Decomposition of a unitary matrix












7












$begingroup$


Following is an excerpt from QCQI:



enter image description here



I can understand that this matrix satisfies a unitary matrix. Also, intuitively, I am able to understand it. However, what is the proof that any given Unitary Matrix can be converted as above? Like why there should be only 4 variables and not more or less? Looking for some formal authentic proof. Thanks.










share|improve this question











$endgroup$

















    7












    $begingroup$


    Following is an excerpt from QCQI:



    enter image description here



    I can understand that this matrix satisfies a unitary matrix. Also, intuitively, I am able to understand it. However, what is the proof that any given Unitary Matrix can be converted as above? Like why there should be only 4 variables and not more or less? Looking for some formal authentic proof. Thanks.










    share|improve this question











    $endgroup$















      7












      7








      7





      $begingroup$


      Following is an excerpt from QCQI:



      enter image description here



      I can understand that this matrix satisfies a unitary matrix. Also, intuitively, I am able to understand it. However, what is the proof that any given Unitary Matrix can be converted as above? Like why there should be only 4 variables and not more or less? Looking for some formal authentic proof. Thanks.










      share|improve this question











      $endgroup$




      Following is an excerpt from QCQI:



      enter image description here



      I can understand that this matrix satisfies a unitary matrix. Also, intuitively, I am able to understand it. However, what is the proof that any given Unitary Matrix can be converted as above? Like why there should be only 4 variables and not more or less? Looking for some formal authentic proof. Thanks.







      quantum-gate mathematics unitarity






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Jan 14 at 15:27









      Blue

      6,00131354




      6,00131354










      asked Jan 14 at 8:32









      user2508039user2508039

      1083




      1083






















          1 Answer
          1






          active

          oldest

          votes


















          10












          $begingroup$


          What is the proof that any given unitary matrix can be converted as above?




          Let $U$ be an arbitrary $2times 2$ unitary matrix. This is equivalent to the rows/columns of $U$ forming an orthonormal system.



          Let us write a generic $U$ as
          $$U=begin{pmatrix}a&b\c&dend{pmatrix}.$$
          The constraints imposed on the coefficients $a,b,c,d$ by the requirement of $U$ being unitary are
          $$|a|^2+|c|^2=1,qquad |b|^2+|d|^2=1,qquad a^* b+c^* d=0.$$
          A pair of complex numbers $a,binmathbb C$ satisfying $|a|^2+|b|^2=1$ can always be parametrized as
          $$a=e^{ialpha_{11}}costheta,qquad b=e^{ialpha_{12}}sintheta,$$
          for some real coefficients $alpha_{ij},thetainmathbb R$.



          It follows that using only the normalization constraint (but without taking into account the orthogonality) we can parametrize $U$ as



          $$U=begin{pmatrix}e^{alpha_{11}}costheta& e^{alpha_{12}}sintheta\
          e^{alpha_{21}}sintheta & e^{alpha_{22}}costheta
          end{pmatrix}.$$



          Requiring the columns to be orthogonal then gives the additional relation



          $$e^{i(alpha_{11}-alpha_{12})}+e^{i(alpha_{21}-alpha_{22})}=0,$$
          that is, $alpha_{11}=alpha_{12}+alpha_{21}-alpha_{22}+pi$.



          We conclude that $U$ is parametrized by three real parameters, here denoted $theta,alpha_{12},alpha_{21},alpha_{22}$.



          To get the form you show you simply need to change variables as follows:



          begin{align}
          theta&=gamma/2, \
          alpha_{12} &= alpha-beta/2+delta/2+pi,\
          alpha_{21} &= alpha+beta/2-delta/2, \
          alpha_{22} &= alpha+beta/2+delta/2.
          end{align}




          Like why there should be only 4 variables and not more or less?




          While the above already proved this, it can be useful to know that this a special case of a more general result.



          A generic unitary $ntimes n$ matrix is specified by $n^2$ real parameters (see wiki page on the unitary group for more details).



          An easy way to see this is again to remember that unitary matrices are characterised by their columns/rows forming an orthonormal system.
          This amounts to $n$ real constraints (imposing each of the $n$ columns to be normalized), plus $binom{n}{2}=n(n-1)/2$ additional complex constraints (imposing each pair of columns to be orthogonal). Each complex constraints amounts to two real constraints, so this sums up to a total of
          $$n+2binom{n}{2}=n+n(n-1)=n^2$$
          real, independent constraints.



