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In the book "Path Integrals in Quantum Mechanics" by J. Zinn-Justin the following integral is evaluated:

In the book he states:

Quite generally we consider integrals of the following form: \begin{equation} I=\int_{\mathbb{R}^2}dxdyf(x,y)=-i\int dzdz'f(x(z,z'),y(z,z')) \end{equation}

makeing use of the following transformation:

\begin{equation} z=(x+iy)/\sqrt{2}, z'=(x-iy)/\sqrt{2} \end{equation}

Then he goes on to add the following statement:

Formal complex conjugation. Notation. We now introduce a standard, convenient but somewhat dangerous notation: we denote the variable $z'$ by $\bar{z}$. We stress that the complex variables $z$ and $\bar{z}$ remain independent integration variables and are complex conjugate only in a formal sense. Indeed, the integration contours over the variables $x$ and $y$ could be deformed in complex space and, then, these variables would also take complex values. The symbol $dzd\bar{z}$ thus corresponds to an integration over a surface of real dimension 2, embedded in $\mathbb{C}^2$.

Now we consider the following expression:

\begin{equation} \mathcal{Z}(\mathbf{A})=\int\left(\prod_{i=1}^n \frac{\mathrm{d} z_i \mathrm{~d} \bar{z}_i}{2 i \pi}\right) \mathrm{e}^{-A(\overline{\mathbf{z}}, \mathbf{z})} \end{equation} Where: \begin{equation} A(\overline{\mathbf{z}}, \mathbf{z})=\sum_{ij}\bar{z_i}A_{ij}z_j \end{equation} $\mathbf{A}$ is a complex matrix with a non-vanishing determinant. The author claims that the integral can be calculated from a purely algebraic viewpoint by an asymmetric change of variables: \begin{equation} z_i\rightarrow z_i'=\sum_{j}A_{ij}z_j \end{equation} leaving the variable $\bar{z}$ unchanged. This transformation has for jacobian $1/\text{det}\mathbf{A}$. What we get is: \begin{equation} \mathcal{Z}(\mathbf{A})=\frac{1}{\text{det}\mathbf{A}}\int\left(\prod_{i=1}^n \frac{\mathrm{d} z_i \mathrm{~d} \bar{z}_i}{2 i \pi}\right) \mathrm{e}^{-(\overline{\mathbf{z}}, \mathbf{z})}=\frac{1}{\text{det}\mathbf{A}}\left[\int\frac{\mathrm{d}z\mathrm{d}\bar{z}}{2i\pi}\mathrm{e}^{-\bar{z}z}\right]^n=\frac{1}{\text{det}\mathbf{A}} \end{equation} My Questions:

$(1)$: Why can this transformation be done in the first place? $z$ and $\bar{z}$ are complex conjugates of each other, how can I ignore one when I'm transforming the other? The statement I added from his book somewhat clarifies things but still confuses me a lot.

$(2)$: I don't understand why in the last integral $\mathrm{d}z_i$ isn't replaced by $\mathrm{d}z_i'$, I think this should be the case. Then the integral: \begin{equation} \int\frac{\mathrm{d}z'\mathrm{d}\bar{z}}{2i\pi}\mathrm{e}^{-\bar{z}z'} \end{equation} is not equal to $1$ because $\bar{z}$ and $z'$ are not complex conjugates. So isn't the last equation written wrong?

