1.II.18F

Quantum Mechanics | Part IB, 2001

Consider a quantum-mechanical particle of mass mm moving in a potential well,

V(x)={0,−a<x<a∞, elsewhere V(x)= \begin{cases}0, & -a<x<a \\ \infty, & \text { elsewhere }\end{cases}

(a) Verify that the set of normalised energy eigenfunctions are

ψn(x)=1asin⁡(nπ(x+a)2a),n=1,2,…\psi_{n}(x)=\sqrt{\frac{1}{a}} \sin \left(\frac{n \pi(x+a)}{2 a}\right), \quad n=1,2, \ldots

and evaluate the corresponding energy eigenvalues EnE_{n}.

(b) At time t=0t=0 the wavefunction for the particle is only nonzero in the positive half of the well,

ψ(x)={2asin⁡(πxa),0<x<a0, elsewhere \psi(x)= \begin{cases}\sqrt{\frac{2}{a}} \sin \left(\frac{\pi x}{a}\right), & 0<x<a \\ 0, & \text { elsewhere }\end{cases}

Evaluate the expectation value of the energy, first using

⟨E⟩=∫−aaψHψdx\langle E\rangle=\int_{-a}^{a} \psi H \psi d x

and secondly using

⟨E⟩=∑n∣an∣2En,\langle E\rangle=\sum_{n}\left|a_{n}\right|^{2} E_{n},

where the ana_{n} are the expansion coefficients in

ψ(x)=∑nanψn(x)\psi(x)=\sum_{n} a_{n} \psi_{n}(x)

Hence, show that

1=12+8π2∑p=0∞(2p+1)2[(2p+1)2−4]21=\frac{1}{2}+\frac{8}{\pi^{2}} \sum_{p=0}^{\infty} \frac{(2 p+1)^{2}}{\left[(2 p+1)^{2}-4\right]^{2}}

Typos? Please submit corrections to this page on GitHub.