          A generic $ntimes n$ matrix is characterised by $n^2$ complex numbers, that is, $2n^2$ real numbers.



          We conclude that the number of free parameters of a generic $ntimes n$ unitary matrix is:
          $$2n^2-n^2=n^2.$$



          Going back to the simple $2times 2$ case above, you can see how we get back the previous result because $2^2=4$ (you might also notice that to count parameters in the special case $2times 2$ I used a different ad-hoc strategy, rather then the one showed here to get the count in the general case).



          Yet another way to count parameters



          Another method I like is to think entirely in terms of orthonormal systems.
          The question is: how many parameters need to be given to specify an orthonormal basis in an $n$-dimensional complex vector space?



          Let us start by the first vector. The only constraint here is that we want the vector to be normalized.
          The number of real parameters needed to specify a normalized vector in $n$-dimensions is $d_1=2n-1$.



          Let us now add another vector. Now we have to impose both the normalization of this additional vector (one real constraint), and the orthogonality of this additional vector to the initial one (two real constraints).
          The additional parameters are therefore $d_2= 2n-3$.



          A third vector will need to be normalized (one real constraint), and orthogonal to the first two vectors ($2times 2$ real constraints), thus $d_3=2n-5$.



          Iterate this reasoning until you get to the last vector, which will be specified by a single real parameter.



          The total number of parameters is therefore:
          $$sum_k d_k=(2n-1)+(2n-3)+cdots+3+1=sum_{k=0}^{n-1} (2k+1).$$
          In other words, the number of parameters is given by the sum of the first $n$ odd integers, which is again readily shown to equal $n^2$.






          share|improve this answer











          $endgroup$













          • $begingroup$
            Awesome. Thanks for the proof. Also, it's OK to use those simultaneous equations because they are linearly independent so we get a 3 variable solution, right?
            $endgroup$
            – user2508039
            Jan 14 at 10:06












          • $begingroup$
            @user2508039 yes. You can also just check that the system can be inverted and therefore you have a bijection between the two sets of parameters
            $endgroup$
            – glS
            Jan 14 at 10:23











          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "694"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: false,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: null,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fquantumcomputing.stackexchange.com%2fquestions%2f5199%2fdecomposition-of-a-unitary-matrix%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          10












          $begingroup$


          What is the proof that any given unitary matrix can be converted as above?




          Let $U$ be an arbitrary $2times 2$ unitary matrix. This is equivalent to the rows/columns of $U$ forming an orthonormal system.



          Let us write a generic $U$ as
          $$U=begin{pmatrix}a&b\c&dend{pmatrix}.$$
          The constraints imposed on the coefficients $a,b,c,d$ by the requirement of $U$ being unitary are
          $$|a|^2+|c|^2=1,qquad |b|^2+|d|^2=1,qquad a^* b+c^* d=0.$$
          A pair of complex numbers $a,binmathbb C$ satisfying $|a|^2+|b|^2=1$ can always be parametrized as
          $$a=e^{ialpha_{11}}costheta,qquad b=e^{ialpha_{12}}sintheta,$$
          for some real coefficients $alpha_{ij},thetainmathbb R$.



          It follows that using only the normalization constraint (but without taking into account the orthogonality) we can parametrize $U$ as



          $$U=begin{pmatrix}e^{alpha_{11}}costheta& e^{alpha_{12}}sintheta\
          e^{alpha_{21}}sintheta & e^{alpha_{22}}costheta
          end{pmatrix}.$$



          Requiring the columns to be orthogonal then gives the additional relation



          $$e^{i(alpha_{11}-alpha_{12})}+e^{i(alpha_{21}-alpha_{22})}=0,$$
          that is, $alpha_{11}=alpha_{12}+alpha_{21}-alpha_{22}+pi$.



          We conclude that $U$ is parametrized by three real parameters, here denoted $theta,alpha_{12},alpha_{21},alpha_{22}$.



          To get the form you show you simply need to change variables as follows:



          begin{align}
          theta&=gamma/2, \
          alpha_{12} &= alpha-beta/2+delta/2+pi,\
          alpha_{21} &= alpha+beta/2-delta/2, \
          alpha_{22} &= alpha+beta/2+delta/2.
          end{align}




          Like why there should be only 4 variables and not more or less?




          While the above already proved this, it can be useful to know that this a special case of a more general result.



          A generic unitary $ntimes n$ matrix is specified by $n^2$ real parameters (see wiki page on the unitary group for more details).



