In Altland and Simons' condensed matter book, complex Gaussian integrals are introduced. Defining $z = x + i y$ and $\bar{z} = x - i y$, the complex integral over $z$ is
$$\int d(\bar{z}, z) = \int_{-\infty}^\infty dx \, dy.$$ In this way, any integral over $z$ can be done by just breaking into real and imaginary parts.

I'm confused about how one would actually use the notation on the left, as it is. It seems it must have some meaning besides just $dx \, dy$, or there should be no point in introducing it.

It is possible to break the double integral $\int d(\bar{z}, z)$ into two single complex integrals and do them individually? For example, if we were to write $$\int d(\bar{z}, z) = \int d \bar{z} \int dz$$ then what would the bounds of integration be? For the inner integral, isn't the value of $z$ determined by the value of $\bar{z}$? Alternatively, if we regard $z$ and $\bar{z}$ as independent, then where does the constraint $\bar{(z)} = \bar{z}$ come in? Should each of these integrals be regarded as regular integrals or contour integrals? If we don't break the integral into two, is $d(\bar{z}, z)$ some kind of area element? In that case, how do you do a complex surface integral?

Overall I don't understand what object $d(\bar{z}, z)$ is. What is it, and how do we integrate over it?

  • 2
    $\begingroup$ Which Mathematics questions? $\endgroup$
    – Qmechanic
    Feb 1, 2017 at 6:29
  • $\begingroup$ May be the following is useful : Under equation (3.17) $$ \int d\left(\mathbf{v}^\dagger,\mathbf{v}\right)e^{-\mathbf{v}^\dagger\mathbf{A}\mathbf{v}}=\pi^{N}\det\mathbf{A}^{-1} \tag{3.17} $$ the authors note : "...where $\:\mathbf{v}\:$ is a complex N-component vector, $\:d\left(\mathbf{v}^\dagger,\mathbf{v}\right)\equiv \prod\limits_{i=1}^{N}d\mathrm{Re}\:v_{i}\,d\mathrm{Im}\:v_{i}\:$, and $\:\mathbf{A}\:$ is a complex matrix with positive definite Hermitian part." $\endgroup$
    – Frobenius
    Feb 1, 2017 at 10:23

3 Answers 3


The complex notations$^1$ $$\int_{\mathbb{C}}\!\mathrm{d}z^{\ast}~\mathrm{d}z, \tag{1}$$ $$\int_{\mathbb{C}}\!\mathrm{d}^2z, \tag{2}$$ and similar notations, mean a real double integration $$N\iint_{\mathbb{R^2}} \!\mathrm{d}x~\mathrm{d}y\tag{3}$$ in the complex plane $\mathbb{C}\cong \mathbb{R}^2$ with coordinates $z=x+iy$, where $N$ is a conventional normalization factor that depends on the author.

$N=1$ convention:

  • A. Altland & B. Simons, Condensed matter field theory, 2nd ed., 2010. See e.g. sentence above eq. (3.11).

$N=2$ convention:

  • J. Polchinski, String Theory, Vol. 1, 1998. See e.g. eq. (2.1.7).

  • R. Blumenhagen, D. Lust & S. Theisen, Basic Concepts of String Theory, 2012. See e.g. footnote on p. 85.

  • D. Tong, Lectures on String Theory. See e.g. subsection 4.0.1.

$N=2i$ convention:

  • J.H. Negele & H. Orland, Quantum Many-Particle Systems, 1998. See e.g. eq. (1.124).


$^1$ Note that $z^{\ast}=x-iy$ denotes the complex conjugate variable. It is not an independent complex variable. In particular, the integration (1) is over $\mathbb{C}$. It is not over $\mathbb{C}^2$. See also e.g. this Math.SE post, this Phys.SE post and links therein.

  • 2
    $\begingroup$ In practice, is there ever a way to "integrate over $z$ and $\bar{z}$ without switching to $x$ and $y$? $\endgroup$
    – knzhou
    Feb 1, 2017 at 18:09
  • $\begingroup$ Yes. E.g. in the complex coherent state method. See e.g. this Math.SE post. $\endgroup$
    – Qmechanic
    Feb 1, 2017 at 19:36
  • $\begingroup$ Could you link to a resource that does an explicit calculation in this formalism? For example, are the calculations here legitimate? $\endgroup$
    – knzhou
    Feb 1, 2017 at 19:39

