I have the following integral :
$$Z(A)=\int \prod_{i=1}^n \frac{dx_i dy_i}{2 \pi} e^{-\sum_{i,j} \bar{z_i}A_{i j} z_j}$$
$$z_i=\frac{x_i+i y_i}{\sqrt{2}}$$
$A$ is an hermitic matrix.
In my course the teacher did a change of variables :
$z=Uz'$ such that $U$ is the unitary matrix that diagonalizes $A$ : $A=UDU^{\dagger}$.
Thus, under such a change of variables
$$\bar{z_i}A_{i j} z_j=\bar{z}'_iD_{ii}z_i$$
And it is written : as the matrix $U$ is unitary the jacobian of the change of variable is 1.
But I don't totally understand this. Indeed, the Jacobian would be the matrix linking the $x_i, y_i$ with $x'_i, y'_i$.
And if I write this I have :
$$ x_i=\frac{z_i+\bar{z}_i}{\sqrt{2}}$$ $$ y_i=\frac{z_i-\bar{z}_i}{\sqrt{2}}$$
Thus :
$$ x=\frac{(U+\bar{U})x'+i(U-\bar{U})y'}{\sqrt{2}}$$ $$ y=\frac{(U-\bar{U})x'+i(U+\bar{U})y'}{\sqrt{2}}$$
And it is not obvious for me that the matrix associated to this change of variable is unitary. [edit] in fact it seems to simplify, the the end of the post.
My questions are :
- Was that obvious from the beginning that the matrix of change of variable will be unitary if the matrix $U$ is ?
- If I remember well, we have $U^{\dagger}=\bar{U}^{t}$ if and only if we write this matrix in an orthogonal basis. Do you agree with me ? So in all the reasoning, we impose that we write everything in an orthonormal basis (we have the choice to do it). Am I right ? Because we need this property to simplify everything.
[edit] Hmmm in fact I tried to simply compute $M^{\dagger} M$ where $M$ is the matrix of change of variables, and it seems it simplifies to a unitary matrix.
$$\frac{1}{4} \begin{bmatrix}U+\bar{U}& U-\bar{U} \\ U-\bar{U} &U+\bar{U}\end{bmatrix} \begin{bmatrix}U^{\dagger}+U^{t}& U^{\dagger}-U^{t} \\ U^{\dagger}-U^{t} &U^{\dagger}+U^{t}\end{bmatrix}$$
For example for the first term we have :
$$\frac{1}{4}( (U+\bar{U})(U^{\dagger}+U^{t})+ (U-\bar{U})(U^{\dagger}-U^{t}) )= \frac{1}{4}( 4*I)=I $$
Consider the change of variables
$$ z_k = U_{kl} z'_l \\ \bar{z}_k = \bar{U}_{lk} \bar{z}'_l $$.
And let $z_k = x_k + i y_k$. Then
$$ x = A x' - B y' \\ y = A y' + B x' $$ with $A = \frac{1}{2}(U + U ^\dagger)$ and $B = \frac{1}{2i}(U - U^\dagger)$. Then the Jacobian is
$$ J = \det \begin{pmatrix} A && B \\ -B && A \end{pmatrix} = \det(A)\det(A + B A^{-1} B) \\ \det(A^2 + B^2) = \det(\frac{1}{2}(U U^\dagger + U^\dagger U)) = 1 $$ where in the second and the last equality it was used that $U$ is unitary (then $[A,B] =0$).
Some remarks: if $U$ is unitary on a complex normed vector space $(V^n,\langle \cdot, \cdot\rangle_\mathbb{C})$, then $U$ is orthogonal on the real normed vector space $(V_\mathbb{R}^{2n},\langle \cdot,\cdot\rangle_\mathbb{R})$, where $\langle \cdot,\cdot\rangle_\mathbb{R} = \Re \langle \cdot, \cdot\rangle_\mathbb{C}$.