Quantum Mechanics | Part IB, 2002

A quantum mechanical particle of mass MM moves in one dimension in the presence of a negative delta function potential

V=22MΔδ(x),V=-\frac{\hbar^{2}}{2 M \Delta} \delta(x),

where Δ\Delta is a parameter with dimensions of length.

(a) Write down the time-independent Schrödinger equation for energy eigenstates χ(x)\chi(x), with energy EE. By integrating this equation across x=0x=0, show that the gradient of the wavefunction jumps across x=0x=0 according to

limϵ0(dχdx(ϵ)dχdx(ϵ))=1Δχ(0)\lim _{\epsilon \rightarrow 0}\left(\frac{d \chi}{d x}(\epsilon)-\frac{d \chi}{d x}(-\epsilon)\right)=-\frac{1}{\Delta} \chi(0)

[You may assume that χ\chi is continuous across x=0.x=0 . ]

(b) Show that there exists a negative energy solution and calculate its energy.

(c) Consider a double delta function potential

V(x)=22MΔ[δ(x+a)+δ(xa)].V(x)=-\frac{\hbar^{2}}{2 M \Delta}[\delta(x+a)+\delta(x-a)] .

For sufficiently small Δ\Delta, this potential yields a negative energy solution of odd parity, i.e. χ(x)=χ(x)\chi(-x)=-\chi(x). Show that its energy is given by

E=22Mλ2, where tanhλa=λΔ1λΔE=-\frac{\hbar^{2}}{2 M} \lambda^{2}, \quad \text { where } \quad \tanh \lambda a=\frac{\lambda \Delta}{1-\lambda \Delta}

[You may again assume χ\chi is continuous across x=±ax=\pm a.]

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