Lecture 1: Introduction, Thermodynamics and Combinatorics, Slides of Thermodynamics

Download Lecture 1: Introduction, Thermodynamics and Combinatorics and more Slides Thermodynamics in PDF only on Docsity! Physics 5C: Introductory Thermodynamics and Quantum Mechanics Fall 2018 Lecture 1: Introduction, Thermodynamics and Combinatorics Lecturer: Feng Wang 23 August Aditya Sengupta Note: LATEXformat adapted from template courtesy of UC Berkeley EECS dept. 1.1 Overview There are two major ways of approaching thermodynamics. The first is a phenomenological view, based on looking at macroscopic systems. These systems are specified by various quantities, such as temperature, energy, work, and pressure. Observing these led to the first and second laws of thermodynamics, as well as the idea of entropy. The second is a microscopic view; the study of particle behaviour to explain the macroscopic effects. This is the basic idea behind statistical mechanics, which treats the actions of particles as essentially random and therefore, on large scales, predictable (random variations average out over a large number of particles). This course will start with a discussion of concepts in thermal physics such as temperature, as well as an introduction to probability, then it will cover the macroscopic and microscopic views of thermodynamics described above. 1.2 Large Number Systems Microscopically, the objects we are considering are atoms and molecules, on size scales of about 10−10m, and macroscopically, the objects are more on the scale of meters and centimeters. This is a considerable gap. Consider a cube with side length 1cm. The number of atoms of size 10−10cm that can fit in it is ( 10−2m 10−10m )3 = 1024 (1.1) The actual quantity that acts as a bridge between macro and micro systems is not far off from this. That is Avogadro’s number, NA ≈ 6.022× 1023. It is defined as the number of carbon atoms in 12g of 12C. An easy unit to use in macroscopic systems is the mole. One mole of something is defined as the quantity of matter that contains NA objects (atoms or molecules). For example, one mole of hydrogen is the quantity required to have 6.022× 1023 atoms of hydrogen. The macroscopic view of thermodynamics is provably correct because it is within the so-called thermodynamic limit of the number of particles, the point after which the random nature of each particle mostly averages out. It is formally defined as a ratio of a number of particles to the volume containing them, but it can be easily understood conceptually. Consider a system consisting of one particle and a pressure sensor on one wall that indicates when a particle collides with it. This system would exhibit sporadic behavior as the sensor is intermittently triggered randomly, because the particle that can collide with it moves randomly. Lecture 1-1 Lecture 1: 23 August Lecture 1-2 Now, consider a similar system with 20 particles. Now, the sensor may average a pressure of 2 or 3 collisions per ∆t time interval, but it will still fluctuate, albeit not quite as much. Increase the number of particles to 106, and the pressure begins to converge. There are still fluctuations, but they are on a much smaller scale; now, the number of collisions averages maybe 100,000. Although it may dip to 99,700 or rise to 100,105, the relative fluctuations will be much smaller. As the number of particles N increases, the absolute fluctuation increases linearly with N , but the relative fluctuation scales only with √ N , to the point that when we reach a macroscopic-sized system of NA particles, the force felt by the sensor is practically a constant. 1.3 Macroscopic Parameters This allows us to define pressure as the ratio of force to the area over which it is applied. Pressure is a thermodynamic state variable, meaning that it is an averaged physical quantity that can be used to describe a macroscopic system in a useful way. The state of a system can be completely defined in terms of state variables such as pressure and temperature. We’ll now try and select other state variables to give a useful description of a macroscopic system. The averaged velocity and momentum, 〈~v〉 and 〈~p〉, are both 0. This is because it is a static system, in which no one direction is preferred over any other. The average kinetic energy can, however, be well-defined and is potentially useful. The quantity that is actually used is U , the total energy of a particle. This is the most important quantity in macro-analysis. Already, with only a vague overview of how state variables define a system, we can use them to understand what will happen to a system under some circumstances. Let there be a box with known pressure, volume, cumulative particle energy, and temperature, which we then split in half. We then want to predict these state variables for one of the smaller boxes thus created. Clearly, the volume is halved, as is the cumulative particle energy because the number of particles is halved. These are called extensive state variables because they depend (usually linearly) on system size. But the temperature remains the same (this is more intuitive but can be more thoroughly shown later) and the pressure is also the same because the number of particles exerting a force is halved, and the area over which that force is distributed is also halved. These two are therefore intensive state variables. 1.4 Ideal Gases We want to find relations between U , P , V , and T that let us predict system behavior in dynamic systems, so that we can analyze cases beyond the above. Consider a system that has a fixed P , V , and T . To experimentally find relations, we can fix one and vary the other two to find a correlation. • If we fix T , we find an inverse relationship between P and V . As volume increases, pressure reduces because of a lower probability of collisions. We find that P ∝ 1 V (1.2) This is called Boyle’s law. Lecture 2: August 28 Lecture 2-1 Physics 5C: Introductory Thermodynamics and Quantum Mechanics Fall 2018 Lecture 2: Statistical and Thermodynamic Concepts Lecturer: Feng Wang August 28 Aditya Sengupta 2.1 Basic Concepts 2.1.1 Probability Previously in physics, everything was deterministic; given a force and a mass, you can predict motion for all times. Or given a charge in an electric field. This is no longer the case, because of the magnitude of the system. It’s not practical to measure all 1023 particles in a system and determinstically find what happens to each of them. Instead, we can use the probability of microscopic states. From there, we can use statistical methods to accurately predict aggregate/average properties such as temperature and pressure, based on the probability distributions of microscopic states (their velocity, energy, etc.) as well as fluctuations around the average. 2.1.1.1 Discrete Probability Distributions These are used when there is a discrete number of outcomes, such as rolling a die. In this case, we can separately write each of P (1) (probability of the first event, e.g. rolling a one), P (2), . . . , P (i). Based on this, we can derive some general quantities and properties: 1. ∑ i P (i) = 1 2. The mean (or average, or expected value) 〈x〉 = ∑ i xiP (i). Similarly, we can say 〈x2〉 = ∑ i x 2 iP (i), or more generally 〈f(x)〉 = ∑ i f(xi)P (i). For example, in a distribution defined as follows: x P(x) 0 1 2 1 1 4 2 1 4 we can find an average value for x and for x2: 〈x〉 = 0× 1/2 + 1× 1/4 + 2× 1/4 = 3/4 (2.1) 〈x2〉 = 02 × 1/2 + 12 × 1/4 + 22 × 1/4 = 5/4 (2.2) 2.1.1.2 Continuous Probability Distributions We use this when we have to define probability in an interval. This is for events that do not have a set/discrete number of potential outcomes. Lecture 2: August 28 Lecture 2-2 Let the interval be (x, x+ ∆x). Then the probability of an event in this interval is P (x)×∆x, which in the limit as ∆x → 0 becomes P (x)dx. We know that the total probability is still 1; therefore if we add all the possible outcomes’ probabilities, we will get 1: lim ∆x→0 ∑ i P (xi)∆x = 1 =⇒ ∫ P (x)dx = 1 (2.3) We can define a mean in a continuous distribution: 〈x〉 = ∑ i xiP (xi)∆xi = ∫ xP (x)dx (2.4) Similarly, a mean square value can be found using ∫ x2P (x)dx, and the mean value for a function using∫ f(x)P (x)dx. Continuous probability also introduces the concept of variance, a measure of the fluctuation or the spread in the value. It may seem difficult to characterize this using only one quantity. To start with, we can see that it must depend upon x − 〈x〉, to see how far away from the mean a value can fall. One idea might be to take the average of this: 〈x− 〈x〉〉 = 0 (2.5) By definition of the mean, this becomes 0. We want a similar quantity that does not go negative, so that deviations from the mean only add to the variance. We could try this: 〈|x− 〈x〉|〉 (2.6) However, this is not an analytic function, so in general it is difficult to deal with. A better solution is to take a square: 〈(x− 〈x〉)2〉 = σ2 x (2.7) This is the variance of a population. To bring the variance into the same units as x, we can take a square root to get the standard deviation: σx = √ 〈(x− 〈x〉)2〉 (2.8) This is the rms (root mean square) deviation from the mean. Lecture 2: August 28 Lecture 2-3 2.1.2 Operations on continuous probability distributions To characterize transformations such as finding the average (rms) momentum from a known velocity, we introduce linear transformations on distributions. Let y = ax + b be a transformation on the continuous variable x. Then, 〈y〉 = 〈ax+ b〉 = a〈x〉+ b (2.9) The standard deviation after a linear transformation can also be calculated: σ2 y = 〈(y − 〈y〉)2〉 = 〈y2 − 2y〈y〉+ 〈y〉2〉 (2.10) σ2 y = 〈y2〉 − 2〈y〉〈y〉+ 〈y〉2 (2.11) σ2 y = 〈y2〉 − 〈y〉2 (2.12) We can calculate these terms in x separately: 〈y2〉 = 〈 (ax+ b)2 〉 = 〈 a2x2 + 2abx+ b2 〉 (2.13)〈 y2 〉 = a2 〈 x2 〉 + 2ab 〈x〉+ b2 (2.14) and 〈y〉2 = (a 〈x〉+ b)2 = a2 〈x〉2 + 2ab 〈x〉+ b2 (2.15) Therefore, we can find σ2 y: σ2 y = a2 (〈 x2 〉 − 〈x〉2 ) = a2σ2 x (2.16) We see that a constant shift of b does not change the spread, but a scaling of a proportionately scales the spread. 