Introduction to Quantum Mechanics - Lecture Slides | PHYS 214, Exams of Quantum Physics

Download Introduction to Quantum Mechanics - Lecture Slides | PHYS 214 and more Exams Quantum Physics in PDF only on Docsity! “‘Quantum mechanics’ is the description of the behavior of matter and light in all its details and, in particular, of the happenings on an atomic scale. Things on a very small scale behave like nothing that you have any direct experience about. They do not behave like waves, they do not behave like particles, they do not behave like clouds, or billiard balls, or weights on springs, or like anything that you have ever seen.” --Richard P. Feynman Introduction to Quantum Mechanics • This week (and the previous week) are critical for the course •Week 4 – Lect. 7,8: Schrödinger Equation -- Definite predictions of quantum behavior -- Examples of particles in infinite wells, finite wells -- Leads up to rest of course • Lab next week -- You will need your “Active Directory” Login {www.ad.uiuc.edu} -- You can save a lot of time by reading the lab ahead of time – it’s a tutorial on how to draw wavefunctions • Midterm Exam – Monday, Feb. 16 - will cover Lectures 1-7 and qualitative aspects of lecture 8 • Week 4 – Discussion and quiz will be on material in lect. 5-6 • Week 4 – Online homework covers material in lects. 7-8. It is due 8 AM Thurs. Feb. 19, but we strongly encourage you to look at the homework before the midterm! • Practice Exams – Additional problems on Lect. 7,8 material • Review – Sunday, Feb. 15, 3-5 pm Lincoln Hall Theater, • Office hours Feb. 15 & 16 (also Feb. 18, for the HW) Overview  Probability distributions  Particle in a “Box” -- matter waves in an infinite square well  quantized energy levels  Wavefunction normalization Good web site for animations http://www.falstad.com/qm1d/ Nice descriptions in the text – Chapter 40 Having established that matter acts qualitatively like a wave, we want to be able to make precise quantitative predictions, under given conditions.  Usually the “conditions” are specified by giving a potential energy U(x,y,z) in which the particle is located. E.g., electron in the coulomb potential of the nucleus electron in a molecule electron in a solid crystal electron in a semiconductor ‘quantum well’ x U(x) For simplicity, here is a 1- dimensional potential energy function: Matter Waves...Quantitative Classically, a particle in the lowest energy state would sit right at the bottom of the well. In QM this is not possible. (Why?) Matter Waves...Quantitative  We will see that we can get good predictions (actually, so far they have never been wrong!!) by assuming that the state of a particle is described by a “wave function” (or “probability amplitude”):  What do we measure? Often it’s: |Ψ(x,y,z,t)|2 --the probability density* for detecting a particle near some place (x,y,z), and at some time t.  We need a “wave equation” describing how Ψ(x,y,z,t) behaves. It should  be as simple as possible  make correct predictions  reduce to the usual classical laws of physics when applied to “classical” objects (e.g., baseballs) *Probability per unit volume = |ψ|2, i.e., ‘psi-squared’ (pronounced “sigh-squared”) Ψ(x,y,z,t)  Start a classical (large) object at some point. At some random time later, what is the probability of finding it at another point? Act 1: Classical probability distributions x P(x) b E Total energy E = KE + U(x) KE x U(x) Ball in a box: Probability is equally distributed x P(x) ? a. b. c. ? E Total energy E = KE + U(x)KE Ball in a valley: x U(x)  Start a classical (large) object at some point. At some random time later, what is the probability of finding it at another point? Act 1: Classical probability distributions More likely to spend time at the edges. x P(x) c E Total energy E = KE + U(x)KE Ball in a valley: x U(x) x P(x) b E Total energy E = KE + U(x) KE x U(x) Ball in a box: Probability is equally distributed To predict a quantum particle’s behavior, we need an equation that tells us exactly how the particle’s wave function, Ψ(x,y,z,t), changes in space and time…  There are two important forms for the SEQ The Schrödinger Equation  In 1926, Erwin Schrödinger proposed an equation that described the time- and space-dependence of the ψ wavefunction for slow matter waves (i.e., electrons, protons,...) The Schrödinger Equation (SEQ) *In this important case, which we’ll be primarily concerned with in this course, the probability density |ψ | 2 associated with the particle does not change with time…. it is in a “stationary state”. First we will focus on a very important special case of the SEQ, the time-independent* SEQ, which is appropriate ONLY when the particle’s wavefunction is associated with a single energy E (we’ll deal with the more general case later). Also simplify ψ(x,y,z)  ψ(x). (1-dimension) Act 2: Particle Wavefunction The three wavefunctions below represent states of a particle in the same potential U(x), and over the same range of x: 1. Which of these wavefunctions represents the particle with the lowest kinetic energy? (Hint: Think “curvature”.) 