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What does it mean to express a gate in Dirac notation?


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}







1












$begingroup$


When discussing the Dirac notation of an operator, for example, let's just say we have the bit flip gate $X$ if we want to write this in the Dirac notation does that just mean writing it as follows?



$$X|psirangle=X(c_0|0rangle+c_1|1rangle)=c_0|1rangle+c_1|0rangle$$










share|improve this question











$endgroup$



















    1












    $begingroup$


    When discussing the Dirac notation of an operator, for example, let's just say we have the bit flip gate $X$ if we want to write this in the Dirac notation does that just mean writing it as follows?



    $$X|psirangle=X(c_0|0rangle+c_1|1rangle)=c_0|1rangle+c_1|0rangle$$










    share|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      When discussing the Dirac notation of an operator, for example, let's just say we have the bit flip gate $X$ if we want to write this in the Dirac notation does that just mean writing it as follows?



      $$X|psirangle=X(c_0|0rangle+c_1|1rangle)=c_0|1rangle+c_1|0rangle$$










      share|improve this question











      $endgroup$




      When discussing the Dirac notation of an operator, for example, let's just say we have the bit flip gate $X$ if we want to write this in the Dirac notation does that just mean writing it as follows?



      $$X|psirangle=X(c_0|0rangle+c_1|1rangle)=c_0|1rangle+c_1|0rangle$$







      quantum-gate notation






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 2 hours ago









      Sanchayan Dutta

      6,75841556




      6,75841556










      asked 2 hours ago









      can'tcauchycan'tcauchy

      2015




      2015






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          This might mean using the ketbra notation:



          $$X = |1rangle langle0| + |1rangle langle0|$$



          This notation describes the effect the operator has on the basis vectors: in this case $X$ converts $|0rangle$ into $|1rangle$ and vice versa.



          A couple of other examples:



          $$Z = |0rangle langle0| - |1rangle langle1|$$



          $$operatorname{CNOT} = |0 ranglelangle0| otimes I + |1 ranglelangle 1| otimes X = |00rangle langle00| + |01rangle langle01| + |11rangle langle10| + |10rangle langle11|$$






          share|improve this answer











          $endgroup$













          • $begingroup$
            For OP: a quick and intuitive way to derive the CNOT's outer product representation is to see what effect it has on the individual qubits. The CNOT flips the second qubit when the control qubit is $|1rangle$. When the control qubit is $|0rangle$ the second qubit remains intact, which is equivalent to applying the identity gate $I$. So we get $|0ranglelangle 0|otimes I$ for this. However, when the control qubit is $|1rangle$ we need to flip the second qubit's state i.e. we apply the $X$ gate to the second qubit. Thus, we get $|1ranglelangle 1|otimes X$ for this.
            $endgroup$
            – Sanchayan Dutta
            1 hour ago





















          0












          $begingroup$

          The Dirac notation for the Pauli-$X$ gate is:



          $$|1rangle langle0| + |1rangle langle0|.$$



          Now you might be wondering where this comes from. The term you're looking for is outer product representation of the $X$ gate. It follows from the spectral decomposition theorem (check Nielsen & Chuang 10th edition, p. 72) which holds for all normal operators. The key point:




          In terms of the outer product representation, this means that $M$ can be written as $M=sum_ilambda_i|iranglelangle i|$,where $lambda_i$ are the eigenvalues of $M$,$|irangle$ is an orthonormal basis for $V$, and each $|irangle$ an eigenvector of $M$ with eigenvalue $lambda_i$.