          An easy way to see this is again to remember that unitary matrices are characterised by their columns/rows forming an orthonormal system.
          This amounts to $n$ real constraints (imposing each of the $n$ columns to be normalized), plus $binom{n}{2}=n(n-1)/2$ additional complex constraints (imposing each pair of columns to be orthogonal). Each complex constraints amounts to two real constraints, so this sums up to a total of
          $$n+2binom{n}{2}=n+n(n-1)=n^2$$
          real, independent constraints.



          A generic $ntimes n$ matrix is characterised by $n^2$ complex numbers, that is, $2n^2$ real numbers.



          We conclude that the number of free parameters of a generic $ntimes n$ unitary matrix is:
          $$2n^2-n^2=n^2.$$



          Going back to the simple $2times 2$ case above, you can see how we get back the previous result because $2^2=4$ (you might also notice that to count parameters in the special case $2times 2$ I used a different ad-hoc strategy, rather then the one showed here to get the count in the general case).



          Yet another way to count parameters



          Another method I like is to think entirely in terms of orthonormal systems.
          The question is: how many parameters need to be given to specify an orthonormal basis in an $n$-dimensional complex vector space?



          Let us start by the first vector. The only constraint here is that we want the vector to be normalized.
          The number of real parameters needed to specify a normalized vector in $n$-dimensions is $d_1=2n-1$.



          Let us now add another vector. Now we have to impose both the normalization of this additional vector (one real constraint), and the orthogonality of this additional vector to the initial one (two real constraints).
          The additional parameters are therefore $d_2= 2n-3$.



          A third vector will need to be normalized (one real constraint), and orthogonal to the first two vectors ($2times 2$ real constraints), thus $d_3=2n-5$.



          Iterate this reasoning until you get to the last vector, which will be specified by a single real parameter.



          The total number of parameters is therefore:
          $$sum_k d_k=(2n-1)+(2n-3)+cdots+3+1=sum_{k=0}^{n-1} (2k+1).$$
          In other words, the number of parameters is given by the sum of the first $n$ odd integers, which is again readily shown to equal $n^2$.






          share|improve this answer











          $endgroup$













          • $begingroup$
            Awesome. Thanks for the proof. Also, it's OK to use those simultaneous equations because they are linearly independent so we get a 3 variable solution, right?
            $endgroup$
            – user2508039
            Jan 14 at 10:06












          • $begingroup$
            @user2508039 yes. You can also just check that the system can be inverted and therefore you have a bijection between the two sets of parameters
            $endgroup$
            – glS
            Jan 14 at 10:23
















          10












          $begingroup$


          What is the proof that any given unitary matrix can be converted as above?




          Let $U$ be an arbitrary $2times 2$ unitary matrix. This is equivalent to the rows/columns of $U$ forming an orthonormal system.



          Let us write a generic $U$ as
          $$U=begin{pmatrix}a&b\c&dend{pmatrix}.$$
          The constraints imposed on the coefficients $a,b,c,d$ by the requirement of $U$ being unitary are
          $$|a|^2+|c|^2=1,qquad |b|^2+|d|^2=1,qquad a^* b+c^* d=0.$$
          A pair of complex numbers $a,binmathbb C$ satisfying $|a|^2+|b|^2=1$ can always be parametrized as
          $$a=e^{ialpha_{11}}costheta,qquad b=e^{ialpha_{12}}sintheta,$$
          for some real coefficients $alpha_{ij},thetainmathbb R$.



          It follows that using only the normalization constraint (but without taking into account the orthogonality) we can parametrize $U$ as



          $$U=begin{pmatrix}e^{alpha_{11}}costheta& e^{alpha_{12}}sintheta\
          e^{alpha_{21}}sintheta & e^{alpha_{22}}costheta
          end{pmatrix}.$$



          Requiring the columns to be orthogonal then gives the additional relation



          $$e^{i(alpha_{11}-alpha_{12})}+e^{i(alpha_{21}-alpha_{22})}=0,$$
          that is, $alpha_{11}=alpha_{12}+alpha_{21}-alpha_{22}+pi$.



          We conclude that $U$ is parametrized by three real parameters, here denoted $theta,alpha_{12},alpha_{21},alpha_{22}$.



          To get the form you show you simply need to change variables as follows:



          begin{align}
          theta&=gamma/2, \
          alpha_{12} &= alpha-beta/2+delta/2+pi,\
          alpha_{21} &= alpha+beta/2-delta/2, \
          alpha_{22} &= alpha+beta/2+delta/2.
          end{align}




          Like why there should be only 4 variables and not more or less?




