Chapter 7: Q12P (page 309)

4} \sqrt{\frac{2 m ω}{h}} x e^{- \frac{m ω}{2 h} x^{2}} (2.62) .$

Short Answer

Expert verified

$(a) <H_{m i n}> \frac{1}{2} h ω \sqrt{\frac{8 n^{2} - 2 n}{8 n^{2} - 2 n - 3}} > \frac{1}{2} h ω (b) <H_{m i n}> \frac{3}{2} h ω \sqrt{\frac{8 n^{2} - 6 n}{8 n^{2} - 6 n - 3}} > \frac{3}{2} h ω (c) ψ_{0} (x) = {(\frac{m ω}{π h})}^{1 / 4 n} e^{- \frac{m ω}{2 h} x^{2}}$

Step by step solution

Given.

We will need the following integral repeatedly:

$ψ (x) = \frac{A}{{(x^{2} + b^{2})}^{n}}, ψ (x) = \frac{B x}{{(x^{2} + b^{2})}^{n}} .$

(a) Generalizing using the trial wave function

$1 = \int_{- \infty}^{\infty} {|ψ|}^{2} d x = 2 {|A|}^{2} \int_{0}^{\infty} \frac{1}{{(x^{2} + b^{2})}^{2 n}} d x = \frac{{|A|}^{2}}{b^{4 n - 1}} \frac{Γ (\frac{1}{2}) Γ (\frac{4 n - 1}{2})}{Γ (2 n)} \Rightarrow A = \sqrt{\frac{b^{4 n - 1} (2 n)}{Γ (\frac{1}{2}) Γ (\frac{4 n - 1}{2})}}$

$<T> = - \frac{h^{2}}{2 m} \int_{- \infty}^{\infty} ψ \frac{d^{2} x}{d x^{2}} d x = - \frac{h^{2}}{2 m} A^{2} \int_{- \infty}^{\infty} \frac{1}{{(x^{2} + b^{2})}^{n}} \frac{d}{d x} [\frac{- 2 n x}{{(x^{2} + b^{2})}^{n + 1}}] d x$

$= \frac{n h^{2}}{m} A^{2} \int_{- \infty}^{\infty} \frac{1}{{(x^{2} + b^{2})}^{n}} [\frac{1}{{(x^{2} + b^{2})}^{n + 1}} - \frac{2 (n + 1) x^{2}}{{(x^{2} + b^{2})}^{n + 2}}] d x = \frac{2 n h^{2}}{m} A^{2} [\int_{0}^{\infty} \frac{1}{{(x^{2} + b^{2})}^{2 n + 1}} d x - 2 (n + 1) \int_{0}^{\infty} \frac{x^{2}}{{(x^{2} + b^{2})}^{2 n + 2}} d x]$

role="math" localid="1658310221074" $= \frac{2 n h^{2}}{m} \frac{b^{4 n - 1} Γ (2 n)}{Γ (\frac{1}{2}) Γ (\frac{4 n - 1}{2})} [\frac{1}{2 b^{4 n + 1}} \frac{Γ (\frac{1}{2}) Γ (\frac{4 n + 1}{2})}{Γ (2 n + 1)} - \frac{2 (n + 1)}{2 b^{4 n - 1}} \frac{Γ (\frac{1}{2}) Γ (\frac{4 n + 1}{2})}{Γ (2 n + 2)}] = \frac{h^{2}}{4 m b^{2}} \frac{n (4 n - 1)}{(2 n + 1)} . <V> = \frac{1}{2} m ω^{2} \int_{- \infty}^{\infty} ψ^{2} x^{2} d x = \frac{1}{2} m ω^{2} 2 A^{2} \int_{0}^{\infty} \frac{x^{2}}{{(x^{2} + b^{2})}^{2 n}} d x . = m ω^{2} \frac{b^{4 n - 1} Γ (2 n)}{Γ (\frac{1}{2}) Γ (\frac{4 n - 1}{2})} \frac{1}{2 b^{4 n - 3}} \frac{Γ (\frac{3}{2}) Γ (\frac{4 n - 3}{2})}{Γ (2 n)} = \frac{m ω^{2} b^{2}}{2 (4 n - 3)} . <H> = \frac{h^{2}}{4 m b^{2}} \frac{n (4 n - 1)}{(2 n + 1)} + \frac{m ω^{2} b^{2}}{(4 n - 3)}; \frac{\partial <H>}{\partial b} = - \frac{h^{2}}{2 m b^{3}} \frac{n (4 n - 1)}{(2 n + 1)} + \frac{m ω^{2} b^{2}}{(4 n - 3)} = 0$

