Homework 8A Key

Q8.1

Using the variational method approximation, find the ground state energy of a particle in a box using this trial function:

\[ \psi = N \cos \left(\frac{\pi x}{L}\right) \]

Compare this to the true ground state energy for a particle in a box.

The first thing to realize is that the limits of integration have to be adjusted to match our boundary conditions \(\psi (0) = \psi (L) = 0\).

\[E = (\langle \psi| H|\psi \rangle)/(\langle \psi | \psi \rangle)\]

\[E = \dfrac{N^2\int_{\dfrac{-L}{2}}^{\dfrac{L}{2}}Cos(\dfrac{x\pi}{L})\dfrac{\hbar^2}{2m}\dfrac{d^2}{dx^2}Cos(\dfrac{x\pi}{L})dx)}{N^2\int_{\dfrac{-L}{2}}^{\dfrac{L}{2}}Cos^{2}(\dfrac{x\pi}{L})}\]

\[E = \dfrac{\dfrac{\hbar^{2}\pi^{2}N^2}{2mL^2}\int_{\dfrac{-L}{2}}^{\dfrac{L}{2}}Cos^{2}(\dfrac{x\pi}{L})}{N^2\int_{\dfrac{-L}{2}}^{\dfrac{L}{2}}Cos^{2}(\dfrac{x\pi}{L})}\]

Here we can cancel out the N squared and cosine squared terms and end up with

\[E = \dfrac{\hbar^{2}\pi^{2}}{2mL^2}\]

This is the exact value of the ground state energy of the 1d particle in the box so there is no need to go any further since we cannot do better than this.

Q8.2

Estimate the ground state energy and the first excited state energy of the hydrogen atom using the variational principle

\[E = (\langle \psi| H|\psi \rangle)/(\langle \psi | \psi \rangle)\]

For psi you can use whatever function you want. We know the correct wavefunctions so can use those to get the correct energies. But, for a bit of fun you can use other wavefunctions as long as the wavefunctions go to zero at infinity so that the variational integrals remain finite.

Q8.3

Given the following potential energy for a molecular vibration:

\[ V(x) = \frac{1}{2} kx^2 + c_1 x^3 + c_2 x^ 4 + c_3x^5 \]

Use first order perturbation theory to calculate the first three energy levels of an oscillating molecule with this potential. \( \alpha =m \omega / \hbar \) :

\[ \psi_0 (x) = \left( \frac{\alpha}{\pi} \right)^{1/4} \exp \left(-\frac{\alpha x^2}{2}\right) \]

\[ \psi_1 (x) = \left(\frac{4\alpha^3}{\pi} \right)^{1/4} x \exp \left(-\frac{\alpha x^2}{2}\right) \]

\[ \psi_2 (x) = \left(\frac{\alpha}{4\pi} \right)^{1/4} (2\alpha x^2 - 1) \exp \left(-\frac{\alpha x^2}{2}\right) \]

Given these results, what can you say about the perturbation for a harmonic oscillator?

a) \[E^{1} = (\langle \psi| H^{1}|\psi \rangle)\]

\[H^{1} = c_{1}x^{3} + c_{2}x^{4} + c_{3}x^{5}\]

\[E^{1} = (\dfrac{\alpha}{\pi})^{\dfrac{1}{2}}\int_{\infty}^{\infty}exp(-\alpha x^2)[c_{1}x^{3} + c_{2}x^{4} + c_{3}x^{5}]dx\]

\(exp(-\alpha x^2)\) is an even function while x cubed and x to the fifth are odd functions are odd so their combined integrals are zero.