Μπαμπης Ποζουκιδης
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  • I'm not familiar with conventions in QM, and I do not own this book, so forgive me if these are stupid questions, but:$\tag{}$ 1. What precisely are you integrating over? Is it a product of specific contours, a hyperrectangles, or something else?$\tag{}$ 2. Are you completely sure $z_i$ and $\overline{z}_i$ are intended to be complex conjugates in this instance? It looks to me like they are being treated as purely separate variables. – Christian E. Ramirez Nov 29 '22 at 03:32
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    I will edit my question to add a part from the book – Μπαμπης Ποζουκιδης Nov 29 '22 at 09:58
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    The notation isnt dangerous at all. $\overline{\frac{x+iy}{\sqrt 2}}=\frac{x-iy}{\sqrt 2}$ – K.defaoite Nov 29 '22 at 10:15
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    It's his words not mine – Μπαμπης Ποζουκιδης Nov 29 '22 at 10:38
  • The $d\bar{z}$ similarly becomes $d\bar{z}'$. – J.G. Nov 29 '22 at 12:29
  • but he doesn't do it in his book. He does not transform $d\bar{z}$ – Μπαμπης Ποζουκιδης Nov 29 '22 at 14:21
  • I find the author is doing more to confuse than to clarify. It would have been easier if he had used $\bar{z}$ from the get go. Then introduced a new variables $z'=\bar{z}$ and $z''i=\sum{j}A_{ij}z_j$. His last integral then becomes $$\frac{1}{\text{det}\mathbf{A}}\left[\int\frac{\mathrm{d}z'\mathrm{d}z''}{2i\pi}\mathrm{e}^{-z''z'}\right]^n$$ and there is no issues. – Raskolnikov Dec 03 '22 at 11:59
  • The trickey thing is that pre-transformation you're integrating over a surface in $C^2$ space where $z$ is conjugates with $\conj{z}$. After the transofrmation, you're integrating over some weird ill-defined surface in $C^2$, so it's not at all clear how to evaluate that simply. – Eric Dec 04 '22 at 16:20
  • @Raskolnikov But why is this true? After this transformation $z$ and $z''$ are not complex conjugates anymore – Μπαμπης Ποζουκιδης Dec 05 '22 at 14:45

1 Answers1

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Notation in this answer: In this answer, let $z,\bar{z}\in \mathbb{C}$ denote two independent complex variables. Let $z^{\ast}$ denote the complex conjugate of $z$.

  • Concerning $n=1$ dimension: When Ref. 1 talks about integration $$\int_{\mathbb{C}}\frac{\mathrm{d}z^{\ast}\wedge\mathrm{d}z}{2\pi\mathrm{i}}\ldots~:=~\int_{\mathbb{R}^2}\frac{\mathrm{dRe}z \wedge\mathrm{dIm}z}{\pi} \ldots \tag{A}$$ in the complex plane $\mathbb{C}\cong \mathbb{R}^2$, it is mainly talking about Gaussian integration (to later in Ref. 1 be applied to perturbation theory in QM). It is a well-known property of Gaussian integrals that $$\forall a,b\in\mathbb{C}:~~ \int_{\mathbb{R}^2}\frac{\mathrm{dRe}z \wedge\mathrm{dIm}z}{\pi} e^{-({\rm Re}z-a)^2-({\rm Im}z-b)^2}~=~1, \tag{B}$$ or equivalently, $$\forall a,b\in\mathbb{C}:~~ \int_{\mathbb{C}}\frac{\mathrm{d}z^{\ast}\wedge\mathrm{d}z}{2\pi\mathrm{i}} e^{-(z^{\ast}-a)(z-b)}~=~1. \tag{C}$$ This fact more or less justifies the following quote from Ref. 1:

    Indeed, we stress that the complex variables $z$ and $\bar{z}$ are independent integration variables and are complex conjugate only in a formal sense: the integration contours over the variables $x$ and $y$ could be deformed in complex space and, then, these variables would also take complex values. The symbol $\mathrm{d}\bar{z}\wedge \mathrm{d}z$ thus corresponds to an integration over a surface of real dimension 2, embedded in $\mathbb{C}^2$.

    See also e.g. this related Phys.SE post.

  • Concerning a positive definite $n\times n$ matrix ${\bf A}$: The Gaussian formula $$ {\cal Z}({\bf A})~=~\frac{1}{\det {\bf A}} \tag{D}$$ can be proven via diagonalization with a unitary matrix, as is done in Ref. 1. Purely algebraic substitution with an asymmetric change of variables in the integral is justified, since it leads to the very same formula. This answers OP's 1st question.

    Ref. 1 argues furthermore that the rational right-hand side can be extended by analytic continuation to any invertible complex matrix (although the Gaussian integral might not be convergent).

    See also e.g. this related Math.SE post.

References:

  1. J. Zinn-Justin, Path Integrals in QM, 2010; p. 136-137.
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