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Simple Harmonic Oscillator and Barriers in Quantum Mechanics

This lecture discusses the concepts of simple harmonic oscillators and barriers in quantum mechanics, including topics such as alpha particle decay and the Schrodinger equation on the hydrogen atom. The lecture also covers the solutions for the Schrodinger equation and the analysis of the parabolic potential well.

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Simple Harmonic Oscillator and Barriers in Quantum Mechanics

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  1. PHYS 3313 – Section 001Lecture # 22 Wednesday, April 22, 2015 Dr. Barry Spurlock • Simple Harmonic Oscillator • Barriers and Tunneling • Alpha Particle Decay • Schrodinger Equation on Hydrogen Atom • Solutions for Schrodinger Equation for Hydrogen Atom PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  2. Announcements • Research paper deadline is Monday, May 4 • Research presentation deadline is Sunday, May 3 • Bring out Homework #5 • Reminder Homework #6 • CH7 end of chapter problems: 7, 8, 9, 12, 17 and 29 • Due on Wednesday, Apr. 29, in class • Reading assignments • CH7.6 and the entire CH8 • Quiz number 5 • At the beginning of the class Wednesday, Apr. 29 • Covers up to what we finish Monday, Apr. 27 PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  3. The Simple Harmonic Oscillator • Simple harmonic oscillators describe many physical situations: springs, diatomic molecules and atomic lattices. • Consider the Taylor expansion of a potential function: The minimum potential at x=x0, so dV/dx=0 and V1=0; and the zero potential V0=0, we have Substituting this into the wave equation: Let and which yields . PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  4. Parabolic Potential Well • If the lowest energy level is zero, this violates the uncertainty principle. • The wave function solutions are where Hn(x) are Hermitepolynomial function of order n. • In contrast to the particle in a box, where the oscillatory wave function is a sinusoidal curve, in this case the oscillatory behavior is due to the polynomial, which dominates at small x. The exponential tail is provided by the Gaussian function, which dominates at large x. PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  5. Analysis of the Parabolic Potential Well • The energy levels are given by • The zero point energy is called the Heisenberg limit: • Classically, the probability of finding the mass is greatest at the ends of motion’s range and smallest at the center (that is, proportional to the amount of time the mass spends at each position). • Contrary to the classical one, the largest probability for this lowest energy state is for the particle to be at the center. PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  6. Ex. 6.12: Harmonic Oscillator stuff • Normalize the ground state wave function ψ0 for the simple harmonic oscillator and find the expectation values <x> and <x2>. PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  7. Barriers and Tunneling • Consider a particle of energy E approaching a potential barrier of height V0 and the potential everywhere else is zero. • We will first consider the case when the energy is greater than the potential barrier. • In regions I and III the wave numbers are: • In the barrier region we have PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  8. Reflection and Transmission • The wave function will consist of an incident wave, a reflected wave, and a transmitted wave. • The potentials and the Schrödinger wave equation for the three regions are as follows: • The corresponding solutions are: • As the wave moves from left to right, we can simplify the wave functions to: PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  9. Probability of Reflection and Transmission • The probability of the particles being reflected R or transmitted T is: • The maximum kinetic energy of the photoelectrons depends on the value of the light frequency f and not on the intensity. • Because the particles must be either reflected or transmitted we have: R + T = 1 • By applying the boundary conditions x→ ±∞, x = 0, and x = L, we arrive at the transmission probability: • When does the transmission probability become 1? PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  10. Tunneling • Now we consider the situation where classically the particle does not have enough energy to surmount the potential barrier, E < V0. • The quantum mechanical result, however, is one of the most remarkable features of modern physics, and there is ample experimental proof of its existence. There is a small, but finite, probability that the particle can penetrate the barrier and even emerge on the other side. • The wave function in region II becomes • The transmission probability that describes the phenomenon of tunneling is PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  11. Uncertainty Explanation • Consider when κL >> 1 then the transmission probability becomes: • This violation allowed by the uncertainty principle is equal to the negative kinetic energy required! The particle is allowed by quantum mechanics and the uncertainty principle to penetrate into a classically forbidden region. The minimum such kinetic energy is: PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  12. Analogy with Wave Optics • If light passing through a glass prism reflects from an internal surface with an angle greater than the critical angle, total internal reflection occurs. The electromagnetic field, however, is not exactly zero just outside the prism. Thus, if we bring another prism very close to the first one, experiments show that the electromagnetic wave (light) appears in the second prism. • The situation is analogous to the tunneling described here. This effect