How to prove that the query oracle is unitary?Why are oracles Hermitian by construction?How is the oracle in Grover's search algorithm implemented?Implementation of the oracle of Grover's algorithm on IBM Q using three qubitsHow would I implement the quantum oracle in Deutsch's algorithm?How is the Deutsch-Jozsa algorithm faster than classical for practical implementation?Balanced vs unbalanced superposition distinguisherWhy doesn't Deutsch-Jozsa Algorithm show that P ≠ BQP?Is it correct to say that we need controlled gates because unitary matrices are reversible?Equivalent unitary transformationsImplementing an oracleDeutsch–Jozsa algorithm: why is $f$ constant?

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How to prove that the query oracle is unitary?


Why are oracles Hermitian by construction?How is the oracle in Grover's search algorithm implemented?Implementation of the oracle of Grover's algorithm on IBM Q using three qubitsHow would I implement the quantum oracle in Deutsch's algorithm?How is the Deutsch-Jozsa algorithm faster than classical for practical implementation?Balanced vs unbalanced superposition distinguisherWhy doesn't Deutsch-Jozsa Algorithm show that P ≠ BQP?Is it correct to say that we need controlled gates because unitary matrices are reversible?Equivalent unitary transformationsImplementing an oracleDeutsch–Jozsa algorithm: why is $f$ constant?













5












$begingroup$


The query oracle: $O_x|irangle|brangle = |irangle|b oplus x_irangle$ used in algorithms like Deutsch Jozsa is unitary. How do I prove it is unitary?










share|improve this question











$endgroup$











  • $begingroup$
    Related: Why are oracles Hermitian by construction?
    $endgroup$
    – Blue
    Mar 26 at 12:25















5












$begingroup$


The query oracle: $O_x|irangle|brangle = |irangle|b oplus x_irangle$ used in algorithms like Deutsch Jozsa is unitary. How do I prove it is unitary?










share|improve this question











$endgroup$











  • $begingroup$
    Related: Why are oracles Hermitian by construction?
    $endgroup$
    – Blue
    Mar 26 at 12:25













5












5








5





$begingroup$


The query oracle: $O_x|irangle|brangle = |irangle|b oplus x_irangle$ used in algorithms like Deutsch Jozsa is unitary. How do I prove it is unitary?










share|improve this question











$endgroup$




The query oracle: $O_x|irangle|brangle = |irangle|b oplus x_irangle$ used in algorithms like Deutsch Jozsa is unitary. How do I prove it is unitary?







algorithm quantum-gate unitarity deutsch-jozsa-algorithm






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Mar 26 at 12:15









Blue

6,63141556




6,63141556










asked Mar 26 at 12:11









DivyatDivyat

283




283











  • $begingroup$
    Related: Why are oracles Hermitian by construction?
    $endgroup$
    – Blue
    Mar 26 at 12:25
















  • $begingroup$
    Related: Why are oracles Hermitian by construction?
    $endgroup$
    – Blue
    Mar 26 at 12:25















$begingroup$
Related: Why are oracles Hermitian by construction?
$endgroup$
– Blue
Mar 26 at 12:25




$begingroup$
Related: Why are oracles Hermitian by construction?
$endgroup$
– Blue
Mar 26 at 12:25










2 Answers
2






active

oldest

votes


















3












$begingroup$

Notice that $mathcal O_x$ is a permutation matrix.



The matrix elements are
$$langle j, crvertmathcal O_xlvert i,brangle
=delta_ijlangle crvert boplus x_irangle
=delta_ijdelta_c,boplus x_i.$$

In other words, $mathcal O_x$ is diagonal with respect to the first register, and, for each block corresponding to a given $i$, connects all and only the indices $b,c$ such that $boplus c=x_i$ (remember that here $b,c,x_iinmathbb Z_2^otimes n$ are length-$n$ bit strings).
Also, notice that for a given $b$ there is no more than one $c$ such that $boplus c=x_i$ (more precisely, there isn't any such $c$ if $x_i=0$, and there is exactly one if $x_ineq 0$).



It follows that $mathcal O_x$ is a (real) permutation matrix, and such matrices are always unitarily diagonalizable with unit eigenvalues (and therefore unitary).
In this case, we have that the eigenvalues of $mathcal O_x$ are $pm1$, and the eigenvectors are, in the case $x_ineq 0$,
$$lvert irangleotimes(lvert branglepmlvert boplus x_irangle)$$
for all $i$ and $b$. If $x_i=0$ then $mathcal O_x$ is the identity, and therefore its spectrum is trivial$^dagger$.



