Why do we have to uncompute rather than simply set registers to zero?












4












$begingroup$


In implementing a quantum subroutine it is important to uncompute temporary registers after use, to ensure the output state of the subroutine is not entangled with them (which would affect its behaviour).



Why is it necessary to perform this through uncomputation rather than simply setting temporary registers to zero afterwards?






quantum-state entanglement







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$endgroup$












  • $begingroup$
    a related question on cstheory
    $endgroup$
    – glS
    14 hours ago
















4












$begingroup$


In implementing a quantum subroutine it is important to uncompute temporary registers after use, to ensure the output state of the subroutine is not entangled with them (which would affect its behaviour).



Why is it necessary to perform this through uncomputation rather than simply setting temporary registers to zero afterwards?






quantum-state entanglement







share|improve this question









$endgroup$












  • $begingroup$
    a related question on cstheory
    $endgroup$
    – glS
    14 hours ago














4












4








4





$begingroup$


In implementing a quantum subroutine it is important to uncompute temporary registers after use, to ensure the output state of the subroutine is not entangled with them (which would affect its behaviour).



Why is it necessary to perform this through uncomputation rather than simply setting temporary registers to zero afterwards?






quantum-state entanglement







share|improve this question









$endgroup$




In implementing a quantum subroutine it is important to uncompute temporary registers after use, to ensure the output state of the subroutine is not entangled with them (which would affect its behaviour).



Why is it necessary to perform this through uncomputation rather than simply setting temporary registers to zero afterwards?






quantum-state entanglement




quantum-state entanglement






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 16 hours ago









Sideshow BobSideshow Bob

1375




1375












  • $begingroup$
    a related question on cstheory
    $endgroup$
    – glS
    14 hours ago


















  • $begingroup$
    a related question on cstheory
    $endgroup$
    – glS
    14 hours ago
















$begingroup$
a related question on cstheory
$endgroup$
– glS
14 hours ago




$begingroup$
a related question on cstheory
$endgroup$
– glS
14 hours ago










2 Answers
2






active

oldest

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2












$begingroup$

Because all operations you carry out (besides measurement) need to be unitary and this one would not.






share|improve this answer









$endgroup$









  • 2




    $begingroup$
    There is one thing you don't address in your answer --- why not use a measurement then, and then flip the bit if the outcome is 1? That would effectively reset the qubit. Not that this is actually a good idea, but could you elaborate your answer to describe why this is not a good idea?
    $endgroup$
    – Niel de Beaudrap
    14 hours ago










  • $begingroup$
    @NieldeBeaudrap If you measure you collapse into one possibility. But say you want to continue applying operations on this output state (in the register of interest)... you see why this is not a good idea?
    $endgroup$
    – cnada
    13 hours ago






  • 1




    $begingroup$
    My comment was that it is something you don't address, which is conspicuous given that 'resetting' the state is exactly what the OP asked about. While I know why it isn't a good idea, I'm not the one who asked the original question.
    $endgroup$
    – Niel de Beaudrap
    12 hours ago



















4












$begingroup$

Uncomputing temporary registers seem to me to be the most efficient way to set temporary registers back to zero.



You can't measure these and classically flip them




  1. because a measurement of $1$ doesn't necessarily mean the state of the temporary register is $|1rangle$


  2. because measuring would cause entangled qubits to collapse into a corresponding state so that would defeat the purpose of uncomputing temporary registers that you mentioned in your question.



