Solution

Put

R = f e^{- γ r}

. Then

R^{'} = e^{- γ r} (f^{'} - γ f), R^{''} = e^{- γ r} (f^{''} - 2 γ f^{'} + γ^{2} f) .

After substitution and cancellation of the

γ^{2} f

terms, the equation becomes

r f^{''} + (2 - 2 γ r) f^{'} + (β - 2 γ) f = 0.

f = \sum_{n = 0}^{\infty} a_{n} r^{n}

, the coefficient of

r^{n - 1}

n (n + 1) a_{n} + (β - 2 γn) a_{n - 1} = 0.

Thus

a_{n} = \frac{2 γn - β}{n ( n + 1 )} a_{n - 1} .

If the series does not terminate, then

a_{n} / a_{n - 1} \sim 2 γ / n

, so

f

acquires the large-

r

behaviour

e^{2 γ r}

and

R

grows like

e^{γ r}

. Normalizability therefore requires termination, which occurs when

2 γ N - β = 0

for some positive integer

N

. Hence

γ_{N} = \frac{β}{2 N}, E_{N} = - \frac{ℏ ^{2} γ _{N}^{2}}{2 m} = - \frac{m q ^{4}}{2 ℏ ^{2} N ^{2}} .

The ground state has

N = 1

E_{1} = - \frac{m q ^{4}}{2 ℏ ^{2}} .

Solved by gpt-5.6-sol high.

 Ancestors (12)