2.1.3 Independent Variables Suppose we have a probability distribution for velocities of a particle in the x and y directions. Consider the probability that vx is in an interval of width dvx, P (vx)dvx, and the same for y. If the two are independent (not correlated), the probability distribution for 2D space can be arrived at just by multiplying the two: P (vx,y) = P (vx)P (vy)dvxdvy (2.17) Lecture 3: 30 August Lecture 3-1 Physics 5C: Introductory Thermodynamics and Quantum Mechanics Fall 2018 Lecture 3: Temperature Lecturer: Feng Wang 30 August Aditya Sengupta 3.1 Thermal Equilibrium Intuitively, we understand that a system comes to thermal equilibrium if a hot object of temperature T1 is connected to a cool object of temperature T2, and over time t they equalize to both attain temperature Tf . At this point, there is no net flow of heat between the objects. Here, an irreversible process has taken place. It intuitively seems like this process should be reversible, as it is only a transfer of energy, but this is not the case (probabilistically). This idea of gaining equilibrium also leads to the 0th law of thermodynamics: if a system A is in equilibrium with B, and A is in equilibrium with some other system C, then B and C are in equilibrium too. (This is Prof. Norman Yao) In principle, a quantum mechanical description of a thermodynamic system would allow for oscillation between the two states (T1, T2) and (Tf , Tf ), but the timescale for oscillation is astronomically large even by astronomically large standards. The 0th law helps us to measure temperature exponentially. We can place a body in an unknown T, and use a calibration system that has a measurable property that exhibits a well-known dependence on temperature. This is the concept behind a thermometer. Note that a thermometer should have a heat capacity much smaller than that of the heat being measured. Recall that heat capacity is the heat required to cause a unit change in temperature, C = dQ dT . This is so that attaining thermal equilibrium does not appreciably change the energy of the system by measuring it. The ideal gas law, PV = nRT , can also be used to measure temperature. Either pressure is held fixed and changes in volume are measured, or vice versa. All of the ways in which we are defining temperature depend on some measurable property. In general, almost all measurable properties are non-linear. For example, the direct pressure/volume relationship does not hold when the liquid being studied freezes. Defining temperature in terms of physically measurable properties would therefore be painful. Instead, we want to find a theoretical - in this case statistical - basis for defining temperature. 3.2 Micro/Macro States Consider a box containing 100 fair coins. We shake the box and look at the configuration. There are 2100 possible configurations, which is about 1030. Each configuration (set of HTTHH...TH) is a microstate, but that probably isn’t how you would characterize the system; you might say something more like “there are 53 heads and 47 tails”. This is a macrostate. The number of microstates corresponding to a macrostate can be very large. In this example, there are 100C50 = 100! 50!50! ≈ 4× 1027 microstates for the one macrostate 50H, 50T. It can also be small; the macrostate 100H, 0T has only one microstate. Lecture 3: 30 August Lecture 3-2 We can describe or measure a system either in terms of many equally likely microstates, or a smaller number of macrostates that are not all equally likely. The most likely macrostate for a system is the one with the most configurations or microstates. A thermal analogy for this would be to specify a microstate for a thermal system (the position/velocity of each atom), which is operationally impossible for any realistic numbe of atoms in a gas. Instead, we generally describe a thermal system via its macrostate. There are several possible ways of going about this, such as total energy, or volume and pressure. To build a statistical notion of temperature, we consider a system of energy E and imagine there exist Ω(E) microstates. This allows us to return to the two systems in thermal contact coming into equilibrium, meaning that some notion of temperature has become the same between the two. Let there be Ω1(E1) microstates of the first system, and Ω2(E2) microstates of the second. For a total energy of E1 + E2, the total number of microstates is Ω1(E1)× Ω2(E2). Systems can exchange energy and eventually come into equilibrium, in a process called thermalization. The key insight here is that a system will equilibrate to a microstate such that the number of microstates Ω1(E1)× Ω2(E2) is maximized. In this model, we assume that each microstate is equally likely, and the dynamics of a system allows us to constantly move between microstates. Additionally, given enough time for the system to equilibrate, the system will explore all its microstates. These assumptions only imply that a system will most likely be found in a configuration with the largest number of microstates. 3.3 Definition of temperature We want to maximize Ω1(E1)× Ω2(E2). Without loss of generality, we choose to shift E1. Then, we know that at the maximum energy, ∂ ∂E1 (Ω1(E1)× Ω2(E2)) = 0 (3.1) We apply the product rule and chain rule: Ω2(E2) ∂Ω1(E1) ∂E1 + Ω1(E1) ∂Ω2(E2) ∂E2 ∂E2 ∂E1 = 0 (3.2) We can also use the fact that E1 + E2 = E, a constant total energy. Taking a derivative tells us that dE1 = −dE2 =⇒ dE2 dE1 = −1, therefore: 1 Ω1 ∂Ω1 ∂E1 − 1 Ω2 Ω2 E2 = 0 (3.3) This can be expressed as the derivative of natural log: d ln Ω1 dE1 = d ln Ω2 dE2 (3.4) Lecture 3: 30 August Lecture 3-3 This serves as a fundamental definition for temperature: 1 kBT = d ln Ω dE (3.5) and the above equation is the condition for thermal equilibrium. This also leads to a fundamental definition for entropy, S: S = kB ln Ω 1 T = ∂S ∂E (3.6) kB , a proportionality constant, is 1.38× 10−23 J K . 3.4 Ensembles The notion of ensembles is introduced to deal with the number of possible states of a system there are; either we repeat an experiment many times, or consider many copies of an identical system all described by the same temperature. There are three ensembles, of which we will only really work with one. 3.4.1 Microcanonical Ensemble This is an ensemble of systems at the same energy. 3.4.2 Canonical Ensemble This is an ensemble of systems, each of which can exchange energy with a large bath. 3.4.3 Grand Canonical Ensemble This is an ensemble of systems, each of which can exchange both energy and particles with a large bath. The exchange of particles allows us to define a chemical potential later in the course. 3.5 The Canonical Ensemble Imagine we have two systems coupled, connected to a bath at temperature T . The whole system has energy E, of which the system has energy ε and the bath has energy E − ε. We assume that the bath is unaffected by the system. The bath has Ω(E − ε) microstates. We want to know the probability that the system has energy ε. This is proportional to the number of microstates of the bath (why?). We can apply the above relation: Lecture 4: September 4 Lecture 4-1 Physics 5C: Introductory Thermodynamics and Quantum Mechanics Fall 2018 Lecture 4: Temperature, Laws of Thermodynamics Lecturer: Feng Wang September 4 Aditya Sengupta We saw last time that a macrostate could be made up of many microstate configurations. Since each microstate has equal probability, thermal equilibrium is reached at the macrostate with the largest number of microstates. We can characterize a macrostate in terms of Ω(V, T,E); this function is the number of microstates. The configuration with the maximum Ω gives thermal equilibrium. Using this concept, we previously derived this relation between a number of microstates at thermal equilibrium and temperature: d ln Ω dE = k = 1 kBT (4.1) kB has units of energy/temperature, or J K , and the numerical value is 1.38× 10−23. Based on the concept of the canonical ensemble, in which a system is connected to a heat reservoir, we can find the probability of each microstate: P (microstate) = e − E kBT∑ i e − Ei kBT (4.2) The numerator is called the Boltzmann factor, and the denominator is Z, the partition function. The whole expression defines the Boltzmann distribution. 