$\Rightarrow b = \sqrt{\frac{h}{m ω}} {[\frac{n (4 n - 1) (4 n - 3)}{2 (2 n + 1)}]}^{1 / 4} . {<H>}_{m i n} = \frac{h^{2}}{4 m} \frac{n (4 n - 1)}{(2 n + 1)} \frac{m ω}{h} \sqrt{\frac{2 (2 n + 1)}{n (4 n - 1) (4 n - 3)}} + \frac{m ω^{2}}{2 (4 n - 3)} \frac{h}{m ω} \sqrt{\frac{n (4 n - 1) (4 n - 3)}{2 (2 n + 1)} .} = \frac{1}{2} h ω \sqrt{\frac{2 n (4 n - 1)}{(2 n + 1) (4 n - 3)}} = \frac{1}{2} h ω \sqrt{\frac{8 n^{2} - 2 n}{8 n^{2} - 2 n - 3}} > \frac{1}{2} h ω$

(b) Finding the least upper bound on the first excited state

$<H> = \frac{3 h^{2}}{4 m b^{2}} \frac{n (4 n - 3)}{(2 n + 1)} + \frac{3}{2} \frac{m ω^{2} b^{2}}{(4 n - 5)}; \frac{\partial <H>}{\partial b} = - \frac{3 h^{2}}{4 m b^{2}} \frac{n (4 n - 3)}{(2 n + 1)} + \frac{m ω^{2} b^{2}}{(4 n - 5)} = 0 b = \sqrt{\frac{h}{m ω}} {[\frac{n (4 n - 3) (4 n - 5)}{2 (2 n + 1)}]}^{1 / 4} . <H> = \frac{3 h^{2}}{4 m} \frac{n (4 n - 3)}{(2 n + 1)} \frac{m ω}{h} \sqrt{\frac{2 (2 n + 1)}{n (4 n - 3) (4 n - 5)}} + \frac{3}{2} \frac{m ω^{2}}{(4 n - 5)} \frac{h}{m ω} \sqrt{\frac{n (4 n - 3) (4 n - 5)}{2 (2 n + 1)}} . = \frac{3}{2} h ω \sqrt{\frac{2 n (4 n - 3)}{(2 n + 1) (4 n - 5)}} = \frac{3}{2} h ω \sqrt{\frac{8 n^{2} - 6 n}{8 n^{2} - 6 n - 5}} > \frac{3}{2} h ω = \frac{3}{2} h ω \sqrt{\frac{2 n (4 n - 3)}{(2 n + 1) (4 n - 5)}} = \frac{3}{2} h ω \sqrt{\frac{8 n^{2} - 6 n}{8 n^{2} - 6 n - 5}} > \frac{3}{2} h ω .$ $1 = 2 {|B|}^{2} \int_{0}^{\infty} \frac{x^{2}}{{(x^{2} + b^{2})}^{2 n}} d x = \frac{{|B|}^{2}}{b^{4 n - 3}} \frac{Γ (\frac{3}{2}) Γ (\frac{4 n - 3}{2})}{Γ (2 n)} \Rightarrow B = \sqrt{\frac{b^{4 n - 3} Γ (2 n)}{Γ (\frac{3}{2}) Γ (\frac{4 n - 3}{2})}} <T> = \frac{h^{2}}{2 m} B^{2} \int_{- \infty}^{\infty} \frac{x}{{(x^{2} + b^{2})}^{n}} \frac{d}{d x} [\frac{x}{{(x^{2} + b^{2})}^{n}} - \frac{2 n x^{2}}{{(x^{2} + b^{2})}^{n + 1}}] d x = - \frac{h^{2} B^{2}}{2 m} \int_{- \infty}^{\infty} \frac{x}{{(x^{2} + b^{2})}^{n}} [\frac{- 2 n x}{{(x^{2} + b^{2})}^{n + 1}} - \frac{4 n x}{{(x^{2} + b^{2})}^{n + 1}} + \frac{4 n (n + 1) x^{3}}{{(x^{2} + b^{2})}^{n + 2}}] d x = \frac{4 n h^{2} B^{2}}{2 m} [3 \int_{0}^{\infty} \frac{x^{2}}{{(x^{2} + b^{2})}^{2 n + 1}} d x - 2 (n + 1) \int_{0}^{\infty} \frac{x^{2}}{{(x^{2} + b^{2})}^{2 n + 2}} d x] = \frac{2 n h^{2}}{m} \frac{b^{4 n - 3} Γ (2 n)}{Γ (\frac{3}{2}) Γ (\frac{4 n - 3}{2})} [\frac{3}{2 b^{4 n - 1}} \frac{Γ (\frac{3}{2}) Γ (\frac{4 n - 3}{2})}{Γ (2 n + 1)} - \frac{2 (2 n + 1)}{2 b^{4 n - 1}} \frac{Γ (\frac{3}{2}) Γ (\frac{4 n - 3}{2})}{Γ (2 n + 2)}] = \frac{3 h^{2}}{4 m b^{2}} \frac{n (4 n - 3)}{(2 n + 1)} .$

