Random Unitary Matrices

Random Unitary Matrices














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So as I understand it, if you choose a random parameter and fill the entries of a matrix with that parameter, you get a random matrix. I was wondering if there are some general conditions under which a matrix can be both unitary and random such that the unitarity condition is now being satisfied.



I am aware that there are already examples of random unitary matrices which can be chosen from the Haar measure, so perhaps this question is related to the concept of the Haar measure?










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    1












    $begingroup$


    So as I understand it, if you choose a random parameter and fill the entries of a matrix with that parameter, you get a random matrix. I was wondering if there are some general conditions under which a matrix can be both unitary and random such that the unitarity condition is now being satisfied.



    I am aware that there are already examples of random unitary matrices which can be chosen from the Haar measure, so perhaps this question is related to the concept of the Haar measure?










    share|cite|improve this question









    $endgroup$














      1












      1








      1





      $begingroup$


      So as I understand it, if you choose a random parameter and fill the entries of a matrix with that parameter, you get a random matrix. I was wondering if there are some general conditions under which a matrix can be both unitary and random such that the unitarity condition is now being satisfied.



      I am aware that there are already examples of random unitary matrices which can be chosen from the Haar measure, so perhaps this question is related to the concept of the Haar measure?










      share|cite|improve this question









      $endgroup$




      So as I understand it, if you choose a random parameter and fill the entries of a matrix with that parameter, you get a random matrix. I was wondering if there are some general conditions under which a matrix can be both unitary and random such that the unitarity condition is now being satisfied.



      I am aware that there are already examples of random unitary matrices which can be chosen from the Haar measure, so perhaps this question is related to the concept of the Haar measure?







      fa.functional-analysis random-matrices






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      asked 9 hours ago









      TomTom

      467514




      467514




















          2 Answers
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          active

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          $begingroup$

          The notion of random unitary matrix usually refers to the Haar measure on the group of unitary matrices; see



          https://case.edu/artsci/math/esmeckes/Meckes_SAMSI_Lecture2.pdf



          http://emis.ams.org/journals/EJP-ECP/article/download/2551/2345.pdf



          ] G. W. Anderson, A. Guionnet, and O. Zeitouni, An introduction to random matrices, Cambridge Studies in Advanced Mathematics, vol. 118, Cambridge University Press, Cambridge,
          2010. MR-2760897






          share|cite|improve this answer









          $endgroup$




















            2












            $begingroup$

            I understand your question as asking for a constructive method to sample uniformly from the unitary group $U(N)$ or orthogonal group $O(N)$, where "uniformly" is understood in the sense of the Haar measure. A simple method starts from an $Ntimes N$ matrix filled with independent Gaussian random variables [complex for $U(N)$ and real for $O(N)$]. Then orthonormalize the columns via Gram-Schmidt and you're done.



            For a more efficient approach, see How to generate random matrices from the classical compact groups by Francesco Mezzadri.



            The construction is implemented in Mathematica in the routines circular unitary ensemble for $U(N)$ and circular real ensemble for $O(N)$.
            Caution: the socalled circular orthogonal ensemble does not sample from $O(N)$ but from the coset $U(N)/O(N)$.






            share|cite|improve this answer











            $endgroup$












            • $begingroup$
              That makes sense but you use independent Gaussian random variables for the matrix, I was wondering if you would take one single random variable as it were (say, all the possible numbers which can be produced by some particular random number generator) and then fill the matrix with every entry being a value which that one variable can take, I take it this is still a random matrix or to be a random matrix does every entry need to correspond to a different independent random variable?
              $endgroup$
              – Tom
              8 hours ago







            • 1




              $begingroup$
              you can use a random number generator to fill the matrix, but it is essential that this random number generator samples from a Gaussian distribution; it your random number generator samples from a uniform distribution, you will have to do some postprocessing to convert that into a Gaussian.
              $endgroup$
              – Carlo Beenakker
              8 hours ago












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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            The notion of random unitary matrix usually refers to the Haar measure on the group of unitary matrices; see



            https://case.edu/artsci/math/esmeckes/Meckes_SAMSI_Lecture2.pdf



            http://emis.ams.org/journals/EJP-ECP/article/download/2551/2345.pdf



            ] G. W. Anderson, A. Guionnet, and O. Zeitouni, An introduction to random matrices, Cambridge Studies in Advanced Mathematics, vol. 118, Cambridge University Press, Cambridge,
            2010. MR-2760897






            share|cite|improve this answer









            $endgroup$

















              2












              $begingroup$

              The notion of random unitary matrix usually refers to the Haar measure on the group of unitary matrices; see



