Multi-dimensional gaussian integral with non-symmetric & non-hermitian coefficient matrix












1














There is a commonly used formula in quantum field theory,
begin{equation}
intprod_{i} frac{dz_{i}^{dagger} dz_{i}}{2pi} e^{-z^{dagger}Az}=frac{1}{det(A)}, (1)
end{equation}

where $z=x+text{i}y$ is a complex vector, and $A$ is a matrix with positive definite hermitian part (We know a matrix can always be written as sumation of hermitian part and anti-hermitian part, $A=frac{A+A^{dagger}}{2}+frac{A-A^{dagger}}{2}$).
The difficulty is coefficient matrix $A$ is neither hermitian nor complex symmetric.



For hermitian matrix $A$, Eq (1) is easy to verify, because a hermitian matrix can be diagonalized by a unitary matrix.



For complex symmetric $A$, the problem is a little bit more tougher although there are no less than two ways to manage.





  1. The first way is analytical continuation from a real symmetric
    matrix. For real symmetric matrix A,



    begin{equation} z^{dagger}A z=[x^{T},y^{T}] begin{bmatrix} A &
    \ & A \ end{bmatrix} begin{bmatrix} x \ y \
    end{bmatrix}, end{equation}



    the integral is
    $sqrt{det(A)}^{-1}timessqrt{det(A)}^{-1}=det(A)^{-1}$.



    If the imaginary part emerges, $Arightarrow A+iB$, where $B$ is
    also a real symmetric matrix. The positive definiteness of the
    original real symmetric $A$ guarantees that there exists a finite
    range on the real axis, i.e., $epsilonin[-alpha,alpha]$, in
    which the matrix $A+epsilon B$ is still positive definite so that
    the integral remains $det(A)^{-1}$.



    When the $epsilon$ becomes complex number the real part of which is in
    the range, $text{Re}({epsilon})in[-alpha,alpha]$, while the
    imaginary part is arbitary, the integral always converges and the
    determinant is not going to touch zero in the range, so the quantities on
    both sides of Eq (1) exist. According to the mathematical theorem
    (equal on piece, equal all over) for analytical functions, when $epsilon = 0+text{i}$, the equation
    still holds, so for complex symmetric $A$ with positive definite real part, Eq (1) is verified. Here we noticed that the original statement is A is a matix with positive definite hermitian part. For complex symmetric matrix, the hermitian part, $frac{A+A^{dagger}}{2}$ is the real part and the anti-hermitian part, $frac{A-A^{dagger}}{2}$ is the imaginary part.



  2. The second way is based on the fact that if the real part of a complex symmetric matrix is positive definite, it can be diagonalized simultaneously with the imaginary part. There is a good description in section 3 and 4 from David K. zhang's note, enter link description here.


I can not figure out the proof for a general complex matrix with positive definite hermitian part.



Looking forward to your reply.










share|cite|improve this question



























    1














    There is a commonly used formula in quantum field theory,
    begin{equation}
    intprod_{i} frac{dz_{i}^{dagger} dz_{i}}{2pi} e^{-z^{dagger}Az}=frac{1}{det(A)}, (1)
    end{equation}

    where $z=x+text{i}y$ is a complex vector, and $A$ is a matrix with positive definite hermitian part (We know a matrix can always be written as sumation of hermitian part and anti-hermitian part, $A=frac{A+A^{dagger}}{2}+frac{A-A^{dagger}}{2}$).
    The difficulty is coefficient matrix $A$ is neither hermitian nor complex symmetric.



    For hermitian matrix $A$, Eq (1) is easy to verify, because a hermitian matrix can be diagonalized by a unitary matrix.



    For complex symmetric $A$, the problem is a little bit more tougher although there are no less than two ways to manage.





    1. The first way is analytical continuation from a real symmetric
      matrix. For real symmetric matrix A,



      begin{equation} z^{dagger}A z=[x^{T},y^{T}] begin{bmatrix} A &
      \ & A \ end{bmatrix} begin{bmatrix} x \ y \
      end{bmatrix}, end{equation}



      the integral is
      $sqrt{det(A)}^{-1}timessqrt{det(A)}^{-1}=det(A)^{-1}$.



