result which implies that it didn't produce again.
answered 2 hours ago
Student404MusStudent404Mus
163
New contributor
Student404Mus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I.e., if we have the entire initial state is written as follows
$| q rangle otimes | beta_{00} rangle =(alpha | 0rangle+beta | 1 rangle ) otimes frac{1}{sqrt{2}}( |00rangle+| 11 rangle )= frac{1}{sqrt{2}}(alpha | 000rangle+alpha | 011 rangle+beta | 100 rangle+beta | 111 rangle )$ ,
then, after the measurement, we obtain the state
$|psi rangle equiv{ |00rangle frac{alpha | 0 rangle+beta | 1rangle}{2}+| 01 rangle frac{alpha | 1 rangle+beta | 0rangle}{2}+| 10 rangle frac{alpha | 0 rangle-beta | 1rangle}{2}+| 11 rangle frac{alpha | 1 rangle-beta | 0rangle}{2} }$ ,
hence we can no longer write
$| q rangle otimes | something rangle$ ,
namely, we can't factorize $| q rangle$ outside the result which implies that it didn't produce again.
answered 2 hours ago
Student404MusStudent404Mus
163
New contributor
Student404Mus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 hours ago
Student404MusStudent404Mus
163
New contributor
Student404Mus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 hours ago
Student404MusStudent404Mus
163
answered 2 hours ago
Student404MusStudent404Mus
163
163
New contributor
Student404Mus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Student404Mus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Student404Mus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
Student404Mus is a new contributor. Be nice, and check out our Code of Conduct.
Student404Mus is a new contributor. Be nice, and check out our Code of Conduct.
Student404Mus is a new contributor. Be nice, and check out our Code of Conduct.
Student404Mus is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Quantum Computing Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fquantumcomputing.stackexchange.com%2fquestions%2f5598%2fhow-to-prove-teleportation-does-not-violate-non-cloning-theorem%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
