A basic question on circuits and matrix representationSWAP gate(s) in the $R(lambda^-1)$ step of the HHL circuit for $4times 4$ systemsDecomposition of arbitrary 2 qubit operatorSolving a circuit implementing a two-level unitary operationMatrix representation and CX gateHow to construct matrix of regular and “flipped” 2-qubit CNOT?Square root of CNOT and spectral decomposition of the Hadamard gateMatrix representation of multiple qubit gates (Hadamard transform on single wire)Tensorial notation for this quantum XOR circuitUnderstanding CNOT gate for indirect measurementHow to create an Ising coupling gate with Qiskit

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A basic question on circuits and matrix representation


SWAP gate(s) in the $R(lambda^-1)$ step of the HHL circuit for $4times 4$ systemsDecomposition of arbitrary 2 qubit operatorSolving a circuit implementing a two-level unitary operationMatrix representation and CX gateHow to construct matrix of regular and “flipped” 2-qubit CNOT?Square root of CNOT and spectral decomposition of the Hadamard gateMatrix representation of multiple qubit gates (Hadamard transform on single wire)Tensorial notation for this quantum XOR circuitUnderstanding CNOT gate for indirect measurementHow to create an Ising coupling gate with Qiskit






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








3












$begingroup$


I have several (rather basic) questions on matrix representation of circuits and I would be very grateful to anyone that could clear up my confusion, thank you in advance.



1) When reading circuit diagrams I know that the input qubit goes in the left hand side. So If we are reading a circuit and it's gates go in the order a,b,c does that mean when we want to write out its matrix representation we multiply the matrices in the order c,a,b . For example if we have a circuit which consists of a swap gate followed by a Hadamard gate on the second qubit , followed by a Hadamard gate on the first qubit , then to calculate it's matrix representation we would have to calculate it in the order $(H_1otimes I)(Iotimes H_2)Swap$, correct ? (as this reflects the order of application of gates on the qubit state).



2)If we are given a gate $S=beginpmatrix1 &0 \0 &i endpmatrix$, but it is on the top line of a two line circuit with a line connecting it to the second (In other words it's a control gate ), I know that the first qubit is the control and the second is the target but I'm unsure of how to write it , should it be $S=beginpmatrix1 & 0 &0&0 \ 0 &-i &0&0\0&0&1&0\0&0&0&-i endpmatrix$,



or should it be



$ S=beginpmatrix1 & 0 &0&0 \ 0 &1 &0&0\0&0&-i&0\0&0&0&-i endpmatrix$,










share|improve this question









$endgroup$




















    3












    $begingroup$


    I have several (rather basic) questions on matrix representation of circuits and I would be very grateful to anyone that could clear up my confusion, thank you in advance.



    1) When reading circuit diagrams I know that the input qubit goes in the left hand side. So If we are reading a circuit and it's gates go in the order a,b,c does that mean when we want to write out its matrix representation we multiply the matrices in the order c,a,b . For example if we have a circuit which consists of a swap gate followed by a Hadamard gate on the second qubit , followed by a Hadamard gate on the first qubit , then to calculate it's matrix representation we would have to calculate it in the order $(H_1otimes I)(Iotimes H_2)Swap$, correct ? (as this reflects the order of application of gates on the qubit state).



    2)If we are given a gate $S=beginpmatrix1 &0 \0 &i endpmatrix$, but it is on the top line of a two line circuit with a line connecting it to the second (In other words it's a control gate ), I know that the first qubit is the control and the second is the target but I'm unsure of how to write it , should it be $S=beginpmatrix1 & 0 &0&0 \ 0 &-i &0&0\0&0&1&0\0&0&0&-i endpmatrix$,



    or should it be



    $ S=beginpmatrix1 & 0 &0&0 \ 0 &1 &0&0\0&0&-i&0\0&0&0&-i endpmatrix$,










    share|improve this question









    $endgroup$
















      3












      3








      3





      $begingroup$


      I have several (rather basic) questions on matrix representation of circuits and I would be very grateful to anyone that could clear up my confusion, thank you in advance.