This issue often arises in conformal field theory, when we may be interested in Euclidean field theory but analytically continue to $\mathbb C^2$. Suppose we have real, Euclidean coordinates $(x,y)$ and form the complex coordinates,

$$z = x+iy, \quad \bar z = x-iy.$$

It is easy to show that the metric $dx^2 + dy^2 = dz d \bar z$, that is, $g_{zz} = g_{\bar z \bar z} = 0$ and $g_{z\bar z} = g_{\bar z z} = \frac12$. From this we can deduce that the measure for integration is,

$$dz d \bar z = 2 dx dy$$

and thus there is a factor of two difference between $\int d^2 z$ and $\int d^2 x$. We can treat $z$ and $\bar z$ as totally independent which then extends us to $\mathbb C^2$. To return to $\mathbb R^2 \subset \mathbb C^2$, we must make the identification that $\bar z = z^\star$, that is, they are related by conjugation and are not independent.

  • 1
    $\begingroup$ Could you give an example of a computation using $dz$ and $d\bar{z}$, that doesn't just switch back to $dx$ and $dy$? $\endgroup$
    – knzhou
    Feb 1, 2017 at 19:37
  • $\begingroup$ @knzhou Well, in general the contour integral $\int dz \int \bar z$ treating $z$ and $\bar z$ as independent means we integrate $z$ over some contour $C_1$ and $\bar z$ over some contour $C_2$. Note that the ordering matters; that is, $\int_{C_1} dz \int_{C_2} d\bar z \neq \int_{C_1} d \bar z \int_{C_2} dz$. There are further subtleties when integrating a multi-variable complex function over multiple contours. $\endgroup$
    – JamalS
    Feb 1, 2017 at 20:01
  • $\begingroup$ But what about the specific case of integrating over the complex plane? That's my issue -- it's confusing what the contours $C_1$ and $C_2$ would be in this case. $\endgroup$
    – knzhou
    Feb 2, 2017 at 1:48

I do not like this notation either, because it implies meaning that isn't there.

The authors define $$\int d(\bar{z}, z) \equiv \int_{-\infty}^\infty dx \, dy.$$ Note the $\equiv$ instead of an $=$ sign.

$d(z,\bar z)$ is literally the area element of the euclidean plane. The example at hand from the book is $$\int d(\bar{z}, z) e^{-\bar z w z} \equiv \int_{-\infty}^\infty dx \, dy e^{-x^2 w - y^2 w} =\sqrt{\frac{\pi}{w}}^2 $$

There are not two independent complex variables in play here. One is integrating functions from $\mathbb{C}\rightarrow\mathbb{C}$ seen as functions from $\mathbb{R}^2\rightarrow\mathbb{C}$.

If I had to guess I'd say they chose the notation because it might feel a little odd to write e.g.

$$\int dxdy\, e^{-\bar z w z}.$$

Maybe a less confusing choice of notation would be $d(z,\bar z) \equiv d\Re(z)d\Im(z) $ indicating integration over real and imaginary part separately.

  • $\begingroup$ I can't agree. If the notation really means nothing but two real integrals, then why spend a page covering complex Gaussian integrals if they're all the same as the real ones? $\endgroup$
    – knzhou
    Feb 1, 2017 at 8:05
  • $\begingroup$ Also, there's this question on Math.SE that demonstrates that manipulations with the $d(\bar{z}, z)$ might actually be possible/useful. $\endgroup$
    – knzhou
    Feb 1, 2017 at 8:06
  • $\begingroup$ Well, one complex Gaussians is just two real ones. If you look at the formulas you'll see they are perfectly analogous to the real case. But because they will be heavily used when the bosonic functional integral is introduced, there is good reason to give a thorough reference. Just glancing over the Math.SE post I'm not convinced, but I will have a look at it. Maybe there is more structure. $\endgroup$
    – Nephente
    Feb 1, 2017 at 8:17
  • $\begingroup$ I think you're right. More exactly the authors note : "Here, $\:\int d(\bar{z}, z) \equiv \int_{-\infty}^\infty dx \, dy\:$ represents the independent integration over the real and imaginary parts of $\:z = x + iy\:$. The identity is easy to prove: owing to the fact that $\: \bar{z}\, z = x^{2}+y^{2}\:$, the integral factorizes into two pieces, each of which is equivalent to Eq. (3.9) with $\:a = 2 w\:$." Eq. (3.9) is $$ \int\limits_{-\infty}^\infty dxe^{-\frac{a}{2}x^{2}} =\sqrt{\dfrac{2\pi}{a}},\quad \mathrm{Re}\:a > 0 \tag{3.9} $$ $\endgroup$
    – Frobenius
    Feb 1, 2017 at 10:15

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