Chemistry 3730 Fall 2000 Assignment 3 Solutions

1.

(a)

The wavefunctions and energies of the unperturbed problem are

> psi0 := (n,x) -> sqrt(2/L)*sin(n*Pi*x/L);

$$\psi 0 := (n, \,x)\rightarrow \sqrt{2\,{\displaystyle \frac {1}{L }} }\,\mathrm{sin}({\displaystyle \frac {n\,\pi \,x}{L}} )$$

> E0 := n -> n^2*h^2/(8*m*L^2);

$$\mathit{E0} := n\rightarrow {\displaystyle \frac {1}{8}} \, {\displaystyle \frac {n^{2}\,h^{2}}{m\,L^{2}}}$$

The values of the parameters are (in SI units)

> L := 1e-9;

$$L := .1\,10^{-8}$$

> h := 6.6261e-34;

$$h := .66261\,10^{-33}$$

> m := 9.1094e-31;

$$m := .91094\,10^{-30}$$

> V0 := 2e-20;

$$\mathit{V0} := .2\,10^{-19}$$

(V₀ is the constant potential in the left side of the box.)

The energy correction is

> E1 := n -> V0*int(psi0(n,x)^2,x=0..L/3);

$$\mathit{E1} := n\rightarrow \mathit{V0}\,{\displaystyle \int _{0} ^{1/3\,L}} \psi 0(n, \,x)^{2}\,dx$$

The total energy is therefore

> En := n -> E0(n) + E1(n);

$$\mathit{En} := n\rightarrow \mathrm{E0}(n) + \mathrm{E1}(n)$$

You can now evaluate the total and zero-order energies (using Maple) and plot them (by hand is easiest).

> E0(1); En(1);

$$.6024710905\,10^{-19}$$

$$.6415713124\,10^{-19}$$

> E0(2); En(2);

$$.2409884361\,10^{-18}$$

$$.2490334250\,10^{-18}$$

> E0(3); En(3);

$$.5422239814\,10^{-18}$$

$$.5488906481\,10^{-18}$$

(b)

Transitions between n=1 and n=2 involve photons of energy E2-E1. For the particle in a box, this difference is

> E0(2)-E0(1);

$$.1807413271\,10^{-18}$$

For the bumpy box problem, we have

> En(2)-En(1);

$$.1848762938\,10^{-18}$$

The latter energy difference is larger so the wavelength (which is inversely proportional to photon energy) is smaller.

(c)

We first need to compute the coefficients. These are (in general)

> c := (n,k) -> V0*int(psi0(k,x)*psi0(n,x),x=0..L/3)/(E0(n)-E0(k));

$$c := (n, \,k)\rightarrow {\displaystyle \frac {\mathit{V0}\, ... ... \,x)\,\psi 0(n, \,x) \,dx}{\mathrm{E0}(n) - \mathrm{E0}(k)}}$$

Here we want n=1 (ground state), so the relevant coefficients are

> c(1,2);

-.03050375384

> c(1,3);

-.008579180767

> c(1,4);

-.001220150156

> c(1,5);

.0009532423062

> c(1,6);

.0008964368480

> c(1,7);

.0002383105776

The only sensible place to cut off the series (satisfying our factor of 5 criterion) is after c_1,3.

> psi_u := x -> psi0(1,x) + c(1,2)*psi0(2,x) + c(1,3)*psi0(3,x);

$$\mathit{psi\_u} := x\rightarrow \psi 0(1, \,x) + \mathrm{c}(1, \, 2)\,\psi 0(2, \,x) + \mathrm{c}(1, \,3)\,\psi 0(3, \,x)$$

Let us now normalize this wavefunction:

> N1 := 1/sqrt(int(psi_u(x)^2,x=0..L));

$$\mathit{N1} := .9994983368$$

> psi_1 := x -> N1*psi_u(x);

$$\mathit{psi\_1} := x\rightarrow \mathit{N1}\,\mathrm{psi\_u}(x)$$

(d)

> plot([psi0(1,x),psi_1(x)],x=0..L,color=[red,blue]);

The probability density is enhanced to the right where the potential energy is zero. This makes perfect sense.

(e)

The average kinetic energy is computed by

> hbar := h/(2*Pi);

$$\mathit{hbar} := .3313050000\,10^{-33}\,{\displaystyle \frac {1}{ \pi }}$$

> evalf(-hbar^2/(2*m)*int(psi_1(x)*diff(psi_1(x),x$2),x=0..L));

$$.6045055541\,10^{-19}$$

The average potential energy is

> V0*int(psi_1(x)^2,x=0..L/3);

$$.3510635072\,10^{-20}$$

The sum of these two quantities is

> %+%%;

$$.6396119048\,10^{-19}$$

The energy computed by perturbation theory is

> En(1);

$$.6415713124\,10^{-19}$$

The difference between these quantities is small.

Bonus:

The difference is probably due to the truncation of the infinite series for the wavefunction.

2.

The energy is associated with the Hamiltonian operator $\hat{H} = \hat{K} + \hat{V}$. If V depends only on x, y and z, then $[\hat{x}^2+\hat{y}^2+\hat{z}^2,\hat{V}] = 0$. There remains only to determine $[\hat{K},\hat{r}^2]$:

$$\begin{eqnarray*}% latex2html id marker 148 \hat{K} & = & \frac{1}{2m}\left(\hat... ...{x}^2] +[\hat{p}_y^2,\hat{y}^2] +[\hat{p}_z^2,\hat{z}^2]\right) \end{eqnarray*}$$

where we used the linearity of commutators to break up the original problem and eliminated several commutators whose values were obviously zero. All three commutators are identical so we will just work one out in detail:

$$\begin{eqnarray*}% latex2html id marker 166 [\hat{p}_x^2,\hat{x}^2] & = & \hat{p... ...\\ & \not= & 0\\ \therefore [\hat{K},\hat{r}^2] & \not= & 0 \end{eqnarray*}$$

r² and the energy are therefore not compatible observables.

3.

$$\begin{eqnarray*}(\hat{L}_x^2 + \hat{L}_y^2)Y_\ell^{m_z}& = & (\hat{L}^2 - \hat... ... & \left[\ell(\ell+1)\hbar^2 - m_z^2\hbar^2\right]Y_\ell^{m_z} \end{eqnarray*}$$

The spherical harmonics are therefore also eigenfunctions of $\hat{L}_x^2 + \hat{L}_y^2$ with eigenvalue $\ell(\ell+1)\hbar^2 - m_z^2\hbar^2$. Note that, in general, any operator which can be expressed as a linear combination of two operators with common eigenfunctions also shares those eigenfunctions.