          The eigenvectors of the Pauli-$X$ gate are $-|0rangle+|1rangle$ and $|0rangle+|1rangle$, and the corresponding eigenvalues are $-1$ and $+1$ cf. Wolfram Alpha. Normalize the eigenvectors to get an orthonormal basis for $X$ i.e. ${frac{-|0rangle+|1rangle}{sqrt{2}},frac{|0rangle+|1rangle}{sqrt{2}}}$. According the spectral decomposition theorem you can represent the $X$ gate as:



          $$-1(frac{-|0rangle+|1rangle}{sqrt{2}})(frac{-langle 0|+langle1|}{sqrt{2}}) + 1(frac{|0rangle+|1rangle}{sqrt{2}})(frac{langle 0|+langle1|}{sqrt{2}})$$
          $$=-frac{1}{2}(|0ranglelangle0|-|0ranglelangle1|-|1ranglelangle0|+|1ranglelangle1|)+frac{1}{2}(|0ranglelangle0|+|0ranglelangle1|+|1ranglelangle0|+|1ranglelangle1|)$$
          $$=|1rangle langle0| + |1rangle langle0|$$



          To convince you that this result is correct let's apply it on an arbitrary qubit state $c_0|0rangle+c_1 |1rangle$:



          $$(|1rangle langle0| + |0rangle langle1|)(c_0|0rangle+c_1|1rangle)$$
          $$=c_0|1ranglelangle0|0rangle+c_1|0ranglelangle 1|1rangle$$
          $$=c_0 |1rangle + c_1 |0rangle$$



          So yes, our result is correct and the bits were indeed flipped upon application of $X=|1rangle langle0| + |1rangle langle0|$ to $c_0|0rangle + c_1|1rangle$.The last step followed from the fact that $langle 0|0rangle$ and $langle 1|1rangle$ are both equal to $1$, as $|0rangle$ and $|1rangle$ are orthonormal vectors i.e. their inner product $langle psi|psirangle=1$ by definition.



          We're done. As an exercise, find the outer product representation of the Pauli-$Z$ gate by yourself. And definitely, do go through the proof of the spectral theorem in Nielsen and Chung if time permits!






          share|improve this answer











          $endgroup$














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            2 Answers
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            active

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            active

            oldest

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            active

            oldest

            votes









            2












            $begingroup$

            This might mean using the ketbra notation:



            $$X = |1rangle langle0| + |1rangle langle0|$$



            This notation describes the effect the operator has on the basis vectors: in this case $X$ converts $|0rangle$ into $|1rangle$ and vice versa.



            A couple of other examples:



            $$Z = |0rangle langle0| - |1rangle langle1|$$



            $$operatorname{CNOT} = |0 ranglelangle0| otimes I + |1 ranglelangle 1| otimes X = |00rangle langle00| + |01rangle langle01| + |11rangle langle10| + |10rangle langle11|$$






            share|improve this answer











            $endgroup$













            • $begingroup$
              For OP: a quick and intuitive way to derive the CNOT's outer product representation is to see what effect it has on the individual qubits. The CNOT flips the second qubit when the control qubit is $|1rangle$. When the control qubit is $|0rangle$ the second qubit remains intact, which is equivalent to applying the identity gate $I$. So we get $|0ranglelangle 0|otimes I$ for this. However, when the control qubit is $|1rangle$ we need to flip the second qubit's state i.e. we apply the $X$ gate to the second qubit. Thus, we get $|1ranglelangle 1|otimes X$ for this.
              $endgroup$
              – Sanchayan Dutta
              1 hour ago


















            2












            $begingroup$

            This might mean using the ketbra notation:



            $$X = |1rangle langle0| + |1rangle langle0|$$



            This notation describes the effect the operator has on the basis vectors: in this case $X$ converts $|0rangle$ into $|1rangle$ and vice versa.



            A couple of other examples:



            $$Z = |0rangle langle0| - |1rangle langle1|$$



            $$operatorname{CNOT} = |0 ranglelangle0| otimes I + |1 ranglelangle 1| otimes X = |00rangle langle00| + |01rangle langle01| + |11rangle langle10| + |10rangle langle11|$$






            share|improve this answer











            $endgroup$













            • $begingroup$
              For OP: a quick and intuitive way to derive the CNOT's outer product representation is to see what effect it has on the individual qubits. The CNOT flips the second qubit when the control qubit is $|1rangle$. When the control qubit is $|0rangle$ the second qubit remains intact, which is equivalent to applying the identity gate $I$. So we get $|0ranglelangle 0|otimes I$ for this. However, when the control qubit is $|1rangle$ we need to flip the second qubit's state i.e. we apply the $X$ gate to the second qubit. Thus, we get $|1ranglelangle 1|otimes X$ for this.
              $endgroup$
              – Sanchayan Dutta
              1 hour ago
















