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Particle in a Box Problem 2 Lecture #10 1 Lecture 10 Objectives: 1. Be able to solve the particle in a box problem 2. Compare the classical and quantum harmonic oscillators 3. Compare the classical and quantum rotors Example: Particle in a box Consider a particle trapped in a one-dimensional box, of length L. The Schr¨odinger equation for this system (considering only the spatial function) is h2 d2ψ − + Uψ = Eψ, 8π2m dx2 where U = 0 when x is inside the box and U = ∞ for x outside the box. This is a second order ordinary differential equation. The general solution to this equation is ψ = α sin γx + β cos γx with γ = (8π2m(E − U)/h2)1/2 We need two boundary conditions to fix α and β. What are they? ψ(0) = 0, and ψ(L) = 0. These BCs give ψ(0) = β = 0 and ψ(L)= α sin γL + β cos γL = 0 hence, either α = 0(trivial solution) or sin γL = 0. Taking the latter we find nπ γL = nπ → γ = L Hence, nπx ψ = α sin n L ∞ ∗ The values of αn can be computed by recalling that 0 ψ ψdx = 1 so that ∞ L L ∗ ∗ ∗ nπxR ψ ψdx = ψ ψdx = α α sin2 = 1 n n L Z0 Z0 Z0 Recalling that sin2udu = u/2 − sin(2u)/4+ C we have that nπ ∗ L R u sin(2u) ∗ L nπ sin(2nπ) αnαn − = αnαn − = 1 nπ 2 4 0 nπ 2 4 1/2 2 Therefore, αn = i L . We can now solve for the energy levels by from the equation L L L ∗ ∗ ∗ ψ Hψdx = ψ Eψdx = E ψ ψdx = E Z0 Z0 Z0 L 2 2 L ∗ h nπ 2 nπx ψ Hψdx = − − sin2 dx = E 8π2m L L L n Z0 ! Z0 2 2 2 nh 1 n h Therefore, En = L 8m = 8mL2 . Lecture #10 2 The Harmonic Oscillator The Classical Harmonic Oscillator A vibrating body subject to a restoring force, which increases in proportion to the displacement from equilibrium, will undergo harmonic motion at constant frequency and is called a harmonic oscillator. Figure 1(a) shows one example of a harmonic oscillator, where a body of mass m is connected to a fixed support by a spring with a force constant, k. We will assume that gravitational forces are absent. Figure 1: (a) Harmonic Oscillator Consisting of a Mass Connected by a Spring to a Fixed Support; (b) Potential Energy, V ,and Kinetic Energy, EK For the Harmonic Oscillator. When the system is at equilibrium, the mass will be at rest, and at this point the displacement, x, from equilibrium has the value zero. As the mass is pulled to the right, there will be a restoring force, f, which is proportional to the displacement. For a spring obeying Hooke’s law, d2x f = −kx = m (1) dt2 The minus sign in equation 1 is related to the fact that the force will be negative, since the mass will tend to be pulled toward the −x direction when the force is positive. From Newton’s second d2x law, the force will be equal to the mass multiplied by the acceleration, dt2 . The equation of motion is a second order ordinary differential equation, obtained by rearranging equation 1 as d2x k + x = 0, (2) dt2 m and has a general solution given by x(t)= A sin ωt + B cos ωt, (3) where ω =(k/m)1/2. (4) The units of ω are radians s−1, and since there are 2π radians/cycle, the frequency ν = ω/2π cycles s−1. [Note: The student should check this solution by substituting equation 3 back into equation 2]. We again require two boundary conditions to specify the constants A and B. We choose the mass to be at x = 0 moving with a velocity v0 at time= 0. The first condition gives x(t =0)= A · 0+ B · 1= B = 0 (5) Lecture #10 3 so that x(t)= A sin ωt. (6) Using this result, the second boundary condition can be written as dx v0 = v(t =0)= = Aω cos ωt|t=0 = Aω, (7) dt t=0 from which we see that A = v0/ω. As the spring stretches, or contracts, when the mass is undergoing harmonic motion, the potential energy of the system will rise and fall, as the kinetic energy of the mass falls and rises. The change in potential energy, dV , is dV = −fdx = kxdx, (8) so that upon integrating, 1 V = kx2 + constant. (9) 2 The constant of integration may be set equal to zero. This potential energy function is shown as the parabolic line in Figure 1(b). The kinetic energy of this harmonic oscillator is given by 1 1 dx 2 m E = mv2 = m = (Aω)2 cos2 ωt (10) K 2 2 dt 2 This function is also plotted in Figure 1(b). The total constant energy, E, of the system is given by 1 m 1 m E = V + E = kx2 + (Aω)2 cos2 ωt = kA2 sin2 ωt + (Aω)2 cos2 ωt, (11) K 2 2 