was observed by Newton and can be demonstrated with two prisms and a laser. The intensity of the second light beam decreases exponentially as the distance between the two prisms increases. PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  13. Potential Well • Consider a particle passing through a potential well region rather than through a potential barrier. • Classically, the particle would speed up passing the well region, because K = mv2 / 2 = E- V0. According to quantum mechanics, reflection and transmission may occur, but the wavelength inside the potential well is shorter than outside. When the width of the potential well is precisely equal to half-integral or integral units of the wavelength, the reflected waves may be out of phase or in phase with the original wave, and cancellations or resonances may occur. The reflection/cancellation effects can lead to almost pure transmission or pure reflection for certain wavelengths. For example, at the second boundary (x= L) for a wave passing to the right, the wave may reflect and be out of phase with the incident wave. The effect would be a cancellation inside the well. PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  14. Alpha-Particle Decay • May nuclei heavier than Pb emits alpha particles (nucleus of He)! The phenomenon of tunneling explains the alpha-particle decay of heavy, radioactive nuclei. • Inside the nucleus, an alpha particle feels the strong, short-range attractive nuclear force as well as the repulsive Coulomb force. • The nuclear force dominates inside the nuclear radius where the potential is approximately a square well. • The Coulomb force dominates outside the nuclear radius. • The potential barrier at the nuclear radius is several times greater than the energy of an alpha particle (~5MeV). • According to quantum mechanics, however, the alpha particle can “tunnel” through the barrier. Hence this is observed as radioactive decay. PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  15. Application of the Schrödinger Equation to the Hydrogen Atom The approximation of the potential energy of the electron-proton system is the Coulomb potential: To solve this problem, we use the three-dimensional time-independent Schrödinger Equation. For Hydrogen-like atoms with one electron (He+ or Li++) Replace e2 with Ze2 (Z is the atomic number) Use appropriate reduced mass μ PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  16. Application of the Schrödinger Equation • Transform to spherical polar coordinates to exploit the radial symmetry. • Insert the Coulomb potential into the transformed Schrödinger equation. • The potential (central force) V(r) depends on the distance r between the proton and electron. PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  17. Application of the Schrödinger Equation The wave function ψ is a function of r, θ and φ. The equation is separable into three equations of independent variables The solution may be a product of three functions. We can separate the Schrodinger equation in polar coordinate into three separate differential equations, each depending only on one coordinate: r, θ, orφ . PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  18. Solution of the Schrödinger Equation -------- azimuthal equation • Only r and θ appear on the left-hand side and only φappears on the right-hand side of the equation • The left-hand side of the equation cannot change asφchanges. • The right-hand side cannot change with either r or θ. • Each side needs to be equal to a constant for the equation to be true in all cases. Set the constant −mℓ2 equal to the right-hand side of the reorganized equation • The sign in this equation must be negative for a valid solution • It is convenient to choose a solution to be . PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  19. Solution of the Schrödinger Equation for Hydrogen • Substitute ψinto the polar Schrodinger equation and separate the resulting equation into three equations: R(r), f(θ), and g(φ). Separation of Variables • The derivatives in Schrodinger eq. can be written as • Substituting them into the polar coord. Schrodinger Eq. • Multiply both sides by r2 sin2θ / Rfg Reorganize PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  20. Solution of the Schrödinger Equation satisfies the previous equation for any value of mℓ. The solution be single valued in order to have a valid solution for anyφ, which requires mℓmust be zero or an integer (positive or negative) for this to work Now, set the remaining equation equal to −mℓ2 and divide either side with sin2θ and rearrange them as Everything depends on r on the left side and θ on the right side of the equation. PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  21. Solution of the Schrödinger Equation • Set each side of the equation equal to constant ℓ(ℓ + 1). • Radial Equation • Angular Equation • Schrödinger equation has been separated into three ordinary second-order differential equations, each containing only one variable. PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  22. Solution of the Radial Equation • The radial equation is called the associated Laguerreequation, and the solutions R that satisfies the appropriate boundary conditions are called associated Laguerre functions. • Assume the ground state has ℓ = 0, and this requires mℓ = 0. We obtain • The derivative of yields two terms, and we obtain PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  23. Solution of the Radial Equation Bohr’s radius Ground state energy of the hydrogen atom Let’s try a solution where Ais a normalization constant, and a0is a constant with the dimension of length. Take derivatives of R, we obtain. To satisfy this equation for any r, each of the two expressions in parentheses must be zero. Set the second parentheses equal to zero and solve for