            2












            2








            2





            $begingroup$

            This might mean using the ketbra notation:



            $$X = |1rangle langle0| + |1rangle langle0|$$



            This notation describes the effect the operator has on the basis vectors: in this case $X$ converts $|0rangle$ into $|1rangle$ and vice versa.



            A couple of other examples:



            $$Z = |0rangle langle0| - |1rangle langle1|$$



            $$operatorname{CNOT} = |0 ranglelangle0| otimes I + |1 ranglelangle 1| otimes X = |00rangle langle00| + |01rangle langle01| + |11rangle langle10| + |10rangle langle11|$$






            share|improve this answer











            $endgroup$



            This might mean using the ketbra notation:



            $$X = |1rangle langle0| + |1rangle langle0|$$



            This notation describes the effect the operator has on the basis vectors: in this case $X$ converts $|0rangle$ into $|1rangle$ and vice versa.



            A couple of other examples:



            $$Z = |0rangle langle0| - |1rangle langle1|$$



            $$operatorname{CNOT} = |0 ranglelangle0| otimes I + |1 ranglelangle 1| otimes X = |00rangle langle00| + |01rangle langle01| + |11rangle langle10| + |10rangle langle11|$$







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 1 hour ago









            Sanchayan Dutta

            6,75841556




            6,75841556










            answered 2 hours ago









            Mariia MykhailovaMariia Mykhailova

            1,9401212




            1,9401212












            • $begingroup$
              For OP: a quick and intuitive way to derive the CNOT's outer product representation is to see what effect it has on the individual qubits. The CNOT flips the second qubit when the control qubit is $|1rangle$. When the control qubit is $|0rangle$ the second qubit remains intact, which is equivalent to applying the identity gate $I$. So we get $|0ranglelangle 0|otimes I$ for this. However, when the control qubit is $|1rangle$ we need to flip the second qubit's state i.e. we apply the $X$ gate to the second qubit. Thus, we get $|1ranglelangle 1|otimes X$ for this.
              $endgroup$
              – Sanchayan Dutta
              1 hour ago




















            • $begingroup$
              For OP: a quick and intuitive way to derive the CNOT's outer product representation is to see what effect it has on the individual qubits. The CNOT flips the second qubit when the control qubit is $|1rangle$. When the control qubit is $|0rangle$ the second qubit remains intact, which is equivalent to applying the identity gate $I$. So we get $|0ranglelangle 0|otimes I$ for this. However, when the control qubit is $|1rangle$ we need to flip the second qubit's state i.e. we apply the $X$ gate to the second qubit. Thus, we get $|1ranglelangle 1|otimes X$ for this.
              $endgroup$
              – Sanchayan Dutta
              1 hour ago


















            $begingroup$
            For OP: a quick and intuitive way to derive the CNOT's outer product representation is to see what effect it has on the individual qubits. The CNOT flips the second qubit when the control qubit is $|1rangle$. When the control qubit is $|0rangle$ the second qubit remains intact, which is equivalent to applying the identity gate $I$. So we get $|0ranglelangle 0|otimes I$ for this. However, when the control qubit is $|1rangle$ we need to flip the second qubit's state i.e. we apply the $X$ gate to the second qubit. Thus, we get $|1ranglelangle 1|otimes X$ for this.
            $endgroup$
            – Sanchayan Dutta
            1 hour ago