This shows explicitly that $mathcal O_x$ is unitarily diagonalizable with unit eigenvalues, and therefore is unitary.




$^dagger$ I'm actually being a bit sloppy for the sake of simplicity here. This analysis holds for each different block of $mathcal O_x$ corresponding to a given $i$. More precisely, I should say that $mathcal O_x$ is block-diagonal as it doesn't connect spaces with $ineq j$ on the first register, and each block is either the identity in the subspace in which it acts if $x_i=0$, or a permutation matrix that connects different pairs of basis states if $x_ineq 0$.






share|improve this answer











$endgroup$




















    7












    $begingroup$

    Apply it twice:
    $$
    O_xO_x|irangle|brangle=O_x|irangle|boplus x_irangle=|irangle|boplus x_ioplus x_irangle=|irangle|brangle
    $$

    Hence, $O_x$ is its own inverse, and therefore reversible.



    To prove unitarity, it makes more sense to prove that $O_x$ has eigenvectors
    $$
    |irangle(|0rangle+|1rangle)quadtextandquad|irangle(|0rangle-|1rangle)
    $$

    for all $i$ with eigenvalues $1$ and $(-1)^x_i$ respectively. These are all orthonormal, and span the full Hilbert space. Consequently, the eigenvalues of $O_x$ are all $pm 1$, and therefore $O_x^star=O_x$ (where $^star$ represents the Hermitian conjugate). Thus,
    $$
    O_xO_x^star=mathbbI,
    $$

    as required for a unitary.






    share|improve this answer











    $endgroup$












    • $begingroup$
      Thanks, I now understand $O_x$ is its own inverse but how does that make it unitary? For a unitary matrix, one would need to show that its inverse is equal to its conjugate transpose, $MM^*=I$. I think we further need to show $O_x$ is Hermitian, then it will be done.
      $endgroup$
      – Divyat
      Mar 26 at 13:56











    • $begingroup$
      One more doubt, I think the answer still assumes that $O_x$ is normal matrix, as then with real eigenvalues we can claim it to be Hermitian. Please tell whether it is okay to assume oracle as normal or some justifications for it?
      $endgroup$
      – Divyat
      Mar 26 at 15:52











    • $begingroup$
      This explanation is not quite correct, because you assumed that $O_x$ is diagonalizable, but it is not follow from $O_x^2=I$. The thing is $O_x$ maps basis vectors to basis vectors. Since $O_x^2=I$ then $Q_x$ is a permutation of basis vectors.
      $endgroup$
      – Danylo Y
      Mar 26 at 16:17











    • $begingroup$
      I actually asked some time ago on math.SE whether one can deduce that $sqrt A$ is diagonalizable from the fact that $A$ is, see here. As far as I understood the answers, $A^2=I$ is not enough to imply that $A$ is unitarily diagonalizable, a counterexample being $beginpmatrixcostheta & 2sintheta \ sintheta/2 & -costhetaendpmatrix$, which is a square root of the identity, but not (unitarily) diagonalizable, and not unitary, even though its eigenvalues are $pm 1$.
      $endgroup$
      – glS
      Mar 26 at 18:01







    • 1




      $begingroup$
      Fair enough, I was being a bit glib, particularly initially with regards to the difference between reversible and unitary, because usually its the reversibility aspect people are interested in.
      $endgroup$
      – DaftWullie
      Mar 27 at 8:44











    Your Answer





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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    Notice that $mathcal O_x$ is a permutation matrix.



    The matrix elements are
    $$langle j, crvertmathcal O_xlvert i,brangle
    =delta_ijlangle crvert boplus x_irangle
    =delta_ijdelta_c,boplus x_i.$$

    In other words, $mathcal O_x$ is diagonal with respect to the first register, and, for each block corresponding to a given $i$, connects all and only the indices $b,c$ such that $boplus c=x_i$ (remember that here $b,c,x_iinmathbb Z_2^otimes n$ are length-$n$ bit strings).
    Also, notice that for a given $b$ there is no more than one $c$ such that $boplus c=x_i$ (more precisely, there isn't any such $c$ if $x_i=0$, and there is exactly one if $x_ineq 0$).



