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I was going through the derivation of Wigner distribution properties, and encountered a certain step in the proof I could not justify.

Namely, the step requires the following equality to be true:

$\int d^2 \lambda \frac{\partial}{\partial \lambda} \left( \exp(-\lambda \alpha^* + \lambda^* \alpha) f(\lambda) \right) \equiv 0$,

where $\alpha$ and $\lambda = x + iy$ are complex numbers, $\int d^2 \lambda \equiv \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} dx dy$, and the differential is a Wirtinger one:

$\frac{\partial}{\partial \lambda} \equiv \frac{1}{2} \left( \frac{\partial}{\partial x} - i \frac{\partial}{\partial y} \right)$.

The function $f$ is bounded, continuous and infinitely differentiable (w.r.t. $x$ and $y$), but not necessarily going to zero on the infinity.

As far as I understand, this is a Fourier transform, and the equality would reduce to a known property (Fourier transform of derivative), if $\lim_{x \rightarrow \infty} f = 0$ and $\lim_{y \rightarrow \infty} f = 0$, which is, unfortunately, not the case. Could anyone point me to the idea behind the proof in such circumstances?

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2 Answers 2

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Returning to this question: it turned out that $f$ actually goes to zero on the infinity. In my case $f(\lambda) \equiv \mathrm{Tr} \{ \hat{A} \hat{D}(\lambda) \}$, where $\hat{A}$ is a Hilbert-Schmidt operator, and $\hat{D}$ is the displacement operator. This trace is square-integrable, as proved in K. Cahill and R. Glauber, Ordered Expansions in Boson Amplitude Operators, Phys. Rev. 177, 1857 (1969). With this amendment, the proof is trivial.

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I have never seen the derivation of this but, if $f$ is entire and bounded (is not it?) then by Liouville's theorem it is constant. This seems to be enough to get the result you are looking for.


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Oh, I completely forgot about this question. No, the problem was that $f$ was not entire (that's why I only used Wirtinger differentials). I have managed to strengthen "$f$ is bounded" to "$f$ goes to zero on the infinity" since then, for which the proof is trivial. I should probably close the question now, as I'm not even sure this equality can be proved with my initial conditions. – fjarri Oct 25 '13 at 10:24
Lioville's theorem only applies to (complex) analytic functions. OP has not specified that $f$ has this property. – Winther Sep 29 at 20:44

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