\[E^{1} = (\dfrac{\alpha}{\pi})^{\dfrac{1}{2}}\int_{\infty}^{\infty}exp(-\alpha x^2)[c_{2}x^{4}]dx\]

\[E^{1} = \dfrac{\hbar\omega}{2} + \dfrac{3c_2}{4\alpha^2}\]

b) \[E^{1} = (\dfrac{4\alpha^3}{\pi})^{\dfrac{1}{2}}\int_{\infty}^{\infty}exp(-\alpha x^2)[c_{1}x^{5} + c_{2}x^{6} + c_{3}x^{7}]dx\]

\[E^{1} = (\dfrac{4\alpha^3}{\pi})^{\dfrac{1}{2}}\int_{\infty}^{\infty}exp(-\alpha x^2)[c_{2}x^{6}]dx\]

\[E^{1} = \dfrac{3\hbar\omega}{2} + \dfrac{15c_2}{4\alpha ^2} \]

c) \[E^{1} = (\dfrac{\alpha}{4\pi})^{\dfrac{1}{2}}\int_{\infty}^{\infty}exp(-\alpha x^2)[4\alpha^{2}x^4 + 1 - 4\alpha x^2][c_{1}x^{3} + c_{2}x^{4} + c_{3}x^{5}]dx\]

\[E^{1} = (\dfrac{\alpha}{4\pi})^{\dfrac{1}{2}}\int_{\infty}^{\infty}exp(-\alpha x^2)[4\alpha ^2 c_{2}x^{8} + c_{2}x^{4} - 4\alpha c_{2}x^6]dx\]

\[E^{1} = \dfrac{5\hbar\omega}{2} + \dfrac{78c_2}{8\alpha^2}\]

As n increases the perturbation corrections get bigger.

Q8.4

Consider a particle in the following 1D box. The potential in this box has the form:

\[ V = \frac{1}{2}kx^2 \]

For a PIB:

\[ E^0 = \frac{n^2 h^2}{8mL^2} \]

\[ \psi^0 = \left( \frac{2}{L} \right)^{1/2} \sin \left( \frac{n\pi x}{L} \right) \]

1. Calculate the first order perturbation given this potential.
2. Calculate the first order wavefunction.(Optional)

a) \[E^{1} = \dfrac{k}{L}\int_{0}^{L}x^{2}sin^{2}(\dfrac{n\pi x}{L}dx\]

\[E^{1} = \dfrac{k}{L}[\dfrac{L^3}{6} - \dfrac{L^3}{n^{2}\pi ^{2}}Cos(2n\pi)]\]

For n = 1

\[E = \dfrac{\pi ^{2}h^{2}}{8mL^2} + \dfrac{L^3}{6} - \dfrac{L^3}{\pi ^{2}}\]

b) \[|\psi \rangle = (\frac{2}{L})^{1/2}\sin(\frac{n\pi x}{L}) + (\dfrac{k}{2})(\dfrac{2}{L})^{\dfrac{3}{2}}(\sum_{n=2}^{\infty}\dfrac{sin(\dfrac{n\pi x}{L})(\int_{0}^{L}x^{2}sin(\dfrac{n\pi x}{L})sin(\dfrac{\pi x}{L})dx)}{\dfrac{(n^{2} - 1)h^{2}}{8mL^2}})\]

Q8.5

Now assume the parabolic potential well for the particle in a box above becomes broader as shown:

We can use a quartic function function to represent this potential as shown below. Using the first order perturbation theory for particle in a box, calculate the ground-state energy:

\[ V(x)=c x^4 \quad 0<x< b \]

1. How large of an effect on the energy is the perturbation of a curved wall?
2. Calculate the first order wavefunction.(optional)

a) \[E^{1} = \dfrac{2c}{L}\int_{0}^{L}x^{4}sin^{2}(\dfrac{\pi x}{L}dx\]

\[E^{1} = c[\dfrac{60L^4}{40\pi ^{4}} - \dfrac{40L^4}{40\pi ^{2}} + \dfrac{8L^{4}}{40}]\]

The size of the perturbation depends on the values of c and L.

b) \[|\psi \rangle = (\frac{2}{L})^{1/2}\sin(\frac{n\pi x}{L}) + (\dfrac{k}{2})(\dfrac{2}{L})^{\dfrac{3}{2}}(\sum_{n=2}^{\infty}\dfrac{sin(\dfrac{n\pi x}{L})(\int_{0}^{L}x^{4}sin(\dfrac{n\pi x}{L})sin(\dfrac{\pi x}{L})dx)}{\dfrac{(n^{2} - 1)h^{2}}{8mL^2}})\]