              https://case.edu/artsci/math/esmeckes/Meckes_SAMSI_Lecture2.pdf



              http://emis.ams.org/journals/EJP-ECP/article/download/2551/2345.pdf



              ] G. W. Anderson, A. Guionnet, and O. Zeitouni, An introduction to random matrices, Cambridge Studies in Advanced Mathematics, vol. 118, Cambridge University Press, Cambridge,
              2010. MR-2760897






              share|cite|improve this answer









              $endgroup$















                2












                2








                2





                $begingroup$

                The notion of random unitary matrix usually refers to the Haar measure on the group of unitary matrices; see



                https://case.edu/artsci/math/esmeckes/Meckes_SAMSI_Lecture2.pdf



                http://emis.ams.org/journals/EJP-ECP/article/download/2551/2345.pdf



                ] G. W. Anderson, A. Guionnet, and O. Zeitouni, An introduction to random matrices, Cambridge Studies in Advanced Mathematics, vol. 118, Cambridge University Press, Cambridge,
                2010. MR-2760897






                share|cite|improve this answer









                $endgroup$



                The notion of random unitary matrix usually refers to the Haar measure on the group of unitary matrices; see



                https://case.edu/artsci/math/esmeckes/Meckes_SAMSI_Lecture2.pdf



                http://emis.ams.org/journals/EJP-ECP/article/download/2551/2345.pdf



                ] G. W. Anderson, A. Guionnet, and O. Zeitouni, An introduction to random matrices, Cambridge Studies in Advanced Mathematics, vol. 118, Cambridge University Press, Cambridge,
                2010. MR-2760897







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 9 hours ago









                Yuval PeresYuval Peres

                1,30177




                1,30177





















                    2












                    $begingroup$

                    I understand your question as asking for a constructive method to sample uniformly from the unitary group $U(N)$ or orthogonal group $O(N)$, where "uniformly" is understood in the sense of the Haar measure. A simple method starts from an $Ntimes N$ matrix filled with independent Gaussian random variables [complex for $U(N)$ and real for $O(N)$]. Then orthonormalize the columns via Gram-Schmidt and you're done.



                    For a more efficient approach, see How to generate random matrices from the classical compact groups by Francesco Mezzadri.



                    The construction is implemented in Mathematica in the routines circular unitary ensemble for $U(N)$ and circular real ensemble for $O(N)$.
                    Caution: the socalled circular orthogonal ensemble does not sample from $O(N)$ but from the coset $U(N)/O(N)$.






                    share|cite|improve this answer











                    $endgroup$












                    • $begingroup$
                      That makes sense but you use independent Gaussian random variables for the matrix, I was wondering if you would take one single random variable as it were (say, all the possible numbers which can be produced by some particular random number generator) and then fill the matrix with every entry being a value which that one variable can take, I take it this is still a random matrix or to be a random matrix does every entry need to correspond to a different independent random variable?
                      $endgroup$
                      – Tom
                      8 hours ago







                    • 1




                      $begingroup$
                      you can use a random number generator to fill the matrix, but it is essential that this random number generator samples from a Gaussian distribution; it your random number generator samples from a uniform distribution, you will have to do some postprocessing to convert that into a Gaussian.
                      $endgroup$
                      – Carlo Beenakker
                      8 hours ago
















                    2












                    $begingroup$

                    I understand your question as asking for a constructive method to sample uniformly from the unitary group $U(N)$ or orthogonal group $O(N)$, where "uniformly" is understood in the sense of the Haar measure. A simple method starts from an $Ntimes N$ matrix filled with independent Gaussian random variables [complex for $U(N)$ and real for $O(N)$]. Then orthonormalize the columns via Gram-Schmidt and you're done.



                    For a more efficient approach, see How to generate random matrices from the classical compact groups by Francesco Mezzadri.



                    The construction is implemented in Mathematica in the routines circular unitary ensemble for $U(N)$ and circular real ensemble for $O(N)$.
                    Caution: the socalled circular orthogonal ensemble does not sample from $O(N)$ but from the coset $U(N)/O(N)$.






                    share|cite|improve this answer











                    $endgroup$












                    • $begingroup$
                      That makes sense but you use independent Gaussian random variables for the matrix, I was wondering if you would take one single random variable as it were (say, all the possible numbers which can be produced by some particular random number generator) and then fill the matrix with every entry being a value which that one variable can take, I take it this is still a random matrix or to be a random matrix does every entry need to correspond to a different independent random variable?
                      $endgroup$
                      – Tom
                      8 hours ago







                    • 1




                      $begingroup$
                      you can use a random number generator to fill the matrix, but it is essential that this random number generator samples from a Gaussian distribution; it your random number generator samples from a uniform distribution, you will have to do some postprocessing to convert that into a Gaussian.
                      $endgroup$
                      – Carlo Beenakker
                      8 hours ago














                    2












                    2








                    2





                    $begingroup$

                    I understand your question as asking for a constructive method to sample uniformly from the unitary group $U(N)$ or orthogonal group $O(N)$, where "uniformly" is understood in the sense of the Haar measure. A simple method starts from an $Ntimes N$ matrix filled with independent Gaussian random variables [complex for $U(N)$ and real for $O(N)$]. Then orthonormalize the columns via Gram-Schmidt and you're done.