            $begingroup$
            For OP: a quick and intuitive way to derive the CNOT's outer product representation is to see what effect it has on the individual qubits. The CNOT flips the second qubit when the control qubit is $|1rangle$. When the control qubit is $|0rangle$ the second qubit remains intact, which is equivalent to applying the identity gate $I$. So we get $|0ranglelangle 0|otimes I$ for this. However, when the control qubit is $|1rangle$ we need to flip the second qubit's state i.e. we apply the $X$ gate to the second qubit. Thus, we get $|1ranglelangle 1|otimes X$ for this.
            $endgroup$
            – Sanchayan Dutta
            1 hour ago















            0












            $begingroup$

            The Dirac notation for the Pauli-$X$ gate is:



            $$|1rangle langle0| + |1rangle langle0|.$$



            Now you might be wondering where this comes from. The term you're looking for is outer product representation of the $X$ gate. It follows from the spectral decomposition theorem (check Nielsen & Chuang 10th edition, p. 72) which holds for all normal operators. The key point:




            In terms of the outer product representation, this means that $M$ can be written as $M=sum_ilambda_i|iranglelangle i|$,where $lambda_i$ are the eigenvalues of $M$,$|irangle$ is an orthonormal basis for $V$, and each $|irangle$ an eigenvector of $M$ with eigenvalue $lambda_i$.




            The eigenvectors of the Pauli-$X$ gate are $-|0rangle+|1rangle$ and $|0rangle+|1rangle$, and the corresponding eigenvalues are $-1$ and $+1$ cf. Wolfram Alpha. Normalize the eigenvectors to get an orthonormal basis for $X$ i.e. ${frac{-|0rangle+|1rangle}{sqrt{2}},frac{|0rangle+|1rangle}{sqrt{2}}}$. According the spectral decomposition theorem you can represent the $X$ gate as:



            $$-1(frac{-|0rangle+|1rangle}{sqrt{2}})(frac{-langle 0|+langle1|}{sqrt{2}}) + 1(frac{|0rangle+|1rangle}{sqrt{2}})(frac{langle 0|+langle1|}{sqrt{2}})$$
            $$=-frac{1}{2}(|0ranglelangle0|-|0ranglelangle1|-|1ranglelangle0|+|1ranglelangle1|)+frac{1}{2}(|0ranglelangle0|+|0ranglelangle1|+|1ranglelangle0|+|1ranglelangle1|)$$
            $$=|1rangle langle0| + |1rangle langle0|$$



            To convince you that this result is correct let's apply it on an arbitrary qubit state $c_0|0rangle+c_1 |1rangle$:



            $$(|1rangle langle0| + |0rangle langle1|)(c_0|0rangle+c_1|1rangle)$$
            $$=c_0|1ranglelangle0|0rangle+c_1|0ranglelangle 1|1rangle$$
            $$=c_0 |1rangle + c_1 |0rangle$$



            So yes, our result is correct and the bits were indeed flipped upon application of $X=|1rangle langle0| + |1rangle langle0|$ to $c_0|0rangle + c_1|1rangle$.The last step followed from the fact that $langle 0|0rangle$ and $langle 1|1rangle$ are both equal to $1$, as $|0rangle$ and $|1rangle$ are orthonormal vectors i.e. their inner product $langle psi|psirangle=1$ by definition.



            We're done. As an exercise, find the outer product representation of the Pauli-$Z$ gate by yourself. And definitely, do go through the proof of the spectral theorem in Nielsen and Chung if time permits!






            share|improve this answer











            $endgroup$


















              0












              $begingroup$

              The Dirac notation for the Pauli-$X$ gate is:



              $$|1rangle langle0| + |1rangle langle0|.$$



              Now you might be wondering where this comes from. The term you're looking for is outer product representation of the $X$ gate. It follows from the spectral decomposition theorem (check Nielsen & Chuang 10th edition, p. 72) which holds for all normal operators. The key point:




              In terms of the outer product representation, this means that $M$ can be written as $M=sum_ilambda_i|iranglelangle i|$,where $lambda_i$ are the eigenvalues of $M$,$|irangle$ is an orthonormal basis for $V$, and each $|irangle$ an eigenvector of $M$ with eigenvalue $lambda_i$.