And somehow implementing a circuit that detected a temporary register's state moved it back to 0 would be much more difficult than simply uncomputing the temporary register.






share|improve this answer











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Because all operations you carry out (besides measurement) need to be unitary and this one would not.






    share|improve this answer









    $endgroup$









    • 2




      $begingroup$
      There is one thing you don't address in your answer --- why not use a measurement then, and then flip the bit if the outcome is 1? That would effectively reset the qubit. Not that this is actually a good idea, but could you elaborate your answer to describe why this is not a good idea?
      $endgroup$
      – Niel de Beaudrap
      14 hours ago










    • $begingroup$
      @NieldeBeaudrap If you measure you collapse into one possibility. But say you want to continue applying operations on this output state (in the register of interest)... you see why this is not a good idea?
      $endgroup$
      – cnada
      13 hours ago






    • 1




      $begingroup$
      My comment was that it is something you don't address, which is conspicuous given that 'resetting' the state is exactly what the OP asked about. While I know why it isn't a good idea, I'm not the one who asked the original question.
      $endgroup$
      – Niel de Beaudrap
      12 hours ago
















    2












    $begingroup$

    Because all operations you carry out (besides measurement) need to be unitary and this one would not.






    share|improve this answer









    $endgroup$









    • 2




      $begingroup$
      There is one thing you don't address in your answer --- why not use a measurement then, and then flip the bit if the outcome is 1? That would effectively reset the qubit. Not that this is actually a good idea, but could you elaborate your answer to describe why this is not a good idea?
      $endgroup$
      – Niel de Beaudrap
      14 hours ago










    • $begingroup$
      @NieldeBeaudrap If you measure you collapse into one possibility. But say you want to continue applying operations on this output state (in the register of interest)... you see why this is not a good idea?
      $endgroup$
      – cnada
      13 hours ago






    • 1




      $begingroup$
      My comment was that it is something you don't address, which is conspicuous given that 'resetting' the state is exactly what the OP asked about. While I know why it isn't a good idea, I'm not the one who asked the original question.
      $endgroup$
      – Niel de Beaudrap
      12 hours ago














    2












    2








    2





    $begingroup$

    Because all operations you carry out (besides measurement) need to be unitary and this one would not.






    share|improve this answer









    $endgroup$



    Because all operations you carry out (besides measurement) need to be unitary and this one would not.







    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered 15 hours ago









    cnadacnada

    1,992212




    1,992212








    • 2




      $begingroup$
      There is one thing you don't address in your answer --- why not use a measurement then, and then flip the bit if the outcome is 1? That would effectively reset the qubit. Not that this is actually a good idea, but could you elaborate your answer to describe why this is not a good idea?
      $endgroup$
      – Niel de Beaudrap
      14 hours ago










    • $begingroup$
      @NieldeBeaudrap If you measure you collapse into one possibility. But say you want to continue applying operations on this output state (in the register of interest)... you see why this is not a good idea?
      $endgroup$
      – cnada
      13 hours ago






    • 1




      $begingroup$
      My comment was that it is something you don't address, which is conspicuous given that 'resetting' the state is exactly what the OP asked about. While I know why it isn't a good idea, I'm not the one who asked the original question.
      $endgroup$
      – Niel de Beaudrap
      12 hours ago














    • 2




      $begingroup$
      There is one thing you don't address in your answer --- why not use a measurement then, and then flip the bit if the outcome is 1? That would effectively reset the qubit. Not that this is actually a good idea, but could you elaborate your answer to describe why this is not a good idea?
      $endgroup$
      – Niel de Beaudrap
      14 hours ago










    • $begingroup$
      @NieldeBeaudrap If you measure you collapse into one possibility. But say you want to continue applying operations on this output state (in the register of interest)... you see why this is not a good idea?
      $endgroup$
      – cnada
      13 hours ago






    • 1




      $begingroup$
      My comment was that it is something you don't address, which is conspicuous given that 'resetting' the state is exactly what the OP asked about. While I know why it isn't a good idea, I'm not the one who asked the original question.
      $endgroup$
      – Niel de Beaudrap
      12 hours ago








    2




    2




    $begingroup$
    There is one thing you don't address in your answer --- why not use a measurement then, and then flip the bit if the outcome is 1? That would effectively reset the qubit. Not that this is actually a good idea, but could you elaborate your answer to describe why this is not a good idea?
    $endgroup$
    – Niel de Beaudrap
    14 hours ago