2 2 where we have substituted equation 6 for x. Substituting equation 4 for ω2, we can write 1 1 E = A2k sin2 ωt + cos2 ωt = A2k (12) 2 2 The total energy is thus a constant and is shown as a horizontal line in Figure 1(b). At the limits of oscillation, where the mass is reversing its direction of motion, its velocity will be momentarily equal to zero, and its momentary kinetic energy will therefore also be zero, meaning that the potential energy will be maximized and equal to the total energy of the system at the two turning points. As the oscillator begins to undergo acceleration away from the turning points, the kinetic energy will increase, and the potential energy will decrease along the curve, V , as shown in Figure 1(b). If the spring constant, k, should not be constant, but should vary slightly from a constant value as x changes, we would be dealing with an anharmonic oscillator. In most cases, an anharmonic oscillator may be closely approximated by the harmonic oscillator equations for small displacements from equilibrium, x. Soon we will be comparing the quantum harmonic oscillator with the classical harmonic oscil- lator, and the probability of finding the mass at various values of x will be of interest. We now calculate this probability for the classical harmonic oscillator. The probability of finding the mass, m, at any given value of x is inversely proportional to the velocity, v, of the mass. This is reasonable, since the faster the mass is moving, the less likely it is to observe the mass. Hence, we expect that the probability of observing the mass will be have Lecture #10 4 a minimum at x = 0, where the velocity is at a maximum, and conversely will exhibit maxima when x = ±A. From equation 7 we see that dx v(x)= = Aω cos ωt, (13) dt so that dx P(x)dx ∝ , (14) Aω cos ωt where P(x)dx is the probability of finding the mass between x and x + dx. Note that since x is a continuous variable that we define P(x)dx as the probability and P(x) is called the probability density, which is the probability per unit length, in this case. We note that at the turning points π 3π of the oscillation, when 1/4 or 3/4 of a cycle has occurred, that t = 2ω or t = 2ω and at these points 1 cos ωt goes to zero, with v going to infinity. We know that the probability of finding the mass at the end points must be a maximum, but must not be infinite. The reason that P(x) remains finite is that dx in equation 14 always has finite width, therefore, P(x) is not defined exactly at a π point. For example, at t = 2ω , P(x) is evaluated over the range −A<x< −A + dx. This probability density function as a function of the x-coordinate, P(x), is plotted along with the velocity, v in Figure 2. The probability density is a smooth function over the range of x available to the oscillator and has exactly one minimum at x = 0. We will soon find that this intuitive classical behavior is not obeyed by the quantized harmonic oscillator. In fact, for the quantum oscillator in the ground state we will find that P(x) has a maximum at x = 0. v(x) P(x) 0 0 -A +A x Figure 2: Probability Density, P(x), for Classical Harmonic Oscillator at Various Displacements, x. P(x) is plotted as the dashed line and the velocity, v(x) is plotted as the solid curve. The two 1 vertical lines give the limits of the oscillator motion. Note that P(x) ∝ v(x) . The Quantum Harmonic Oscillator The quantum harmonic oscillator is a very important problem in quantum mechanics. For example, it serves as a first-order approximation for the bond vibrational problem in diatomic and (with Lecture #10 5 coupling) polyatomic molecules. We will examine the quantum harmonic oscillator in some detail, comparing it with what we know about the classical harmonic oscillator from the previous section. The potential energy function for the quantum harmonic function is the same as for the classical harmonic oscillator, namely, V = 1/2kx2. Thus, in the quantum Hamiltonian is Hop = EKop + Vop (15) and we may write the Schr¨odinger equation as −¯h2 ∂2ψ 1 + kx2ψ = Eψ (16) 2m ∂x2 2 The general solution to this problem (which we will not derive) can be written as 1/4 1/2 α 1 −y2/2 ψn(x)= n Hn(y)e , (17) π 2 n! mω 1/2 where n = 0, 1, 2,..
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