      1) When reading circuit diagrams I know that the input qubit goes in the left hand side. So If we are reading a circuit and it's gates go in the order a,b,c does that mean when we want to write out its matrix representation we multiply the matrices in the order c,a,b . For example if we have a circuit which consists of a swap gate followed by a Hadamard gate on the second qubit , followed by a Hadamard gate on the first qubit , then to calculate it's matrix representation we would have to calculate it in the order $(H_1otimes I)(Iotimes H_2)Swap$, correct ? (as this reflects the order of application of gates on the qubit state).



      2)If we are given a gate $S=beginpmatrix1 &0 \0 &i endpmatrix$, but it is on the top line of a two line circuit with a line connecting it to the second (In other words it's a control gate ), I know that the first qubit is the control and the second is the target but I'm unsure of how to write it , should it be $S=beginpmatrix1 & 0 &0&0 \ 0 &-i &0&0\0&0&1&0\0&0&0&-i endpmatrix$,



      or should it be



      $ S=beginpmatrix1 & 0 &0&0 \ 0 &1 &0&0\0&0&-i&0\0&0&0&-i endpmatrix$,










      share|improve this question









      $endgroup$




      I have several (rather basic) questions on matrix representation of circuits and I would be very grateful to anyone that could clear up my confusion, thank you in advance.



      1) When reading circuit diagrams I know that the input qubit goes in the left hand side. So If we are reading a circuit and it's gates go in the order a,b,c does that mean when we want to write out its matrix representation we multiply the matrices in the order c,a,b . For example if we have a circuit which consists of a swap gate followed by a Hadamard gate on the second qubit , followed by a Hadamard gate on the first qubit , then to calculate it's matrix representation we would have to calculate it in the order $(H_1otimes I)(Iotimes H_2)Swap$, correct ? (as this reflects the order of application of gates on the qubit state).



      2)If we are given a gate $S=beginpmatrix1 &0 \0 &i endpmatrix$, but it is on the top line of a two line circuit with a line connecting it to the second (In other words it's a control gate ), I know that the first qubit is the control and the second is the target but I'm unsure of how to write it , should it be $S=beginpmatrix1 & 0 &0&0 \ 0 &-i &0&0\0&0&1&0\0&0&0&-i endpmatrix$,



      or should it be



      $ S=beginpmatrix1 & 0 &0&0 \ 0 &1 &0&0\0&0&-i&0\0&0&0&-i endpmatrix$,







      circuit-construction matrix-representation






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked 8 hours ago









      bhapibhapi

      3578 bronze badges




      3578 bronze badges























          1 Answer
          1






          active

          oldest

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          3












          $begingroup$

          1. Yes, in the circuit the qubit "enters" to the left, and exits to the right, but when applying the gates to a state you must apply the one on the far left first, then the next and so on, so concretely you do write them down right to left, but it's just a consequence of writing the operator that we want to apply on the left of the vector, while our natural way of writing dictates a "time direction" left to right. As an example the following circuit

          enter image description here



          implements the operation



          $$(Iotimes mathrmCNOT)(mathrmCNOTotimes I)(Hotimes Iotimes I)|000rangle $$



          1. A controlled gate does nothing if the first qubit is $0$ and performs the gate if it is $1$, you can represent it this way

          $$mathrmCS=|0ranglelangle0|otimes I + |1ranglelangle1|otimes S $$
          Hence in matrix form it would be



          $$mathrmCS=beginpmatrix
          I&0\0&S
          endpmatrix =beginpmatrix
          1&0&0&0\0&1&0&0\0&0&1&0\0&0&0&i
          endpmatrix$$






          share|improve this answer









          $endgroup$














          • $begingroup$
            Thank you for succinctly answering all my questions :)
            $endgroup$
            – bhapi
            5 hours ago













          Your Answer








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          $begingroup$

          1. Yes, in the circuit the qubit "enters" to the left, and exits to the right, but when applying the gates to a state you must apply the one on the far left first, then the next and so on, so concretely you do write them down right to left, but it's just a consequence of writing the operator that we want to apply on the left of the vector, while our natural way of writing dictates a "time direction" left to right. As an example the following circuit

          enter image description here



          implements the operation



          $$(Iotimes mathrmCNOT)(mathrmCNOTotimes I)(Hotimes Iotimes I)|000rangle $$



          1. A controlled gate does nothing if the first qubit is $0$ and performs the gate if it is $1$, you can represent it this way

          $$mathrmCS=|0ranglelangle0|otimes I + |1ranglelangle1|otimes S $$
          Hence in matrix form it would be



          $$mathrmCS=beginpmatrix
          I&0\0&S
          endpmatrix =beginpmatrix
          1&0&0&0\0&1&0&0\0&0&1&0\0&0&0&i
          endpmatrix$$






          share|improve this answer









          $endgroup$














          • $begingroup$
            Thank you for succinctly answering all my questions :)
            $endgroup$
            – bhapi
            5 hours ago