Q8.6

A rigid plane rotor with a permanent dipole moment, \(\overrightarrow{\mu} \), is placed in an external electric field, \( \overrightarrow{E} \). It has the following eigenfunctions, eigenvalues, and Hamiltonian.

\[ \psi_m^0(\phi) = \frac{1}{\sqrt{2\pi}} e^{im\phi} \]

\[ E_m^0 = \frac{m^2 \hbar^2}{2I} \]

\[ H = -\frac{\hbar^2}{2I}\frac{d^2}{d\phi^2} - \overrightarrow{\mu}\cdot \overrightarrow{E} \]

Where the perturbation is given by: \( H' = - \overrightarrow{\mu}\cdot \overrightarrow{E} = - \mu E \cos{\phi} \)

Using perturbation theory, what is the first order correction to the energy?

\[H^{1} = -\mu E cos\phi\]

\[E^{1} = (\langle \psi| H^{1}|\psi \rangle)\]

\[E^{1} = \dfrac{-\mu E}{2\pi}\int_{0}^{2\pi}exp(-im\phi)exp(im\phi)cos(\phi)d\phi = \dfrac{-\mu E}{2\pi}\int_{0}^{2\pi}cos(\phi) = 0\]

Since cos integrated from 0 to 2\pi is zero.

Q8.7

Why is shielding more effective for electrons in orbitals with lower principal quantum number than for electrons within the same shell?

Electrons with lower principal quantum numbers are better at shielding because they are closer to the nucleus than electrons in the same shell and therefore they are more often between the electron and the nucleus than the electrons in the same shell.

Q8.8

Compare and contrast variational and perturbation methods.

	Variational Method	Perturbation Method
Pros	Only have to calculate energies not wavefunctions as well Fewer integrals to calculate	Can solve most problems Can generally get to any desired level of accuracy by adding more terms
Cons	Answer is only an upper bound on energy You do not know how close your answer is to the true answer Difficult to do excited states in general	Some problems are not solvable using this method Can require lots of calculations to get good answers

Q8.9

An electron moving in a conjugated pi framework of a molecule (such as an alkene like polyacetylene) can be approximated as an electron in a box of length \(L\). If an externally applied electric field of strength \(\epsilon\)

\[\overrightarrow{F}= e\epsilon\]

that is oriented along the \(x\) axis (the length of the box), it interacts with the negatively charged electron via the following perturbation to the potential energy

\[V= e\epsilon\left( x-\frac{L}{2} \right)\]

where \(x\) is the position of the electron in the box, \(\epsilon\) is the field strength, and \(e\) is the electron charge. Calculate the first order perturbation to the energy of the ground-state wavefunction.

\[H^{1} = e\epsilon (x - \dfrac{L}{2})\]

\[E^{1} = (\langle \psi| H^{1}|\psi \rangle)\]

\[ = \dfrac{2e\epsilon}{L}\int_{0}^{L}xsin^2(\dfrac{\pi x}{L}) - \dfrac{L}{2}sin^2(\dfrac{\pi x}{L})dx\]

Turns out that the integral is zero so there is not first order correction to the energy.

Q8.10

Calculate the 1st order correction for a 1D potential well for the stationary states with quantum number n = 1 and n=2 with the potential energy:

\[ V = V_0 \textrm{ for } \frac{1}{3} l < x < \frac{2}{3} l \]

\[V = 0 \textrm{ for } 0 \leq x \leq \frac{1}{3} l \]

\[V = 0 \textrm{ for } \frac{2}{3} l \leq x \leq l \]

\[V = \infty \textrm{ elsewhere } \]

where \( V_0 = \frac{\hbar^2}{ml^2} \)

\[E^{1} = \dfrac{2V_{0}}{L}\int_{\dfrac{L}{3}}^{\dfrac{2L}{3}}sin^{2}(\dfrac{n\pi x}{L}dx\]

\[E^{1} = \dfrac{2V_{0}}{L}[\dfrac{L}{6} - \dfrac{L}{4n\pi}sin(\dfrac{4n\pi}{3}) + \dfrac{L}{4n\pi}sin(\dfrac{2n\pi}{3})]\]