      If the imaginary part emerges, $Arightarrow A+iB$, where $B$ is
      also a real symmetric matrix. The positive definiteness of the
      original real symmetric $A$ guarantees that there exists a finite
      range on the real axis, i.e., $epsilonin[-alpha,alpha]$, in
      which the matrix $A+epsilon B$ is still positive definite so that
      the integral remains $det(A)^{-1}$.



      When the $epsilon$ becomes complex number the real part of which is in
      the range, $text{Re}({epsilon})in[-alpha,alpha]$, while the
      imaginary part is arbitary, the integral always converges and the
      determinant is not going to touch zero in the range, so the quantities on
      both sides of Eq (1) exist. According to the mathematical theorem
      (equal on piece, equal all over) for analytical functions, when $epsilon = 0+text{i}$, the equation
      still holds, so for complex symmetric $A$ with positive definite real part, Eq (1) is verified. Here we noticed that the original statement is A is a matix with positive definite hermitian part. For complex symmetric matrix, the hermitian part, $frac{A+A^{dagger}}{2}$ is the real part and the anti-hermitian part, $frac{A-A^{dagger}}{2}$ is the imaginary part.



    2. The second way is based on the fact that if the real part of a complex symmetric matrix is positive definite, it can be diagonalized simultaneously with the imaginary part. There is a good description in section 3 and 4 from David K. zhang's note, enter link description here.


    I can not figure out the proof for a general complex matrix with positive definite hermitian part.



    Looking forward to your reply.










    share|cite|improve this question

























      1












      1








      1







      There is a commonly used formula in quantum field theory,
      begin{equation}
      intprod_{i} frac{dz_{i}^{dagger} dz_{i}}{2pi} e^{-z^{dagger}Az}=frac{1}{det(A)}, (1)
      end{equation}

      where $z=x+text{i}y$ is a complex vector, and $A$ is a matrix with positive definite hermitian part (We know a matrix can always be written as sumation of hermitian part and anti-hermitian part, $A=frac{A+A^{dagger}}{2}+frac{A-A^{dagger}}{2}$).
      The difficulty is coefficient matrix $A$ is neither hermitian nor complex symmetric.



      For hermitian matrix $A$, Eq (1) is easy to verify, because a hermitian matrix can be diagonalized by a unitary matrix.



      For complex symmetric $A$, the problem is a little bit more tougher although there are no less than two ways to manage.





      1. The first way is analytical continuation from a real symmetric
        matrix. For real symmetric matrix A,



        begin{equation} z^{dagger}A z=[x^{T},y^{T}] begin{bmatrix} A &
        \ & A \ end{bmatrix} begin{bmatrix} x \ y \
        end{bmatrix}, end{equation}



        the integral is
        $sqrt{det(A)}^{-1}timessqrt{det(A)}^{-1}=det(A)^{-1}$.



        If the imaginary part emerges, $Arightarrow A+iB$, where $B$ is
        also a real symmetric matrix. The positive definiteness of the
        original real symmetric $A$ guarantees that there exists a finite
        range on the real axis, i.e., $epsilonin[-alpha,alpha]$, in
        which the matrix $A+epsilon B$ is still positive definite so that
        the integral remains $det(A)^{-1}$.



        When the $epsilon$ becomes complex number the real part of which is in
        the range, $text{Re}({epsilon})in[-alpha,alpha]$, while the
        imaginary part is arbitary, the integral always converges and the
        determinant is not going to touch zero in the range, so the quantities on
        both sides of Eq (1) exist. According to the mathematical theorem
        (equal on piece, equal all over) for analytical functions, when $epsilon = 0+text{i}$, the equation
        still holds, so for complex symmetric $A$ with positive definite real part, Eq (1) is verified. Here we noticed that the original statement is A is a matix with positive definite hermitian part. For complex symmetric matrix, the hermitian part, $frac{A+A^{dagger}}{2}$ is the real part and the anti-hermitian part, $frac{A-A^{dagger}}{2}$ is the imaginary part.



