# 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

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

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

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