2. Which corresponds to the highest kinetic energy? ψ(x) x ψ(x) x (a) (b) (c)ψ(x) x Act 2: Particle Wavefunction - Solution The three wavefunctions below represent states of a particle in the same potential U(x), and over the same range of x: 1. Which of these wavefunctions represents the particle with the lowest kinetic energy? (Hint: Think “curvature”.) 2. Which corresponds to the highest kinetic energy? ψ(x) x ψ(x) x (a) (b) (c)ψ(x) x Highest KE Lowest KE The curvature of the wavefunction represents kinetic energy: Since (b) clearly has the least curvature, that particle has lowest KE. (a) has highest curvature  highest KE Probability distribution Difference between classical and quantum cases E x U(x) x P(x) Classical (particle with same energy as in qunatum case) In classical mechanics, the particle is most likely to be found where its speed is slowest Quantum (lowest energy state state) E x U(x) x P(x) = ψ2 In quantum mechanics, the particle can be found where it is “forbidden in classical mechanics . In classical mechanics, the particle moves back and forth coming to rest at each “turning point” In quantum mechanics, the particle is most likely to be found at the center.  Notice that if U(x) = constant, this equation has the simple form: For positive C, what is the form of the solution? For negative C, what is the form of the solution? where is a constant that might be positive or negative. a) sin kx b) cos kx c) eax d) e-ax a) sin kx b) cos kx c) eax d) e-ax KE term PE term Total E term Last lecture: The time-independent SEQ (in 1D) Particle in a Box  The solution of the Schrodinger Eq. discussed today is NOT mysterious!  We choose simple cases to illustrate quantum waves – just as we did for classical waves  The waves have EXACTLY the same form as standing waves on a string, sound waves in a pipe, etc.  The wavelength is determined by the condition that it “fits in the box”  The ONLY difference is the relation of the frequency (energy) of the wave to its wavelength!  On a string the wave is a displacement y(x) and the square is the intensity, etc.  For an electron in a well, the wave is the probability amplitude Ψ(x) ond the square |Ψ(x)|2 is the probability of finding the electron at point x Constraints on the form of ψ(x)  |ψ(x)|2 corresponds to a physically meaningful quantity – the probability of finding the particle near x. Physically meaningful states must have: ψ(x) must be single-valued, and finite (finite to avoid infinite probability density) Ψ(x ) must be continuous, with finite dψ/dx (because dψ/dx is related to the momentum density) In regions with finite potential, dψ/dx must be continuous, with finite d2ψ/dx2, to avoid infinite energies. There is usually no significance to the overall sign of ψ(x). (it goes away when we take the absolute square) {In fact, we will see that ψ(x,t) is usually complex!} Act 3 - Solution (a) (b) 2. Which of the following wavefunctions corresponds to a particle more likely to be found on the left side? ψ(x) 0 x ψ(x) 0 x (c) ψ(x) 0 x None of them! (a) is clearly completely symmetric. (b) might seem to be “higher” on the left than on the right, but it is only the absolute square that determines the probability. |ψ|2 0 x (www.micro.uiuc.edu) This is a basic problem in “Nano-science”. It’s a simplified (1D) model for an electron confined in a quantum structure (e.g., “quantum dot”), which scientists/engineers make, e.g., at the UIUC Microelectronics Laboratory! Application of SEQ: “Particle in a Box” KE term PE term Total E term U = 0 for 0 < x < L U = ∞ everywhere else (www.kfa-juelich.de/isi/) (newt.phys.unsw.edu.au) ‘Quantum dots’ U(x) 0 L ∞ ∞  As a specific important example, consider a quantum particle confined to a small region, 0 < x < L, by infinite potential walls. We call this a “one-dimensional (1D) box”.  