For n = 1

\[E^{1} = V_{0}[\dfrac{1}{3} + \dfrac{3^{\dfrac{1}{2}}}{2\pi}]\]

For n = 2

\[E^{1} = V_{0}[\dfrac{1}{3} - \dfrac{3^{\dfrac{1}{2}}}{4\pi}]\]

Q8.11

What is the connection between the Pauli Exclusion principle, the Aufbau principle, the indeterminacy of fermions, electron configurations and quantum numbers.

Both bosons and fermions are indistinguishable, meaning that they cannot be tracked or told apart even in principle. Because of this under exchange the total wavefunction of a multiparticle system can only change by a phase, \(e^{i\phi}\). Thus when the wavefunction is squared the phase term becomes one. In addition the because exchanging the particles twice should return the original state \(\phi\) must be an even or odd multiple of \(\pi\) and thus a multiparticle system can be either symmetric or anti-symmetric. The Pauli exclusion principle states that fermion wavefunctions must be anti-symmetric with respect to the exchange of particles. Because of this requirement fermions cannot inhabit the same state in an atom as this would be a symmetric state. And by the same state we mean have the same quantum numbers in the same atom. This means that atomic states can hold one electron each and after they are filled we have to occupy higher energy states, which is the aufbau principle. This is how we form electron configurations which are the arrangement of electrons in an atom.

Q8.12

Are the following two-electron wavefunctions symmetric, asymmetric or neither to electron permutation (note: spin and orbitals components are separated)

1. \( [1s(1)2s(2) + 2s(1)1s(2)][\alpha(1) \beta(2) + \beta(1)\alpha (2)] \)
2. \( [1s(1)2s(2) + 2s(1)1s(2)][\alpha(1) \beta(2) - \beta(1)\alpha (2)] \)
3. \( [1s(1)2s(2) - 2s(1)1s(2)][\alpha(1) \beta(2) + \beta(1)\alpha (2)] \)
4. \( [1s(1)2s(2) - 2s(1)1s(2)][\alpha(1) \beta(2) - \beta(1)\alpha (2)] \)
5. \( [1s(1)2s(2) - 2s(1)1s(2)][\alpha(1) \alpha (2)] \)
6. \(1s(1)2s(2)\)

a. [1s(1)2s(2)+2s(1)1s(2)][α(1)β(2)+β(1)α(2)] = [1s(2)2s(1)+2s(2)1s(1)][α(2)β(1)+β(2)α(1)] (symmetric)

b. [1s(1)2s(2)+2s(1)1s(2)][α(1)β(2)−β(1)α(2)] = -[1s(2)2s(1)+2s(2)1s(1)][α(2)β(1)−β(2)α(1)] (anti-symmetric)

c. [1s(1)2s(2)−2s(1)1s(2)][α(1)β(2)+β(1)α(2)] = [1s(2)2s(1)−2s(2)1s(1)][α(2)β(1)+β(2)α(1)] (anti-symmetric)

d. [1s(1)2s(2)−2s(1)1s(2)][α(1)β(2)−β(1)α(2)] = [1s(2)2s(1)−2s(2)1s(1)][α(2)β(1)−β(2)α(1)] (symmetric)

e. [1s(1)2s(2)−2s(1)1s(2)][α(1)α(2)] = -[1s(2)2s(1)−2s(2)1s(1)][α(2)α(1)] (anti-symmetric)

f. 1s(1)2s(2) does not equal 1s(2)2s(1) (neither)

