How to prove teleportation does not violate non-cloning theorem?












1












$begingroup$


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?enter image description here






algorithm entanglement teleportation no-cloning-theorem







share|improve this question









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.







$endgroup$

















    1












    $begingroup$


    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?enter image description here






    algorithm entanglement teleportation no-cloning-theorem







    share|improve this question









    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.







    $endgroup$















      1












      1








      1





      $begingroup$


      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?enter image description here






      algorithm entanglement teleportation no-cloning-theorem







      share|improve this question









      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?enter image description here






      algorithm entanglement teleportation no-cloning-theorem




      algorithm entanglement teleportation no-cloning-theorem






      share|improve this question









      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 question









      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 question




      share|improve this question








      edited 2 hours ago









      Blue♦

      6,29541355




      6,29541355






      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




      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.






















          2 Answers
          2






          active

          oldest

          votes


















          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











          $endgroup$





















            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.






            $endgroup$













              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.










              draft saved

              draft discarded


















              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









              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












                $begingroup$

                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 1 hour ago

























                  answered 3 hours 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




                          Student404Mus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.









                          answered 1 hour 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.






















                              Student404Mus is a new contributor. Be nice, and check out our Code of Conduct.










                              draft saved

                              draft discarded


















                              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.




                              draft saved


                              draft discarded














                              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





















































                              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







                              -

                              Popular posts from this blog

                              Ponta tanko

                              Image
                              Usona ponta tanko M60A1 Ponta tanko estas, simile kiel kirasita savveturilo, inter specialaj kirasitaj transportiloj. Kiel pontaj tankoj eblas konsideri ĉiujn memmovigaj rimedoj ebligantaj superi obstaklojn ĝis larĝeco 24 metroj rekte en spaco de bataloperacoj. Ties ĉefa tasko estas ebligi al la propraj taĉmentoj daŭrigi en rapida antaŭenirado sen senbezona reteniĝo. El tio pli-malpli rezultas taktikaj-teknikaj postuloj, kiujn ili devas plenumi. Precipe temas pri kapablo operacii komune kun unue-sinsekvaj taĉmentoj kaj transpontigi obstaklojn dum kelke da minutoj. This page is only for reference, If you need detailed information, please check here
                              Read more

                              Tantalo (mitologio)

                              Image
                              Ĉi tiu artikolo temas pri mitologio. Por aliaj signifoj vidu la paĝon Tantalo (apartigilo) . Tantalo , laŭ Gioacchino Assereto. Laŭ la helena mitologio, Tantalo (el la greka Τάνταλος , ŝancelanta aŭ tre malfeliĉa ) estis filo de Zeŭso kaj la oceanidino Pluto, laŭ la plej divastigata versio. Li estis reĝo de Frigio [1] aŭ de la monto Sipilo, en Lidio (Malgrandazio) [2] kaj post sia morto li suferis eternan torturadon ĉe la Tartaro, la plej profunda parto de Hadeso, rezervita por la plej perversaj homoj. Tamen, kelkaj rakontoj asertas ke li estis reĝo de Argo aŭ de Korinto [3] , aŭ ke li migris norden el la monto Sipilo por akiri la tronon de Paflagonio, el kie Ilo forpelis lin pro tio, ke Tantalo forkaptis kaj seksperfortis lian fraton Ganimedon [4] , kaj tiel ofendis la diojn. Verŝajne li ankaŭ regis urbon en okcidenta Anatolio, kiun oni nomis Tantalis [5] pro li, aŭ Sipilo , pro la monto sur kies flanko ankoraŭ staras ĝiaj ruinaĵoj [6] . Laŭ la geografiisto Strabo,...
                              Read more

                              Erzsébet Schaár

                              Image
                              Erzsébet Schaár Memportreto de Erzsébet Schaár en Pécs Personaj informoj Naskiĝo 27-an de julio 1908 ( 1908-07-27 ) en Budafok Morto 29-an de aŭgusto 1975 ( 1975-08-29 ) (67-jara) en Budapeŝto Ŝtataneco Hungario Profesio skulptisto Familianoj Edz(in)o Tibor Vilt v   •   d   •   r Erzsébet Schaár [erĵEbet ŝAr], laŭ hungarlingve kutima sinsekvo Schaár Erzsébet estis hungara skulptisto. Ŝia edzo estis Tibor Vilt. Déryné Széppataki Róza en Miskolc fare de Erzsébet Schaár Erzsébet Schaár [1] naskiĝis la 27-an de julio 1908 en Budafok, ŝi mortis la 29-an de aŭgusto 1975 en Budapeŝto. Enhavo 1 Biografio 2 Elektitaj solaj ekspozicioj 3 Elektitaj kolektivaj ekspozicioj 4 Elektitaj skulptaĵoj 5 Elektitaj disĉiploj 6 Fontoj 7 Referencoj Biografio | Erzsébet Schaár studis en Industriarta Lernejo, poste en Reĝa Belarta Altlernejo (1924-1926). Ŝia ĉefinstruisto estis Zsig...
                              Read more