                    For a more efficient approach, see How to generate random matrices from the classical compact groups by Francesco Mezzadri.



                    The construction is implemented in Mathematica in the routines circular unitary ensemble for $U(N)$ and circular real ensemble for $O(N)$.
                    Caution: the socalled circular orthogonal ensemble does not sample from $O(N)$ but from the coset $U(N)/O(N)$.






                    share|cite|improve this answer











                    $endgroup$



                    I understand your question as asking for a constructive method to sample uniformly from the unitary group $U(N)$ or orthogonal group $O(N)$, where "uniformly" is understood in the sense of the Haar measure. A simple method starts from an $Ntimes N$ matrix filled with independent Gaussian random variables [complex for $U(N)$ and real for $O(N)$]. Then orthonormalize the columns via Gram-Schmidt and you're done.



                    For a more efficient approach, see How to generate random matrices from the classical compact groups by Francesco Mezzadri.



                    The construction is implemented in Mathematica in the routines circular unitary ensemble for $U(N)$ and circular real ensemble for $O(N)$.
                    Caution: the socalled circular orthogonal ensemble does not sample from $O(N)$ but from the coset $U(N)/O(N)$.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 8 hours ago

























                    answered 9 hours ago









                    Carlo BeenakkerCarlo Beenakker

                    83.2k9196301




                    83.2k9196301











                    • $begingroup$
                      That makes sense but you use independent Gaussian random variables for the matrix, I was wondering if you would take one single random variable as it were (say, all the possible numbers which can be produced by some particular random number generator) and then fill the matrix with every entry being a value which that one variable can take, I take it this is still a random matrix or to be a random matrix does every entry need to correspond to a different independent random variable?
                      $endgroup$
                      – Tom
                      8 hours ago







                    • 1




                      $begingroup$
                      you can use a random number generator to fill the matrix, but it is essential that this random number generator samples from a Gaussian distribution; it your random number generator samples from a uniform distribution, you will have to do some postprocessing to convert that into a Gaussian.
                      $endgroup$
                      – Carlo Beenakker
                      8 hours ago

















                    • $begingroup$
                      That makes sense but you use independent Gaussian random variables for the matrix, I was wondering if you would take one single random variable as it were (say, all the possible numbers which can be produced by some particular random number generator) and then fill the matrix with every entry being a value which that one variable can take, I take it this is still a random matrix or to be a random matrix does every entry need to correspond to a different independent random variable?
                      $endgroup$
                      – Tom
                      8 hours ago







                    • 1




                      $begingroup$
                      you can use a random number generator to fill the matrix, but it is essential that this random number generator samples from a Gaussian distribution; it your random number generator samples from a uniform distribution, you will have to do some postprocessing to convert that into a Gaussian.
                      $endgroup$
                      – Carlo Beenakker
                      8 hours ago
















                    $begingroup$
                    That makes sense but you use independent Gaussian random variables for the matrix, I was wondering if you would take one single random variable as it were (say, all the possible numbers which can be produced by some particular random number generator) and then fill the matrix with every entry being a value which that one variable can take, I take it this is still a random matrix or to be a random matrix does every entry need to correspond to a different independent random variable?
                    $endgroup$
                    – Tom
                    8 hours ago





                    $begingroup$
                    That makes sense but you use independent Gaussian random variables for the matrix, I was wondering if you would take one single random variable as it were (say, all the possible numbers which can be produced by some particular random number generator) and then fill the matrix with every entry being a value which that one variable can take, I take it this is still a random matrix or to be a random matrix does every entry need to correspond to a different independent random variable?
                    $endgroup$
                    – Tom
                    8 hours ago





                    1




                    1




                    $begingroup$
                    you can use a random number generator to fill the matrix, but it is essential that this random number generator samples from a Gaussian distribution; it your random number generator samples from a uniform distribution, you will have to do some postprocessing to convert that into a Gaussian.
                    $endgroup$
                    – Carlo Beenakker
                    8 hours ago





                    $begingroup$
                    you can use a random number generator to fill the matrix, but it is essential that this random number generator samples from a Gaussian distribution; it your random number generator samples from a uniform distribution, you will have to do some postprocessing to convert that into a Gaussian.
                    $endgroup$
                    – Carlo Beenakker
                    8 hours ago


















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