How to superpose two composite qubit states? Planned maintenance scheduled April 17/18, 2019...

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How to superpose two composite qubit states?



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$begingroup$


Assuming we have two sets of $n$ qubits. The first set of $n$ qubits is in state $|arangle$ and second set in $|brangle$. Is there a fixed procedure that generates a superposed state of the two $|arangle + |brangle$ ?






quantum-state circuit-construction superposition







share|improve this question











$endgroup$



















    3












    $begingroup$


    Assuming we have two sets of $n$ qubits. The first set of $n$ qubits is in state $|arangle$ and second set in $|brangle$. Is there a fixed procedure that generates a superposed state of the two $|arangle + |brangle$ ?






    quantum-state circuit-construction superposition







    share|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      Assuming we have two sets of $n$ qubits. The first set of $n$ qubits is in state $|arangle$ and second set in $|brangle$. Is there a fixed procedure that generates a superposed state of the two $|arangle + |brangle$ ?






      quantum-state circuit-construction superposition







      share|improve this question











      $endgroup$




      Assuming we have two sets of $n$ qubits. The first set of $n$ qubits is in state $|arangle$ and second set in $|brangle$. Is there a fixed procedure that generates a superposed state of the two $|arangle + |brangle$ ?






      quantum-state circuit-construction superposition




      quantum-state circuit-construction superposition






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 14 hours ago









      Sanchayan Dutta♦

      6,69641556




      6,69641556










      asked 19 hours ago









      wang1908wang1908

      313




      313






















          2 Answers
          2






          active

          oldest

          votes


















          5












          $begingroup$

          Depending on what precisely your assumptions are about a, b, I think this is essentially impossible, and is something called the "no superposing theorem". Please see this paper.






          share|improve this answer











          $endgroup$





















            3












            $begingroup$

            Your question is not quite correctly defined.



            First of all, $|arangle + |brangle$ is not a state. You need to normalize it by considering $frac{1}{|||arangle + |brangle||}(|arangle + |brangle)$.

            Secondly, in fact, you don't have access to the states $|arangle$ and $|brangle$ but to the states up to some global phase, i.e. you can think that the first register is in the vector-state $e^{iphi}|arangle$ and the second register is in the vector-state $e^{ipsi}|brangle$ with inaccessible $phi, psi$. Since you don't have access to $phi, psi$, you can't define sum $|arangle + |brangle$. But you can ask to construct normalized state $|arangle + e^{ it}|brangle$ for some $t$. This question is correct. Though, as DaftWullie pointed out in his answer, it is impossible to construct such a state.






            share|improve this answer











            $endgroup$













            • $begingroup$
              Can't you just set $t = 0 mod 2pi$? Or is $t$ something out of your control?
              $endgroup$
              – wizzwizz4
              14 hours ago








            • 2




              $begingroup$
              @wizzwizz4 Yes, $t$ is unknown. Again, you can't define sum $e^{iphi}|arangle + e^{ipsi}|brangle$ to be equal to some exact state (that depends only on $|arangle$ and $|brangle$). But you can define this sum to be of some type of state. This type of state can be defined as $|arangle + e^{ it}|brangle$ since $e^{iphi}|arangle + e^{ipsi}|brangle = e^{iphi} (|arangle + e^{ i(psi - phi)}|brangle) propto |arangle + e^{ i(psi - phi)}|brangle$.
              $endgroup$
              – Danylo Y
              13 hours ago














            Your Answer








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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            5












            $begingroup$

            Depending on what precisely your assumptions are about a, b, I think this is essentially impossible, and is something called the "no superposing theorem". Please see this paper.






            share|improve this answer











            $endgroup$


















              5












              $begingroup$

              Depending on what precisely your assumptions are about a, b, I think this is essentially impossible, and is something called the "no superposing theorem". Please see this paper.






              share|improve this answer











              $endgroup$
















                5












                5








                5





                $begingroup$

                Depending on what precisely your assumptions are about a, b, I think this is essentially impossible, and is something called the "no superposing theorem". Please see this paper.






                share|improve this answer











                $endgroup$



                Depending on what precisely your assumptions are about a, b, I think this is essentially impossible, and is something called the "no superposing theorem". Please see this paper.







                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited 14 hours ago









                Sanchayan Dutta♦

                6,69641556




                6,69641556










                answered 18 hours ago









                DaftWullieDaftWullie

                15.5k1642




                15.5k1642

























                    3












                    $begingroup$

                    Your question is not quite correctly defined.



