Schrödinger Equation for Particle in Ring

The Schrödinger Equation for particle on a ring:We take the Potential energy to be zero for a particle at a distance r from thecenter and to be infinitely high at all other positions.The eigenfunctions will be functions of the angular variable ø, so we willrepresent the eigenfunctions by Φ instead of the general symbol ψ.So we write the general Schrödinger EquationHψ = εψAsHΦ= εψfor a particle on a ring.Thus we can write Schrödinger Equation- h 2 d 2 Φ8π 2 I dx 2= εΦ --------- (1)The function that satisfy the equation (1) and is shown asΦ = A e imøwhere im is an imaginary numberThe first order derivative with respect to ø isdΦ / dø = im A e imøand the second order derivative with respect to ø isd 2 Φ / dø 2 = im (im) A e imø

d 2 Φ / dø 2 = -m 2 A e imø= - m 2 ΦThus, the equation (1) becomes- h 2 (- m 2 Φ ) = εΦ8π 2 Ih 2 m 2 = ε --------- (2)8π 2 Iwhere m = 0, ±1, ±2, ±3 ……This is the same result, which was shown by fitting de Broglie waves intocircumference with a general symbol n for quantum number.The pattern of the energies of the allowed rotational states is shown asfollows.

For a particle-on-a-ring, we can write the probability as2π∫ Φ 2 dø = 10When solutions to the Schrödinger equation satisfy such equation, thesolutions functions are said to be normalized.The normalization of function Φ, gives the value for A.2π∫ Φ 2 dø = 102π∫ Φ * Φ dø = 102π∫ (Ae -imø ) (Ae imø ) dø = 102π∫ A 2 dø = 102πA 2 [ ø ] = 10A 2 [ 2π ] = 1

A = 1 / 2π ------- (3)Thus the normalized eigenfunction Φ can be written asΦ = A e imøΦ = (1 / 2π )( e imø ) -------- (4)where m = 0, ±1, ±2, ±3 ……Angular-Momentum values:We can use this function to get the values of angular momentum operatorand deduce the angular momentum associated with each quantum stateL z Φ = (h / 2 π i) ( dΦ /dø)From equation (4)h d 1L z Φ = ( e imø )2 π i dø 2πh 1L z Φ = (im) ( e imø )2 π i 2πFrom equation (4)hL z Φ = (im) Φ2 π i

hL z Φ = m Φ2 πThus, the angular momentum ishL z = m ------ (5)2 π where m = 0, ±1, ±2, ±3 ……Thus, the two states at each energy other than for m = 0, have angularmomenta which correspond to the particle moving clockwise and anticlockwise,depending upon the sign of m and are uniformly distributedaround the ring.The Heisenberg Uncertainty Principle and the Particle on the ring:According to Heisenberg, there is an inherent limitations to the related orconjugate properties of particles.The momentum p and the position x are the conjugate properties.According to Heisenberg, we cannot know the precise values of p and x.Any calculated values of p and x will be uncertain to the extent of ∆p and∆x.And thery are related to each other as∆p ∆x = hwhere h is planck’s constant.At the lowest energy state, i.e. m = 0The angular momemtum Lz

From equation 5hL z = m = 02 πRotation in three dimesions:The Schrödinger equation for a particle in a three dimensional system iswritten as follows if the Potential energy experienced by a particle is zero:- h 2 d 2 ψ d 2 ψ d 2 ψ+ + = εψ -------- (6)8π 2 m dx 2 dy 2 dz 2The Schrödinger equation can be expressed in terms of polar co-ordinatesand moment of inertia I = mr 2 as follows:8π 2 I sinθ dθ sin 2 θ 2- h 2 1 d dψ 1 d 2 ψsinθ + = εψdθ dø------- (7)

Angular Momentum:The angular momentum operator along z direction is given asL z = (h / 2 π i) [x (d/ dy) - y (d/dx)]SimilarlyThe angular momentum operator along x direction is given asL x = (h / 2 π i) [y (d/ dz) - z (d/dy)]AndThe angular momentum operator along y direction is given asL y = (h / 2 π i) [z (d/ dx) - x (d/dz)]Thus, the square of the total angular momentum operator for threedimesional system is given asL 2 = L x2+ L y2+ L z2Using Polar co-ordinates, the angular momentum operator in z-directionisgiven asL z= (h / 2 π i) ( d /dø)And the square of the total angular-momentum operator ash 2 1 d d 1 d 2L 2 = (sinθ) +4π 2 I sinθ dθ dθ sin 2 θ dø 2------- (8)

As we know that from equation (5), the angular momentum in z direction isquantized in units of (h/2π) and is given ashL z = m2 π where m = 0, ±1, ±2, ±3 ……Then, the total angular momentum is given ashL = l ( l + 1) ------ (9)2 πwhere l = 0, 1, 2, 3 ……andwhere m = 0, ±1, ±2, ±3 …… , ±l.