I. INTRODUCTION II. THE SOLUTIONS OF THE SCHR ¨ ODINGER EQUATION FOR A CENTRAL POTENTIAL The Schr¨ odinger equation for a central potential is a ”classic ” example in undergraduate and first year graduate quantum mechanics courses (see [1], for example). It is a good exercise in the separation of variables; a good example of the conservation of angular momentum, and a starting point for treating the Hydrogen atom problem. An important concept that is introduced along with this problem is that of the ”effective radial potential”, where the central potential that appears in the radial equation - after separation of variables-acquires and additional term that depends on the total angular momentum quantum number l (see below). One then has a one-dimensional Schr¨ odinger equation with an effective potential The behavior of the radial wave function is interpreted in view of this effective potential. In particular, the new repulsive part, is interpreted as a centripetal barrier that attempts to throw the particle away from the origin, thus determining its shape for different values of l . The present paper analyzes the angular equation along similar lines (for the polar variable θ ) and demonstrate that, viewed as a one-dimensional Schr¨ odinger equation, we can identify an effective polar potential that shapes the angular part of the solution. The aim is to give the students of quantum mechanics a ”feeling” of how the spherical harmonics, that constitute the angular part of the wave function for any radial potential, assume the distribution they have about the z -axis. We will identify a polar potential - more precisely, a family of polar potentials that depend on the magnetic quantum number m - that squeezes the particle towards the xy -plane with a strength that increases with m . The physical picture introduced, and the intuition employed, we believe, will also provide the students with a simple prototype of the intuitive thinking that is employed by practicing physicists in understanding more complex real-world systems. V$_{eff}$ ( r ) = V ( r ) + ℏ 2 2 m l ( l + 1) r 2 (1) − ℏ 2 2 m ∇ $^{2}$ψ + V ψ = Eψ The Schr¨ odinger equation 2