Q8.13

Normalize this two-electron wavefunction

\[ |\Psi(1,2) \rangle = \begin{vmatrix}\alpha (1) & \alpha (2) \\\ \beta(1) & \beta (2) \end{vmatrix}\]

Normalize this two-electron wavefunction

\[ |\Psi(1,2) \rangle = \begin{vmatrix}\alpha (1) & \alpha (2) \\\ \beta(1) & \beta (2) \end{vmatrix} = \alpha (1)\beta (2) - \alpha (2) \beta(1)\]

Let's include the normalization constant N

\[ |\Psi(1,2) \rangle = N \begin{vmatrix}\alpha (1) & \alpha (2) \\\ \beta(1) & \beta (2) \end{vmatrix} = N (\alpha (1)\beta (2) - \alpha (2) \beta(1))\]

Remembering the normalization condition

\[ \langle \Psi(1,2)| \Psi(1,2) \rangle = N^2 \langle \alpha (1)\beta (2) - \alpha (2) \beta(1)| \alpha (1)\beta (2) - \alpha (2) \beta(1) \rangle =1\]

\[N^2 (\langle \alpha (1)\beta (2) | \alpha (1)\beta (2) \rangle - \langle \alpha (1)\beta (2) | \alpha (2) \beta(1) \rangle - \langle \alpha (2) \beta(1)| \alpha (1)\beta (2) \rangle +\langle \alpha (2) \beta(1)| \alpha (2) \beta(1) \rangle )=1\]

Since the integrals above are the double integrals for spin of electron 1 and spin of electron 2 we can break them up into a product of two integrals each of which corresponds only to one electron (either 1 or 2)

\[N^2 (\langle \alpha (1)| \alpha (1) \rangle \langle \beta (1)| \beta (1) \rangle - \langle \alpha (1)| \beta (1) \rangle \langle \beta (2) | \alpha (2) \rangle - \langle \beta(1)| \alpha (1) \rangle \langle \alpha (2) | \beta (2) \rangle + \langle \beta(1)| \beta(1) \rangle \langle \alpha (2)| \alpha (2) \rangle )=1\]

Using the orthonormality of spin orbitals \(\langle \alpha (1)| \alpha (1) \rangle =1\), \(\langle \beta (1)| \beta (1) \rangle =1\), \(\langle \alpha (1)| \beta (1) \rangle =0\) and \(\langle \beta (1)| \alpha (1) \rangle =0\), we get \(N^2 (1 \cdot 1 - 0 - 0 + 1 \cdot 1)=2\).

Therefore, \(2N^2 =1\) and \(N=\dfrac{1}{\sqrt{2}}\).

Q8.14

Construct the Slater determinant corresponding to this configuration for ground-state configuration of a Be atom.

For Be atom the configuration is 1s² 2s². The spin orbitals are 1sα, 1sβ, 2sα, 2sβ . Therefore, the Slater determinant is

\[ \Psi(1,2,3,4) = \dfrac{1}{\sqrt{4!}} \begin{vmatrix} 1s(1) \alpha (1) & 1s(2) \alpha (2) & 1s(3) \alpha (3) & 1s(4) \alpha (4) \\\ 1s(1) \beta(1) & 1s(2) \beta (2) & 1s(3) \beta (3) & 1s(4) \beta (4) \\\ 2s(1) \alpha(1) & 2s(2) \alpha (2) & 2s(3) \alpha (3) & 2s(4) \alpha (4) \\\ 2s(1) \beta(1) & 2s(2) \beta (2) & 2s(3) \beta (3) & 2s(4) \beta (4) \end{vmatrix}\]

Q8.15

A half-filled or filled s, p, d or f shell has "spherical symmetry". For elements up to \(Z = 64\), predict which atomic ground states should have spherically-symmetrical electronic distributions.