                    First of all, $|arangle + |brangle$ is not a state. You need to normalize it by considering $frac{1}{|||arangle + |brangle||}(|arangle + |brangle)$.

                    Secondly, in fact, you don't have access to the states $|arangle$ and $|brangle$ but to the states up to some global phase, i.e. you can think that the first register is in the vector-state $e^{iphi}|arangle$ and the second register is in the vector-state $e^{ipsi}|brangle$ with inaccessible $phi, psi$. Since you don't have access to $phi, psi$, you can't define sum $|arangle + |brangle$. But you can ask to construct normalized state $|arangle + e^{ it}|brangle$ for some $t$. This question is correct. Though, as DaftWullie pointed out in his answer, it is impossible to construct such a state.






                    share|improve this answer











                    $endgroup$













                    • $begingroup$
                      Can't you just set $t = 0 mod 2pi$? Or is $t$ something out of your control?
                      $endgroup$
                      – wizzwizz4
                      14 hours ago








                    • 2




                      $begingroup$
                      @wizzwizz4 Yes, $t$ is unknown. Again, you can't define sum $e^{iphi}|arangle + e^{ipsi}|brangle$ to be equal to some exact state (that depends only on $|arangle$ and $|brangle$). But you can define this sum to be of some type of state. This type of state can be defined as $|arangle + e^{ it}|brangle$ since $e^{iphi}|arangle + e^{ipsi}|brangle = e^{iphi} (|arangle + e^{ i(psi - phi)}|brangle) propto |arangle + e^{ i(psi - phi)}|brangle$.
                      $endgroup$
                      – Danylo Y
                      13 hours ago


















                    3












                    $begingroup$

                    Your question is not quite correctly defined.



                    First of all, $|arangle + |brangle$ is not a state. You need to normalize it by considering $frac{1}{|||arangle + |brangle||}(|arangle + |brangle)$.

                    Secondly, in fact, you don't have access to the states $|arangle$ and $|brangle$ but to the states up to some global phase, i.e. you can think that the first register is in the vector-state $e^{iphi}|arangle$ and the second register is in the vector-state $e^{ipsi}|brangle$ with inaccessible $phi, psi$. Since you don't have access to $phi, psi$, you can't define sum $|arangle + |brangle$. But you can ask to construct normalized state $|arangle + e^{ it}|brangle$ for some $t$. This question is correct. Though, as DaftWullie pointed out in his answer, it is impossible to construct such a state.






                    share|improve this answer











                    $endgroup$













                    • $begingroup$
                      Can't you just set $t = 0 mod 2pi$? Or is $t$ something out of your control?
                      $endgroup$
                      – wizzwizz4
                      14 hours ago








                    • 2




                      $begingroup$
                      @wizzwizz4 Yes, $t$ is unknown. Again, you can't define sum $e^{iphi}|arangle + e^{ipsi}|brangle$ to be equal to some exact state (that depends only on $|arangle$ and $|brangle$). But you can define this sum to be of some type of state. This type of state can be defined as $|arangle + e^{ it}|brangle$ since $e^{iphi}|arangle + e^{ipsi}|brangle = e^{iphi} (|arangle + e^{ i(psi - phi)}|brangle) propto |arangle + e^{ i(psi - phi)}|brangle$.
                      $endgroup$
                      – Danylo Y
                      13 hours ago
















                    3












                    3








                    3





                    $begingroup$

                    Your question is not quite correctly defined.



                    First of all, $|arangle + |brangle$ is not a state. You need to normalize it by considering $frac{1}{|||arangle + |brangle||}(|arangle + |brangle)$.

                    Secondly, in fact, you don't have access to the states $|arangle$ and $|brangle$ but to the states up to some global phase, i.e. you can think that the first register is in the vector-state $e^{iphi}|arangle$ and the second register is in the vector-state $e^{ipsi}|brangle$ with inaccessible $phi, psi$. Since you don't have access to $phi, psi$, you can't define sum $|arangle + |brangle$. But you can ask to construct normalized state $|arangle + e^{ it}|brangle$ for some $t$. This question is correct. Though, as DaftWullie pointed out in his answer, it is impossible to construct such a state.






                    share|improve this answer











                    $endgroup$



                    Your question is not quite correctly defined.



