Introduction to Quantum Mechanics · Problem 10.22

Scattering — David J. Griffiths

Physics — Quantum Mechanics · two-dimensional-scattering-theory

Problem

Two-dimensional scattering theory. By analogy with Section 10.2, develop partial wave analysis for two dimensions. (a) In polar coordinates (r,θ)(r, \theta) the Laplacian is ∇2=∂2∂x2+∂2∂y2=∂2∂r2+1r∂∂r+1r2∂2∂θ2.(10.105)\nabla^2 = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} = \frac{\partial^2}{\partial r^2} + \frac{1}{r}\frac{\partial}{\partial r} + \frac{1}{r^2}\frac{\partial^2}{\partial\theta^2}. \tag{10.105} Find the separable solutions to the (time-independent) Schrödinger equation, for a potential with azimuthal symmetry (V(r,θ)→V(r)V(r,\theta) \to V(r)). Answer: ψ(r,θ)=R(r) eijθ,(10.106)\psi(r,\theta) = R(r)\,e^{ij\theta}, \tag{10.106} where jj is an integer, and u≡r Ru \equiv \sqrt{r}\,R satisfies the radial equation −ℏ22md2udr2+[V(r)+ℏ22m(j2−1/4)r2]u=Eu.(10.107)-\frac{\hbar^2}{2m}\frac{d^2u}{dr^2} + \left[V(r) + \frac{\hbar^2}{2m}\frac{(j^2 - 1/4)}{r^2}\right]u = Eu. \tag{10.107} (b) By solving the radial equation for very large rr (where both V(r)V(r) and the centrifugal term go to zero), show that an outgoing radial wave has the asymptotic form R(r)∼eikrr,(10.108)R(r) \sim \frac{e^{ikr}}{\sqrt{r}}, \tag{10.108} where k≡2mE/ℏk \equiv \sqrt{2mE}/\hbar. Check that an incident wave of the form AeikxAe^{ikx} satisfies the Schrödinger equation, for V(r)=0V(r) = 0 (this is trivial, if you use cartesian coordinates). Write down the two-dimensional analog to Equation 10.12, and compare your result to Problem 10.2. Answer: ψ(r,θ)≈A[eikx+f(θ)eikrr],for large r.(10.109)\psi(r,\theta) \approx A\left[e^{ikx} + f(\theta)\frac{e^{ikr}}{\sqrt{r}}\right], \quad \text{for large } r. \tag{10.109} (c) Construct the analog to Equation 10.21 (the wave function in the region where V(r)=0V(r) = 0 but the centrifugal term cannot be ignored). Answer: ψ(r,θ)=A{eikx+∑j=−∞∞cjHj(1)(kr) eijθ},(10.110)\psi(r,\theta) = A\left\{e^{ikx} + \sum_{j=-\infty}^{\infty} c_j H_j^{(1)}(kr)\,e^{ij\theta}\right\}, \tag{10.110} where Hj(1)H_j^{(1)} is the Hankel function (not the spherical Hankel function!) of order jj. (d) For large zz, Hj(1)(z)∼2/π e−iπ/4(−i)jeizz.(10.111)H_j^{(1)}(z) \sim \sqrt{2/\pi}\,e^{-i\pi/4}(-i)^j\frac{e^{iz}}{\sqrt{z}}. \tag{10.111} Use this to show that f(θ)=2/πk e−iπ/4∑j=−∞∞(−i)jcj eijθ.(10.112)f(\theta) = \sqrt{2/\pi k}\,e^{-i\pi/4}\sum_{j=-\infty}^{\infty}(-i)^j c_j\,e^{ij\theta}. \tag{10.112} (e) Adapt the argument of Section 10.1.2 to this two-dimensional geometry. Instead of the area dσd\sigma, we have a length, dbdb, and in place of the solid angle dΩd\Omega we have the increment of scattering angle ∣dθ∣|d\theta|; the role of the differential cross-section is played by D(θ)≡∣dbdθ∣,(10.113)D(\theta) \equiv \left|\frac{db}{d\theta}\right|, \tag{10.113} and the effective "width" of the target (analogous to the total cross-section) is B≡∫02πD(θ) dθ.(10.114)B \equiv \int_0^{2\pi} D(\theta)\,d\theta. \tag{10.114} Show that D(θ)=∣f(θ)∣2,andB=4k∑j=−∞∞∣cj∣2.(10.115)D(\theta) = |f(\theta)|^2, \quad \text{and} \quad B = \frac{4}{k}\sum_{j=-\infty}^{\infty}|c_j|^2. \tag{10.115} (f) Consider the case of scattering from a hard disk (or, in three dimensions, an infinite cylinder) of radius aa: V(r)={∞,(r≤a),0,(r>a).(10.116)V(r) = \begin{cases} \infty, & (r \leq a), \\ 0, & (r > a). \end{cases} \tag{10.116} By imposing appropriate boundary conditions at r=ar = a, determine BB. You'll need the analog to Rayleigh's formula: eikx=∑j=−∞∞(i)jJj(kr) eijθ(10.117)e^{ikx} = \sum_{j=-\infty}^{\infty}(i)^j J_j(kr)\,e^{ij\theta} \tag{10.117} (where JjJ_j is the Bessel function of order JJ). Plot BB as a function of kaka, for 0<ka<20 < ka < 2.

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