    It follows that $mathcal O_x$ is a (real) permutation matrix, and such matrices are always unitarily diagonalizable with unit eigenvalues (and therefore unitary).
    In this case, we have that the eigenvalues of $mathcal O_x$ are $pm1$, and the eigenvectors are, in the case $x_ineq 0$,
    $$lvert irangleotimes(lvert branglepmlvert boplus x_irangle)$$
    for all $i$ and $b$. If $x_i=0$ then $mathcal O_x$ is the identity, and therefore its spectrum is trivial$^dagger$.



    This shows explicitly that $mathcal O_x$ is unitarily diagonalizable with unit eigenvalues, and therefore is unitary.




    $^dagger$ I'm actually being a bit sloppy for the sake of simplicity here. This analysis holds for each different block of $mathcal O_x$ corresponding to a given $i$. More precisely, I should say that $mathcal O_x$ is block-diagonal as it doesn't connect spaces with $ineq j$ on the first register, and each block is either the identity in the subspace in which it acts if $x_i=0$, or a permutation matrix that connects different pairs of basis states if $x_ineq 0$.






    share|improve this answer











    $endgroup$

















      3












      $begingroup$

      Notice that $mathcal O_x$ is a permutation matrix.



      The matrix elements are
      $$langle j, crvertmathcal O_xlvert i,brangle
      =delta_ijlangle crvert boplus x_irangle
      =delta_ijdelta_c,boplus x_i.$$

      In other words, $mathcal O_x$ is diagonal with respect to the first register, and, for each block corresponding to a given $i$, connects all and only the indices $b,c$ such that $boplus c=x_i$ (remember that here $b,c,x_iinmathbb Z_2^otimes n$ are length-$n$ bit strings).
      Also, notice that for a given $b$ there is no more than one $c$ such that $boplus c=x_i$ (more precisely, there isn't any such $c$ if $x_i=0$, and there is exactly one if $x_ineq 0$).



      It follows that $mathcal O_x$ is a (real) permutation matrix, and such matrices are always unitarily diagonalizable with unit eigenvalues (and therefore unitary).
      In this case, we have that the eigenvalues of $mathcal O_x$ are $pm1$, and the eigenvectors are, in the case $x_ineq 0$,
      $$lvert irangleotimes(lvert branglepmlvert boplus x_irangle)$$
      for all $i$ and $b$. If $x_i=0$ then $mathcal O_x$ is the identity, and therefore its spectrum is trivial$^dagger$.



      This shows explicitly that $mathcal O_x$ is unitarily diagonalizable with unit eigenvalues, and therefore is unitary.




      $^dagger$ I'm actually being a bit sloppy for the sake of simplicity here. This analysis holds for each different block of $mathcal O_x$ corresponding to a given $i$. More precisely, I should say that $mathcal O_x$ is block-diagonal as it doesn't connect spaces with $ineq j$ on the first register, and each block is either the identity in the subspace in which it acts if $x_i=0$, or a permutation matrix that connects different pairs of basis states if $x_ineq 0$.






      share|improve this answer











      $endgroup$















        3












        3








        3





        $begingroup$

        Notice that $mathcal O_x$ is a permutation matrix.



        The matrix elements are
        $$langle j, crvertmathcal O_xlvert i,brangle
        =delta_ijlangle crvert boplus x_irangle
        =delta_ijdelta_c,boplus x_i.$$

        In other words, $mathcal O_x$ is diagonal with respect to the first register, and, for each block corresponding to a given $i$, connects all and only the indices $b,c$ such that $boplus c=x_i$ (remember that here $b,c,x_iinmathbb Z_2^otimes n$ are length-$n$ bit strings).
        Also, notice that for a given $b$ there is no more than one $c$ such that $boplus c=x_i$ (more precisely, there isn't any such $c$ if $x_i=0$, and there is exactly one if $x_ineq 0$).



        It follows that $mathcal O_x$ is a (real) permutation matrix, and such matrices are always unitarily diagonalizable with unit eigenvalues (and therefore unitary).
        In this case, we have that the eigenvalues of $mathcal O_x$ are $pm1$, and the eigenvectors are, in the case $x_ineq 0$,
        $$lvert irangleotimes(lvert branglepmlvert boplus x_irangle)$$
        for all $i$ and $b$. If $x_i=0$ then $mathcal O_x$ is the identity, and therefore its spectrum is trivial$^dagger$.