          While the above already proved this, it can be useful to know that this a special case of a more general result.



          A generic unitary $ntimes n$ matrix is specified by $n^2$ real parameters (see wiki page on the unitary group for more details).



          An easy way to see this is again to remember that unitary matrices are characterised by their columns/rows forming an orthonormal system.
          This amounts to $n$ real constraints (imposing each of the $n$ columns to be normalized), plus $binom{n}{2}=n(n-1)/2$ additional complex constraints (imposing each pair of columns to be orthogonal). Each complex constraints amounts to two real constraints, so this sums up to a total of
          $$n+2binom{n}{2}=n+n(n-1)=n^2$$
          real, independent constraints.



          A generic $ntimes n$ matrix is characterised by $n^2$ complex numbers, that is, $2n^2$ real numbers.



          We conclude that the number of free parameters of a generic $ntimes n$ unitary matrix is:
          $$2n^2-n^2=n^2.$$



          Going back to the simple $2times 2$ case above, you can see how we get back the previous result because $2^2=4$ (you might also notice that to count parameters in the special case $2times 2$ I used a different ad-hoc strategy, rather then the one showed here to get the count in the general case).



          Yet another way to count parameters



          Another method I like is to think entirely in terms of orthonormal systems.
          The question is: how many parameters need to be given to specify an orthonormal basis in an $n$-dimensional complex vector space?



          Let us start by the first vector. The only constraint here is that we want the vector to be normalized.
          The number of real parameters needed to specify a normalized vector in $n$-dimensions is $d_1=2n-1$.



          Let us now add another vector. Now we have to impose both the normalization of this additional vector (one real constraint), and the orthogonality of this additional vector to the initial one (two real constraints).
          The additional parameters are therefore $d_2= 2n-3$.



          A third vector will need to be normalized (one real constraint), and orthogonal to the first two vectors ($2times 2$ real constraints), thus $d_3=2n-5$.



          Iterate this reasoning until you get to the last vector, which will be specified by a single real parameter.



          The total number of parameters is therefore:
          $$sum_k d_k=(2n-1)+(2n-3)+cdots+3+1=sum_{k=0}^{n-1} (2k+1).$$
          In other words, the number of parameters is given by the sum of the first $n$ odd integers, which is again readily shown to equal $n^2$.






          share|improve this answer











          $endgroup$













          • $begingroup$
            Awesome. Thanks for the proof. Also, it's OK to use those simultaneous equations because they are linearly independent so we get a 3 variable solution, right?
            $endgroup$
            – user2508039
            Jan 14 at 10:06












          • $begingroup$
            @user2508039 yes. You can also just check that the system can be inverted and therefore you have a bijection between the two sets of parameters
            $endgroup$
            – glS
            Jan 14 at 10:23














          10












          10








          10





          $begingroup$


          What is the proof that any given unitary matrix can be converted as above?




          Let $U$ be an arbitrary $2times 2$ unitary matrix. This is equivalent to the rows/columns of $U$ forming an orthonormal system.



          Let us write a generic $U$ as
          $$U=begin{pmatrix}a&b\c&dend{pmatrix}.$$
          The constraints imposed on the coefficients $a,b,c,d$ by the requirement of $U$ being unitary are
          $$|a|^2+|c|^2=1,qquad |b|^2+|d|^2=1,qquad a^* b+c^* d=0.$$
          A pair of complex numbers $a,binmathbb C$ satisfying $|a|^2+|b|^2=1$ can always be parametrized as
          $$a=e^{ialpha_{11}}costheta,qquad b=e^{ialpha_{12}}sintheta,$$
          for some real coefficients $alpha_{ij},thetainmathbb R$.



          It follows that using only the normalization constraint (but without taking into account the orthogonality) we can parametrize $U$ as



          $$U=begin{pmatrix}e^{alpha_{11}}costheta& e^{alpha_{12}}sintheta\
          e^{alpha_{21}}sintheta & e^{alpha_{22}}costheta
          end{pmatrix}.$$



          Requiring the columns to be orthogonal then gives the additional relation



          $$e^{i(alpha_{11}-alpha_{12})}+e^{i(alpha_{21}-alpha_{22})}=0,$$
          that is, $alpha_{11}=alpha_{12}+alpha_{21}-alpha_{22}+pi$.



          We conclude that $U$ is parametrized by three real parameters, here denoted $theta,alpha_{12},alpha_{21},alpha_{22}$.