4.1 Examples of the Boltzmann distribution 4.1.1 Atmosphere Consider an isothermal (equal temperature) atmosphere, with a number of molecules h. We want to find the distribution of air density in this atmosphere. This can be done with classical mechanics, but the Boltzmann distribution presents an easier way. At a height z, the (potential) energy is E = mgz. Therefore P (z) ∝ e− mgz kBT (4.3) This is the probability of finding a molecule at height z. We can make a similar function for density that is proportional to this probability, n(z) = Ae − mgz kBT n(0) = A =⇒ n(z) = n(0)e − mgz kBT (4.4) Lecture 4: September 4 Lecture 4-2 4.1.2 Chemical Reactions For H2 and O2 to combine to form H2O, there is an energy barrier to be crossed, on the order of 0.5 eV. The kBT at room temperature is 25× 10−3 eV, therefore thermal energy usually needs to be added to the system to cross the barrier. However, there is a very small probability that the system will naturally attain that energy, which is described by the Boltzmann distribution. The probability is proportional to r0 = e −EactkBT (4.5) Consider a fridge that lowers the temperature of its contents by ∆T . Then, the activation energy becomes r′ = e − Eact kB(T−∆T ) (4.6) We can rearrange and Taylor expand the denominator of r′ to get: r′ = e − Ea kBT (1+ ∆T T ) r ′ r0 = e − Ea kBT ∆T T (4.7) With some reasonable assumptions, we get r0 ≈ 8r′, meaning reactions (e.g. food being spoiled) take place 8 times more slowly in a fridge. If something would have spoiled in 1 day outside, it will spoil in 8 days inside. 4.2 State Variables Let a system be anything we want to study, and let its surroundings be everything else (treated as a thermal reservoir). We will consider the state of thermal equilibrium for now, which we can characterize through macroscopic observable vaues. Functions of state (that do not depend on the system’s history) include temperature, pressure, energy, and volume; these are ways of indirectly measuring heat or work. Consider a system consisting of state variables xn = (x1, x2, · · · ). Let f(x) be a function of state. Then, if a state variable changes xi → xf , then ∆f = ∫ xf xi df = f(xf )− f(xi) (4.8) For example, let there be two state variables x and y that completely define a system, and let f(x, y) = xy. Then, any path between (0, 0) and (1, 1) wil cause this change in f : ∆f = ∫ (1,1) (0,0) df = [xy] (1,1) (0,0) = 1 (4.9) Lecture 4: September 4 Lecture 4-3 Consider another variable defined in terms of a differential, dg = ydx. Here, the path can be a factor. Integrate between the same two points along the line y = x: ∫ (1,1) (0,0) ydx = ∫ 1 0 xdx = 1 2 (4.10) whereas along y = 0 followed by x = 1: ∫ x=1 x=0 ydx+ ∫ y=1 y=0 ydx = 0 + 0 = 0 (4.11) These cannot be used to define state variables, as they are not conservative. 4.3 First Law of Thermodynamics We want to find a relation between work/heat and intrinsic state variables. The law provides this; it states that heat and work are both forms of energy, or that total energy is conserved. We can define an internal energy U , the sum of all the energies in the system. Then, we can say that ∆U = ∆Q+ ∆W : any change in energy must come only from the changes in heat or work. Consider a thermally isolated system in which ∆Q = 0, so dU = dW (the d in dW should have a bar across it to indicate W isn’t a conservative state variable, but eh), and we compress the system, changing the volume and therefore the pressure. We know dW = ~F · d~x, and therefore dW = −F A (Adx) = −PA A dV = −PdV (4.12) (with a negative because volume is decreasing). Assuming no friction and everything in equilibrium, this relation holds. 4.4 Heat capacity of a gas system Consider a gas system in which energy is dependent on temperature and volume: U = U(T, V ). Then dU = ( ∂U ∂T ) V dT = ( ∂U ∂V ) T dV (4.13) Then, a change in the system from U(T1, V1) to U(T2, V2) causes a change in energy ∆U , which can be described by the above equation in terms of the changes in T and U . To find dQ dT , we can use the First Law and the equation for dW derived above: Lecture 5: 6 September Lecture 5-2 Cv lnT ∣∣∣T2 T1 = −R lnV ∣∣∣V 2 V1 (5.7) ln T2 T1 = − R Cv ln V2 V1 (5.8) R = Cp − Cv =⇒ R Cv = Cp Cv − 1 = γ − 1 (5.9) ∴ ln T2 T1 = −(γ − 1) ln V2 V1 (5.10) lnT2 = (γ − 1) lnV2 = lnT1 + (γ − 1) lnV1 (5.11) (5.12) This tells us that the quantity TV γ−1 is a constant for a system. We can apply the ideal gas law again to give us a constant relation in volume and pressure: PV = nRT (5.13) PV nR · V γ−1 = PV γ nR = constant (5.14) ∴ PV γ = constant. (5.15) Lecture 5: 6 September Lecture 5-3 5.1.3 Adiabatic atmosphere The previous assumption that the atmosphere is isothermic was not very good. The atmosphere is much closer to adiabatic. We want to use the hydrostatic equation to find an expression for atmospheric pressure as a function of z, which can easily be found via balancing forces: ∆PA = − (ρV ) g = −ρgA∆z (5.16) dP = −ρgdz (5.17) (5.18) Let ρ = N V ·m0, the mass in a unit volume due to one mole multiplied by the number of moles. The ideal gas law tells us that this is ρ = PV kBT ·m0 (5.19) (5.20) Therefore, we can substitute this back into the hydrostatic equation to get dP = − P kBT mgdz (5.21) T dP P = −mg kB dz (5.22) Note that if we hold temperature constant, we find that pressure decays exponentially with z, which is the result we got in the isothermal case. We can now apply the adiabatic condition, P ( RT P )γ = constant (5.23) P 1−γ · T γ = constant (5.24) (5.25) We can take a derivative, (1− γ) dP P + γ dT T = 0 (5.26) dP P = − γ 1− γ dT T = γ γ − 1 dT T (5.27) (5.28) Having isolated dP P , we can substitute this back into the equation with a differential in z from above: Lecture 5: 6 September Lecture 5-4 T γ γ − 1 dT T = −mg kB dz (5.29) γ γ − 1 dT = −mg kB dz (5.30) That is, dT dz is constant, and negative. 5.2 Heat Engines and the Second Law The Second Law of Thermodynamics is the first concept in physics that is time-dependent. There are a couple of different ways of stating it; Clausius’s statement is no process whose sole result is to transfer heat from cold to hot is possible, and Kelvin’s is no process whose sole result is complete conversion of heat to work is possible. A Carnot engine is a simplification of a real heat engine, retaining only the essential parts. This makes intuitively studying it much easier. It is a system operating a cyclic process that converts heat to work. Because dW = PdV , the work done by a Carnot engine is the area enclosed in the above P/V diagram. Let the four corners of the quadrilateral describing the Carnot engine be A through D. From A to B (at T = Th) and from C to D (at T = Tl), an isothermal process takes place; from B to C and from D to A, it is adiabatic. Lecture 6: 11 September Lecture 6-2 ηR = Qh W = Qh Qh −Ql = Th Th − Tl (6.3) 6.3 Clausius’s theorem In a Carnot engine, we know that Qh Ql = Th Tl (6.4) Qh Th − Ql Tl = 0 (6.5)∑ cycle ∆Qi Ti = Qh Th + (−Ql) Tl = 0 (6.6)∮ dQ T = 0 (6.7) The above is true for a Carnot cyle. Consider a large number of Carnot engines placed one after another, each making an infinitesimal change in temperature to produce a net ∆W consisting of many dW components. Let the initial working temperature be T , then the change in heat at step i is dQi Ti = dQh,i T = dQi + dWi T =⇒ dWi = dQi ( T Ti − 1 ) (6.8) Then, by the second law, our total work is ≤ 0, so Wtotal = ∑ i dWi + ∆W ≤ 0 (6.9) By the first law, ∆W = ∑ i dQi (6.10) Therefore ∑ (dWi + dQi) < 0 ∑[( T Ti − 1 ) dQi + dQi ] T ∑ i dQi Ti = T ∮ dQi Ti ≤ 0 (6.11) 6.4 Entropy ∮ dQrev T = 0 =⇒ ∫ B A dQrev T is path independent. Therefore it is an exact differential. So, we introduce a new function of state S: dS = dQrev T (6.12)∫ B A dQrev T = S(B)− S(A) (6.13) 6.5 Irreversible Processes An irreversible process has the characteristic ∮ dQ T ≤ 0 (6.14) Therefore ∫ B A dQ T + ∫ A B dQrev T ≤ 0 (6.15)∫ B A dQ T ≤ − ∫ A B dQrev T = ∫ B A dQrev T (6.16) 0 = ∫ B A dQ T (isolated system) ≤ ∫ B A dQrev T = ∫ B A dS = S(B)− S(A) (6.17) We can now revisit the first law to find: dU = dQ+ dW (6.18) dU = TdS − PdV (6.19) 1 kBT = d ln Ω dU =⇒ 1 kB ( ∂S ∂U ) V (6.20) which gives us something about microstates. (Revisit) Lecture 7-3 Lecture 8: September 18 Lecture 8-1 Physics 5C: Introductory Thermodynamics and Quantum Mechanics Fall 2018 Lecture 8: Micro and Macro Definitions of Entropy Lecturer: Feng Wang September 18 Aditya Sengupta 8.1 Mathematical Relations between Thermodynamic Quantities We’ve seen the first law of thermodynamics many times by now: dU = dQ+ dW (8.1) In a reversible process, we have expressions for dQ and dW : dU = TdS − PdV (8.2) Since these are all exact differentials, this expression holds true even for irreversible processes. This lets us express U as a function of entropy and volume U(S, V ). This lets us rewrite dU by differentiating: dU = ( ∂U ∂S ) V dS + ( ∂U ∂V ) S dV (8.3) Comparing coefficients tells us that ( ∂U ∂S ) V = T (8.4)( ∂U ∂V ) S = −P (8.5) (8.6) and therefore P T = − ( ∂U ∂V ) S ( ∂S ∂U ) V (8.7) Also, we can rewrite the original equation coming from the First Law, dS = dU T + P T dV (8.8) At thermal equilibrium, ∆S = 0. 