          3












          $begingroup$

          1. Yes, in the circuit the qubit "enters" to the left, and exits to the right, but when applying the gates to a state you must apply the one on the far left first, then the next and so on, so concretely you do write them down right to left, but it's just a consequence of writing the operator that we want to apply on the left of the vector, while our natural way of writing dictates a "time direction" left to right. As an example the following circuit

          enter image description here



          implements the operation



          $$(Iotimes mathrmCNOT)(mathrmCNOTotimes I)(Hotimes Iotimes I)|000rangle $$



          1. A controlled gate does nothing if the first qubit is $0$ and performs the gate if it is $1$, you can represent it this way

          $$mathrmCS=|0ranglelangle0|otimes I + |1ranglelangle1|otimes S $$
          Hence in matrix form it would be



          $$mathrmCS=beginpmatrix
          I&0\0&S
          endpmatrix =beginpmatrix
          1&0&0&0\0&1&0&0\0&0&1&0\0&0&0&i
          endpmatrix$$






          share|improve this answer









          $endgroup$














          • $begingroup$
            Thank you for succinctly answering all my questions :)
            $endgroup$
            – bhapi
            5 hours ago













          3












          3








          3





          $begingroup$

          1. Yes, in the circuit the qubit "enters" to the left, and exits to the right, but when applying the gates to a state you must apply the one on the far left first, then the next and so on, so concretely you do write them down right to left, but it's just a consequence of writing the operator that we want to apply on the left of the vector, while our natural way of writing dictates a "time direction" left to right. As an example the following circuit

          enter image description here



          implements the operation



          $$(Iotimes mathrmCNOT)(mathrmCNOTotimes I)(Hotimes Iotimes I)|000rangle $$



          1. A controlled gate does nothing if the first qubit is $0$ and performs the gate if it is $1$, you can represent it this way

          $$mathrmCS=|0ranglelangle0|otimes I + |1ranglelangle1|otimes S $$
          Hence in matrix form it would be



          $$mathrmCS=beginpmatrix
          I&0\0&S
          endpmatrix =beginpmatrix
          1&0&0&0\0&1&0&0\0&0&1&0\0&0&0&i
          endpmatrix$$






          share|improve this answer









          $endgroup$



          1. Yes, in the circuit the qubit "enters" to the left, and exits to the right, but when applying the gates to a state you must apply the one on the far left first, then the next and so on, so concretely you do write them down right to left, but it's just a consequence of writing the operator that we want to apply on the left of the vector, while our natural way of writing dictates a "time direction" left to right. As an example the following circuit

          enter image description here



          implements the operation



          $$(Iotimes mathrmCNOT)(mathrmCNOTotimes I)(Hotimes Iotimes I)|000rangle $$



          1. A controlled gate does nothing if the first qubit is $0$ and performs the gate if it is $1$, you can represent it this way

          $$mathrmCS=|0ranglelangle0|otimes I + |1ranglelangle1|otimes S $$
          Hence in matrix form it would be



          $$mathrmCS=beginpmatrix
          I&0\0&S
          endpmatrix =beginpmatrix
          1&0&0&0\0&1&0&0\0&0&1&0\0&0&0&i
          endpmatrix$$







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 6 hours ago









          user2723984user2723984

          3158 bronze badges




          3158 bronze badges














          • $begingroup$
            Thank you for succinctly answering all my questions :)
            $endgroup$
            – bhapi
            5 hours ago
















          • $begingroup$
            Thank you for succinctly answering all my questions :)
            $endgroup$
            – bhapi
            5 hours ago















          $begingroup$
          Thank you for succinctly answering all my questions :)
          $endgroup$
          – bhapi
          5 hours ago




          $begingroup$
          Thank you for succinctly answering all my questions :)
          $endgroup$
          – bhapi
          5 hours ago

















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