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Hydrogen and the Central Force Problem

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Hydrogen and the Central Force Problem Hydrogen and the Central Force Problem Our treatment of quantum mechanics is getting increasingly more realistic. In the last chapter, we made the leap from 1 to 3 dimensions. In this chapter we solve our first problem with a physically realistic potential– the attractive electrostatic (Coulomb) potential between two oppositely charged particles. In the case of atomic hydrogen these particles are an electron and a proton, but the theory can be easily extended to one electron ions such as Li++, muonic atoms where a muon is captured in orbit about a proton, or positronium (an electron in orbit about an antielectron). In this chapter we will ignore the complications of electron spin. Heretofore we have been discussing the motion of a single particle in an “external” potential of mysterious origin. Because the proton is about 2000 times more massive than the electron and doesn’t move very much, we could think of it as providing an external potential to the “orbiting” electron and still get fairly accurate results. However positronium gives us a great excuse to discuss two particle wave functions which interact in a potential mutually created by both particles. The Two Particle Reduction Lets consider two particles in one dimension. The first particle has mass m1 and is located at x1 ; the second has mass m2 and is located at x2. The wave func- tion becomes a function of both x1 and x2 : ψ(x1,x2). It is natural to generalize 2 the probabilistic interpretation of the wave function PDF(x1,x2)= |ψ(x1,x2)| , where PDF(x1,x2) gives the probability density of finding particle 1 at position x1 and simultaneously finding particle 2 at position x2. To clarify this last point, the probability of finding particle 2 within a distance ∆ of particle 1 would be written as: +∞ x1+∆ ∗ P(|x1 − x2| < ∆) = dx1 dx2 ψ (x1,x2)ψ(x1,x2) −∞ x1−∆ 1 Thewavefunctionψ(x1,x2)obeysaSchr¨odinger equation which is built on 2 2 the total energy of the two particle system: E = p1/(2m1)+p1/(2m1)+V (x1,x2). 2 2 −¯h ∂2 −¯h ∂2 2 ψ(x1,x2)+ 2 ψ(x1,x2)+V (x1,x2) ψ(x1,x2)=Eψ(x1,x2)(1) 2m1 ∂x1 2m2 ∂x2 We thus have a partial differential equation which is traditionally solved using separation of variables. The separation of variables would be trivial if V (x1,x2)=V1(x1)+V2(x2) since we just have a diffiQ in x1 added to a diffiQ in x2. This would be the situation if particle 1 and 2 moved in external potentials but felt no mutual interaction. In the case of positronium, we usually have the opposite problem of zero external potential but the electron and positron (antielectron) attract each other with a potential related to their separation. The one dimensional analogy would be V (x1 −x2). It is very difficult to separate variables for this potential in the original (x1,x2) coordinates but there is a classic change of variables trick which allows for a separation of variables solutions. The clue comes from classical mechanics. In many classical situations motion about center of mass is essentially decoupled from motion of the center of mass. An example would be if one were to throw a system consising of two masses coupled with a spring from the surface of the moon. The masses will oscillate following simple harmonic motion about their center of gravity; while the center of mass follows the same parabolic trajectory as a point like particle in the grav- itational field of the moon. Guided by the classical problem, we tranform the variables of Eq. (1) into the position of the center of mass X and the position of x1 relative to x2: m1 m2 X = x1 + x2 where M = m1 + m2 M M x = x1 − x2 (2) 2 We next transform the derivatives using the chain rule. ∂ ∂X ∂ ∂x ∂ ∂ m1 ∂ ∂ = + , = + ∂x1 ∂x1 ∂X ∂x1 ∂x ∂x1 M ∂X ∂x ∂ ∂X ∂ ∂x ∂ ∂ m2 ∂ ∂ = + , = − (3) ∂x2 ∂x2 ∂X ∂x2 ∂x ∂x2 M ∂X ∂x 2 2 The kinetic energy operator Tˇ = p1/(2m1)+p2/(2m2) can be written in terms of the derivatives of Eq. (3) 2 −¯h m1 ∂ ∂ m1 ∂ ∂ Tˇ = + + 2m1 M ∂X ∂x M ∂X ∂x 2 −¯h m2 ∂ ∂ m2 ∂ ∂ + − − (4) 2m2 M ∂X ∂x M ∂X ∂x We note that the cross terms proportional to ∂2/(∂x∂X) cancel leaving the terms: − 2 2 − 2 2 ˇ ¯h m1 + m2 ∂ ¯h 1 1 ∂ T = 2 2 + + 2 2 M ∂X 2 m1 m2 ∂x − 2 2 − 2 2 ˇ ¯h ∂ ¯h ∂ 1 1 1 T = 2 + 2 where = + (5) 2M ∂X 2µ ∂x µ m1 m2 We see that the kinetic energy has a contribution