Notes on Quantum Harmonic Oscillator: Brute Force Methods, Lecture notes of Quantum Mechanics

Download Notes on Quantum Harmonic Oscillator: Brute Force Methods and more Lecture notes Quantum Mechanics in PDF only on Docsity! 8.04: Quantum Mechanics Professor Allan Adams Massachusetts Institute of Technology 2013 March 5 Lecture 8 Quantum Harmonic Oscillator: Brute Force Methods Assigned Reading: E&R 5all, 61,2,8 Li. 3all, 41, 51, 6all Ga. 24, 3all Sh. 4all, 51,2 We will now continue our journey of exploring various systems in quantum mechanics for which we have now laid down the rules. Roughly speaking, there are two sorts of states in quantum mechanics: 1. Bound states: the particle is somewhat localized and cannot escape the potential: 2. Unbound states: the particle can escape the potential. Note that for the same potential, whether something is a bound state or an unbound state depends on the energy considered. Figure 1: For the finite well, the energy represented by the lower black line is for a bound state, while the energy represented by the upper black line is for an unbound state But note that in qunatum mechanics, because of the possibility of tunneling as seen before, the definition of whether a state is bound or not differs between classical and quantum mechanics. The point is that we need to compare E with limx→±∞ V (x) to determine if a state is bound or not. Why do we split our cases that way? Why do we study bound and unbound states separately if they obey the same equations? After all, in classical mechanics, both obey F = ṗ. In quantum mechanics, both obey ∂ψ(x, t) Êψ(x, t) = iJ . ∂t 2 8.04: Lecture 8 Figure 2: The same energy denoted by the black line is a bound classical and quantum state for the potential on the left, while the classical bound state is a quantum unbound state for the potential on the right The distinction is worth making because the bound and unbound states exhibit qualitatively different behaviors: Mechanics Bound States Unbound States Classical periodic motion aperiodic motion Quantum discrete energy spectrum continuous energy spectrum For now, we will focus on bound states, with discussions of unbound states coming later. Let us remind ourselves of some of the properties of bound states. 1. Infinite square well: • Ground state: no nodes • nth excited state: n nodes (by the node theorem) • Energy eigenfunctions chosen to be real − iEt• Time evolution of energy eigenfunctions through complex phase e � . – Different states evolve at different rates – Energy eigenstates have no time evolution in observables as p(x) for such states is independent of t – Time evolution of expectation values for observables comes only through in­ terference terms between energy eigenfunctions 2. Symmetric finite square well • Node theorem still holds • V (x) is symmetric – Leads to symmetry or antisymmetry of φ(x; E) – Antisymmetric φ(x; E) are fine as |φ(x; E)|2 is symmetric • Exponetial tails in classically forbidden regions leads to discrete energy spectrum (picture of shooting for finite square well?) 3. Asymmetric finite square well 5 8.04: Lecture 8 This is certainly much neater, but it is not any easier. To make things easier, we need to develop some more mathematical techniques. As a mathematical detour, we need to discuss asymptotic analysis. Let us suppose that we are solving the equation ∂f(x) 1 + 1 + f(x) = 0. (0.6) ∂x x In asymptotic analysis, we look at the different limits. Let us consider that 1 1 lim 1 + ≈ . x→0 x x This means that lim x→0 ∂f(x) ∂x ≈ f(x) x (0.7) which is solved by f(x) = f0x −1 . But this only considers the behavior as x → 0. Let us say that f(x) = x −1 g(x) where g(x) is finite and well-behaved as x → 0. Plugging this in yields ∂g(x) = −g(x), ∂x which is solved by g(x) = g0e −x . This means that the full solution is g0 f(x) = (0.8) xex which is exact! The first approximation was just to make our lives easier, and there ended up being no accuracy sacrificed anywhere. Furthermore, e−x is indeed relatively constant compared to x−1 as x → 0. To summarize, we looked at the extreme behavior of the differential equation to peel off a part of the solution. We then plugged that back in to get a simpler differential equation to solve. ∂g(x)But what if we did not know how to solve ∂x + g(x) = 0? How would we figure it out? � � � � � � � � 6 8.04: Lecture 8 This is where the series method comes in. We can expand the function as a power series ∞ g(x) = aj x j (0.9) j=0 and perhaps figure out what the coefficients need to be. Now, we know that ∞ ∞ ∞ ∂g(x) j−1 s = jaj x j−1 = jaj x = (s + 1)as+1x (0.10) ∂x j=0 j=1 s=0 first by removing the j = 0 term and second by redefining s ≡ j − 1. But s and j are just labels, so we might as well just say that ∂g(x) ∞ = (j + 1)aj+1x j . (0.11) ∂x j=0 This means that our differential equation is now ∞ ((j + 1)aj+1 + aj )x j = 0. (0.12) j=0 The left side is a polynomial in x. If this is zero for all x, then the overall coefficients are identically zero: (j + 1)aj+1 + aj = 0 (0.13) so aj aj+1 = − (0.14) j + 1 implying (−1)j aj = a0 (0.15) j! so ∞ ∞ j (−x)j −x g(x) = aj x = a0 = a0e (0.16) j! j=0 j=0 as expected! To summarize, we just expanded the function as a power series, found the recursion relation for its coefficients, and then plugged in the initial conditions. Let us get back into the physics of this. We want to solve the equation ∂2φ = (u 2 − ε)φ. ∂u2 ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ � 7 8.04: Lecture 8 First, we apply asymptotic analysis: lim (u 2 − ε) ≈ u 2 (0.17) u→±∞ so ∂2φ lim ≈ u 2φ. (0.18) u→±∞ ∂u2 We can try αu2 2φ(u) = φ0e . We find that ∂φ = αuφ ∂u and ∂2φ = (α + α2 u 2)φ ≈ α2 u 2φ ∂u2 in the same limit. Comparing this to our differential equation, we see that we need α2 = 1 so α = ±1. We then get 2 2 2 2φ(u) = φAe u + φB e − u . The first part is not normalizable, though, so we do not want that. We keep only the second part and generalize the constant to a function of u that is relatively constant and well-behaved as u → ±∞, so we say that 2φ(u) = s(u)e − u 2 . Now recall that the last step in asymptotic analysis is to plug this form into the original differential equation to yield a differential equation that can be solved for s(u): ∂2s ∂s − 2u + (ε − 1) s = 0. (0.19) ∂u2 ∂u Before we solve this, we need to figure out the behavior of s(u). We know that it should grow 2 2less rapidly than e+ u as u → ±∞. Also, there should be multiple solutions corresponding to discrete bound states. Finally, the nth solution should have n nodes. Let us try the power series expansion ∞ s(u) = aj u j (0.20) j=0 ∑