        This shows explicitly that $mathcal O_x$ is unitarily diagonalizable with unit eigenvalues, and therefore is unitary.




        $^dagger$ I'm actually being a bit sloppy for the sake of simplicity here. This analysis holds for each different block of $mathcal O_x$ corresponding to a given $i$. More precisely, I should say that $mathcal O_x$ is block-diagonal as it doesn't connect spaces with $ineq j$ on the first register, and each block is either the identity in the subspace in which it acts if $x_i=0$, or a permutation matrix that connects different pairs of basis states if $x_ineq 0$.






        share|improve this answer











        $endgroup$



        Notice that $mathcal O_x$ is a permutation matrix.



        The matrix elements are
        $$langle j, crvertmathcal O_xlvert i,brangle
        =delta_ijlangle crvert boplus x_irangle
        =delta_ijdelta_c,boplus x_i.$$

        In other words, $mathcal O_x$ is diagonal with respect to the first register, and, for each block corresponding to a given $i$, connects all and only the indices $b,c$ such that $boplus c=x_i$ (remember that here $b,c,x_iinmathbb Z_2^otimes n$ are length-$n$ bit strings).
        Also, notice that for a given $b$ there is no more than one $c$ such that $boplus c=x_i$ (more precisely, there isn't any such $c$ if $x_i=0$, and there is exactly one if $x_ineq 0$).



        It follows that $mathcal O_x$ is a (real) permutation matrix, and such matrices are always unitarily diagonalizable with unit eigenvalues (and therefore unitary).
        In this case, we have that the eigenvalues of $mathcal O_x$ are $pm1$, and the eigenvectors are, in the case $x_ineq 0$,
        $$lvert irangleotimes(lvert branglepmlvert boplus x_irangle)$$
        for all $i$ and $b$. If $x_i=0$ then $mathcal O_x$ is the identity, and therefore its spectrum is trivial$^dagger$.



        This shows explicitly that $mathcal O_x$ is unitarily diagonalizable with unit eigenvalues, and therefore is unitary.




        $^dagger$ I'm actually being a bit sloppy for the sake of simplicity here. This analysis holds for each different block of $mathcal O_x$ corresponding to a given $i$. More precisely, I should say that $mathcal O_x$ is block-diagonal as it doesn't connect spaces with $ineq j$ on the first register, and each block is either the identity in the subspace in which it acts if $x_i=0$, or a permutation matrix that connects different pairs of basis states if $x_ineq 0$.







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Mar 26 at 18:05

























        answered Mar 26 at 17:49









        glSglS

        4,333740




        4,333740























            7












            $begingroup$

            Apply it twice:
            $$
            O_xO_x|irangle|brangle=O_x|irangle|boplus x_irangle=|irangle|boplus x_ioplus x_irangle=|irangle|brangle
            $$

            Hence, $O_x$ is its own inverse, and therefore reversible.



            To prove unitarity, it makes more sense to prove that $O_x$ has eigenvectors
            $$
            |irangle(|0rangle+|1rangle)quadtextandquad|irangle(|0rangle-|1rangle)
            $$

            for all $i$ with eigenvalues $1$ and $(-1)^x_i$ respectively. These are all orthonormal, and span the full Hilbert space. Consequently, the eigenvalues of $O_x$ are all $pm 1$, and therefore $O_x^star=O_x$ (where $^star$ represents the Hermitian conjugate). Thus,
            $$
            O_xO_x^star=mathbbI,
            $$

            as required for a unitary.






            share|improve this answer











            $endgroup$












            • $begingroup$
              Thanks, I now understand $O_x$ is its own inverse but how does that make it unitary? For a unitary matrix, one would need to show that its inverse is equal to its conjugate transpose, $MM^*=I$. I think we further need to show $O_x$ is Hermitian, then it will be done.
              $endgroup$
              – Divyat
              Mar 26 at 13:56











            • $begingroup$
              One more doubt, I think the answer still assumes that $O_x$ is normal matrix, as then with real eigenvalues we can claim it to be Hermitian. Please tell whether it is okay to assume oracle as normal or some justifications for it?
              $endgroup$
              – Divyat
              Mar 26 at 15:52











