QuantumPhysics.dvi

The norms ofJ±|j,m±〉were already computed earlier, and thus require that

λ(j)−m+(m++ 1) = 0

λ(j)−m−(m−−1) = 0 (5.97)

Eliminatingλ(j) gives a relation directly betweenm+andm−, (m++m−)(m+−m−+1) = 0.

Since we have m+−m− ≥ 0, the second parenthesis never vanishes, and we must have

m−=−m+. We assumed that the quantum system was irreducible, so that forgivenj, all

states with differentmare mapped into one another underJ±. Thus, successively applying

J−to|j,m+〉must ultimately yield the state|j,m−〉. Since the number of timesJ−is being

applied is of clearly an integer, we must have thatm+−m−= 2m+is a positive or zero

integer. We definejto be this integer or half-integerj≡m+, so that

λ(j) =j(j+ 1) 2 j+ 1∈N (5.98)

The integer 2j+ 1 is the dimension of the Hilbert space Hj for this irreducible quantum

system, since the number of basis vectors|j,m〉withm=−j,−j+ 1,−j+ 2,···,j− 1 ,jis

precisely 2j+ 1. For later convenience, we list the matrix elements of all the operators,

J^2 |j,m〉 = j(j+ 1) ̄h^2 |j,m〉

J 3 |j,m〉 = mh ̄|j,m〉

J+|j,m〉 =

√

(j−m)(j+m+ 1) ̄h|j,m+ 1〉

J−|j,m〉 =

√

(j−m+ 1)(j+m) ̄h|j,m− 1 〉 (5.99)

5.8 The Coulomb problem

The Coulomb problem involves a single mobile charged particle in the electro-static poten-

tial of a fixed central charge. In Hydrogen-like atoms, for example, the fixed charge is a

nucleus and the mobile charge is an electron. The Coulomb potential isthen attractive and

produces bound states. But it also covers systems in which the mobile charge is a proton or

a positron, so that no bound states exist. For the time being, effects of spin are neglected.

The Schr ̈odinger equation for the Coulomb problem with chargesq 1 andq 2 is given by,

−

̄h^2

2 m

∆ψE(r) +

q 1 q 2

r

ψE(r) =EψE(r) (5.100)

As for any spherically symmetric problem, we work in spherical coordinatesr,θ,φ, in terms

of which the Laplace operator takes the form,

∆ =

1

r^2

∂

∂r

(

r^2

∂

∂r

)

−

L^2

̄h^2 r^2

(5.101)