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1) How to prove that $N\times N$ matrix integral over complex matrices $Z$ $$ \int d Z d Z^\dagger e^{-Tr Z Z^\dagger} \frac{x_1\det e^Z -x_2 \det e^{AZ^\dagger}}{\det(1-x_1e^Z)\det(1-x_2e^{AZ^\dagger})} $$ does not depend on the external Hermitian matrix $A$? $x_1$ and $x_2$ are numbers. The statement is trivial for $1\times1$ case.

2)The same for

$$ \int d Z d Z^\dagger e^{-Tr Z Z^\dagger} \frac{x_1\det e^Z -x_2 \det e^{AZ^\dagger}}{\det(1-x_1e^Zg)\det(1-x_2e^{AZ^\dagger}g)} $$

where g - arbitrary $GL(N)$ matrix.

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  • $\begingroup$ I understand that $dZ$ is the Lebesgue measure on $N\times N$ complex matrices, that is $dZ=\prod_{i,j}d\Re z_{ij}d\Im z_{ij}$, but what $dZ^\dagger$ stands for ? $\endgroup$ Jul 20, 2011 at 6:14
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    $\begingroup$ This is just a notation sometimes used in the mathphys literature to show that you integrate over $2N^2$ real variables contrary to $N^2$ for the Hermitian model. $\endgroup$
    – Sasha
    Oct 20, 2011 at 9:14

1 Answer 1

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Let us first slightly generalize OP's first integral to

$$\tag{1} I(x,y,A,B) ~:=~ \int_{{\rm Mat}_{n\times n}(\mathbb{C})} \! dZ~dZ^{\dagger} \frac{e^{-{\rm tr}(ZZ^{\dagger})}\left(xe^{{\rm tr}(ZA)}-ye^{{\rm tr}(BZ^{\dagger})}\right)}{\det \left({\bf 1}-xe^{ZA}\right)\det \left({\bf 1}-ye^{BZ^{\dagger}}\right)}, $$ where $A,B,Z\in {\rm Mat}_{n\times n}(\mathbb{C})$ and $x,y\in\mathbb{C}$.

I) Case $n=1$: As OP observes, it is straightforward to confirm that the integral (1)

$$I(x,y,a,b)~:=~ \int_{\mathbb{C}} \! dz~d\bar{z} ~e^{-|z|^2}\frac{xe^{az}-ye^{b\bar{z}}}{(1-xe^{az})(1-ye^{b\bar{z}})}$$ $$\tag{2} ~=~\int_{\mathbb{C}} \! dz~~d\bar{z}~e^{-|z|^2}\left(\frac{1}{1-xe^{az}}-\frac{1}{1-ye^{b\bar{z}}}\right)~=~\frac{\pi}{1-x}-\frac{\pi}{1-y}$$

is indeed independent of $a$ and $b$.

II) Case $n=2$: In two dimensions we have

$$\tag{3}\Delta(x,M)~:=~ \det \left({\bf 1}-xe^{M}\right) ~=~1-x{\rm tr}(e^{M})+x^2e^{{\rm tr}(M)},$$ $$\tag{4}{\rm tr}(e^{M}) ~=~e^{{\rm tr}(MP_+)}+e^{{\rm tr}(MP_-)} ,\qquad M~\in~ {\rm Mat}_{2\times 2}(\mathbb{C}),$$

where $P_{\pm}=\frac{1}{2}({\bf 1}_{2\times 2}\pm \sigma_{3})$ are projections.

The integral (1) now becomes

$$\tag{5} \int_{{\rm Mat}_{2\times 2}(\mathbb{C})} \! dZ~dZ^{\dagger}~ e^{-{\rm tr}(ZZ^{\dagger})} \frac{xe^{{\rm tr}(ZA)}-ye^{{\rm tr}(BZ^{\dagger})}}{\Delta(x,ZA)\Delta(y,BZ^{\dagger})}. $$

Let us view the integral (5) as a formal power series in the indeterminates $x$ and $y$. Let us consider just the coefficient in front of $x^2y$. It is$^1$

$$ \int_{{\rm Mat}_{2\times 2}(\mathbb{C})} \! dZ~dZ^{\dagger}~e^{-{\rm tr}(ZZ^{\dagger})} $$ $$\times \left\{e^{{\rm tr}(ZA)} {\rm tr}(e^{ZA}){\rm tr}(e^{BZ^{\dagger}})- e^{{\rm tr}(BZ^{\dagger})}\left( \left({\rm tr}(e^{ZA})\right)^2- e^{{\rm tr}(ZA)} \right)\right\}, $$ $$ ~=~\int_{{\rm Mat}_{2\times 2}(\mathbb{C})} \! dZ~dZ^{\dagger}~e^{-{\rm tr}(ZZ^{\dagger})} $$ $$\times \left\{ \left(e^{{\rm tr}(ZA(1+P_+))}+e^{{\rm tr}(ZA(1+P_-))}\right)\left(e^{{\rm tr}(P_+BZ^{\dagger})}+e^{{\rm tr}(P_-BZ^{\dagger})}\right)- e^{{\rm tr}(BZ^{\dagger})}\left( e^{2{\rm tr}(ZAP_+)}+e^{2{\rm tr}(ZAP_-)} +e^{{\rm tr}(ZA)} \right)\right\} $$ $$~\stackrel{(7)}{=}~ \pi^4 \left( {\rm tr}(e^{BA})-e^{{\rm tr}(BA)}\right), \tag{6} $$ which depends on the matrices $A$ and $B$, even if $A={\bf 1}_{2\times 2}$ and $B$ is Hermitian as in OP's case. So OP's conjecture is false.

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$^1$ Recall the Gaussian integral formula

$$\tag{7} I(A,B)~:=~ \int_{{\rm Mat}_{n\times n}(\mathbb{C})} \! dZ~dZ^{\dagger} ~e^{{\rm tr}(-ZZ^{\dagger}+ZA+BZ^{\dagger})}~=~~\pi^{n^2}e^{{\rm tr}(BA)}. $$

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  • $\begingroup$ Purely by pattern matching, shouldn't the "Gaussian" intergral have a determinant in it and not a exp-tr? In that case, it may be possible for $A$ to disappear as conjectured... $\endgroup$
    – Suvrit
    Oct 9, 2014 at 17:24

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