    $begingroup$
    There is one thing you don't address in your answer --- why not use a measurement then, and then flip the bit if the outcome is 1? That would effectively reset the qubit. Not that this is actually a good idea, but could you elaborate your answer to describe why this is not a good idea?
    $endgroup$
    – Niel de Beaudrap
    14 hours ago












    $begingroup$
    @NieldeBeaudrap If you measure you collapse into one possibility. But say you want to continue applying operations on this output state (in the register of interest)... you see why this is not a good idea?
    $endgroup$
    – cnada
    13 hours ago




    $begingroup$
    @NieldeBeaudrap If you measure you collapse into one possibility. But say you want to continue applying operations on this output state (in the register of interest)... you see why this is not a good idea?
    $endgroup$
    – cnada
    13 hours ago




    1




    1




    $begingroup$
    My comment was that it is something you don't address, which is conspicuous given that 'resetting' the state is exactly what the OP asked about. While I know why it isn't a good idea, I'm not the one who asked the original question.
    $endgroup$
    – Niel de Beaudrap
    12 hours ago




    $begingroup$
    My comment was that it is something you don't address, which is conspicuous given that 'resetting' the state is exactly what the OP asked about. While I know why it isn't a good idea, I'm not the one who asked the original question.
    $endgroup$
    – Niel de Beaudrap
    12 hours ago













    4












    $begingroup$

    Uncomputing temporary registers seem to me to be the most efficient way to set temporary registers back to zero.



    You can't measure these and classically flip them




    1. because a measurement of $1$ doesn't necessarily mean the state of the temporary register is $|1rangle$


    2. because measuring would cause entangled qubits to collapse into a corresponding state so that would defeat the purpose of uncomputing temporary registers that you mentioned in your question.



    And somehow implementing a circuit that detected a temporary register's state moved it back to 0 would be much more difficult than simply uncomputing the temporary register.






    share|improve this answer











    $endgroup$


















      4












      $begingroup$

      Uncomputing temporary registers seem to me to be the most efficient way to set temporary registers back to zero.



      You can't measure these and classically flip them




      1. because a measurement of $1$ doesn't necessarily mean the state of the temporary register is $|1rangle$


      2. because measuring would cause entangled qubits to collapse into a corresponding state so that would defeat the purpose of uncomputing temporary registers that you mentioned in your question.



      And somehow implementing a circuit that detected a temporary register's state moved it back to 0 would be much more difficult than simply uncomputing the temporary register.






      share|improve this answer











      $endgroup$
















        4












        4








        4





        $begingroup$

        Uncomputing temporary registers seem to me to be the most efficient way to set temporary registers back to zero.



        You can't measure these and classically flip them




        1. because a measurement of $1$ doesn't necessarily mean the state of the temporary register is $|1rangle$


        2. because measuring would cause entangled qubits to collapse into a corresponding state so that would defeat the purpose of uncomputing temporary registers that you mentioned in your question.



        And somehow implementing a circuit that detected a temporary register's state moved it back to 0 would be much more difficult than simply uncomputing the temporary register.






        share|improve this answer











        $endgroup$



        Uncomputing temporary registers seem to me to be the most efficient way to set temporary registers back to zero.



        You can't measure these and classically flip them




        1. because a measurement of $1$ doesn't necessarily mean the state of the temporary register is $|1rangle$


        2. because measuring would cause entangled qubits to collapse into a corresponding state so that would defeat the purpose of uncomputing temporary registers that you mentioned in your question.



        And somehow implementing a circuit that detected a temporary register's state moved it back to 0 would be much more difficult than simply uncomputing the temporary register.







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited 10 hours ago









        Blue♦

        5,70721354




        5,70721354










        answered 14 hours ago









        Malcolm ReganMalcolm Regan

        1797




        1797






























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