Recall, from last lecture, the time-independent SEQ in one dimension: Real standing waves of electron density in a “quantum corral” IBM Almaden Single atoms (Fe) Cu Observation of an electron wave “in a box” Image taken with a scanning tunneling microscope (more later) (Note: the color is not real! – it is a representation of the electrical current observed in the experiment) Particle in a Box (2) Region II: When U = 0, what is ψ(x)? The general solution to this is: U(x) 0 L ∞ ∞ II Plugging ψ(x) into the SEQ, we find the requirement: Consistent with DeBroglie’s hypothesis: p = h/λ = ħk and E = p2/2m B1 and B2 are coefficients to be determined by the boundary conditions. Oscillatory solutions! ψ Particle in a Box (3)  Matching ψ at the first boundary (x = 0) The wavefunction takes on the following form: At x = 0: because cos(0) = 1 and sin(0) = 0 So this “boundary condition” requires that there be no cos(kx) term! Region I: Region II: Key point: The total wavefunction ψ(x) must be continuous at all boundaries ! U(x) 0 L ∞ ∞ II I I ψΙ ψΙ ψΙΙ Particle in a Box (4)  Matching ψ at the second boundary (x = L) At x = L: This constraint forces k to have special values ! This is precisely the condition we found for confined waves, e.g., waves on a string, EM waves in a laser cavity, etc.: For matter waves, the wavelength is related to the particle energy: En = h2/2mλ2 Therefore n λ (= c/f) 4 L/2 3 2L/3 2 L 1 2L U(x) 0 L ∞ ∞ II I I ψΙ ψΙ ψΙΙ Particle-in-Box: Solution  Calculate ground state energy and energy for a transition. An electron is trapped in a “quantum wire” that is L = 4 nm long. Assuming that the potential seen by the electron is approximately that of an infinite square well, estimate the ground (lowest) state energy of the electron. Allowed energies for electron in a 1D box: What photon energy is required to excite the trapped electron to the next available energy level (i.e., n = 2)? So, energy difference between n = 2 and n = 1 levels: = 0.071 eV E1 = ground state energy (n = 1) U=∞ U=∞ 0 xL En n=1 n=2 n=3 2. If we decrease the size of the dot, the difference between two energy levels (e.g., between n = 2 and 3) will a) decrease b) increase c) stay the same 1. An electron is in a quantum “dot”. If we decrease the size of the dot, the ground state energy of the electron will a) decrease b) increase c) stay the same Lecture 7, Act 4 U=∞ U=∞ 0 xL En n=1 n=2 n=3 2. If we decrease the size of the dot, the difference between two energy levels (e.g., between n = 2 and 3) will a) decrease b) increase c) stay the same 1. An electron is in a quantum “dot”. If we decrease the size of the dot, the ground state energy of the electron will a) decrease b) increase c) stay the same Lecture 7, Act 4 - Solution En = n2 E1 E3 - E2 = (9 – 4)E1 = 5E1 Since E1 increases, so does ΔE U=∞ U=∞ 0 xL En n=1 n=2 n=3 Probability Density Reflects INTERFERENCE effects caused by the wave-like character of quantum particles! For a classical particle bouncing back and forth in a well: The probability of finding the particle is equally likely throughout the well. For a quantum particle in a stationary state, the probability distribution is NOT uniform…there are “nodes” where the probability is ZERO! The “normalization factor” N is determined by setting the total probability of finding the particle in the well equal to one: Probability per unit length (in 1-dimension) |ψ|2 0 xLn=3 N2 Integral under the curve = 1 N = Note: N is a “normalization factor” N is not an integer Probability Density – Example problem Consider an electron trapped in a 1D well with L = 5 nm. Let’s say that the electron is in the following state: |ψ|2 0 x5 nm N2 Discussed in section. a) What is the energy of the electron in this state (in eV)? b) What is the value of the normalization factor squared N2? c) Estimate the probability of finding the electron within ±0.1 nm of the center of the well? (No integral required. Do it graphically.) Probability Density – Example problem Consider an electron trapped in a 1D well with L = 5 nm. Let’s say that the electron is in the following state: |ψ|2 0 x5 nm N2 Answer = 0.135 eV a) What is the energy of the electron in this state (in eV)? b) What is the value of the normalization factor squared N2? c) Estimate the probability of finding the electron within ±0.1 nm of the center of the well? (No integral required. Do it graphically.) Answer = 0.08 = 8% Answer = 0.4 nm-1 Since (sin(3πx/L))2 ≈ 1 for x ≈ L/2