          To get the form you show you simply need to change variables as follows:



          begin{align}
          theta&=gamma/2, \
          alpha_{12} &= alpha-beta/2+delta/2+pi,\
          alpha_{21} &= alpha+beta/2-delta/2, \
          alpha_{22} &= alpha+beta/2+delta/2.
          end{align}




          Like why there should be only 4 variables and not more or less?




          While the above already proved this, it can be useful to know that this a special case of a more general result.



          A generic unitary $ntimes n$ matrix is specified by $n^2$ real parameters (see wiki page on the unitary group for more details).



          An easy way to see this is again to remember that unitary matrices are characterised by their columns/rows forming an orthonormal system.
          This amounts to $n$ real constraints (imposing each of the $n$ columns to be normalized), plus $binom{n}{2}=n(n-1)/2$ additional complex constraints (imposing each pair of columns to be orthogonal). Each complex constraints amounts to two real constraints, so this sums up to a total of
          $$n+2binom{n}{2}=n+n(n-1)=n^2$$
          real, independent constraints.



          A generic $ntimes n$ matrix is characterised by $n^2$ complex numbers, that is, $2n^2$ real numbers.



          We conclude that the number of free parameters of a generic $ntimes n$ unitary matrix is:
          $$2n^2-n^2=n^2.$$



          Going back to the simple $2times 2$ case above, you can see how we get back the previous result because $2^2=4$ (you might also notice that to count parameters in the special case $2times 2$ I used a different ad-hoc strategy, rather then the one showed here to get the count in the general case).



          Yet another way to count parameters



          Another method I like is to think entirely in terms of orthonormal systems.
          The question is: how many parameters need to be given to specify an orthonormal basis in an $n$-dimensional complex vector space?



          Let us start by the first vector. The only constraint here is that we want the vector to be normalized.
          The number of real parameters needed to specify a normalized vector in $n$-dimensions is $d_1=2n-1$.



          Let us now add another vector. Now we have to impose both the normalization of this additional vector (one real constraint), and the orthogonality of this additional vector to the initial one (two real constraints).
          The additional parameters are therefore $d_2= 2n-3$.



          A third vector will need to be normalized (one real constraint), and orthogonal to the first two vectors ($2times 2$ real constraints), thus $d_3=2n-5$.



          Iterate this reasoning until you get to the last vector, which will be specified by a single real parameter.



          The total number of parameters is therefore:
          $$sum_k d_k=(2n-1)+(2n-3)+cdots+3+1=sum_{k=0}^{n-1} (2k+1).$$
          In other words, the number of parameters is given by the sum of the first $n$ odd integers, which is again readily shown to equal $n^2$.






          share|improve this answer











          $endgroup$




          What is the proof that any given unitary matrix can be converted as above?




          Let $U$ be an arbitrary $2times 2$ unitary matrix. This is equivalent to the rows/columns of $U$ forming an orthonormal system.



          Let us write a generic $U$ as
          $$U=begin{pmatrix}a&b\c&dend{pmatrix}.$$
          The constraints imposed on the coefficients $a,b,c,d$ by the requirement of $U$ being unitary are
          $$|a|^2+|c|^2=1,qquad |b|^2+|d|^2=1,qquad a^* b+c^* d=0.$$
          A pair of complex numbers $a,binmathbb C$ satisfying $|a|^2+|b|^2=1$ can always be parametrized as
          $$a=e^{ialpha_{11}}costheta,qquad b=e^{ialpha_{12}}sintheta,$$
          for some real coefficients $alpha_{ij},thetainmathbb R$.



          It follows that using only the normalization constraint (but without taking into account the orthogonality) we can parametrize $U$ as



          $$U=begin{pmatrix}e^{alpha_{11}}costheta& e^{alpha_{12}}sintheta\
          e^{alpha_{21}}sintheta & e^{alpha_{22}}costheta
          end{pmatrix}.$$



          Requiring the columns to be orthogonal then gives the additional relation



          $$e^{i(alpha_{11}-alpha_{12})}+e^{i(alpha_{21}-alpha_{22})}=0,$$
          that is, $alpha_{11}=alpha_{12}+alpha_{21}-alpha_{22}+pi$.



          We conclude that $U$ is parametrized by three real parameters, here denoted $theta,alpha_{12},alpha_{21},alpha_{22}$.



          To get the form you show you simply need to change variables as follows:



          begin{align}
          theta&=gamma/2, \
          alpha_{12} &= alpha-beta/2+delta/2+pi,\
          alpha_{21} &= alpha+beta/2-delta/2, \
          alpha_{22} &= alpha+beta/2+delta/2.
          end{align}




          Like why there should be only 4 variables and not more or less?




