# The variational principle and the quantum mechanical particle-in-a-box

#### Question P8.4.4

Consider a one-dimensional quantum mechanical particle in a box ($-1\le x \le 1$) described by the SchrÃ¶dinger equation: $$-\frac{\mathrm{d}^2\psi}{\mathrm{d}x^2} = E\psi,$$ in energy units for which $\hbar^2/(2m) = 1$ with $m$ the mass of the particle. The exact solution for the ground state of this system is given by $$\psi = \cos\left(\frac{\pi x}{2}\right), \quad E = \frac{\pi^2}{4}.$$ An approximate solution may be arrived at using the variational principle by minimizing the expectation value of the energy of a trial wavefunction, $$\psi_\mathrm{trial} = \sum_{n=0}^N a_n\phi_n(x)$$ with respect to the coefficients $a_n$. Taking the basis functions to have the following symmetrized polynomial form, $$\phi_n = (1-x)^{N-n+1}(x+1)^{n+1},$$ use scipy.optimize.minimize and scipy.integrate.quad to find the optimum value of the expectation value (Rayleigh-Ritz ratio): $$\mathcal{E} = \frac{\langle \psi_\mathrm{trial} | \hat{H} | \psi_\mathrm{trial} \rangle}{\langle \psi_\mathrm{trial} | \psi_\mathrm{trial} \rangle} = -\frac{\int_{-1}^1\psi_\mathrm{trial}\frac{\mathrm{d}^2}{\mathrm{d}x^2}\psi_\mathrm{trial}\;\mathrm{d}x}{\int_{-1}^1\psi_\mathrm{trial}\psi_\mathrm{trial}\;\mathrm{d}x}.$$ Compare the estimated energy, $\mathcal{E}$, with the exact answer for $N=1,2,3,4$.

[Hint: Use np.polynomial.Polynomial objects to represent the basis and trial wavefunctions].

#### Solution

To access solutions, please obtain an access code from Cambridge University Press at the Lecturer Resources page for my book (registration required) and then sign up to scipython.com providing this code.