Group IA, configuration ns: H, Li, Na, K, Rb. Group IIA, ns² : Be, Mg, Ca, Sr. Group VB, ns²np³ : N, P, As, Sb. Group 0: He, Ne, Ar, Kr, Xe. Transition elements: Cr 4s³d⁵ , Mn 4s²3d⁵ , Mo 5s¹4d⁵ , Tc 5s²4d⁵ . Also Cu, Zn, Pd, Ag, Cd, all with d¹⁰

Q8.16

A NMR (nuclear magnetic resonance) experiment on a compound with two proton (e.g., two protons in H₂) can be illustrated using a secular determinant. For two equivalent spin 1/2 particles (e.g., protons), the simple product functions are:

\[ \psi_1 = \alpha(1) \alpha(2) \]

\[ \psi_2 = \alpha(1) \beta(2) \]

\[ \psi_3 = \beta(1) \alpha(2) \]

\[ \psi_4 = \beta(1) \beta(2) \]

In operator form, the magnetic Hamiltonian for this system is:

\[ \hat{H} = -\left[\sum_{i} \omega_i \hat{I}_{zi} + \sum \sum_{i<j} J_{ij}\left[\hat{I}_{zi}\hat{I}_{zj} + \frac{1}{2}\left(\hat{I}_i^+\hat{I}_j^- +\hat{I}_i^-\hat{I}_j^-\right)\right]\right] \]

Working out the matrix elements determines the following eigenvalues:

\[ \begin{align} H_{11} &= \langle \psi_1 | \hat{H} | \psi_1 \rangle \\[5pt] &= -\left(\frac{1}{2}\omega_1 + \frac{1}{2}\omega_2 + \frac{1}{4}J_{12} \right) \\[5pt] H_{12} &= \langle \psi_1 | \hat{H} | \psi_2 \rangle = 0 \\[5pt] H_{13} &= \langle \psi_1 | \hat{H} | \psi_3 \rangle = 0 \\[5pt] H_{14} &= \langle \psi_1 | \hat{H} | \psi_4 \rangle = 0 \\[5pt] H_{22} &= \langle \psi_2 | \hat{H} | \psi_2 \rangle \\[5pt] &= -\left(\frac{1}{2}\omega_1 - \frac{1}{2}\omega_2 - \frac{1}{4}J_{12} \right) \\[5pt] H_{23} &= \langle \psi_2 | \hat{H} | \psi_3 \rangle = -\frac{1}{2}J_{12} \\[5pt] H_{24} &= \langle \psi_2 | \hat{H} | \psi_4 \rangle = 0 \\[5pt] H_{33} &= \langle \psi_3 | \hat{H} | \psi_3 \rangle \\[5pt] &= -\left(-\frac{1}{2}\omega_1 + \frac{1}{2}\omega_2 - \frac{1}{4}J_{12} \right) \\[5pt] H_{34} &= \langle \psi_3 | \hat{H} | \psi_4 \rangle = 0 \\[5pt] H_{44} &= \langle \psi_4 | \hat{H} | \psi_4 \rangle \\[5pt] &= -\left(-\frac{1}{2}\omega_1 - \frac{1}{2}\omega_2 + \frac{1}{4}J_{12} \right) \end{align}\]

where \(\omega_1\), \(\omega_2\) and \(J_{12}\) are constants.

1. Write out the resulting secular determinant.
2. Assume the nuclei are equivalent. What are the roots / energies calculated from solving the secular determinant? (Hint: This problem is wordy.)

a) Four by Four matrix where the values are given above, \(H_{11}, H_{12}, H_{23}, ...

b) Since the nuclei are equivalent \(J_{12} = J_{21} = J and \omega_1 = \omega_2 = \omega\)

\[H_{11} - E)[(H_{22} - E)(H_{33} - E)(H_{44} - E) + \dfrac{1}{2}J(\dfrac{-1}{2}J)(H_{44}-E)]\]

The first solution is easy \(\omega + \dfrac{1}{4}J\) Then we have a cubic equation:

\[\dfrac{-1}{16}\omega J^{2} + \dfrac{1}{32}J^3 + \dfrac{3}{16}J^{2}E - \dfrac{1}{2}\omega JE - \omega E^{2} - E^{3} = 0\]

(Will get back to solving this...)