Chapter 6: Q1P (page 254)

Suppose we put a delta-function bump in the center of the infinite square well:
$H^{'} = αδ (x - a / 2)$
where $a$ is a constant.
(a) Find the first-order correction to the allowed energies. Explain why the energies are not perturbed for even $n$ .
(b) Find the first three nonzero terms in the expansion (Equation 6.13) of the correction to the ground state, $Ψ_{1}^{1}$ .

Short Answer

Expert verified

The first-order correction to the allowed energies $= \frac{2 α}{a} \sin^{2} \frac{nπ}{2}$

The the first three nonzero terms in the expansion of the correction to the ground state, $Ψ_{1}^{1} is = \sqrt{\frac{a}{2}} \frac{mα}{π^{2} ħ^{2}} [\sin (\frac{3 πx}{a}) - \frac{1}{3} \sin (\frac{5 πx}{a}) + \frac{1}{6} \sin (\frac{7 πx}{a})]$

Step by step solution

Stationary state of a one-dimensional infinite square well.

The stationary state of a one-dimensional infinite square wellis:

$Ψ_{n}^{0} = \sqrt{\frac{2}{a}} \sin (\frac{nπ}{a} x)$

Step 2: The first-order correction to the allowed energies.

For the infinite square well:

$\hat{H}' = α δ (x - \frac{a}{2}), α = con st$

Solve the problem by considering the stationary state of a one-dimensional infinite square well, that is: $Ψ_{n}^{0} = \sqrt{\frac{2}{a}} \sin (\frac{nπ}{a} x) E_{n}^{'} = <Ψ_{n}^{0} |\hat{H}'| Ψ_{n}^{0}> = \int_{0}^{a} \hat{H}' |{\hat{Ψ}}_{n}^{0}| dX = \frac{2 α}{a} \int_{0}^{a} δ (x - \frac{a}{2}) \sin^{2} (\frac{nπx}{a}) dx = \frac{2 {αa}^{2}}{\sin} (\frac{nπ}{a} . \frac{a}{2}) = \frac{2 α}{a} \sin^{2} \frac{nπ}{2} - \{for n \equiv odd \Rightarrow E_{n}^{,} = \frac{2 α}{a} \sin^{2} (\frac{nπ}{2}) = \frac{2 α}{a}\} - \{for n \equiv even \Rightarrow E_{n}^{'} = 0\}$

Step 3: The first three nonzero terms in the expansion.

Use the formula and substitute each value.

$Ψ_{n}^{1} = \sum_{m \neq n} \frac{<Ψ_{n}^{0} |\hat{H}'| Ψ_{n}^{0}>}{E_{n}^{0} - E_{m}^{0}} Ψ_{m}^{0}$

For n=1

$<Ψ_{n}^{0} |\hat{H}'| Ψ_{n}^{0}> = \frac{2 α}{a} \int_{0}^{a} dx \sin (\frac{mπx}{a}) δ (x - \frac{a}{2}) \sin (\frac{πx}{a}) = \frac{2 α}{a} \sin (\frac{mπ}{a} . \frac{a}{2}) \sin (\frac{π}{a} . \frac{a}{2}) = \frac{2 α}{a} \sin (\frac{mπ}{2})$

Note that: $m \neq 1, (n = 1, m \neq n)$

for $m = 0 \Rightarrow \sin (0) = 0$

for $m = 0 \Rightarrow \sin (mπ) = 0$

The first three non- zero terms (odd)

$m = 3, 5, 7; n = 1 E_{n}^{0} = \frac{n^{2} π^{2} ħ}{2 {ma}^{2}} \Rightarrow E_{1}^{0} = \frac{π^{2} ħ^{2}}{2 {ma}^{2}} \Rightarrow Ψ_{1}^{1} = \frac{2 a}{a} \sqrt{\frac{2}{a}} [\frac{\sin (\frac{3 π}{2})}{\frac{π^{2} ħ^{2}}{2 {ma}^{2}} (1 - 9)} \sin (\frac{3 πx}{a}) + \frac{\sin (\frac{5 π}{2}) \sin (\frac{5 πx}{a})}{\frac{π^{2} ħ^{2}}{2 {ma}^{2}} (1 - 25)} + \frac{sinin (\frac{5 π}{2}) \sin (\frac{5 πx}{a})}{\frac{π^{2} ħ^{2}}{2 {ma}^{2}} (1 - 49)}] = \frac{2 a}{a} \sqrt{\frac{2}{a}} \frac{2 {ma}^{2}}{π^{2}} \frac{2 {ma}^{2}}{π^{2} ħ^{2}} [\frac{1}{8} \sin (\frac{3 πx}{a}) - \frac{1}{24} \sin (\frac{5 πx}{a}) + \frac{1}{48} \sin (\frac{7 πx}{a})]$