      2. The second way is based on the fact that if the real part of a complex symmetric matrix is positive definite, it can be diagonalized simultaneously with the imaginary part. There is a good description in section 3 and 4 from David K. zhang's note, enter link description here.


      I can not figure out the proof for a general complex matrix with positive definite hermitian part.



      Looking forward to your reply.










      share|cite|improve this question













      There is a commonly used formula in quantum field theory,
      begin{equation}
      intprod_{i} frac{dz_{i}^{dagger} dz_{i}}{2pi} e^{-z^{dagger}Az}=frac{1}{det(A)}, (1)
      end{equation}

      where $z=x+text{i}y$ is a complex vector, and $A$ is a matrix with positive definite hermitian part (We know a matrix can always be written as sumation of hermitian part and anti-hermitian part, $A=frac{A+A^{dagger}}{2}+frac{A-A^{dagger}}{2}$).
      The difficulty is coefficient matrix $A$ is neither hermitian nor complex symmetric.



      For hermitian matrix $A$, Eq (1) is easy to verify, because a hermitian matrix can be diagonalized by a unitary matrix.



      For complex symmetric $A$, the problem is a little bit more tougher although there are no less than two ways to manage.





      1. The first way is analytical continuation from a real symmetric
        matrix. For real symmetric matrix A,



        begin{equation} z^{dagger}A z=[x^{T},y^{T}] begin{bmatrix} A &
        \ & A \ end{bmatrix} begin{bmatrix} x \ y \
        end{bmatrix}, end{equation}



        the integral is
        $sqrt{det(A)}^{-1}timessqrt{det(A)}^{-1}=det(A)^{-1}$.



        If the imaginary part emerges, $Arightarrow A+iB$, where $B$ is
        also a real symmetric matrix. The positive definiteness of the
        original real symmetric $A$ guarantees that there exists a finite
        range on the real axis, i.e., $epsilonin[-alpha,alpha]$, in
        which the matrix $A+epsilon B$ is still positive definite so that
        the integral remains $det(A)^{-1}$.



        When the $epsilon$ becomes complex number the real part of which is in
        the range, $text{Re}({epsilon})in[-alpha,alpha]$, while the
        imaginary part is arbitary, the integral always converges and the
        determinant is not going to touch zero in the range, so the quantities on
        both sides of Eq (1) exist. According to the mathematical theorem
        (equal on piece, equal all over) for analytical functions, when $epsilon = 0+text{i}$, the equation
        still holds, so for complex symmetric $A$ with positive definite real part, Eq (1) is verified. Here we noticed that the original statement is A is a matix with positive definite hermitian part. For complex symmetric matrix, the hermitian part, $frac{A+A^{dagger}}{2}$ is the real part and the anti-hermitian part, $frac{A-A^{dagger}}{2}$ is the imaginary part.



      2. The second way is based on the fact that if the real part of a complex symmetric matrix is positive definite, it can be diagonalized simultaneously with the imaginary part. There is a good description in section 3 and 4 from David K. zhang's note, enter link description here.


      I can not figure out the proof for a general complex matrix with positive definite hermitian part.



      Looking forward to your reply.







      gaussian-integral






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 28 '18 at 17:36









      Ye CaoYe Cao

      61




      61






















          0






          active

          oldest

          votes











          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: "69"
          };
          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: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          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%2fmath.stackexchange.com%2fquestions%2f3017439%2fmulti-dimensional-gaussian-integral-with-non-symmetric-non-hermitian-coefficie%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          0






          active

          oldest

          votes








          0






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes
















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics 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.





          Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


          Please pay close attention to the following guidance:


          • 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.


          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%2fmath.stackexchange.com%2fquestions%2f3017439%2fmulti-dimensional-gaussian-integral-with-non-symmetric-non-hermitian-coefficie%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

          Quarter-circle Tiles

          build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

          Mont Emei