8.5 Entropy and probability Given a two-part system that we can characterize by 〈U〉 , 〈S〉 , 〈P 〉, we want to find the entropy in part 1 and 2 (〈S〉). S = kB ln Ω (8.28) 〈S2〉 = ∑ i PiSi = ∑ i PikB lnni (8.29) Let the whole system (1 + 2) have N microstates, and let (2) have ni microstates. Then the probability of a particle being in 2 is Pi = ni N . This allows us to define an average entropy for part 1: 〈S1〉 = Stotal − S2 = kB( ∑ i Pi) lnN − ∑ i PikB lnni = kB ∑ i Pi ln 1 Pi (8.30) Lecture 8-4 Lecture 9: September 20 Lecture 9-1 Physics 5C: Introductory Thermodynamics and Quantum Mechanics Fall 2018 Lecture 9: Democracy doesn’t work (microstate distributions, equipartition) Lecturer: Feng Wang September 20 Aditya Sengupta We saw previously that S = kB ln Ω (9.1) However, this assumes that every microstate has equal probability, which is not the case. This definition of entropy was built on a binomial distribution of microstates, assuming each one was equally likely (analogous to flipping a coin, etc.), but that isn’t always true. For example, the distribution of particles in a box with a gravitational field attracting it to one side is not binomial. We saw previously that the entropy of a system can be expressed in terms of the probability of microstate i, SI = −kB ∑ i Pi lnPi (9.2) which is a more general expression, as it works for any probability distribution. We verify this by considering it for the equal distribution, in which Pi = 1 Ω . Then, S = −kB Ω∑ i=1 = 1 Ω ln 1 Ω = −kBΩ 1 Ω − ln Ω = kB ln Ω (9.3) as expected. We don’t want to count the microstates that end up not conserving energy (we can give those zero proba- bility). So, we have these two constraints: 〈E〉 = ∑ i PiEi = U (9.4)∑ i Pi = 1 (9.5) Since entropy wants to maximize itself, we can find the equilibrium state by finding the state with the maximum entropy. ∂S ∂Pi = 0 (9.6) Using Lagrangian multipliers, we get Lecture 9: September 20 Lecture 9-2 ∂ ∂Pi ( S kB − α(constraint 1)− β((constraint2)) ) = 0 (9.7) ∂ ∂Pi (∑ i (−Pi lnPi − αPi − βPiEi) ) = 0 (9.8) Taking derivatives: 0 = − lnPi − Pi 1 Pi − α− βEi (9.9) Pi = e−1−α−βEi = e−βEi z (9.10) (defining z = e1+α) This looks a lot like the Boltzmann distribution. Applying the constraint that ∑ i Pi = 1, and that ∑ i PiEi = U : 1 = ∑ i e−βEi z =⇒ z = ∑ i e−βEi (9.11) ∑ i PiEi = U =⇒ U = ∑ i Eie −βEi z (9.12) β = 1 kBT =⇒ Pi = e − Ei kBT z (9.13) 9.1 Equipartition theorem Analogous to how energy is proportional to velocity squared or momentum squared, or a length squared (for potential energy), we can say that energy is proportional to efficiency squared. We can find this by taking a continuous version of the Boltzmann distribution, as a function of v, the velocity of a particle in this microstate (cite?) 〈E〉 = ∑∞ −∞ P (v)E(v)dv = 1 β ∫∞ ∞ (β 1 2mv 2)e−β 1 2 mv2 dv∫∞ −∞ e−β 1 2 mv2 dv Use the change of variable x = √ mβ 2 v, and cancel a lot of things to get 〈E〉 = 1 β ∫∞ −∞ x2e−x 2 dx∫∞ −∞ e−x2dx (9.14) Lecture 10: 25 September Lecture 10-1 Physics 5C: Introductory Thermodynamics and Quantum Mechanics Fall 2018 Lecture 10: Finding System Parameters from the Probability Distribution Lecturer: Feng Wang 25 September Aditya Sengupta 10.1 The Partition Function We previously covered that the probability distribution of a system governed by statistical mechanics is p = e−βEi∑ i e −βEi (10.1) We can define z = ∑ i e −βEi , the partition function. Based on this, all of statistical mechanics can be summarized as follows: 1. Calculate z 2. To obtain any parameter (U, S,Cv, P ), follow a set procedure based on the probability distribution. 10.2 Single-Particle Function In a system in which no particles interact with each other, the partition function can be found by multiplying the partition functions of all the particles. Consider a single particle interacting with a reservoir, with fluctuations. Let the system have two discrete energy levels in which the particle can be; an energy of ∆ 2 or of −∆ 2 . z = ∑ α e−βEα = e−β(−∆ 2 ) + e−β( ∆ 2 ) = 2 cosh β∆ 2 (10.2) For N equally spaced energy levels numbered 0 to N − 1, with separation ∆ between each, z = N−1∑ α=0 e−β(α∆) = 1− e−Nβ∆ 1− e−β∆ (10.3) If N →∞, this simplifies to z = 1 1− e−β∆ (10.4) Lecture 10: 25 September Lecture 10-2 These systems can be described by modelling them as harmonic oscillators. Using a common result from analysis of harmonic oscillators, in which En = n~ω + ~ω 2 , we can say zharmonic oscillator = ∑ α e−β ~ω 2 e−β(α~ω) (10.5) zharmonic oscillator = e−β ~ω 2 1 1− e−β~ω (10.6) 10.3 Obtaining functions of state 10.3.1 Internal energy U We know that 〈U〉 = ∑ i EiPi = ∑ iEie −βEi∑ i e −βEi (10.7) 〈U〉 = − ∑ i ∂ ∂β e −βEi z = − ∂z ∂β z (10.8) 〈U〉 = d ln z dβ (10.9) 10.3.2 Entropy S S = −kB ∑ i Pi lnPi = −kB ∑ i ( e−βEi z ln ( e−βEi z )) (10.10) S = −kB z ∑ i ( e−βEi(−βEi)− e−βEi ln z ) (10.11) S = kB ( βU + ln z ∑ i Pi ) = kB (βU + ln z) = U T + kB ln z (10.12) 10.3.3 Heat capacity at constant volume Cv Cv = ( ∂U ∂T ) V = ∂ ∂T ( −d ln z dβ ) |V (10.13) dβ can be rewritten as d 1 kBT = −dT kBT 2 , so Cv = ∂ ∂T ( d ln z dT kBT 2 ) = kBT ( z ( ∂ ln z ∂T ) + T ( ∂2 ln z ∂T 2 ) V ) (10.14) Lecture 10: 25 September Lecture 10-3 10.4 Free Energy / the Helmholz function We know that dU = TdS−PdV , which allows us to write U in terms of natural variables (S, V). However, it may be desirable to use a derivative-based representation of U , in terms of (T, V). This leads to the following definition: F = U − TS (10.15) dF = dU − TdS − SdT = TdS − PdV − TdS − SdT = −PdV − SdT (10.16) This is also referred to as the Helmholz free energy. 10.5 Gibbs Function This comes out of wanting to express U in terms of (T, P). G = F + PV (10.17) dG = dF + PdV + V dP = −SdT + V dP (10.18) This is also referred to as the Gibbs free energy. We can define the partition function in terms of these free energies: F = U − TS = U − T ( U T + kB ln z ) = −kBT ln z =⇒ z = e − F kBT (10.19) These also allow us to redefine pressure and entropy in terms of the Helmholz free energy: P = −∂F ∂V T = kBT ( ∂ ln z ∂V ) T (10.20) S = −∂F ∂T V (10.21) 10.6 Two-Level System Parameters We previously derived z = 2 cosh β∆ 2 for a two-level system. We can find U based on this: U = −d ln z dβ = −1 z dz dβ = − 1 eβ∆/2 + e−β∆/2 ( ∆ 2 eβ ∆ 2 − ∆ 2 e−β ∆ 2 ) (10.22) U = −∆ 2 tanh β∆ 2 (10.23) Lecture 11: September 27 Lecture 11-2 11.2 Multiple-Particle Partition Functions So far, we have assumed all partition functions deal with a single particle, each with its own independent energy levels. We have assumed no interaction between these particles that would affect their probability distributions. We can make a model for multiparticulate systems based on this. Let a system consist of particles i, j, k, . . . . Then we can say the energy is equal to the sum of individual particulate energies: Eijk = E (a) i + E (b) j + E (c) k + . . . (11.9) which would make the multi-particle partition function equal to Z = ∑ {i,j,k,... } e−βEtotal = ∑ {i,j,k,... } e−(Ei+Ej+Ek) = ∑ i ∑ j ∑ k e−βEie−βEje−βEk (11.10) Z = ZiZjZk(splitting the summations and interpreting as partition functions) (11.11) lnZ = ∑ {m=i,j,k,... } lnZm (11.12) This is a good approximation if there is no particle interaction. With N independent particles, for example, Z = (Zone) N . Also, for diatomic molecules with translational, vibrational, and rotational components to their energy, we can say Ztotal = ZtZvZr. 11.3 Statistical Mechanics of Light In classical mechanics, light is described by electromagnetic waves. We can describe their thermal energy. Consider a metal cube with side length L. This allows us to define boundary conditions on the cube. In the x-direction, E|| = 0 at x = 0 and x = L. If we take E(x) = sin(kxx), then we apply the E(L) = 0 boundary condition, we find that kxL = nπ. By symmetry, we can extend this to E(x, y, z) = sin(kxx) sin(kyy) sin(kzz) (11.13) kx = n π L , ky = m π L , kz = l π L (11.14) n, m, l define different EM modes. Each mode is a different degree of freedom. Let ~k = (kx, ky, kz), then ω = c|k|. We want to know how many modes there are with frequency less than some ω. We can graph this in 3D space; we get integer-valued (scaled by π L ) nodes on a cube. We can restrict this by requiring that the distance from the origin to the point on the cube is less than ω c . Then, we can find the number of modes: N = Total volume Element volume = 1 8 4 3π ( ω c )3( π L )3 (11.15) N = ω3L3 6π2c3 = ω3V 6π2c3 (11.16) We can double this due to polarization (?). We can also derive a density of state per unit frequency dN dω = ω2 π2c3V = g(ω), and based on this an energy of light: U = ∫ ∞ 0 g(ω)dω × 〈E0〉 = ∫ ∞ 0 ( kBT ω2V π2c3 ) dω (11.17) since we know the average energy of a state is kBT (half of this each due to the electric field and magnetic field). We can express this as the product of a spectral energy density and an energy per unit frequency, U(ω) = kBT ω2 π2c3 (11.18) This is Rayleigh-Jeans’ law. Note that it scales with increasing