16.5: Exercises - Variational Method

An asterisk denotes a harder problem, which you are nevertheless encouraged to try!

1. Estimate the ground-state energy of a 1-dimensional simple harmonic oscillator using as trial function

(a) \(\psi_a(x) = \cos \alpha x\) for \(|\alpha x| < \pi /2\), zero elsewhere,

(b) \(\psi_b(x) = \alpha^2 − x^2\) for \(|x| < \alpha\), zero elsewhere,

(c) \(\psi_c(x) = C \text{ exp}(−\alpha x^2 )\)

(d) * \(\psi_d(x) = C(\alpha − |x|)\) (for \(|x| < \alpha\), zero elsewhere)

(e) \(\psi_e(x) = C \sin \alpha x\) (for \(|\alpha x| < \pi\), zero elsewhere)

In each case, \(\alpha\) is the variational parameter. Don’t forget the normalisation. Sketch the wavefunctions and compare them with the actual ground-state wavefunction.

Numerical answers for checking (a) \(0.568\hbar \omega\) (b) \(0.598\hbar \omega \) (c) \(0.5\hbar \omega\) (d) \(0.5477\hbar \omega \) (e) \(1.67\hbar \omega\)

Where does the kinetic energy come from in (d)? You will have to consider carefully how to calculate it.

Write down the overlap integral \(\langle \psi_i(x)|\psi_j (x)\rangle\) between the wavefunction in (e) and each of the other four: Why does (e) represent an estimate of the first excited state?

You may use the results

\[\int^{\infty}_{−\infty} \text{ exp}(−\alpha x^2 )dx = \sqrt{\frac{\pi}{\alpha}} \text{ and } \int^{\infty}_{−\infty} x^2 \text{ exp }−\alpha x^2 dx = \frac{\sqrt{\pi}}{2\alpha^{3/2}} \nonumber\]

\[\int^{\pi /2\alpha}_{-\pi / 2 \alpha} x^2 \cos^2 \alpha x dx = \pi (\pi^2 − 6)/24\alpha^3 \nonumber\]

2. * A particle moves in one dimension in the potential

\[V (x) = \infty , \quad |x| > a, \quad V (x) = 0, \quad |x| \leq a \nonumber\]

Use a trial function of the form

\[\psi_T (x) = \begin{cases} (a^2 − x^2 )(1 + cx^2 ), & |x| \leq a \\ 0, & |x| > a \end{cases} \nonumber\]

and show that the energy is

\[E(c) = \left( \frac{3\hbar^2}{4ma^2} \right) \frac{11a^4 c^2 + 14a^2 c + 35}{a^4c^2 + 6a^2c + 21} \nonumber\]

Now, treating \(c\) as a variational parameter, obtain an upper bound on the ground-state energy. (You may wish to use maple).

\[E(c^{(1)}) = 1.23372 \frac{\hbar^2}{ma^2}, \quad E(c^{(2)}) \nonumber\]

How does your bound compare with the exact ground-state energy?

3. * Repeat the previous problem taking

\[\psi_T (x) = \begin{cases} (a^2 − x^2 )(x + cx^3 ), & |x| \leq a \\ 0, & |x| > a \end{cases} \nonumber\]

as the trial function. Why does this give an upper bound for the first excited energy level? Compare your variational result with the exact eigenvalue of the \(n = 2\) level.

4. A “1d atom” has ground state wavefunction \(u_1(x) = \text{ exp }−\alpha |x|\).

Consider a ring of N such atoms, one centerd on \(x = 0\) separated by a distance \(d\). Using the single site wavefunctions \(u_{1j} (x + jd)\) as LCAO basis functions, what are the eigenfunctions according to Bloch’s theorum in 1D.

write down the single-particle ground state wavefunction which is an eigenstate of the displacement operator assuming that

\[\langle u_1(x)|u_1(x + d)\rangle << 1 \nonumber\]

What is the normalisation for this wavefunction?

A computational physicist solves for this wavefunction using the variational method, with a trial wavefunction \(\psi_T (r)\) and a set of variational parameters \(c_k, k = 2\pi /N d\)

\[\psi_T (x) = \sum_k c_k \cos kx \nonumber\]

Using your knowledge of the symmetry of the exact atomic solution \(u_1\), what can you say without calculation about the coefficients \(c_k\), and allowed values of \(k\)?

What would you get from a trial wavefunction of the form

\[\psi_T (r) = \sum_k c_k \sin kx \nonumber\]

5. Obtain a variational estimate of the ground-state energy of the hydrogen atom by taking as trial function

\[\psi_T (r) = \text{ exp}(−\alpha r^2 ) \nonumber\]

How does your result compare with the exact result? Sketch the trial wavefunction and the actual wavefunction on the same graph.

You may use the integrals in Question 1 and

\[\int^{\infty}_0 r^4 \text{ exp}(−2ar)dr = \frac{3 \sqrt{2\pi}}{64a^{5/2}} \nonumber\]

\[\int e^{−\alpha r^2} (4\alpha^2 r^2 − 6\alpha )e^{−\alpha r^2} 4\pi r^2 dr = \frac{−3}{16} \sqrt{\frac{2\pi}{a}} \nonumber\]