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






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      asked Nov 28 '18 at 17:36









      Ye CaoYe Cao

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