chapter4 - Chapter 4 BOUND STATES OF A CENTRAL POTENTIAL...

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Chapter 4 BOUND STATES OF A CENTRAL POTENTIAL 4.1 General Considerations As an application of some of these ideas we brie f y investigate the properties of a particle of mass m subject to a force deriving from a spherically symmetric potential V ( r ) . Obviously, in many of the cases where this problem is of interest, the mass m referred to is the reduced mass m = m 1 m 2 m 1 + m 2 (4.1) of an interacting two-particle system and the radial coordinate r corresponds to the mag- nitude of the relative position vector ~ r = ~ r 2 ~ r 1 , although these details need not concern us here. The corresponding quantum system is governed by a Hamiltonian operator H = P 2 2 m + V ( R ) (4.2) which in the position representation takes the usual form ~ 2 2 2 m + V ( r ) , (4.3) and our goal is information about the bound state solutions | ψ i to the energy eigenvalue equation ( H ε ) | ψ i =0 , (4.4) whereweassumethat V ( r ) 0 as r →∞ , so that bound state solutions are identi F ed as those square normalizeable solutions h ψ | ψ i = Z d 3 r | ψ ( ~ r ) | 2 =1 (4.5) for which ε 0 . The spherical symmetry of V ( r ) ( and thus of H ) suggest the use of spherical coordinates for which the identity 2 = 1 r 2 r 2 r + 1 r 2 · 2 ∂θ 2 +cot θ + 1 sin 2 θ 2 ∂φ 2 ¸ (4.6) holds for all points except at r , which is a singular point of the transformation. This last expression can be written in the form P 2 = ~ 2 2 = ~ 2 r 2 r 2 r + ~ 2 L 2 r 2 (4.7) where L 2 = · 2 2 θ + 1 sin 2 θ 2 2 ¸ (4.8)
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124 Bound States of a Central Potential is the operator, in this representation, associated with the square of the (dimensionless) orbital angular momentum ~ L = ~ R × ~ K. Thus, as in classical mechanics, the kinetic energy H 0 separates into a radial part and a rotational part H 0 = P 2 r 2 m + ~ 2 L 2 2 mr 2 , (4.9) where P 2 r = ~ 2 1 r 2 r 2 r = · ~ i 1 r r r ¸ 2 . (4.10) This last form suggests that the operator P r = ~ i 1 r r r (4.11) can be viewed as the radial component of the momentum. This is a legitimate inference, and indeed P r as de f ned above does correspond to the Hermitian part P r = 1 2 h ~ P · ˆ R + ˆ R · ~ P i . (4.12) of the operator ~ P · ˆ R . Some care must be taken, however, because (it can be shown that) P r is only Hermitian on the space of wave functions ψ ( ~ r ) such that lim r 0 r ψ ( ~ r )=0 , while the eigenfunctions of P r all diverge at r =0 as r 1 . Thus, P r is an example of a Hermitian operator that is not an observable. None of this is too important for the task at hand, however. Indeed, with the aforementioned formulae we can write the energy eigenvalue equation of interest in the form ½ ~ 2 2 m 1 r 2 r 2 r + ~ 2 L 2 2 mr 2 + V ( r ) ε ¾ ψ ( ~ r (4.13) which is valid at all points except r . We now make the observation that L 2 and L z = i / ∂φ (or any other component of ~ L ) commute with the each other and with any function of r or P r .I t i so b v i o u s , therefore, that they commute with H .Thu s ,be cau se £ H,L 2 ¤ = £ L 2 ,L z ¤ =[ z ]=0 (4.14) we know that there exists a basis of eigenstates common to the three operators 2 , and L z .
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This note was uploaded on 12/19/2010 for the course PHYSICS ph 463 taught by Professor Paule. during the Fall '10 term at Missouri S&T.

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chapter4 - Chapter 4 BOUND STATES OF A CENTRAL POTENTIAL...

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