            • $begingroup$
              This explanation is not quite correct, because you assumed that $O_x$ is diagonalizable, but it is not follow from $O_x^2=I$. The thing is $O_x$ maps basis vectors to basis vectors. Since $O_x^2=I$ then $Q_x$ is a permutation of basis vectors.
              $endgroup$
              – Danylo Y
              Mar 26 at 16:17











            • $begingroup$
              I actually asked some time ago on math.SE whether one can deduce that $sqrt A$ is diagonalizable from the fact that $A$ is, see here. As far as I understood the answers, $A^2=I$ is not enough to imply that $A$ is unitarily diagonalizable, a counterexample being $beginpmatrixcostheta & 2sintheta \ sintheta/2 & -costhetaendpmatrix$, which is a square root of the identity, but not (unitarily) diagonalizable, and not unitary, even though its eigenvalues are $pm 1$.
              $endgroup$
              – glS
              Mar 26 at 18:01







            • 1




              $begingroup$
              Fair enough, I was being a bit glib, particularly initially with regards to the difference between reversible and unitary, because usually its the reversibility aspect people are interested in.
              $endgroup$
              – DaftWullie
              Mar 27 at 8:44















            7












            $begingroup$

            Apply it twice:
            $$
            O_xO_x|irangle|brangle=O_x|irangle|boplus x_irangle=|irangle|boplus x_ioplus x_irangle=|irangle|brangle
            $$

            Hence, $O_x$ is its own inverse, and therefore reversible.



            To prove unitarity, it makes more sense to prove that $O_x$ has eigenvectors
            $$
            |irangle(|0rangle+|1rangle)quadtextandquad|irangle(|0rangle-|1rangle)
            $$

            for all $i$ with eigenvalues $1$ and $(-1)^x_i$ respectively. These are all orthonormal, and span the full Hilbert space. Consequently, the eigenvalues of $O_x$ are all $pm 1$, and therefore $O_x^star=O_x$ (where $^star$ represents the Hermitian conjugate). Thus,
            $$
            O_xO_x^star=mathbbI,
            $$

            as required for a unitary.






            share|improve this answer











            $endgroup$












            • $begingroup$
              Thanks, I now understand $O_x$ is its own inverse but how does that make it unitary? For a unitary matrix, one would need to show that its inverse is equal to its conjugate transpose, $MM^*=I$. I think we further need to show $O_x$ is Hermitian, then it will be done.
              $endgroup$
              – Divyat
              Mar 26 at 13:56











            • $begingroup$
              One more doubt, I think the answer still assumes that $O_x$ is normal matrix, as then with real eigenvalues we can claim it to be Hermitian. Please tell whether it is okay to assume oracle as normal or some justifications for it?
              $endgroup$
              – Divyat
              Mar 26 at 15:52











            • $begingroup$
              This explanation is not quite correct, because you assumed that $O_x$ is diagonalizable, but it is not follow from $O_x^2=I$. The thing is $O_x$ maps basis vectors to basis vectors. Since $O_x^2=I$ then $Q_x$ is a permutation of basis vectors.
              $endgroup$
              – Danylo Y
              Mar 26 at 16:17











            • $begingroup$
              I actually asked some time ago on math.SE whether one can deduce that $sqrt A$ is diagonalizable from the fact that $A$ is, see here. As far as I understood the answers, $A^2=I$ is not enough to imply that $A$ is unitarily diagonalizable, a counterexample being $beginpmatrixcostheta & 2sintheta \ sintheta/2 & -costhetaendpmatrix$, which is a square root of the identity, but not (unitarily) diagonalizable, and not unitary, even though its eigenvalues are $pm 1$.
              $endgroup$
              – glS
              Mar 26 at 18:01







            • 1




              $begingroup$
              Fair enough, I was being a bit glib, particularly initially with regards to the difference between reversible and unitary, because usually its the reversibility aspect people are interested in.
              $endgroup$
              – DaftWullie
              Mar 27 at 8:44













            7












            7








            7





            $begingroup$

            Apply it twice:
            $$
            O_xO_x|irangle|brangle=O_x|irangle|boplus x_irangle=|irangle|boplus x_ioplus x_irangle=|irangle|brangle
            $$

            Hence, $O_x$ is its own inverse, and therefore reversible.



