Past exam of the quantum theory course of the University of Oxford / 2015 / 3 / i

Solution

Introduce the reduced radial wavefunction

u (r) = r ψ (r)

- 2 m r / ℏ^{2}

gives

u^{''} - \frac{ℓ ( ℓ + 1 )}{r ^{2}} u + \frac{2 m K Z}{ℏ ^{2} r} u + \frac{2 m E}{ℏ ^{2}} u = 0.

For the decaying bound state, take

α > 0

and put

u = f e^{- α r}

u^{''} = e^{- α r} (f^{''} - 2 α f^{'} + α^{2} f) .

Since

2 m E / ℏ^{2} = - α^{2}

, the two constant multiples of

f

cancel. Dividing by the nonzero exponential gives

f^{''} - 2 α f^{'} - \frac{ℓ ( ℓ + 1 )}{r ^{2}} f + \frac{2 m K Z}{ℏ ^{2} r} f = 0.

There are no discussions about this article yet.

There are currently no matching articles.