a0. Set the first parentheses equal to zero and solve for E. Both equal to the Bohr’s results PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  24. Principal Quantum Number n • The principal quantum number, n, results from the solution of R(r) in the separate Schrodinger Eq. since R(r) includes the potential energy V(r). The result for this quantized energy is • The negative sign of the energy E indicates that the electron and proton are bound together. PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  25. Quantum Numbers • The full solution of the radial equation requires an introduction of a quantum number, n, which is a non-zero positive integer. • The three quantum numbers: • n Principal quantum number • ℓ Orbital angular momentum quantum number • mℓ Magnetic quantum number • The boundary conditions put restrictions on these • n= 1, 2, 3, 4, . . . (n>0) Integer • ℓ = 0, 1, 2, 3, . . . , n − 1 (ℓ < n) Integer • mℓ = −ℓ, −ℓ + 1, . . . , 0, 1, . . . , ℓ − 1, ℓ (|mℓ| ≤ ℓ) Integer • The predicted energy level is PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  26. Ex 7.3: Quantum Numbers & Degeneracy What are the possible quantum numbers for the state n=4 in atomic hydrogen? How many degenerate states are there? n ℓmℓ 4 0 0 4 1 -1, 0, +1 4 2 -2, -1, 0, +1, +2 4 3 -3, -2, -1, 0, +1, +2, +3 The energy of a atomic hydrogen state is determined only by the primary quantum number, thus, all these quantum states, 1+3+5+7 = 16, are in the same energy state. Thus, there are 16 degenerate states for the state n=4. PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  27. Hydrogen Atom Radial Wave Functions The radial solution is specified by the values of n and ℓ First few radial wave functions Rnℓ PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  28. Solution of the Angular and Azimuthal Equations • The solutions forazimuthal eq. are or • Solutions to the angular and azimuthal equations are linked because both have mℓ • Group these solutions together into functions ---- spherical harmonics PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  29. Normalized Spherical Harmonics PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  30. Ex 7.1: Spherical Harmonic Function Show that the spherical harmonic function Y11(θ,φ) satisfies the angular Schrodinger equation. PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  31. Solution of the Angular and Azimuthal Equations The radial wave function R and the spherical harmonics Y determine the probability density for the various quantum states. Thus the total wave function ψ(r,θ,φ) depends on n, ℓ, and mℓ. The wave function can be written as PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  32. Orbital Angular Momentum Quantum Number ℓ • It is associated with the R(r) and f(θ) parts of the wave function. • Classically, the orbital angular momentum with L = mvorbitalr. • ℓis related to the magnitude of L by . • In an ℓ = 0 state, . It disagrees with Bohr’s semi-classical “planetary” model of electrons orbiting a nucleus L = nħ. PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  33. Orbital Angular Momentum Quantum Number ℓ • Certain energy level is degenerate with respect to ℓ when the energy is independent of ℓ. • Use letter names for the various ℓ values • ℓ = 0 1 2 3 4 5 . . . • Letter = spdfgh. . . • Atomic states are referred by their n and ℓ • s=sharp, p=principal, d=diffuse, f =fundamental, then alphabetical • A state with n = 2 and ℓ= 1 is called the 2pstate • Is 2d state possible? • The boundary conditions require n > ℓ PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  34. Magnetic Quantum Number mℓ • The angleφis a measure of the rotation about the z axis. • The solution for specifies that mℓ is an integer and related to the z component of L. • The relationship of L, Lz, ℓ, and mℓ for ℓ = 2. • is fixed. • Because Lz is quantized, only certain orientations of are possible and this is called space quantization. • mℓ is called the magnetic moment since z axis is chosen customarily along the direction of magnetic field. PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  35. Magnetic Quantum Number mℓ • Quantum mechanics allows to be quantized along only one direction in space and because of the relationship L2 = Lx2 + Ly2 + Lz2, once a second component is known, the third component will also be known.  violation of uncertainty principle • One of the three components, such as Lz, can be known clearly but the other components will not be precisely known • Now, since we know there is no preferred direction, • We expect the average of the angular momentum components squared to be: PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  36. Magnetic Effects on Atomic Spectra—The Normal Zeeman Effect • A Dutch physicist Pieter Zeeman showed as early as 1896 that the spectral lines emitted by atoms in a magnetic field split into multiple energy levels. It is called the Zeeman effect. The Normal Zeeman effect: • A spectral line of an atom is split into three lines. • Consider the atom to behave like a small magnet. • The current loop has a magnetic moment μ = IA and the period T = 2πr / v.If an electron can be considered as orbiting a circular current loop of I = dq / dt around the nucleus, we obtain • where L = mvr is the magnitude of the orbital angular momentum PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  37. The Normal Zeeman Effect • Since there is no magnetic field to align them, points in random directions. • The dipole has a potential energy The angular momentum is aligned with the magnetic moment, and the torque between and causes a precession of . Where μB = eħ / 2mis called the Bohr magneton. cannot align exactly in the z direction and has only certain allowed quantized orientations. PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  38. The Normal Zeeman Effect • The potential energy is quantized due to the magnetic quantum number mℓ. • When a magnetic field is applied, the 2p level of atomic hydrogen is split into three different energy states with the electron energy difference of ΔE = μBB Δmℓ. • So split is into a total of 2ℓ+1 energy states PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  39. The Normal Zeeman Effect A transition from 1s to 2p A transition from 2p to 1s PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  40. The Normal Zeeman Effect An atomic beam of particles in the ℓ= 1 state pass through a magnetic field along the z direction. (Stern-Gerlach experiment) The mℓ = +1 state will be deflected down, the mℓ = −1 state up, and the mℓ = 0 state will be undeflected.  