              The eigenvectors of the Pauli-$X$ gate are $-|0rangle+|1rangle$ and $|0rangle+|1rangle$, and the corresponding eigenvalues are $-1$ and $+1$ cf. Wolfram Alpha. Normalize the eigenvectors to get an orthonormal basis for $X$ i.e. ${frac{-|0rangle+|1rangle}{sqrt{2}},frac{|0rangle+|1rangle}{sqrt{2}}}$. According the spectral decomposition theorem you can represent the $X$ gate as:



              $$-1(frac{-|0rangle+|1rangle}{sqrt{2}})(frac{-langle 0|+langle1|}{sqrt{2}}) + 1(frac{|0rangle+|1rangle}{sqrt{2}})(frac{langle 0|+langle1|}{sqrt{2}})$$
              $$=-frac{1}{2}(|0ranglelangle0|-|0ranglelangle1|-|1ranglelangle0|+|1ranglelangle1|)+frac{1}{2}(|0ranglelangle0|+|0ranglelangle1|+|1ranglelangle0|+|1ranglelangle1|)$$
              $$=|1rangle langle0| + |1rangle langle0|$$



              To convince you that this result is correct let's apply it on an arbitrary qubit state $c_0|0rangle+c_1 |1rangle$:



              $$(|1rangle langle0| + |0rangle langle1|)(c_0|0rangle+c_1|1rangle)$$
              $$=c_0|1ranglelangle0|0rangle+c_1|0ranglelangle 1|1rangle$$
              $$=c_0 |1rangle + c_1 |0rangle$$



              So yes, our result is correct and the bits were indeed flipped upon application of $X=|1rangle langle0| + |1rangle langle0|$ to $c_0|0rangle + c_1|1rangle$.The last step followed from the fact that $langle 0|0rangle$ and $langle 1|1rangle$ are both equal to $1$, as $|0rangle$ and $|1rangle$ are orthonormal vectors i.e. their inner product $langle psi|psirangle=1$ by definition.



              We're done. As an exercise, find the outer product representation of the Pauli-$Z$ gate by yourself. And definitely, do go through the proof of the spectral theorem in Nielsen and Chung if time permits!






              share|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                The Dirac notation for the Pauli-$X$ gate is:



                $$|1rangle langle0| + |1rangle langle0|.$$



                Now you might be wondering where this comes from. The term you're looking for is outer product representation of the $X$ gate. It follows from the spectral decomposition theorem (check Nielsen & Chuang 10th edition, p. 72) which holds for all normal operators. The key point:




                In terms of the outer product representation, this means that $M$ can be written as $M=sum_ilambda_i|iranglelangle i|$,where $lambda_i$ are the eigenvalues of $M$,$|irangle$ is an orthonormal basis for $V$, and each $|irangle$ an eigenvector of $M$ with eigenvalue $lambda_i$.




                The eigenvectors of the Pauli-$X$ gate are $-|0rangle+|1rangle$ and $|0rangle+|1rangle$, and the corresponding eigenvalues are $-1$ and $+1$ cf. Wolfram Alpha. Normalize the eigenvectors to get an orthonormal basis for $X$ i.e. ${frac{-|0rangle+|1rangle}{sqrt{2}},frac{|0rangle+|1rangle}{sqrt{2}}}$. According the spectral decomposition theorem you can represent the $X$ gate as:



                $$-1(frac{-|0rangle+|1rangle}{sqrt{2}})(frac{-langle 0|+langle1|}{sqrt{2}}) + 1(frac{|0rangle+|1rangle}{sqrt{2}})(frac{langle 0|+langle1|}{sqrt{2}})$$
                $$=-frac{1}{2}(|0ranglelangle0|-|0ranglelangle1|-|1ranglelangle0|+|1ranglelangle1|)+frac{1}{2}(|0ranglelangle0|+|0ranglelangle1|+|1ranglelangle0|+|1ranglelangle1|)$$
                $$=|1rangle langle0| + |1rangle langle0|$$



