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?enter image description here






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?enter image description here






    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?enter image description here






      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?enter image description here






      algorithm entanglement teleportation no-cloning-theorem




      algorithm entanglement teleportation no-cloning-theorem






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      edited 1 hour ago









      Blue♦

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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!






          share|improve this answer











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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.






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              2 Answers
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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!






              share|improve this answer











              $endgroup$


















                2












                $begingroup$

                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!






                share|improve this answer











                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  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!






                  share|improve this answer











                  $endgroup$



                  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!







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited 29 mins ago

























                  answered 1 hour ago









                  Blue♦Blue

                  6,29541355




                  6,29541355

























                      0












                      $begingroup$

                      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.






                      share|improve this answer








                      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.






                      $endgroup$


















                        0












                        $begingroup$

                        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.






                        share|improve this answer








                        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.






                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          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.






                          share|improve this answer








                          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.







                          share|improve this answer








                          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.









                          share|improve this answer



                          share|improve this answer






                          New contributor




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                          answered 21 mins ago









                          Student404MusStudent404Mus

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                          163




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