from the center of mass motion X which enters with a mass M = m1 + m2 and a contribution from the relative −1 −1 −1 motion x = x1 − x2 which enters with an “reduced mass” µ = m1 + m2 . Lets now assume a potential of the form V (x1,x2)=Vr(x1 − x2)+Vcg(X) which might be applicable, for example, to two masses connected with a spring 2 which experience a potential Vr(x1 − x2)=κ(x1 − x2) /2, as the mass spring system falls in a gravitational field Vcg(X)=(m1 gx1 +m2 gx2)=MgX.The 3 Schr¨odinger equation reads: 2 2 −¯h ∂2 −¯h ∂2 + V (x) ψ(x, X)+ + V (X) ψ(x, X)=Eψ(x, X) 2µ ∂x2 r 2M ∂X2 cg r cg (6) We now have a classic separation of variables situation. We factor the wave function as ψ(x, X)=ψr(x) × ψcg(X). Inserting this into Eq. (6) and dividing both sides by ψr(x) × ψcg(X) we get two separate Schr¨odinger equations: 2 −¯h d2 ψ (x)+V (x)ψ (x)=E ψ (x) 2µ dx2 r r r r r 2 −¯h d2 ψ (X)+V (X)ψ (X)=E ψ (X) ,E+ E = E (7) 2M dX2 cg g cg cg cg r cg To gain insight into Eq. (7) lets consider the specific example of two masses m1 and m2 undergoing simple harmonic motion, while their center of gravity travels to the right with a total momentumhK ¯ . If the two masses are tied together with a spring with spring constant κ, the oscillation frequency will be −1 −1 −1 ω = κ/µ where the “reduced mass” is µ = m1 + m2 . The ψcg(X) portion of the wave function which describes the center of gravity is just a free particle traveling wave solution of the Schr¨odinger equation of the form: ψcg(X)=exp(iKX) ; while the ψr(x) portion of the wave function which describes the relative coordinate x1−x2 is constructed out of Hermite polynomials √ 2 and given by ψr(x)=Hn( βx)exp(−βx/2). The complete wave function is: 2 µω ψ(x, X)=N H ( βx) e−βx /2 eiKX where β = (8) n ¯h The energy of the system, or E in Eq. (6), is just equal to the sum of Er (the harmonic oscillator energy) and Ecg (the kinetic energy of the center of mass 4 which acts as a particle of mass M traveling with a momentumhK ¯ . 2 1 ¯h K2 E = n + ¯hω + (9) 2 2M A brief summary might be in order. Quite often in reality, we have the problem of describing a situation where two particles interact with each other through a potential which depends on their separation. In such cases, one can affect a separation of variable solution by writing the Schr¨odinger equation in terms of the relative displacement of two particles and location of their center of mass. We’ve shown that the equation describing the center of mass, becomes totally independent of the equation describing the relative coordinates. The Schr¨odinger equation for the relative displacement between the two bodies is exactly the same as the equation for a single particle of reduced mass µ interacting in an external potential V (x)=V (x1 − x2). Hence the fact that we have two particles rather than one creates no additional mathematical complications at all! The Angular Reduction Besides the reduction of a two particle system down to a one particle system with a reduced mass µ, the conservation of angular momentum creates additional simplifications in the central force problem in both classical mechanics and quan- tum mechanics. A few words about the classical orbit problem (illustrated below) wouldn’t hurt. r V(r) r φ r φ Perigee Apogee 2 L = µ r φ r = µ r φ 5 We have described the elliptical orbit in terms of the polar coordinates r and ˙ 1 φ. The tangential velocity is vφ = r φ; while the radial velocity isr ˙. The energy expression is given by: 1 E = µ r˙2 +(rφ˙)2 + V (r) (10) 2 2 ˙ The angular momentum, given by L = µvφ r = µr φ, is a constant of motion. We can express φ˙ in terms of L and insert it in Eq. (10) to obtain an expression in r only: 2 L 1 2 L φ˙ = ,E= µr˙ + + V (r) (11) µr2 2 2 µr2 Eq. (11) is very analogous to the conservation of energy in a one dimensional 1 2 potential well: E = 2 m x˙ + V (x). To complete the analogy we combine the angular momentum term with the potential actual potential energy to form an “effective potential”. L2 Veff = + V (r) (12) 2 µr2 The L2/(2 µr2) term is often called the “centrifugal potential”. We illustrate radial motion of the orbit for a V = −K/r potential in an effective potential well below: V eff 2 -K L V = + 2 eff r 2µ r dominated by centrifugal potential r E Perigee Apogee dominated by Coulomb potential 1 In classical mechanics, one often writesr ˙ to stand for dr/dt.
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