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266 7 QUANTUM THEORY
�e �rst term on the le� hand side depends only on r, but the second term
depends on r and the angles. Hence, to make this equation separable,
multiply through by r2, and rearrange to give
− ħ2r
2mR
∂2(rR)
∂r2
− Er2 = ħ2Λ̂2Y
2mY
�e le� side is dependent only on r and the right side depends only on
the angles. Hence, this equation is only true for all r and all angles if both
sides are equal to the same constant, arbitrarily called λ. �is gives two
equations
− ħ2r
2mR
∂2(rR)
∂r2
− Er2 = λ
ħ2Λ̂2Y
2mY
= λ
�e solutions to the second equation are the spherical harmonics, which
are eigenfunctions of the operator Λ̂2, with eigenvalue −l(l + 1), which
gives −ħ2 l(l +1)Y/2mY = λ and so λ = −ħ2 l(l +1)/2m.�is makes the
equation in r
− ħ2r
2mR
∂2(rR)
∂r2
− Er2 = −ħ
2 l(l + 1)
2m
which is rearranged to
− ħ2
2mr2
∂2(rR)
∂r
+ ħ2 l(l + 1)
2mr2
R = ER
(b) For the case when l = 0, this becomes
− ħ2
2mr
∂2(rR)
∂r2
= ER
For the trial solution R(r) = N sin(nπr/a)/r, and so rR = N sin(nπr/a)
− ħ2
2mr
∂2(N sin(nπr/a))
∂r2
= E N sin(nπr/a)
r
Note that d2 sin ax/dx2 = da cos ax/dx = −a2 sin ax
− ħ2
2mr
× −(nπ
a
)
2
N sin(nπr/a) = EN sin(nπr/a)/r
Which is solved if
E = ħ2(nπ)2
2ma2
= (h/2)2n2
2ma2
= h
2n2
8ma2
where ħ = h/2π has been used
(c) As the wavefunction must be zero outside the cavity, and be continuous,
the wavefunction must equal zero at r = a, for all angles, and so the
boundary condition is that R(a) = 0. Hence sin(nπa/a)/a = 0, which
is true if sin(nπ) = 0, which implies that n must be integral. Hence, the
allowed energies are En = n2h2/8ma2, n = 1, 2, 3...