            To prove unitarity, it makes more sense to prove that $O_x$ has eigenvectors
            $$
            |irangle(|0rangle+|1rangle)quadtextandquad|irangle(|0rangle-|1rangle)
            $$

            for all $i$ with eigenvalues $1$ and $(-1)^x_i$ respectively. These are all orthonormal, and span the full Hilbert space. Consequently, the eigenvalues of $O_x$ are all $pm 1$, and therefore $O_x^star=O_x$ (where $^star$ represents the Hermitian conjugate). Thus,
            $$
            O_xO_x^star=mathbbI,
            $$

            as required for a unitary.






            share|improve this answer











            $endgroup$



            Apply it twice:
            $$
            O_xO_x|irangle|brangle=O_x|irangle|boplus x_irangle=|irangle|boplus x_ioplus x_irangle=|irangle|brangle
            $$

            Hence, $O_x$ is its own inverse, and therefore reversible.



            To prove unitarity, it makes more sense to prove that $O_x$ has eigenvectors
            $$
            |irangle(|0rangle+|1rangle)quadtextandquad|irangle(|0rangle-|1rangle)
            $$

            for all $i$ with eigenvalues $1$ and $(-1)^x_i$ respectively. These are all orthonormal, and span the full Hilbert space. Consequently, the eigenvalues of $O_x$ are all $pm 1$, and therefore $O_x^star=O_x$ (where $^star$ represents the Hermitian conjugate). Thus,
            $$
            O_xO_x^star=mathbbI,
            $$

            as required for a unitary.







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited Mar 27 at 8:51

























            answered Mar 26 at 12:18









            DaftWullieDaftWullie

            15.2k1542




            15.2k1542











            • $begingroup$
              Thanks, I now understand $O_x$ is its own inverse but how does that make it unitary? For a unitary matrix, one would need to show that its inverse is equal to its conjugate transpose, $MM^*=I$. I think we further need to show $O_x$ is Hermitian, then it will be done.
              $endgroup$
              – Divyat
              Mar 26 at 13:56











            • $begingroup$
              One more doubt, I think the answer still assumes that $O_x$ is normal matrix, as then with real eigenvalues we can claim it to be Hermitian. Please tell whether it is okay to assume oracle as normal or some justifications for it?
              $endgroup$
              – Divyat
              Mar 26 at 15:52











            • $begingroup$
              This explanation is not quite correct, because you assumed that $O_x$ is diagonalizable, but it is not follow from $O_x^2=I$. The thing is $O_x$ maps basis vectors to basis vectors. Since $O_x^2=I$ then $Q_x$ is a permutation of basis vectors.
              $endgroup$
              – Danylo Y
              Mar 26 at 16:17











            • $begingroup$
              I actually asked some time ago on math.SE whether one can deduce that $sqrt A$ is diagonalizable from the fact that $A$ is, see here. As far as I understood the answers, $A^2=I$ is not enough to imply that $A$ is unitarily diagonalizable, a counterexample being $beginpmatrixcostheta & 2sintheta \ sintheta/2 & -costhetaendpmatrix$, which is a square root of the identity, but not (unitarily) diagonalizable, and not unitary, even though its eigenvalues are $pm 1$.
              $endgroup$
              – glS
              Mar 26 at 18:01







            • 1




              $begingroup$
              Fair enough, I was being a bit glib, particularly initially with regards to the difference between reversible and unitary, because usually its the reversibility aspect people are interested in.
              $endgroup$
              – DaftWullie
              Mar 27 at 8:44
















            • $begingroup$
              Thanks, I now understand $O_x$ is its own inverse but how does that make it unitary? For a unitary matrix, one would need to show that its inverse is equal to its conjugate transpose, $MM^*=I$. I think we further need to show $O_x$ is Hermitian, then it will be done.
              $endgroup$
              – Divyat
              Mar 26 at 13:56











            • $begingroup$
              One more doubt, I think the answer still assumes that $O_x$ is normal matrix, as then with real eigenvalues we can claim it to be Hermitian. Please tell whether it is okay to assume oracle as normal or some justifications for it?
              $endgroup$
              – Divyat
              Mar 26 at 15:52











            • $begingroup$
              This explanation is not quite correct, because you assumed that $O_x$ is diagonalizable, but it is not follow from $O_x^2=I$. The thing is $O_x$ maps basis vectors to basis vectors. Since $O_x^2=I$ then $Q_x$ is a permutation of basis vectors.
              $endgroup$
              – Danylo Y
              Mar 26 at 16:17











