A theorem concerning radial symmetric elements under the Bargmann transform is stated and proved. It is shown that for any \(f\in S^{'}_{1/2}(\mathbb R^d)\) the following conditions are equivalent. \(1^{\circ}\) \(f\) is radial symmetric; \(2^{\circ}\) if \(U\) is unitary on \(\mathbb R^d\), then \((\mathfrak{V}f)(Uz)=(\mathfrak{V}f)(z)\); \(3^{\circ}\) \((\mathfrak{V}f)(z)=F_0(\langle z, z\rangle)\), for some entire function \(F_0\) on \(C\); \(4^{\circ}\) \((f,h_{\alpha})=0\) for every \(\alpha=(\alpha_1,\ldots,\alpha_d)\in N^d\) such that at least one of \(\alpha_j\) is odd, and \(\dfrac{\alpha !}{\sqrt{(2\alpha)!}}(f,h_{2\alpha})=\dfrac{\beta !}{\sqrt{(2\beta)!}}(f,h_{2\beta})\) for any \(\beta\in\mathbb N^d\) with \(| \alpha| =| \beta| \).
Here \(\mathfrak{V}f\) is the Bargmann transform. The element \(f\in\mathcal S^{'}_{1/2}(\mathbb R^d)\), where \(\mathcal S^{'}_{1/2}(\mathbb R^d)\) is the Gelfand-Shilov distribution space, is called symmetric, if the pullback \(U^{*}f\) is equal to \(f\), for every unitary transformation \(U\) on \(\mathbb R^d\).