ω, meaning that a high frequency would correspond to essentially infinite energy. This was referred to as the Rayleigh-Jeans Catastrophe. Since it’s unlikely that infinite energy would actually be produced, the experimental scientists involved tried a lot of different mathematical models till they found one that fit the data they had. Experimentally, they found U(ω) = ω2 π2c3 ( ~ω e ~ω kBT − 1 ) (11.19) which is the expression that was derived from the harmonic oscillator model before. We can get this analytically from integrating this expression: U = ∫ ∞ 0 g(ω) V dω ~ω eβ~ω − 1 = ~ π2c3 ∫ ∞ 0 ω3dω eβ~ω − 1 (11.20) U = ~ π2c3 ( 1 ~β )4 ∫ ∞ 0 x3dx ex − 1 = π2k4 B 15c3~3 T 4 = AT 4 (11.21) which is the Stefan-Boltzmann Law. Lecture 11-3 Lecture 12: 2 October Lecture 12-1 Physics 5C: Introductory Thermodynamics and Quantum Mechanics Fall 2018 Lecture 12: Quantum Mechanics Lecturer: Feng Wang 2 October Aditya Sengupta The theoretical foundation for resolving the Rayleigh-Jeans Catastrophe was Planck’s seemingly odd step of assuming energy was quantized, i.e. that E = hν (12.1) Based on this, we can define the electron-volt as a microscopic unit of energy: the energy required to move one electron through one volt of potential difference. Thermal energy at 300K is on the order of 25 MeV. A consequence of this quantization is that the photoelectric effect, in which light of a certain freque1ncy causes electronic emission, can be explained; the light is delivering quanta of energy due to their frequency. We can more accurately measure the effect; consider a photoelectric surface at which light is incident at an arbitrary frequency, and a voltage is applied between this and a second plate at voltage −Va. Electrons can cross this barrier if they are given energy at least eVa. We can measure this by measuring the current in the circuit between the plates, as a function of frequency. At some critical frequency at which electrons are first moved, Ek = 0, after which frequency and Ek scale directly. This effect cannot be purely described by electromagnetism; if it were, then with increased intensity of light at the same frequency, more energy would be delivered and therefore there would be more electronic emission. If we believe that particles can carry quanta of energy based on their frequency, we have an easy explanation for the photoelectric effect; after a work function is overcome at some critical frequency Lecture 13: 4 October Lecture 13-1 Physics 5C: Introductory Thermodynamics and Quantum Mechanics Fall 2018 Lecture 13: QM Is Still Weird Lecturer: Feng Wang 4 October Aditya Sengupta 13.1 Bohr’s Model Experimentally, we have found out that the hydrogen atom has all of its positive charges at the center, that it is stable (i.e. despite electrons accelerating, the system is in equilibrium), and that lines with frequency ν = A ( 1 22 − 1 n2 ) are produced. From these conclusions, Bohr was able to make a model of the atom that unified classical and quantum mechanics. The first postulate of Bohr’s model is that electrons can have stable orbitals, with discrete energy levels: Ei − Ef = hν (13.1) Energy levels are discrete, but we cannot say En ∝ n for any n, because then a transition between energy states would not produce multiple frequency lines as per the Balmer spectrum. Instead, we can try letting angular momentum take on discrete values. L = n~ (13.2) where ~ is pretty much randomly chosen as a proportionality constant. In classical mechanics, we know that the Coulombic attraction balances the centripetal force: e2 4πε0r2 = mv2 r (13.3) We also know the angular momentum is discretized: L = n~ = mvr (13.4) Therefore we can find a radius for the nth orbital: rn = 4πε0mv 2r2 e2 = 4πε0(n~)2 me2 = n2aB (13.5) where aB is the Bohr radius (the radius for n = 1). We can find the energy of the electron, Lecture 13: 4 October Lecture 13-2 En = 1 2 mv2 n − e2 4πε0rn = 1 2 e2 4πε0rn − e2 4πε0rn (13.6) En = − 1 n2 e2 8πε0aB = −E0 n2 (13.7) Based on this and the energy level difference equation before, we can say that going from lower to higher n corresponds to light absorption, and higher to lower n corresponds to light emission, at certain frequencies. This matches with the hydrogen line emission observations. Look at the energy difference between two states, ∆E = Ei − Ef = E0 ( 1 n2 f − 1 n2 i ) (13.8) which matches with the Balmer line expression exactly as long as nf = 2. Therefore we can see that this model matches up well with real observations. 13.2 Wave Properties Quantization of angular momentum helps to explain observed properties, but it was not known why energy should be quantized. It was thought that the wave nature of particles could explain this. Enter De Broglie’s Hypothesis, in which light could be explained both by properties of electromagnetic waves (λ, ν) and those of photon particles (E, p). We can make relations between these: E = hν and p = h ν . We can connect these expressions via momentum. In the non-relativistic limit, Ek = p2 2m , therefore p = √ 2m0Ek. We also know that p = hλ, so λ = h√ 2m0Ek (13.9) In general, E2 tot = E2 0 + c2p2 (13.10) p = 1 c √ E2 tot − E2 0 (13.11) E0 = m0c 2;Etot = E0 + Ek (13.12) ∴ λ = hc√ E2 tot − E2 0 = h/(m0c)√( Ek E0 )2 + 2 ( Ek E0 ) (13.13) We can see that the wavelength scales with 1√ m . 13.3 Davisson-Germer Experiment Davisson and Germer essentially scattered light off a crystal and measured its intensity at various angles from 0o to 90o, and they got a weird plot, peaking at some angle around 30o. It is not obvious what this tells us about wave diffraction and interference. The optical path length difference in light diffracting off crystals is D sinφ. When this is equal to nλ, constructive interference is created. At n = 1, the wavelength is 2.15 × sin50A = 1.65A (get the angstrom symbol). We can also find the wavelength based on the equation derived from particle properties: λ = h (2m0e× 56V )1/2 = 1.66A (13.14) (where 56V is also experimentally determined). Once again, the model follows reality quite well. Bragg diffraction: ∆ = 2D cos θ = nλ. θ is the complement of φ, so they both work. Apparently. Idk. Lecture 13-3 Eψ = ( p2 2m + V (x) ) ψ (14.10) ~ωψ = ( ~2k2 2m + V (x) ) ψ (14.11) ~i ∂ψ ∂t = ( ~2 2m ( −i ∂ ∂x ) + V (x) ) ψ (14.12) i~ ∂ψ ∂t = ( − ~2 2m ∂2 ∂x2 + V (x) ) ψ (14.13) “So this is the Schrodinger equation, based on our educated guesses. In retrospect, it seems so natural.” entire room bursts out laughing. “There are many parts of this at which you may ask, why would you do that.” Often, we separate the Schrodinger equation into two parts: i~ ∂ψ ∂t = Ĥψ (14.14) Ĥ = − ~2 2m ∂2 ∂x2 + V (14.15) where Ĥ is the Hamiltonian, an operator that acts on a wavefunction. The rest of this class is basically going to be learning how to do things with the Schrodinger equation. For example, we can characterize wavefunction evolution: ψ(x, 0)→ Hψ(x, 0) = ( − h2 2m ∂2 ∂x2 + V (x) ) ψ(x0) (14.16) ψ(x,∆t) = ψ(x, 0) + ∂ψ ∂t ∆t = ψ(x, 0) + ∆t · 1 i~ (Hψ(x, 0)) (14.17) Essentially, this just repeatedly applies a linear approximation. This is fine for computers, but not for humans. 14.5 Analytical solutions to the Schrodinger equation If psi1, ψ2, . . . are solutions to the Schrodinger equation, then ∑ n Cnψn is also a solution. Fortunately, we can decompose the energy to get this sort of expansion: E(r, t) = ∑ n En(r)e−iωnt (14.18) ψ(x, t) = ∑ n ψn(x)e−iωnt = ∑ n ψn(x)ei En ~ t (14.19) Each term in this summation is referred to as an energy eigenfunction. Lecture 14-3 Lecture 15: 11 October Lecture 15-1 Physics 5C: Introductory Thermodynamics and Quantum Mechanics Fall 2018 Lecture 15: Understanding and Solving Schrodinger’s Equation Lecturer: Feng Wang 11 October Aditya Sengupta (not in that order, probably) 15.1 Solving the 1D case (kinda) Restating Schrodinger’s equation in one dimension: i~ ∂ ∂t ψ(x, t) = ( − ~2 2m ∂2 ∂x2 + V (x) ) ψ(x, t) = Hψ(x, t) (15.1) To find the general solution, we want to take a superposition of many different linearly independent solutions. These solutions will have well defined energies of the form E = ~ω; to convert the wavefunction to frequency space so that we can find this form of energy, we take a Fourier transform: ψ(x, t) = ∑ n ψn(x)e−iωnt (15.2) where ψn(x)e−iωnt is a state with fixed frequency, or fixed energy. It is often referred to as an energy eigenstate. This gives us i~ ∂ ∂t ( ψn(x)e−iωnt ) = ~ωψn(x)e−iωnt = ( ~2 2m ∂2 ∂x2 + V (x) ) ψn(x)e−iωt (15.3) Eψn(x) = ~ωψn(x) = ( ~2 2m ∂2 ∂x2 + V (x) ) ψ(x) (15.4) (15.5) We’ve managed to change the Schrodinger equation into a differential equation only in space. Solving this for ψn allows us to characterize the time evolution of a one-state wavefunction: ψ(x, t) = ψn(x)e−i En ~ t (15.6) and the probability distribution of the particle: p(x, t) = |ψ(x, t)|2 = ∣∣∣ψn(x)e−i En h t ∣∣∣2 = |ψn(x)|2 (15.7) This tells us that the probability distribution is independent of time. Lecture 15: 11 October Lecture 15-2 15.2 Is this where cats die? (Spoiler alert: no) Consider a particle in a box (an infinite square well) whose potential is characterized by V (x) = { 0 0 < x < L ∞ otherwise (15.8) This allows us to split the box into three regions. In the first and third regions, where x < 0 or x > L respectively, the potential is infinite therefore there is no chance of finding the particle there. ψ(x) = 0. In the second region, we get − ~2 2m ∂2 ∂x2 ψn(x) = Enψn(x) (15.9) ∂2 ∂x2 ψn(x) = −2mEn ~2 ψn(x) = −k2 nψn(x) (15.10) where kn is a constant defined for convenience. This has the general solution ψn(x) = C1 sin knx+ C2 cos knx (15.11) We can refine our solution with boundary conditions. One is that ψ(x = 0) = ψ(x = L) = 0 for continuity of the wavefunction. At x = 0, ψn(0) = C1 · 0 + C2 · 1 = C2 = 0 (15.12) So the solution can be simplified to ψn(x) = C1 sin knx. At x = L, ψn(L) = C1 sin knL = 0 (15.13) C1 = 0or sin knL = 0 (15.14) If C1 were 0, then the wavefunction would be 0 everywhere. This is technically valid but physically useless. Therefore sin knL = 0 (15.15) kn = nπ L (15.16) En = n2~2π2 2mL2 (15.17) Lecture 16: 18 October Lecture 16-2 Using quantum mechanics, however, we can analyze the wave behaviour of a particle that could not classically exist in the forbidden region energy state. By the Schrodinger equation, we know Eψ = ( − ~2 2m ∂2 ∂x2 + V ) ψ (16.3) ~2 2m ∂2 ∂x2 ψ = (V − E)ψ (16.4) ∂2ψ ∂x2 = 2m(V − E) ~2 ψ = α2ψ (16.5) The solutions to this are ψ = e±αx, of which only e−αx makes physical sense. So the wavefunction has overall exponential decay; it peaks in the classically permitted region, but it is nonzero in the classically forbidden region. Quantum mechanically, it can take on classically forbidden energies. The wave that exits the potential well is called an evanescent wave (citation needed; all I can find online is E&M related), and this process is called quantum tunneling. 16.3 Qualitative features of bound state wavefunctions 16.3.1 Curvature and oscillatins We know that Lecture 16: 18 October Lecture 16-3 ∂2ψ ∂x2 = 2m(V − E) ~2 ψ (16.6) We can qualitatively analyze this equation to find some trends. If V > E, the equation describes exponential decay with positive curvature. If V >> E then the curvature will be very large, so ψ will decrease faster. If V < E, the wavefunction will oscillate and have negative curvature. If V << E, the oscillations get faster. Based on this, what might the wavefunction of a particle with bound energy states look like? At the ground state, V = 0, therefore |0−Emin| is the smallest energy level, given that there are many possible energy states in the potential well, which will be oscillating at different rates. Because there are many possible energy states, we may ask whether there is a ground state with no oscillations. To find this, we apply the boundary condition that ψ is continuous and differentiable, i.e. that ψ and ∂ψ ∂x are both continuous. Otherwise, the second derivative becomes weird to work with. We know that in a bound energy state, the wavefunction has an exponential decay with nonzero, but small and finite, slope. The continuity of the first derivative, coupled with the small and finite slope of the wavefunction at either end, means no energy state can be flat: E1 6= 0. There must be at least half an oscillation of the wavefunction within the bounding box; this is E1. Higher and higher energy states are characterized by more oscillations within the box. E2 has one complete period of a sine wave, and so on; roughly, ψn ∝ sin nx L (I think). I’m not yet sure why the oscillations in this graphic seem to be in an offset sinc envelope, where the middle oscillations are smaller, rather than all being equal; I’m sure we’ll get to that later. For now: higher energies correspond to more oscillations within the bounding box. The ground state has no node (point at which the energy is the averaged energy over the whole box), and after that, the nth eigenstate will have n− 1 nodes. Lecture 16: 18 October Lecture 16-4 16.3.2 Amplitudes Suppose we have a potential well with two minimum points, first a lower value of V then a higher one. As we saw in the qualitative analysis, the initial larger value of E − V corresponds to faster oscillations and smaller wavelengths (of what?), and the later smaller value of E − V corresponds to slower oscillations and larger wavelengths. How does the amplitude (of the wavefunction?) change? Since large values of E − V correspond to large kinetic energies, v is large, so the particle spends less time in these regions, so the probability is low. Amplitude scales with probability (of finding the particle at a particular point described by the wavefunction), so high energies have low amplitudes. By counting the number of nodes, we can find which eigenstate of this potential well is represented; in the above image, there are six, so this is the seventh eigenstate. 16.3.3 Symmetry Symmetry arguments may also be useful in analyzing wavefunctions. If a potential well bottoms out (reaches a minima) at x = 0, and has the characteristic V (x) = V (−x), then we can say |ψ(x)|2 = |ψ(−x)|2. For real-valued ψ(x), we can say that either ψ(x) = ψ(−x) or ψ(x) = −ψ(−x); the first is the general definition for an even wavefunction, and the second is that for an odd wavefunction. Lecture 17: 23 October Lecture 17-1 Physics 5C: Introductory Thermodynamics and Quantum Mechanics Fall 2018 Lecture 17: Fun with the Schrodinger equation Lecturer: Feng Wang 23 October Aditya Sengupta 17.1 Finite square well wavefunctions By definition, k = (2mE)1/2 ~ and α = (2m(V0−E))1/2 ~ . We can apply this to the constraint equations from before. Since three unknowns are to be determined (E is also an unknown), a third constraint is applied, which is the normalization of the wavefunction; the sum of all probabilities is equal to 1. (Probabilities of finding the particle?) For the even case, we know that tan kL 2 = α k (17.1) tan (√ 2mE ~ L 2 ) = ( V0 E − 1 )1/2 (17.2) and for the odd case, cot kL 2 = −α k (17.3) cot (√ 2mE ~ L 2 ) = − ( V0 E − 1 )1/2 (17.4) This is difficult to analytically solve, so we can introduce some graphical methods. First, we make some algebraic replacements: √ 2mE ~ L 2 = θ (17.5) √ 2mV0 ~ L 2 = θ0 (17.6) V0 E = ( θ0 θ )2 (17.7) Then, for the even case, tan θ = √( θ0 θ )2 − 1 (17.8) Lecture 17: 23 October Lecture 17-2 and for the odd case, cot θ = − √( θ0 θ )2 − 1 (17.9) We can see on a graph of either case where the decaying square root curve meets the continuously repeating tangent lines, and can numerically estimate the solutions. For the even case, the first solution is close to but less than π 2 , the second is close to but less than 3π 2 , and so on incrementing by π in each case until θ = θ0. The odd solutions are smaller than but close to integral multiples of π. In a finite square well, the waves are more spread out than in the infinite case, so the energy in a finite square well is lower. We get this from the graphic solution (how). There is a finite number of bound states in both even and odd wavefunctions. There are more states with larger values of θ0, leading to a deeper well. There will be at least one bound state (in a one-dimensional case? what’s that mean) 17.2 The harmonic oscillator V = 1 2 cx2 (17.10) Substituting this into the Schrodinger equation, − ~2 2m d2ψ dx2 + 1 2 cx2ψ = Eψ (17.11) We examine the limiting behaviour, in which ψ e−αx = e− √ 2m(V0−E) ~ x. The harmonic with number 0 acts as ψ e− 2m( c 2 x2−E) ~ x (17.12) which as x→∞ acts as a Gaussian distribution, e− √ mc ~ x2 . At large x, for the general Gaussian expression, we can take x derivatives: dψ dx = −x a2 ψ (17.13) d2ψ dx2 = − 1 a2 ψ + x2 a4 ψ (17.14) Therefore, substituting back into the Schrodinger equation gives us − ~2 2m x2 a4 + 1 2 cx2 = 0 =⇒ a = ( ~2 mc ) 1 4 (17.15) All the eigenstates decay in almost the same way (?) as x→∞. The ψns are very different, but in general we can say ψn(x) = fn(x)e− x2 2a2 (17.16) We want to choose an fn(x) in general such that it is alternating between even and odd. A convenient choice may be fn(x) = xn. Instead of just large x, we want to ensure this choice fulfils the Schrodinger equation for all x. ψ0(x) = e− x2 2a2 (17.17) We just did this one, actually. The only difference is we can no longer set E = 0 because we’re not in the very-large-x regime, so solving in much the same way as the above yields E = ca2 2 . ψ1(x) = xe− x2 2a2 (17.18) Substituting this into the Schrodinger equation gives us −a4 d 2ψ dx2 + x2ψ = 2E1 c ψ (17.19) Derivatives suck, but they tell us that E1 = 3 2 ca2 = 3E0 = 3 · 1 2 ~ω0 (17.20) In general, we see that state n has energy 1 2~ω0 + (n− 1)~ω0. Lecture 17-3 Lecture 19: 30 October Lecture 19-1 Physics 5C: Introductory Thermodynamics and Quantum Mechanics Fall 2018 Lecture 19: Quantum Amplitudes and Other Made-Up Words Lecturer: Feng Wang 30 October Aditya Sengupta 19.1 Quantum Amplitudes The probability distribution of a particle is important, but is not sufficient to describe everything. We introduce the amplitude, which is a complex number directly related to the wavefunction. One of the key ways of examining photon polarization is an analyzer loop. Consider a polarization analyzer which splits polarized light into basis