Modern Physics (PHY 3305) - Lecture Notes. Modern Physics (PHY 3305) Lecture Notes

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1 Modern Physics (PHY 3305) Lecture Notes Modern Physics (PHY 3305) Lecture Notes Bound States II (Ch ) SteveSekula, 25 February 2010 (created 13 December 2009) tags: lecture 1 of 10 02/25/ :56 AM

2 Review We started to dig into the mechanics of solving the Shroedinger Wave Equation How to solve in four easy steps: Identify the player(s) in the problem (a particle, multiple particles, etc.) Identify constraints and boundary conditions in the problem (define your potential(s)) Solve for the wave function by (1) applying differential equation solving, (2) enforcing smoothness, and (3) enforcing normalization Determine outcomes in the problem by applying the wave function We applied the first 3 of these to the infinite square well, and in spirit (by graphing wave functions and probability densities) step 4 Today We get a little more realistic: the finite square-well We get a lot more realistic: the harmonic oscillator We expand on Step 4: we start solving for outcomes using the wave function The Finite Square Well Need to write down the SWE in three different regions. In Regions II and III: Öh 2 2 d ï(x) À + U ï(x) ï(x) 2m 2 0 = E and in Region I: Öh 2 2 d ï(x) À 2m 2 = Eï(x) 2 of 10 02/25/ :56 AM

3 which is just like in the infinite square well. Let's begin by solving the equations in Regions II and III. Rewrite as: 2 d ï(x) 2 = 2m(U 0 À E) ï(x) Öh 2 Solutions will be of the form: ï(x) = Ce Ëx + De ÀËx where Ë = p 2m(U o À E)=Öh 2. But in order to find appropriate solutions for this problem, we need to apply physicality considerations, specifically normalization. No matter what the value of A, at x = À1 the above solution shoots off to infinite normalization, as it does at x = + 1. To achieve a finite normalization for these wave functions, we need to remove one of the two halves of the general solution. Which half depends on whether x < 0 or x > 0. We find: ï(x) II = Ce Ëx and ï(x) III = Ge ÀËx We don't yet know what C and G are. What about in Region I? Previously, we wrote down the general solution: ï(x) I = A sin(kx) + Bcos(kx) and tossed the cosine part because we needed the function to be 0 at x = 0. Now, however, that requirement is relaxed and we have to leave the solution in this more general form. 3 of 10 02/25/ :56 AM

4 DISCUSSION: How do we solve for A, B, C, and G? ANSWER: there are conditions we have not yet applied: smoothness (continuity). Imposing continuity on the problem means that: ï(x = 0) II = ï (x = 0) I ï(x = L) I = ï (x = L) III dï(x) II dï(x) Ì = I Ì (x=0) dï(x) I dï(x) Ì = III Ì (x=l) Ì (x=0) Ì (x=l) ï(x = 0) II = ï (x = 0) I leads to C = B dï(x) II dï(x) Ì = I Ì leads to ËC = ka. (x=0) (x=0) ï(x = L) I = ï (x = L ) III leads to Asin(kL) + B cos(kl) = Ge ÀËL dï(x) I dï(x) Ì = III ÀËL Ì leads to kacos(kl) À k Bsin(kL) = ÀËGe (x=l) (x=l) It's not immediately obvious, but these four equations impose a QUANTIZATION CONDITION on the solutions. Working through the algebra, we find: 2cot(kL) = k Ë À Ë k This holds only for discrete values of E, which can be seen from a graph of energy vs. kl (and choosing a horizontal line on the graph representing U 0). (see slides) In addition, the wavelengths of these waves are longer, and so the energies of each allowed wave are necessarily lower, than in the infinite square well. 4 of 10 02/25/ :56 AM

5 You can't solve for the exact wave functions without specifying E (e.g. all of those coefficients are not well-defined without a well-defined problem) - that is, normalization to determine the coefficients requires specific E, and thus k and Ë, values. You can then plot the wave functions and their probability densities (see slides) and compare to the infinite square well. Entering the Forbidden Region Given the wave nature of matter, it's not completely surprising that solutions can penetrate into the classically forbidden region. However, this stands as the most striking contrast with the infinite square well. You can compute the "penetration depth" of the wave function into the forbidden region by evaluating the wave functions in regions II or III for the case where x = 1 =Ë = Î (the "penetration depth"). This is the place on the wave function where the amplitude falls by 1/e, and is a standard definition for progress along an exponential. Î = Öh p 2m(U ) 0 À E The penetration into this region should become more and more significant as the energy approaches U, and indeed the graphs bear that out. 0 Can you see particles in the forbidden region? This is a tricky question. Let's think about this carefully. DISCUSSION: what is needed to "see" particles in the forbidden region? ANSWER: You need to conduct an experiment whose spatial precision is better than or comparable to Î. That meants Î = Áx. By the Uncertainty Principle, Áp Õ Öh =(2Î), which means that the momentum uncertainty is Áp Õ Öh = p 2mU 0 À E This would imply that KE = p 2 =2m Õ Öh 2 =2mÎ 2 = U 0 À E So the kinetic energy can be no smaller than U 0 À E, which is a NEGATIVE kinetic energy ( KE = E À U 0 ) 5 of 10 02/25/ :56 AM

