Wigner functions of $s$ waves

J. P. Dahl, S. Varro, A. Wolf, and W. P. Schleich

Phys. Rev. A 75, 052107 – Published 11 May 2007

Abstract

We derive explicit expressions for the Wigner function of wave functions in $D$ dimensions which depend on the hyperradius—that is, of $s$ waves. They are based either on the position or the momentum representation of the $s$ wave. The corresponding Wigner function depends on three variables: the absolute value of the $D$ -dimensional position and momentum vectors and the angle between them. We illustrate these expressions by calculating and discussing the Wigner functions of an elementary $s$ wave and the energy eigenfunction of a free particle.

Received 15 January 2007

DOI:https://doi.org/10.1103/PhysRevA.75.052107

Authors & Affiliations

J. P. Dahl^1,2, S. Varro^3,2, A. Wolf², and W. P. Schleich²

¹Chemical Physics, Department of Chemistry, Technical University of Denmark, DTU 207, DK-2800 Lyngby, Denmark
²Institut für Quantenphysik, Universität Ulm, D-89069 Ulm, Germany
³Research Institute for Solid State Physics and Optics, H-1525 Budapest, P.O. Box 49, Hungary

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Issue

Vol. 75, Iss. 5 — May 2007

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Images

Figure 1
Wigner function $W_{S}$ of the shell wave function corresponding to the limit $a \to \pm \infty$ in Eq. (32) in its dependence on the angle $θ$ between $\vec{r}$ and $\vec{k}$ (horizontal) and the number $D$ of dimensions (vertical). Here we do not show the Wigner function $W_{S}$ but $W_{S} {(r k)}^{D - 1} (\sin^{D - 2} θ) S^{(D)} S^{(D - 1)}$ . In the bottom of each figure we display contour lines of $W_{S}$ . The thick line marks the curve where the Wigner function vanishes, separating positive domains from negative domains. The horizontal axes $r$ and $k$ are identical in all figures. However, the vertical axis changes with increasing angles—that is, going from one row to the next—but is identical in each column.Reuse & Permissions
Figure 2
Wigner function $W_{\max}$ of the maximally entangled state corresponding to the choice $a = - 2 ∕ D$ in Eq. (32). The arrangement of figures is as in Fig. 1.Reuse & Permissions
Figure 3
Wigner function $W_{E}$ corresponding to the free nonrelativistic particle in an eigenstate of energy $E \equiv {(ℏ k_{0})}^{2} ∕ (2 M)$ with $k_{0} = 1$ in its dependence on the absolute value $r$ of the position vector (horizontal) and the number $D$ of dimensions (vertical). Here we do not show the Wigner function $W_{E}$ but $V^{(D)} W_{E} = {(r k)}^{D - 1} (\sin^{D - 2} θ) S^{(D)} S^{(D - 1)} W_{E}$ . Along the thick line the Wigner function vanishes, separating positive domains from negative domains. The horizontal axis $θ$ and $k$ are identical in all figures. However, the vertical axis changes with increasing angles—that is, going from one row to the next—but is identical in each column.Reuse & Permissions
Figure 4
Integration over the wave vector $\vec{q}$ in the Fourier ansatz, Eq. (54), for the Wigner function $W_{E}$ of an energy eigenstate. Since the Schrödinger equation (52) for $W_{E}$ determines the length $q_{0}$ of $\vec{q}$ , we are left with an integration over a sphere in $D$ dimensions. Due to the transversality condition, Eq. (57), this integration takes place in the space orthogonal to $\vec{k}$ , leading to a reduction of the number of dimensions. For the case of $D = 3$ depicted here we find a circle.Reuse & Permissions
Figure 5
Orientation of the $\vec{ξ}$ -coordinate system employed to perform the integration over the $D$ Cartesian variables $ξ_{1}, \dots, ξ_{D}$ relative to the given vectors $\vec{r}$ and $\vec{k}$ . The ${\vec{e}}_{1}$ axis defining the component $ξ_{1}$ of $\vec{ξ}$ is along $\vec{r}$ whereas the ${\vec{e}}_{2}$ axis defining the component $ξ_{2}$ of $\vec{ξ}$ is orthogonal to $\vec{r}$ and in the plane defined by $\vec{r}$ and $\vec{k}$ . Here we have only depicted the case $D = 3$ .Reuse & Permissions
Figure 6
Orientation of $\vec{q}$ -coordinate system employed to perform the integration over the $D$ Cartesian variables $q_{1}, \dots, q_{D}$ relative to the given vectors $\vec{r}$ and $\vec{k}$ . In contrast to Fig. 5 the ${\vec{e}}_{1}$ axis defining the component $q_{1}$ of $\vec{q}$ is along the wave vector $\vec{k}$ and the ${\vec{e}}_{2}$ axis is orthogonal to $\vec{k}$ and in the plane defined by $\vec{r}$ and $\vec{k}$ . Here we have only depicted the case $D = 3$ .Reuse & Permissions

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