                    First of all, $|arangle + |brangle$ is not a state. You need to normalize it by considering $frac{1}{|||arangle + |brangle||}(|arangle + |brangle)$.

                    Secondly, in fact, you don't have access to the states $|arangle$ and $|brangle$ but to the states up to some global phase, i.e. you can think that the first register is in the vector-state $e^{iphi}|arangle$ and the second register is in the vector-state $e^{ipsi}|brangle$ with inaccessible $phi, psi$. Since you don't have access to $phi, psi$, you can't define sum $|arangle + |brangle$. But you can ask to construct normalized state $|arangle + e^{ it}|brangle$ for some $t$. This question is correct. Though, as DaftWullie pointed out in his answer, it is impossible to construct such a state.







                    share|improve this answer














                    share|improve this answer



                    share|improve this answer








                    edited 14 hours ago









                    Sanchayan Dutta♦

                    6,69641556




                    6,69641556










                    answered 15 hours ago









                    Danylo YDanylo Y

                    58116




                    58116












                    • $begingroup$
                      Can't you just set $t = 0 mod 2pi$? Or is $t$ something out of your control?
                      $endgroup$
                      – wizzwizz4
                      14 hours ago








                    • 2




                      $begingroup$
                      @wizzwizz4 Yes, $t$ is unknown. Again, you can't define sum $e^{iphi}|arangle + e^{ipsi}|brangle$ to be equal to some exact state (that depends only on $|arangle$ and $|brangle$). But you can define this sum to be of some type of state. This type of state can be defined as $|arangle + e^{ it}|brangle$ since $e^{iphi}|arangle + e^{ipsi}|brangle = e^{iphi} (|arangle + e^{ i(psi - phi)}|brangle) propto |arangle + e^{ i(psi - phi)}|brangle$.
                      $endgroup$
                      – Danylo Y
                      13 hours ago




















                    • $begingroup$
                      Can't you just set $t = 0 mod 2pi$? Or is $t$ something out of your control?
                      $endgroup$
                      – wizzwizz4
                      14 hours ago








                    • 2




                      $begingroup$
                      @wizzwizz4 Yes, $t$ is unknown. Again, you can't define sum $e^{iphi}|arangle + e^{ipsi}|brangle$ to be equal to some exact state (that depends only on $|arangle$ and $|brangle$). But you can define this sum to be of some type of state. This type of state can be defined as $|arangle + e^{ it}|brangle$ since $e^{iphi}|arangle + e^{ipsi}|brangle = e^{iphi} (|arangle + e^{ i(psi - phi)}|brangle) propto |arangle + e^{ i(psi - phi)}|brangle$.
                      $endgroup$
                      – Danylo Y
                      13 hours ago


















                    $begingroup$
                    Can't you just set $t = 0 mod 2pi$? Or is $t$ something out of your control?
                    $endgroup$
                    – wizzwizz4
                    14 hours ago






                    $begingroup$
                    Can't you just set $t = 0 mod 2pi$? Or is $t$ something out of your control?
                    $endgroup$
                    – wizzwizz4
                    14 hours ago






                    2




                    2




                    $begingroup$
                    @wizzwizz4 Yes, $t$ is unknown. Again, you can't define sum $e^{iphi}|arangle + e^{ipsi}|brangle$ to be equal to some exact state (that depends only on $|arangle$ and $|brangle$). But you can define this sum to be of some type of state. This type of state can be defined as $|arangle + e^{ it}|brangle$ since $e^{iphi}|arangle + e^{ipsi}|brangle = e^{iphi} (|arangle + e^{ i(psi - phi)}|brangle) propto |arangle + e^{ i(psi - phi)}|brangle$.
                    $endgroup$
                    – Danylo Y
                    13 hours ago






                    $begingroup$
                    @wizzwizz4 Yes, $t$ is unknown. Again, you can't define sum $e^{iphi}|arangle + e^{ipsi}|brangle$ to be equal to some exact state (that depends only on $|arangle$ and $|brangle$). But you can define this sum to be of some type of state. This type of state can be defined as $|arangle + e^{ it}|brangle$ since $e^{iphi}|arangle + e^{ipsi}|brangle = e^{iphi} (|arangle + e^{ i(psi - phi)}|brangle) propto |arangle + e^{ i(psi - phi)}|brangle$.
                    $endgroup$
                    – Danylo Y
                    13 hours ago




















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