6 Thus no experiment can be certain that the particle has negative kinetic energy, which is what is required to be sure that the particle has entered the classically forbidden region So all we can say for sure is the following: Until we make an observation of the location of the particle, the particle itself is a standing wave whose physical extent penetrates into the classically forbidden region (we cannot "watch" the particle continuously to see where it goes) Once we make a measurement, there is a finite probability of finding it outside the box The Harmonic Oscillator See slides for graphical discussions of the harmonic oscillator. We have discussed the: classical "mass on a spring" - a harmonic oscillator we have discussed atomic bound states Not that the deepest part of the atomic bound state potential has a shape quite similar to the harmonic oscillator of the mass on a spring. Therefore, this is really our first "physical" example that can be solved using the SWE. What does the SWE look like? There are no longer "regions" where the potential takes a step - instead, it increases steadily in both the negative and positive directions, and its minimum is taken to be at x = 0. The SWE then looks like: Öh 2 2 d ï(x) 1 2 À + Ôx ï(x) = Eï(x) 2m 2 2 DISCUSSION: What might the solutions to this equation look like? ANSWER: you need a solution that, passed through the derivative twice, 6 of 10 02/25/ :56 AM

7 yields an 2 x out in front of the original wave function. A good candidate is: ï(x) = Ae Àax2 This is the Gaussian Function. It turns up A LOT in physics, math, and especially statistics. If we try this solution, we arrive at: Ò Öh 2 a m Ó À E À Ò 2Öh 2 a 2 1 Ó À Ô x 2 = 0 m 2 We can choose to solve this by requiring that: a = p mô=2ö h which forces the right-term on the left-hand-side of the equation to be zero. E = 2 Öh r Ô m We then have to solve for the remaining unknown in the wave function. DISCUSSION: what unused property of the wave function can we apply to do this? ANSWER: Normalization! Z +1 À1 Z r +1 2 Ù jï(x)j = A 2 e À2ax2 = A 2 2a À1 which yields A = mô Ñ 1=8 Ù 2 Öh 2 7 of 10 02/25/ :56 AM

8 Thus: E = 2 Öh r Ô m and ï(x) = mô Ñ 1=8 p e À( mô =2Ö)x h 2 Ù 2 Öh 2 In fact, the rigorous solution to this problem teaches us that: 1 E = ( n + )Ö! h 2 0 where q! 0 = Ô=m Features of the Solution to the Harmonic Oscillator The solutions are equally spaced (see slides and consider the energies given above) - this is characteristic of many oscillating systems For each energy, there is a unique wave function (see table 5.1 in Harris) The simple Gaussian we considered above is the GROUND STATE of the system, but there are obviously other more general functions that work. There are the product of Gaussian functions with Hermite polynomials and a normalization constant for each energy. Contrary to the classical case, the particle is most likely to be found at the center of the well (in the ground state) and not at the sides of the well. As n increases, however, we regain the classical expectation: that the particle is most likely to be found near the edges. 8 of 10 02/25/ :56 AM

9 Applying the Harmonic Oscillator: the thermal behavior of a diatomic gas Diatomic gases are very common - you are breathing a few of them right now ( O 2 and N 2). A diatomic molecule is a bound state of two identical atoms, and under normal room conditions the atoms remain bound. What can we learn about such gases from the harmonic oscillator approximation? In classical thermodynamics, we learn that different gases have different molar heat capacities - basically, their ability to store energy as they are heated. Monatomic gases have just one value for that heat capacity, independent of the element, while diatomic gases have varied molar heat capacities depending on the element. Heat capacity is the change in energy per unit change in temperature. You have different places you can "stick" energy in a diatomic molecule as you heat it. Think of heat capacity as being defined by the number of buckets into which you can stick energy. More buckets = more heat capacity. Theoretically, the motions of a diatomic molecule under application of heat energy corresponds to translation, rotation, and vibration of the molecule. The latter is due to the "interatomic springiness" of the system. At room temperature, predictions for the molar heat capacity from classical considerations well with data, but ONLY as long at the molecules cannot vibrate. Why might this be? The consideration of the quantum harmonic oscillator suggests that applying classical HO-thinking to the problem will lead to incorrect assumptions about the ground state, the behavior of probability versus separation of the molecules, etc. Unlike a classical spring, only certain frequencies are allowed in the quantum case. A quantum oscillator CANNOT drop below its ground state, so even if it's not translating or rotating, it's vibrating. But this also means that if it cannot absorb enough energy from translation or rotation, it will remain stuck in its vibrational ground state. Given the definition of heat capacity, if over a certain range of temperatures you don't add enough energy to bump the system up from its ground state, you'll never add anything to its vibrational energy, and in turn vibration will add nothing to the heat capacity. You don't get to put energy in the vibration "bucket" until the temperature gets high enough, and then suddenly the heat capacity changes since you can stick energy into vibration. Harris works through the case for Hydrogen ( H ), and demonstrates that at 2 9 of 10 02/25/ :56 AM

10 room temperature (300K) the amount of energy available to hydrogen gas from translation is 10 times smaller than what is needed to get hydrogen out of its ground state and into the next state. This means that the heat capacity does not obtain a contribution from vibration until you reach 1000s of Kelvin, which is exactly what is observed. Classical physics could not explain this observation, but quantum physics readily offers an explanation from even "meager" considerations. Absolute Zero An important consequence of the non-zero ground state is that even when the temperature of an oscillator is reduced to zero, it possesses energy. Solids are a good example. SOlids are nothing but atoms in bound states, free to vibrate around points in space. But being an oscillator, they can never have zero energy. Removing all energy only leaves the system in its ground state, NOT at a state of zero energy. But if T=0, something must be zero! What is it? DISCUSSION: what IS zero at absolute zero? ANSWER: entropy. A system with T=0 has some energy, but ZERO disorder. It's as perfectly ordered as it can be, with all available states neatly filled. Next Time mid-term! Then: We will study wave and particle propagation We will discuss scattering off of a boundary We will discuss the quantum phenomenon called "tunneling" and its applications 10 of 10 02/25/ :56 AM