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Past exam of the quantum physics course of the University of Oxford / 2016 / 9 / c / Solution by

Codex  0 2026-10-25

The linear confinement correction to Coulomb bound states uses

H^{'} = b r

, and its first-order energy correction in a normalized Coulomb eigenstate is

b ⟨ r ⟩

. Replace

a_{0}

by

a_{q}

in the hydrogen wavefunctions. The normalized spherical harmonics remove the angular integral, leaving

⟨ r ⟩_{n l} = \int_{0}^{\infty} r^{3} ∣ R_{n l} (r) ∣^{2} d r .

For

1 s

, the integral of

r^{3} e^{- 2 r / a_{q}}

yields

⟨ r ⟩_{1 s} = \frac{4}{a _{q}^{3}} \frac{3 !}{( 2/ a _{q} ) ^{4}} = \frac{3}{2} a_{q} .

For

2 s

, put

u = r / a_{q}

and square the full radial polynomial, including its cross term:

⟨ r ⟩_{2 s} = \frac{a _{q}}{2} \int_{0}^{\infty} u^{3} (1 - u + \frac{u ^{2}}{4}) e^{- u} d u = \frac{a _{q}}{2} (3! - 4! + \frac{5 !}{4}) = 6 a_{q} .

For

2 p

, the same calculation gives

⟨ r ⟩_{2 p} = \frac{a _{q}}{24} \int_{0}^{\infty} u^{5} e^{- u} d u = 5 a_{q} .

Consequently the first two Coulomb levels become, to first order,

E_{1 s} E_{2 s} E_{2 p, m} = - \frac{m _{q} a ^{2}}{4 ℏ ^{2}} + \frac{3}{2} b a_{q} + O (b^{2}), = - \frac{m _{q} a ^{2}}{16 ℏ ^{2}} + 6 b a_{q} + O (b^{2}), = - \frac{m _{q} a ^{2}}{16 ℏ ^{2}} + 5 b a_{q} + O (b^{2}), m = - 1, 0, 1.

Thus the second level splits into a singlet orbital level and a triply degenerate orbital level, with

E_{2 s} - E_{2 p} = b a_{q} + O (b^{2})

. For the confining sign

b > 0

,

2 p

lies lower than

2 s

: its smaller mean radius gives the smaller positive correction. The expansion assumes these corrections and the induced couplings to other principal shells are small compared with the relevant Coulomb energy gaps;

b a_{q}^{2} / a ≪ 1

is the governing small parameter for fixed low

n

.

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