components (using the X/Y basis, the X’/Y’ basis, or the L/R basis). To make light interfere with itself, add a second analyzer that makes the direct lines of light collinear. This is a reversed analyzer. Here, the output is the same thing as the input to the first analyzer, which doesn’t seem to serve much of a purpose. But doing this helps to verify that we can split light into its polarization basis elements correctly, and that opens the door to all kinds of analysis. These two analyzers form an analyzer loop. On diagrams, an analyzer loop is denoted by a box in which an oval is drawn, and the two basis elements indicated on the two arcs of the oval. Consider Y-polarized light incident on an X’-Y’ analyzer loop, with an angle of θ = 30o between Y’ and Y. We feed the output of this loop into an X-Y analyzer. We can block any arm of any analyzer in this system; say we block the Y’ channel on the analyzer loop (block it between the analyzer and the reverse analyzer). The light is forced to be polarized in the X’ state with some probability, so an X’ polarization is projected onto the X-Y analyzer. The probability of projection is cos2 θ; for Y to X’, this is 60o, so the probability of this transmission is ( 1 2 )2 = 1 4 , and for X’ to X, this is 30o, so the probability of this transmission is(√ 3 2 )2 = 3 4 . Therefore the overall probability of transmission is 3 16 . Suppose now that we block the X’ channel. The light is forced to be polarized in the Y’ state; the probability of transmission of Y to Y’ is 3 4 , and the probability of transmission Y’ to X is 1 4 . Therefore, again, the probability of transmission overall again is 3 16 . If both channels are open, the probability becomes 0. This is because the loop analyzer now does nothing, so the setup is just measuring the X polarization of Y-polarized light, which is 0 by definition. Why do two nonzero probabilities combine to make a zero probability? This is because of destructive interference; when only one source of light at a time goes through thin slits, there is some intensity at each point, but it is possible for the two sources to destructively add at certain points where there were nonzero intensities before. More concretely, this comes from I ∝ |E2 total| = |E1 + E2|2; the magnitudes of the electric fields individually may be nonzero and create nonzero intensities, but their vector sum may end up being zero. In general, for quantum amplitudes, we know: 1. Probability = |Quantum amplitude|2 2. Quantum amplitude total = ∑ i (amplitude)i. 19.2 Formalism for projection amplitudes We represent a projection between state i and state f by 〈f |i〉, for example 〈Y |Y ′〉. Projections are applied in series by multiplying them together, e.g. 〈X|X ′〉 〈X|Y 〉. This lets us write the quantum amplitude of the previous system, QA = 〈X|X ′〉 〈X ′|Y 〉+ 〈X|Y ′〉 〈Y ′|Y 〉 = 0 (19.1) P = | 〈X|X ′〉 |2 = cos2 θ =⇒ | 〈X|X ′〉 | = cos θ =⇒ 〈X|X ′〉 = cos θeiφ (19.2) and we can derive the other polarization projections, | 〈X ′|Y 〉 | = ∣∣∣cos (π 2 − θ )∣∣∣ (19.3) | 〈Y ′|Y 〉 | = |cos θ| (19.4) | 〈X|Y ′〉 = ∣∣∣cos (π 2 + θ )∣∣∣ (19.5) Choose the convention where everything is real and positive, we can just drop all the absolute value signs above. In general, the cosine of the angle between each pair of arrows is the probability projection of one onto the other. Lecture 19-2 Lecture 20: 1 November Lecture 20-1 Physics 5C: Introductory Thermodynamics and Quantum Mechanics Fall 2018 Lecture 20: Linear and Circular Polarization Projections Lecturer: Feng Wang 1 November Aditya Sengupta 20.1 Linear to Circular Probability Projections Recall 〈X|X ′〉 = 〈Y ′|Y 〉 = cos θ (20.1) 〈X ′|Y 〉 = 〈Y |X ′〉 = sin θ (20.2) 〈X|Y ′〉 = 〈Y ′|X〉 = − sin θ (20.3) We can use the same logic to analyze left and right circular polarization. Consider an R-L analyzer loop, consisting of an LR analyzer followed by an inverted LR analyzer. Place this analyzer loop before an X’-Y’ analyzer. Then, consider what happens when R is open, when L is open, and when both are open. When R is open, we get 〈Y ′|R〉 〈R|Y ′〉. We know that the probability of projecting linearly polarized light to circularly polarized light is 1 2 , that is, | 〈Y |R〉 |2 = | 〈R|Y 〉2 | = 1 2 , and similarly for the left polarization. Therefore we get a probability of 1 4 in both polarizations, which gives us a quantum amplitude of 〈Y ′|L〉 〈L|Y 〉+ 〈Y ′|R〉 〈R|Y 〉 = cos θ (20.4) and a transmission probability of cos2 θ, the amplitude squared. Also, due to symmetry, all the components are equal: | 〈Y ′|R〉 | = | 〈R|Y 〉 | = | 〈Y |L〉 | = | 〈L|Y 〉 | = 1√ 2 (20.5) It is not possible for all the amplitudes to be real, so in general we give all of them complex phases: 〈Y ′|R〉 = 1√ 2 eiα (20.6) 〈R|Y 〉 = 1√ 2 eiβ (20.7) 〈Y ′|L〉 = 1√ 2 eiγ (20.8) 〈L|Y 〉 = 1√ 2 eiδ (20.9) Lecture 21: 6 November Lecture 21-1 Physics 5C: Introductory Thermodynamics and Quantum Mechanics Fall 2018 Lecture 21: State vectors Lecturer: Feng Wang 6 November Aditya Sengupta A state vector can be expressed as a linear combination of basis vectors; since the basis vectors are orthonor- mal, the coefficients are equal to the dot products between the state vector and each basis element: |ψ〉 = ∑ i |ψi〉 〈ψi|ψ〉 (21.1) Position representation uses the position state as the basis (invoking the particle nature of the thing), where |xi〉 is a position eigenstate. The position states have the projection on one another 〈xi|xj〉 = δ(xi − xj) (21.2) which invokes the Dirac delta. We can translate between the wavefunction ket and the position representation by projection, |ψ〉 = ∑ i |xi〉 〈xi|ψ〉 =⇒ ψ(xi) = 〈xi|ψ〉 (21.3) The probability of finding ψ at xi is therefore | 〈xi|ψ〉 |2 = |ψ(xi)|2 (21.4) which turns out to basically be a different formulation of the Schrodinger equation. 21.1 Time dependence of quantum states Till now we have only discussed particles in a steady state. We can extend this theory to include time dependence as well, starting with a simple case as an example. We previously covered the energy eigenstates to describe the bound states in an infinite square well, which we can revisit to include time dependence, ψn(x, t) = ψn(x)e−i En ~ t (21.5) We know that Pn(x, t) = |ψn(x, t)|2 (21.6) Lecture 21: 6 November Lecture 21-2 which in the stationary state becomes |ψn(x)|2. We can construct a superposition of energy eigenstates. Considering the first two energy eigenstates and adding time dependence, ψ(x, t) = A sin (πx L ) e−iω1t +A sin ( 2πx L ) eiω2t (21.7) The probability is no longer fixed, as there is a phase change in the time dependent factor (we know that ω2 = 4ω1). We can find the time dependent factor at certain time intervals, e−iω1t e−iω2t t = 0 1 1 t = π ω1 e−iω1 π ω1 = −1 e−iω2 π ω1 = 1 t = 2π ω1 1 1 Based on these, we can plot the wavefunction at these fixed times. Say we have ψ(x, t) = f1(x)e−iω1t + f2(x)e−iω2t (21.8) |ψ(x, t)|2 = ψ(x, t)ψ∗(x, t) = f1(x)2 + f2(x)2 + 2f1(x)f2(x) cos(ω1 − ω2)t (21.9) which is demonstrably independent of the reference zero energy or frequency. Although the particle exists in a well-defined single state, a particle with a superposition of quantum states will have equal probability of an energy measurement showing each of its constituent energy states. The probability split for the current distribution is | 〈ψ1|ψ〉 |2 + | 〈ψ2|ψ〉 |2 = ∣∣∣Ae−i(ω1t+∆t) ∣∣∣2 +A2 = 2A2 = 1 (21.10) The average energy is E1+E2 2 , although this cannot be measured directly. To experimentally determine the time-dependent probability distribution of a particle, it is necessary to prepare many copies of the same state, and to do the same measurement at exactly the same time, otherwise interference from the cos((ω2 − ω1)t) term would invalidate the equivalence of the states. Consider the behaviour of a wave packet in a square well. The wavefunction in this state is a superposition of energy eigenstates, ψ(x, t) = ∞∑ n=1 Bn sin (nπx L ) e−i En ~ t (21.11) At t = 0, ψ(x, t) = ∞∑ n=1 Bn sin (nπx L ) (21.12) which is the expression for a Fourier series expansion. We can analyze this further using state orthogonality, the condition that 〈ψj |ψi〉 = 0. Therefore, in a projection from ψ to ψj , we get 〈ψj |ψ〉 = ∫ sin jπx L ψdx (21.13)∫ sin jπx L ψ(x, 0)dx = ∞∑ n=1 Bn ∫ sin nπx L sin jπx L dx = Bj ∫ L 0 sin jπx L sin jπx L dx = Bj L 2 (21.14) ∴ Bj = 2 L ∫ L 0 sin ( jπx L ) ψ(x, 0)dx (21.15) Consider a simple example, with ψ(x) = A where L 2 − b 2 < x < L 2 + b 2 and 0 elsewhere. The coefficients are Bn = 2 L ∫ L 0 sin nπx L ψ(x, 0)dx = 2A L · L nπ ∫ L 2 + b 2 L 2 − b 2 (21.16) which can, through a LOT of algebra, be found: Bn = 2Ab L sin nπb 2L nπb 2L sin nπ 2 (21.17) Lecture 21-3 Lecture 22: 8 November Lecture 22-3 ψ(x, t) ∼ ∫ Bke i(kx−E~ t)dk ∼ ∫ Bke i(kx− ~k2 2m t)dk (22.14) This allows us to define a group velocity of a wave packet. We know that ω k = λ T = vp, the phase velocity. We also know that ω = ~k2 2m , so dω dk = ~k m = p m = vgroup (22.15) Consider a specific case where k is restricted to the range k0 ±∆k. This makes the general integral for ψ equal to ∫ ∆k k′=−∆k ei(k0+k′)x− ~ 2m (k0+k′)2tdk (22.16) This is an ugly integral, so assuming ∆k << k0, we get ψ = ei(k0x− ~k2 0t 2m ) ∫ ∆k −∆k ei(x− ~k0 m )k′dk′ (22.17) which integrates to something kind of ugly. We can take the magnitude and it looks like a sinc, |ψ(x, t)|2 = ∣∣∣∣∣ sin (( (x− ~k0 m t)∆k )) (x− ~k0 m t)∆k ∣∣∣∣∣ 2 (22.18) This looks terrible. Lecture 22-4 Lecture 23: November 13 Lecture 23-1 Physics 5C: Introductory Thermodynamics and Quantum Mechanics Fall 2018 Lecture 23: Analysis of Uncertainty Lecturer: Feng Wang November 13 Aditya Sengupta It is impossible to accurately measure both position and momentum; there is always some spread in position, which can be Fourier transformed to find the uncerainty in momentum. The condition is ∆x ·∆p ≥ ~. In measurements, this comes from counting uncertainty, as derived in the previous lecture. We found that ∆λ ≥ λ2 ∆L . Defining ∆λ/λ2 = ∆k, we can restate the uncertainty principle as ∆k ·∆l ≥ 1. The time period of a wave T = ∆τ N , where ∆τ is the width of a larger wave packet, can be used to find the time-domain uncertainty: ∆T = ( ∆τ N − ∆τ N + 1 ) ≈ ∆τ ·∆τ N2 1 ∆τ (23.1) ∆T T 2 ·∆τ ≥ 1 =⇒ ∆ω ·∆τ ≥ 1 (23.2) Time-frequency uncertainty. Worlds are colliding. Since energy and frequency are directly proportional with proportionality constant ~, therefore ∆E ·∆τ ≥ ~. 