saw only 2 with silver atom If the space quantization were due to the magnetic quantum number mℓ, the number of mℓstates is always odd at (2ℓ+ 1) and should have produced an odd number of lines. PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  41. Intrinsic Spin • In 1920, to explain spectral line splitting of Stern-Gerlach experiment, Wolfgang Pauli proposed the forth quantum number assigned to electrons • In 1925, Samuel Goudsmit and George Uhlenbeck in Holland proposed that the electron must have an intrinsic angular momentumand therefore a magnetic moment. • Paul Ehrenfest showed that the surface of the spinning electron should be moving faster than the speed of lightto obtain the needed angular momentum!! • In order to explain experimental data, Goudsmit and Uhlenbeck proposed that the electron must have an intrinsic spin quantum numbers = ½. PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  42. Intrinsic Spin The electron’sspin will be either “up” or “down” and can never be spinning with its magnetic moment μs exactly along the z axis. For each state of the other quantum numbers, there are two spins values The intrinsic spin angular momentum vector . The spinning electron reacts similarly to the orbiting electron in a magnetic field. (Dirac showed that this is necessary due to special relativity..) We should try to find L, Lz, ℓ, and mℓ. The magnetic spin quantum numberms has only two values, ms = ±½. PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  43. Energy Levels and Electron Probabilities • In ground state an atom cannot emit radiation. It can absorb electromagnetic radiation, or gain energy through inelastic bombardment by particles. • For hydrogen, the energy level depends on the principle quantum number n. PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  44. Selection Rules • We can use the wave functions to calculate transition probabilities for the electron to change from one state to another. Allowed transitions: Electrons absorbing or emitting photons can change states when Δℓ = ±1. (Evidence for the photon carrying one unit of angular momentum!) Δn=anything Δℓ = ±1 Δmℓ = 0, ±1 Forbidden transitions: Other transitions possible but occur with much smaller probabilities when Δℓ ≠ ±1. PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  45. Probability Distribution Functions • We must use wave functions to calculate the probability distributions of the electrons. • The “position” of the electron is spread over space and is not well defined. • We may use the radial wave function R(r) to calculate radial probability distributions of the electron. • The probability of finding the electron in a differential volume element dτis PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  46. Equipartition Theorem • The formula for average kinetic energy 3kT/2 works for monoatomic molecule what is it for diatomic molecule? • Consider oxygen molecule as two oxygen atoms connected by a massless rod  This will have both translational and rotational energy • How much rotational energy is there and how is it related to temperature? • EquipartitionTheorem: • In equilibrium a mean energy of ½ kT per molecule is associated with each independent quadratic term in the molecule’s energy. • Each independent phase space coordinate: degree of freedom • Essentially the mean energy of a molecule is ½ kT*NDoF PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  47. Equipartition Theorem • In a monoatomic ideal gas, each molecule has • There are three degrees of freedom. • Mean kinetic energy is • In a gas of N helium molecules, the total internal energy is • The heat capacity at constant volume is • For the heat capacity for 1 mole, • using the ideal gas constant R = 8.31 J/K. PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  48. The Rigid Rotator Model • For diatomic gases, consider the rigid rotator model. • The molecule has rotational E only when it rotates about x or y axis. • The corresponding rotational energies are • There are five degrees of freedom (three translational and two rotational) resulting in mean energy of 5kT/2 per molecule according to equi-partition principle (CV=5R/2) PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  49. Table of Measured Gas Heat Capacities PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

  50. Equipartition Theorem • Most the mass of an atom is confined to a nucleus whose magnitude is smaller than the whole atom. • Izis smaller than Ix and Iy. • Only rotations about x and ycontributes to the energy • In some circumstances it is better to think of atoms connected to each other by a massless spring. • The vibrational kinetic energy is • There are seven degrees of freedom (three translational, two rotational, and two vibrational).  7kT/2 per molecule • While it works pretty well, the simple assumptions made for equi-partition principle, such as massless connecting rod, is not quite sufficient for detailed molecular behaviors PHYS 3313-001, Spring 2015 Dr. Jaehoon Yu

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