# Quantum Chemistry

Chapter06 - Home
Exercises: 1 - 2 - 3 - 4 - 5 - 6 - 7 - 8 - 9 - 10 - 11 - 12 - 13 - 14 - 15 - 16 - 17 - 18 - 19 - 20 - 21 - 22 - 23 - 24 - 25 - 26 - 27 - 28 - 29 - 30 - 31 - 32 - 33 - 34 - 35 - 36 - 37 - 38 - 39 - 40 - 41 - 42 - 43 - 44 - 45 - 46 - 47 - 48 - 49

Exercise. This is the solution to exercise 6.1 in the book.

(a) If the three force constants are equal ${k}_{x}={k}_{y}={k}_{z}=k$ then the vibration frequency is the same along all directions:

${\nu }_{x}={\nu }_{y}={\nu }_{z}=\frac{1}{2\pi }\sqrt{\frac{k}{m}}$

Hence, according to exercise 4.18  the potential energy operator is:

$\stackrel{̂}{V}=2{\pi }^{2}{\nu }^{2}m\left({x}^{2}+{y}^{2}+{z}^{2}\right)=2{\pi }^{2}{\nu }^{2}m{r}^{2}$

Thus in this case the potential energy is spherically symmetric. It was shown in the text (following the derivation of equations up to 6.16) that in this case the wave function can be written as a product of a radial function and another function dependent on $\theta$ and $\varphi$.

(b) Again, following the arguments leading to equation 6.16 in the text we get that:

$G\left(\theta ,\varphi \right)={Y}_{l}^{m}\left(\theta ,\varphi \right)$

(c) $f\left(r\right)$ must satisfy equation 6.17 in the text.

(d) In exercise 4.18  we found that:

$\psi \left(x,y,z\right)={\psi }_{x}\left(x\right){\psi }_{y}\left(y\right){\psi }_{z}\left(z\right)$

For the isotropic harmonic oscillator in the ground state ${\nu }_{x}={\nu }_{y}={\nu }_{z}=\nu$ and ${\psi }_{x}={\psi }_{y}={\psi }_{z}={\psi }_{0}\left(x\right)$, where ${\psi }_{0}\left(x\right)$ is given by (according to 4.49 in the text):

${\psi }_{0}\left(x\right)={c}_{0}{e}^{-\alpha ∕2{x}^{2}}\phantom{\rule{1em}{0ex}}\alpha =\frac{2\pi \nu m}{\hslash }$

The energy is:

$E=\frac{3}{2}h\nu$

${c}_{0}$ is given by the normalization condition:

${\int }_{-\infty }^{\infty }{\psi }_{0}^{2}\left(x\right)dx=1$

or

${c}_{0}^{2}{\int }_{-\infty }^{\infty }{e}^{-\alpha {x}^{2}}dx=1$

${c}_{0}^{2}\sqrt{\frac{\pi }{\alpha }}=1$

or

${c}_{0}={\left(\frac{\alpha }{\pi }\right)}^{1∕4}$

Hence,

$\psi \left(x,y,z\right)={\left(\frac{\alpha }{\pi }\right)}^{3∕4}{e}^{-\alpha ∕2\left({x}^{2}+{y}^{2}+{z}^{2}\right)}$

or

$\psi ={\left(\frac{\alpha }{\pi }\right)}^{3∕4}{e}^{-\alpha ∕2{r}^{2}}$

Hence, in this case:

$G\left(\theta ,\varphi \right)=1$

and

$f\left(r\right)={\left(\frac{\alpha }{\pi }\right)}^{3∕4}{e}^{-\alpha ∕2{r}^{2}}$

We can show that $f\left(r\right)$ verifies equation 6.17 in the text for:

$E=\frac{3}{2}h\nu \phantom{\rule{1em}{0ex}}\stackrel{̂}{V}=2{\pi }^{2}{\nu }^{2}m{r}^{2}$

when $l=0$ (i.e., the ground state).

It is easy to show that:

${f}^{\prime }\left(r\right)=-\alpha rf\left(r\right)$

and

$f”\left(r\right)=\left({\alpha }^{2}{r}^{2}-\alpha \right)f\left(r\right)$

Hence:

$f”\left(r\right)+\frac{2}{r}{f}^{\prime }\left(r\right)=\left({\alpha }^{2}{r}^{2}-3\alpha \right)f\left(r\right)$

and then the left hand side of the equation is:

$LHS=\frac{3\alpha {\hslash }^{2}}{2m}f\left(r\right)$

With energy:

$E=\frac{3}{2}h\nu =\frac{3}{2}\frac{\alpha {\hslash }^{2}}{m}$

then indeed equation 6.17 is satisfied.