23.1 Consequences of Uncertainty The energy-time uncertainty means finding the exact energy of a particle would take infinite time. More concrete examples of this uncertainty can be stated. Consider the infinite square well with energies E1, E2, E3. Two electrons in the energy states E1 and E2 can be said to form a dipole, which in classical models has decaying oscillations. In quantum mechanics, the excited state decays suddenly according to a probability distribution. Consider the interval from time t to time t+ ∆t. The probability of decay ∆P should depend linearly on ∆t, a rate constant γ, and on the overall probability P (if the state has decayed the probability of the state decaying in the next time interval is zero). Therefore dP P = −γdt =⇒ P (t) = Poe −γt. We know that P (t) ∝ |ψ(t)|2, therefore |ψ(t)| = Poe − t τ (23.3) where τ = 1 γ , the decay time. This characterizes the uncertainty in time of the excited state. If the atomic transition has a time uncertainty of 10−8s, then the uncertainty in energy turns out to be ∆E ∼ 10−7 eV. Something about forbidden transitions. Consider the case of gamma ray radiation from unstable nuclei. These have a time uncertainty of 10−19s, using the same kind of argument as above. 23.2 Shape of the Spectrum As derived previously, the wavefunction acts as a decaying exponential, which lets us define amplitude via a Fourier transform, Lecture 24: 15 November Lecture 24-2 ∂ψI ∂x = ∂ψII ∂x |x=0 =⇒ A0ik −Aik = Bik2 =⇒ A0k −Ak = Bk2 (24.9) (A0 −A) √ 2mE ~ = B √ 2m(E − V0 ~ (24.10) Algebra gives us A = A0 k1 − k2 k1 + k2 (24.11) B = A0 2k1 k1 + k2 (24.12) From this, we can define reflection and transmission coefficients. R is defined below, R = ∣∣∣∣ AA0 ∣∣∣∣2 = ( k1 − k2 k1 + k2 )2 (24.13) Consider the case where E − V0 << V0, i.e. the energy is just above the barrier. In this case, there is little transmission (R ≈ 1), and k2 << k1. In the case where E − V0 >> V0, we get k1 ≈ k2 and R ≈ 0, so there is little reflection. The reflection coefficient can be graphed as a function of energy. The transmission coefficient is T = ∣∣∣∣ BA0 ∣∣∣∣2 · v2 v1 (24.14) This is slightly different because of the change in potential corresponding to a change in velocity. The transmission coefficient should represent the probability of passing an electron past the potential barrier per unit time. Therefore it should be based on a flux of electrons; the density of electrons that could cross the barrier, multiplied by the ratio of velocities in the old and new potentials. If the electrons have zero velocity, then the transmission probability should be zero. T = ∣∣∣∣ BA0 ∣∣∣∣2 · v2 v1 = ( 2k1 k1 + k2 )2 · k2 k1 = 4k1k2 (k1 + k2)2 (24.15) We can verify this by noting that R+ T = (k1 − k2)2 (k1 + k2)2 + 4k1k2 (k1 + k2)2 = 1 (24.16) as it should be. Lecture 24: 15 November Lecture 24-3 24.2 Conservation of current From 5B, we know that changes in local charge density over time come only from an inflow or outflow of current: dρ dt = −∇~j = − ∂j ∂x (24.17) Consider a box from x0 to x0 + ∆x, with inflow of current jin and outflow jout. In this case, the total ρ can be found by integrating a constant ρ over that length, d ∫ x0+∆x x0 ρdx dt = j(x0)− j(x0 + ∆x) (24.18) Physically, the charge density comes from the amplitude of the wavefunction (?), therefore we can write ρ = |ψ|2 = ψ∗ψ (24.19) so on the left hand side of the above expression, we get d dt ∫ x0+∆x x0 ψ∗(x, t)ψ(x, t)dx (24.20) Differentiating under the integral gives us d dt (ψ∗ψ) = ψ∗ dψ dt + ψ dψ∗ dt (24.21) Since ψ and ψ∗ both satisfy the Schrodinger equation, we know that i~ dψ dt = ( − ~2 2m ∂2 ∂x2 + V (x) ) ψ (24.22) −i~dψ ∗ dt = ( − ~2 2m ∂2 ∂x2 + V (x) ) ψ∗ (24.23) Therefore we can use the first time derivatives of ψ and ψ∗ in the charge-density expression: d dt (ψ∗ψ) = 1 i~ ψ∗ ( − ~2 2m ∂2 ∂x2 + V (x) ) ψ − 1 i~ ψ ( − ~2 2m ∂2 ∂x2 + V (x) ) ψ (24.24) Lecture 24: 15 November Lecture 24-4 d dt (ψ∗ψ) = i~ 2m ( ψ∗ ∂2ψ ∂x2 − ψ∂ 2ψ∗ ∂x2 ) (24.25) Therefore, the integral becomes ∫ x0+∆x x0 i~ 2m ( ψ∗ ∂2ψ ∂x2 − ψ∂ 2ψ∗ ∂x2 ) dx (24.26) Note that using the Fundamental Theorem of Calculus, that is, ∫ b a ∂F ∂x dx = F |ba (24.27) we can write this as the evaluation of i~ 2m ( ψ∗ ∂ψ∂x − ψ ∂ψ∗ ∂x ) at x0 + ∆x, minus that at x0. Since the right-hand side is the difference between the evaluation of j at x0 and x0 + ∆x, we can set j equal to the negative of the above expression, j ≡ − i~ 2m ( ψ∗ ∂ψ ∂x − ψ∂ψ ∗ ∂x ) (24.28) For ψ = Aeikx, this expression gives us j = − i~ 2m · 2ik · |A|2 = ~k m |A|2 = v|A|2 (24.29) Note that if the wavefunction is real, the charge density is zero as the wavefunction is its own complex conjugate. 24.3 Scattering by a 1D potential well Consider a potential well in which V = 0 for x < 0 and x > L, and V = −V0 elsewhere. The wavefunction has the form ψI = A0e ik1x +Ae−ik1x (24.30) ψII = Beik2x + Ce−ik2x (24.31) ψIII = Deik1x (24.32) Lecture 25: 27 November Lecture 25-2 ikA0 + (−ik)A = αB (25.6) and so A A0 = ik + α ik − α (25.7) B A0 = 2ik ik − α (25.8) This gives us the reflection coefficient, R = ∣∣∣∣ AA0 ∣∣∣∣2 = α2 + k2 α2 + k2 = 1 (25.9) There is complete reflection, and no transmission. However, with a finite-width square barrier, this is different. The wavefunction here has three components, ψI = A0e ikx +Ae−ikx (25.10) ψII = Be−αx + Ceαx (25.11) ψIII = Deikx (25.12) There are four boundary conditions here. Continuity of the wavefunction at x = 0, A0 +A = B + C (25.13) Continuity of the derivative at x = 0, ikA0 + (−ik)A = (α)B + αC (25.14) Continuity of the wavefunction at x = L, Be−αL + CeαL = DeikL (25.15) and continuity of the derivative at x = L, (−α)Be−αL + αCeαL = ikDeikL (25.16) Lecture 25: 27 November Lecture 25-3 Therefore, algebra gives us D A0 = 4ikα eikL ((α+ ik)2e−αL − (α− ik)2eαL) (25.17) The tunneling probability is then given by vt|D|2 vi|A0|2 = |D|2 |A0|2 = 16k2α2 |(α+ ik)2e−αL − (α− ik)2eαL|2 (25.18) With the assumption that αL >> 1, the decaying exponential with that exponent can be violently murdered, and as we hold a towel over its mouth we can write the simplified expression T = 16k2α2 (α2 + k2)2e2αL = 16 E V0 ( 1− E V0 ) e−2αL (25.19) The exponential decay term tends to dominate here: T ∼ e−2αL = e− 2 ~ √ 2m(V0−E)L (25.20) With a large mass, this transmission becomes negligible. Even an atom qualifies as “large” here. Consider an electron in a metal trying to tunnel through to another metal with an energy gap. For Ek ≈ 3eV and V ≈ 10eV, and a thickness L ≈ 1Å, the exponential term is 1.4×1010m−1×10−10m, and so the transmission is e−2×1.4 ≈ 1 10 This is used in the scanning tunneling microscope. 25.2 Probability current The probability current of the wavefunction in the first region is JI = v0|A0|2 − v0|A|2 (25.21) and in the third is JIII = v0|D|2 (25.22) Forcing these two to be equal gives |A0|2 − |A|2 = |D|2. In the middle, we can calculate the probability current using the first-principles equation based on the form of the wavefunction, JII = − i~ 2m ( (B∗e−αx + C∗eαx)(−αBe−αx + Cαeαx)− (Be−αx + Ceαx)(B∗(−α)e−αx + C∗αeαx) ) (25.23) JII = − i~α m (B∗C −BC∗) (25.24) For this to be equal to the probability current in the other two regions, B and C have to be finite, and complex with different phases. 25.3 Approximation for barrier penetration In an arbitrary potential rectangular barrier, we know that ψ(L) ψ(0) ∼ e− 2m(V0−E)1/2 ~ L (25.25) and α(x) = √ 2m ~ (V (x)− E) (25.26) characterizes the exponential decay. Over a small slice ∆x, the ratio of wavefunctions is ψ(x+ ∆x) ψ(x) ∼ e−α(x)∆x (25.27) With a linear approximation, we get ψ(x) + ∂ψ ∂x ∆x ∼ (x− α(x)∆x)ψ(x) (25.28) and therefore ∂ψ(x) ψ(x) = −α(x)dx (25.29) This gives us a complex exponential in the integral of −αx, which is a function of V (x), to characterize the wavefunction ratios and therefore the transmission, T = ∣∣∣∣ψ(x2) ψ(x1) ∣∣∣∣2 =∼ exp ( −2 ∫ x2 x1 √ 2m(V (x)− E) ~ dx ) (25.30) Lecture 25-4