            • $begingroup$
              I actually asked some time ago on math.SE whether one can deduce that $sqrt A$ is diagonalizable from the fact that $A$ is, see here. As far as I understood the answers, $A^2=I$ is not enough to imply that $A$ is unitarily diagonalizable, a counterexample being $beginpmatrixcostheta & 2sintheta \ sintheta/2 & -costhetaendpmatrix$, which is a square root of the identity, but not (unitarily) diagonalizable, and not unitary, even though its eigenvalues are $pm 1$.
              $endgroup$
              – glS
              Mar 26 at 18:01







            • 1




              $begingroup$
              Fair enough, I was being a bit glib, particularly initially with regards to the difference between reversible and unitary, because usually its the reversibility aspect people are interested in.
              $endgroup$
              – DaftWullie
              Mar 27 at 8:44















            $begingroup$
            Thanks, I now understand $O_x$ is its own inverse but how does that make it unitary? For a unitary matrix, one would need to show that its inverse is equal to its conjugate transpose, $MM^*=I$. I think we further need to show $O_x$ is Hermitian, then it will be done.
            $endgroup$
            – Divyat
            Mar 26 at 13:56





            $begingroup$
            Thanks, I now understand $O_x$ is its own inverse but how does that make it unitary? For a unitary matrix, one would need to show that its inverse is equal to its conjugate transpose, $MM^*=I$. I think we further need to show $O_x$ is Hermitian, then it will be done.
            $endgroup$
            – Divyat
            Mar 26 at 13:56













            $begingroup$
            One more doubt, I think the answer still assumes that $O_x$ is normal matrix, as then with real eigenvalues we can claim it to be Hermitian. Please tell whether it is okay to assume oracle as normal or some justifications for it?
            $endgroup$
            – Divyat
            Mar 26 at 15:52





            $begingroup$
            One more doubt, I think the answer still assumes that $O_x$ is normal matrix, as then with real eigenvalues we can claim it to be Hermitian. Please tell whether it is okay to assume oracle as normal or some justifications for it?
            $endgroup$
            – Divyat
            Mar 26 at 15:52













            $begingroup$
            This explanation is not quite correct, because you assumed that $O_x$ is diagonalizable, but it is not follow from $O_x^2=I$. The thing is $O_x$ maps basis vectors to basis vectors. Since $O_x^2=I$ then $Q_x$ is a permutation of basis vectors.
            $endgroup$
            – Danylo Y
            Mar 26 at 16:17





            $begingroup$
            This explanation is not quite correct, because you assumed that $O_x$ is diagonalizable, but it is not follow from $O_x^2=I$. The thing is $O_x$ maps basis vectors to basis vectors. Since $O_x^2=I$ then $Q_x$ is a permutation of basis vectors.
            $endgroup$
            – Danylo Y
            Mar 26 at 16:17













            $begingroup$
            I actually asked some time ago on math.SE whether one can deduce that $sqrt A$ is diagonalizable from the fact that $A$ is, see here. As far as I understood the answers, $A^2=I$ is not enough to imply that $A$ is unitarily diagonalizable, a counterexample being $beginpmatrixcostheta & 2sintheta \ sintheta/2 & -costhetaendpmatrix$, which is a square root of the identity, but not (unitarily) diagonalizable, and not unitary, even though its eigenvalues are $pm 1$.
            $endgroup$
            – glS
            Mar 26 at 18:01





            $begingroup$
            I actually asked some time ago on math.SE whether one can deduce that $sqrt A$ is diagonalizable from the fact that $A$ is, see here. As far as I understood the answers, $A^2=I$ is not enough to imply that $A$ is unitarily diagonalizable, a counterexample being $beginpmatrixcostheta & 2sintheta \ sintheta/2 & -costhetaendpmatrix$, which is a square root of the identity, but not (unitarily) diagonalizable, and not unitary, even though its eigenvalues are $pm 1$.
            $endgroup$
            – glS
            Mar 26 at 18:01





            1




            1




            $begingroup$
            Fair enough, I was being a bit glib, particularly initially with regards to the difference between reversible and unitary, because usually its the reversibility aspect people are interested in.
            $endgroup$
            – DaftWullie
            Mar 27 at 8:44




            $begingroup$
            Fair enough, I was being a bit glib, particularly initially with regards to the difference between reversible and unitary, because usually its the reversibility aspect people are interested in.
            $endgroup$
            – DaftWullie
            Mar 27 at 8:44

















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