$<V> = \frac{1}{2} m ω^{2} 2 B^{2} \int_{0}^{\infty} \frac{x^{4}}{{(x^{2} + b^{2})}^{2 n}} d x = \frac{1}{2} m ω^{2} \frac{b^{4 n - 3} Γ (2 n)}{Γ (\frac{3}{2}) Γ (\frac{4 n - 3}{2})} \frac{2}{2 b^{4 n - 5}} \frac{Γ (\frac{5}{2}) Γ (\frac{4 n - 5}{2})}{Γ (2 n)} = \frac{3}{2} \frac{m ω^{2} b^{2}}{(4 n - 5)} .$

Step4: (c) The bounds approach the exact energy as n→∞

As $n \to \infty,$ ψ becomes more and more Gaussian”.

Analytically, for large n $b \approx \sqrt{\frac{h}{m ω}} {(\frac{n . 4 n . 4 n}{2.2 n})}^{1 / 4} = \sqrt{\frac{2 n h}{m ω}},$ so

${(x^{2} + b^{2})}^{n} = b^{2 n} (1 + x)$ ²b²n≈b2n1+mωx22hnn→b2nemωx2/h

Meanwhile, using Sterling’s approximation in the form $Γ (z + 1) \approx z^{z} e^{- z} :$

$A^{2} = \frac{b^{4 n - 1} Γ (2 n)}{Γ (\frac{1}{2}) Γ (2 n - \frac{1}{2})} \approx \frac{b^{4 n - 1}}{\sqrt{π}} \frac{{(2 n - 1)}^{2 n - 1} e^{- (2 n - 1)}}{{(2 n - \frac{3}{2})}^{2 n - 3 / 2} e^{- (2 n - 3 / 2)}} \approx \frac{b^{4 n - 1}}{\sqrt{π}} \frac{1}{\sqrt{e}} (\frac{2 n - 1}{2 n - \frac{3}{2}})$

localid="1658317073153" $But (\frac{1 - \frac{1}{2 n}}{1 - \frac{3}{4 n}}) \approx (1 - \frac{1}{2 n}) (1 - \frac{3}{4 n}) \approx 1 + \frac{3}{4 n} - \frac{1}{2 n} = 1 + \frac{1}{4 n}; s o {(\frac{2 n - 1}{2 n - \frac{3}{2}})}^{2 n - 1} \approx {[{(1 + \frac{1}{4 n})}^{n}]}^{2} \frac{1}{1 + 1 / 4 n} \to {(e^{1 / 4})}^{2} = \sqrt{e} . = \frac{b^{4 n - 1}}{\sqrt{πe}} \sqrt{e} \sqrt{2 n} = \sqrt{\frac{2 n}{π}} b^{4 n - 1} \Rightarrow A \approx {(\frac{2 n}{π})}^{1 / 4} b^{2 n - 1 / 2}$

So,

$ψ \approx {(\frac{2 n}{π})}^{1 / 4} b^{2 n - 1 / 2} \frac{1}{b^{2 n}} e^{- m ω x^{2} / 2 h} = {(\frac{2 n}{π})}^{1 / 4} {(\frac{m ω}{2 n h})}^{1 / 4} e^{- m ω x^{2} / 2 h} = {(\frac{m ω}{πh})}^{1 / 4} e^{- m ω x^{2} / 2 h}$

Which is precisely the ground state of the harmonic oscillator (Eq. 2.60). So it’s no accident that we get the exact energies, in the limit n → ∞.

$ψ_{0} (x) = {(\frac{m ω}{π h})}^{1 / 4} e^{- \frac{m ω}{2 h} x^{2}} .$

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Short Answer

Step by step solution

Given.

(a) Generalizing using the trial wave function

(b) Finding the least upper bound on the first excited state

Step4: (c) The bounds approach the exact energy as n→∞

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