          While the above already proved this, it can be useful to know that this a special case of a more general result.



          A generic unitary $ntimes n$ matrix is specified by $n^2$ real parameters (see wiki page on the unitary group for more details).



          An easy way to see this is again to remember that unitary matrices are characterised by their columns/rows forming an orthonormal system.
          This amounts to $n$ real constraints (imposing each of the $n$ columns to be normalized), plus $binom{n}{2}=n(n-1)/2$ additional complex constraints (imposing each pair of columns to be orthogonal). Each complex constraints amounts to two real constraints, so this sums up to a total of
          $$n+2binom{n}{2}=n+n(n-1)=n^2$$
          real, independent constraints.



          A generic $ntimes n$ matrix is characterised by $n^2$ complex numbers, that is, $2n^2$ real numbers.



          We conclude that the number of free parameters of a generic $ntimes n$ unitary matrix is:
          $$2n^2-n^2=n^2.$$



          Going back to the simple $2times 2$ case above, you can see how we get back the previous result because $2^2=4$ (you might also notice that to count parameters in the special case $2times 2$ I used a different ad-hoc strategy, rather then the one showed here to get the count in the general case).



          Yet another way to count parameters



          Another method I like is to think entirely in terms of orthonormal systems.
          The question is: how many parameters need to be given to specify an orthonormal basis in an $n$-dimensional complex vector space?



          Let us start by the first vector. The only constraint here is that we want the vector to be normalized.
          The number of real parameters needed to specify a normalized vector in $n$-dimensions is $d_1=2n-1$.



          Let us now add another vector. Now we have to impose both the normalization of this additional vector (one real constraint), and the orthogonality of this additional vector to the initial one (two real constraints).
          The additional parameters are therefore $d_2= 2n-3$.



          A third vector will need to be normalized (one real constraint), and orthogonal to the first two vectors ($2times 2$ real constraints), thus $d_3=2n-5$.



          Iterate this reasoning until you get to the last vector, which will be specified by a single real parameter.



          The total number of parameters is therefore:
          $$sum_k d_k=(2n-1)+(2n-3)+cdots+3+1=sum_{k=0}^{n-1} (2k+1).$$
          In other words, the number of parameters is given by the sum of the first $n$ odd integers, which is again readily shown to equal $n^2$.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Jan 14 at 17:45

























          answered Jan 14 at 9:35









          glSglS

          3,946638




          3,946638












          • $begingroup$
            Awesome. Thanks for the proof. Also, it's OK to use those simultaneous equations because they are linearly independent so we get a 3 variable solution, right?
            $endgroup$
            – user2508039
            Jan 14 at 10:06












          • $begingroup$
            @user2508039 yes. You can also just check that the system can be inverted and therefore you have a bijection between the two sets of parameters
            $endgroup$
            – glS
            Jan 14 at 10:23


















          • $begingroup$
            Awesome. Thanks for the proof. Also, it's OK to use those simultaneous equations because they are linearly independent so we get a 3 variable solution, right?
            $endgroup$
            – user2508039
            Jan 14 at 10:06












          • $begingroup$
            @user2508039 yes. You can also just check that the system can be inverted and therefore you have a bijection between the two sets of parameters
            $endgroup$
            – glS
            Jan 14 at 10:23
















          $begingroup$
          Awesome. Thanks for the proof. Also, it's OK to use those simultaneous equations because they are linearly independent so we get a 3 variable solution, right?
          $endgroup$
          – user2508039
          Jan 14 at 10:06






          $begingroup$
          Awesome. Thanks for the proof. Also, it's OK to use those simultaneous equations because they are linearly independent so we get a 3 variable solution, right?
          $endgroup$
          – user2508039
          Jan 14 at 10:06














          $begingroup$
          @user2508039 yes. You can also just check that the system can be inverted and therefore you have a bijection between the two sets of parameters
          $endgroup$
          – glS
          Jan 14 at 10:23




          $begingroup$
          @user2508039 yes. You can also just check that the system can be inverted and therefore you have a bijection between the two sets of parameters
          $endgroup$
          – glS
          Jan 14 at 10:23


















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Quantum Computing Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fquantumcomputing.stackexchange.com%2fquestions%2f5199%2fdecomposition-of-a-unitary-matrix%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          Popular posts from this blog

          Aardman Animations

          Are they similar matrix

          “minimization” problem in Euclidean space related to orthonormal basis