                To convince you that this result is correct let's apply it on an arbitrary qubit state $c_0|0rangle+c_1 |1rangle$:



                $$(|1rangle langle0| + |0rangle langle1|)(c_0|0rangle+c_1|1rangle)$$
                $$=c_0|1ranglelangle0|0rangle+c_1|0ranglelangle 1|1rangle$$
                $$=c_0 |1rangle + c_1 |0rangle$$



                So yes, our result is correct and the bits were indeed flipped upon application of $X=|1rangle langle0| + |1rangle langle0|$ to $c_0|0rangle + c_1|1rangle$.The last step followed from the fact that $langle 0|0rangle$ and $langle 1|1rangle$ are both equal to $1$, as $|0rangle$ and $|1rangle$ are orthonormal vectors i.e. their inner product $langle psi|psirangle=1$ by definition.



                We're done. As an exercise, find the outer product representation of the Pauli-$Z$ gate by yourself. And definitely, do go through the proof of the spectral theorem in Nielsen and Chung if time permits!






                share|improve this answer











                $endgroup$



                The Dirac notation for the Pauli-$X$ gate is:



                $$|1rangle langle0| + |1rangle langle0|.$$



                Now you might be wondering where this comes from. The term you're looking for is outer product representation of the $X$ gate. It follows from the spectral decomposition theorem (check Nielsen & Chuang 10th edition, p. 72) which holds for all normal operators. The key point:




                In terms of the outer product representation, this means that $M$ can be written as $M=sum_ilambda_i|iranglelangle i|$,where $lambda_i$ are the eigenvalues of $M$,$|irangle$ is an orthonormal basis for $V$, and each $|irangle$ an eigenvector of $M$ with eigenvalue $lambda_i$.




                The eigenvectors of the Pauli-$X$ gate are $-|0rangle+|1rangle$ and $|0rangle+|1rangle$, and the corresponding eigenvalues are $-1$ and $+1$ cf. Wolfram Alpha. Normalize the eigenvectors to get an orthonormal basis for $X$ i.e. ${frac{-|0rangle+|1rangle}{sqrt{2}},frac{|0rangle+|1rangle}{sqrt{2}}}$. According the spectral decomposition theorem you can represent the $X$ gate as:



                $$-1(frac{-|0rangle+|1rangle}{sqrt{2}})(frac{-langle 0|+langle1|}{sqrt{2}}) + 1(frac{|0rangle+|1rangle}{sqrt{2}})(frac{langle 0|+langle1|}{sqrt{2}})$$
                $$=-frac{1}{2}(|0ranglelangle0|-|0ranglelangle1|-|1ranglelangle0|+|1ranglelangle1|)+frac{1}{2}(|0ranglelangle0|+|0ranglelangle1|+|1ranglelangle0|+|1ranglelangle1|)$$
                $$=|1rangle langle0| + |1rangle langle0|$$



                To convince you that this result is correct let's apply it on an arbitrary qubit state $c_0|0rangle+c_1 |1rangle$:



                $$(|1rangle langle0| + |0rangle langle1|)(c_0|0rangle+c_1|1rangle)$$
                $$=c_0|1ranglelangle0|0rangle+c_1|0ranglelangle 1|1rangle$$
                $$=c_0 |1rangle + c_1 |0rangle$$



                So yes, our result is correct and the bits were indeed flipped upon application of $X=|1rangle langle0| + |1rangle langle0|$ to $c_0|0rangle + c_1|1rangle$.The last step followed from the fact that $langle 0|0rangle$ and $langle 1|1rangle$ are both equal to $1$, as $|0rangle$ and $|1rangle$ are orthonormal vectors i.e. their inner product $langle psi|psirangle=1$ by definition.



                We're done. As an exercise, find the outer product representation of the Pauli-$Z$ gate by yourself. And definitely, do go through the proof of the spectral theorem in Nielsen and Chung if time permits!







                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited 1 hour ago

























                answered 2 hours ago









                Sanchayan DuttaSanchayan Dutta

                6,75841556




                6,75841556






























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