How to prove teleportation does not violate non-cloning theorem?
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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
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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 1 hour ago
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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 1 hour ago
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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 1 hour ago
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edited 1 hour ago
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2 Answers
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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 1 hour ago
Blue♦Blue
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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 27 mins ago
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$begingroup$
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 1 hour 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 1 hour 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 1 hour 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 1 hour ago
Blue♦Blue
6,29541355
edited 35 mins ago
answered 1 hour ago
Blue♦Blue
6,29541355
answered 1 hour ago
Blue♦Blue
6,29541355
answered 1 hour 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 27 mins ago
Student404MusStudent404Mus
163
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Student404Mus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
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$begingroup$
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 27 mins 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 27 mins 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 27 mins ago
Student404MusStudent404Mus
163
New contributor
Student404Mus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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answered 27 mins ago
Student404MusStudent404Mus
163
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answered 27 mins ago
Student404MusStudent404Mus
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answered 27 mins ago
Student404MusStudent404Mus
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163
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Student404Mus is a new contributor. Be nice, and check out our Code of Conduct.
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