Schrodinger Equation - Quantum Mechanics - Solved Past Paper, Exams of Physics

Download Schrodinger Equation - Quantum Mechanics - Solved Past Paper and more Exams Physics in PDF only on Docsity! PHY 137A (D. Budker) Midterm 1 Solutions TA: Uday Varadarajan 1. Derive the general solution of the Schrödinger Equation for a free particle. Solution: The Schrödinger Equation for a free particle is given by −
2 2m ∂2ψ(x, t) ∂x2 = i
∂ψ(x) ∂t (1.1) We use the seperation of variables trick, considering a solution of the form ψ(x, t) = ψ(x)χ(t). Then, as in the general case, this reduces the time dependent Schrödinger equation to the time independent Schrödinger Equation for every positive value of energy E, ψE(x, t) = ψE(x)e−iEt/
. That is, ψE(x) is a solution to −
2 2m ∂2ψE(x) ∂x2 = EψE(x) (1.2) We make the ansatz that the general solution to this equation has the form ψE(x) = Aeikx +Be−ikx, (1.3) where k2 = 2mE
2 . Plugging this into the time independent Schrödinger equation, we get that −
2 2m ∂2ψE(x) ∂x2 = −
2 2m (A(ik)2eikx + (−ik)2Be−ikx) =
2k2 2m (Aeikx +Be−ikx) = E(Aeikx +Be−ikx) = EψE(x), (1.4) which shows that our ansatz is indeed a solution. That this is the most general solution follows from the fact that a second order linear differential equation has precisely two linearly independent solutions, and we indeed have exactly two free parameters. Now, since the time dependent Schrödinger equation is a linear differential equation, a general solution for a free particle is given by some normalizable superposition of the ψE(x, t) for possibly all values of E, ψ(x, t) = ∫ ∞ 0 c(E)ψE(x, t)dE (1.5) It is more convenient to use k instead of E as the variable of integration, so one instead writes the most general solution as ψ(x, t) = 1√ 2π ∫ ∞ −∞ φ(k)eikx−i
k2t 2m dk. (1.6) where φ(k) can be determined from initial conditions and the requirement that ψ(x, t) is properly normalized. 2. Capillary waves on the surface of water are due to surface tension. They prevail over gravity waves when the wavelengths are smaller or on the order of 1cm. For such waves, the frequency ω scales with the wavelength according to: ω ∝ λ−3/2 (1.7) (a) How does the phase velocity scale with wavelength for such waves? (b) How does the group velocity scale with wavelength? 1 (c) Find the ratio of the group and phase velocities. Explain the meaning of your result. Solution: We can use the relation between wavelength and wave vectors, k = 2πλ to rewrite the dispersion relation as ω = Ck 3 2 , (1.8) The phase velocity and group velocity are given by the expressions Phase Velocity = ω k = Ck 1 2 ∝ λ− 12 (1.9) Group Velocity = ∂ω ∂k = 3 2 Ck 1 2 ∝ λ− 12 . (1.10) Thus, the ratio of the group velocity to the phase velocity is clearly 32 . This means that the group of waves forming a particular packet is moving faster than any individual crest of the wave. So, if we were far away, we would see the swell of the wave moving past us rather quickly, but each crest would seem to lag behind the group, falling further and further behind till it died away. (d) Why does this question appear on a quantum mechanics test? TA’s take: Well, understanding the behaviour of waves is after all one of the best ways to gain intuition for the wave like behaviour of quantum particles, and nonrelativistic particles have a dispersion relation not too different from the above relation ω ∝ λ−2. In particular, the qualitative behaviour of matter waves is much the same as what you’ve found for capillary waves, but with some altered constants. 3. Consider the scattering of a particle with E  mα2
2 , from a triple δ-function potential: V (x) = α [δ(x− a) + δ(x− b) + δ(x − c)] (1.11) (a) What are the relative positions of the potential spikes (a, b, c) that maximize the reflection coefficient. Solution: The amplitude for reflection off any given well is given by M = iβ 1− iβ , (1.12) where β = mα 2 E
2 . Now, in the limit that E  mα 2 2
2 , β  1, and thus, we see that the reflection amplitudes are very small, and we can approximate the reflection amplitude for scattering off all three delta functions as the sum of the amplitudes for scattering off each delta function seperately, along with a phase factor which accounts for the extra distance travelled by the corresponding waves. In particular, assuming that the potential spikes are in alphabetical order from left to right, the amplitude for reflection is given by M3 =M +Me2ik(b−a) +Me2ik(c−a). (1.13) We will get maximal constructive interference if the reflected waves off all three potential spikes all have the same phase, which happens if we satisfy the conditions 2k(b− a) = 2πn (1.14) 2k(c− a) = 2πm. (1.15) This will happen, for example, if (b−a) = π/k and (c−a) = 2π/k. In terms of the DeBroigle wavelength of the particle λ = 2πk , this just translates into the requirement that the potential spikes each be seperated by half integral multiples of its DeBroigle wavelength. (b) How does the reflection coefficient in the arrangement of part (a) compare to the reflection coefficient from a single δ-function potential? Solution: In this case, M3 = 3M so R = |M3|2 = 9|M |2, is nine times what we would have found for the single delta function potential. 2