Quantum Mechanics in One Dimension - Introduction to Modern Physics - Lecture Notes, Study notes of Physics

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Quantum mechanics in one dimension Schrödinger’s equation is the analogue to the wave equation of sound, light, water, 2 2 22 2 1 t y vx y      , which works for all classical waves, that have either a photon associated with it or a pseudo-particle (such as a phonon) harmonic wave, plane wave is solution to this equation moving to the right y(x,t) = y0 cos 2 π (x/λ - t/T) = y0 cos 2π/λ (x – vt) where v = λ/T these functions describe something with physical significance, e.g. the E vector, the amplitude on a water wave, the air pressure in a sound wave the square of that function y(x,t)2 ~ η energy per unit volume, intensity (I) of waves is energy density times wave speed, so I ~ y(x,t)2 ~ η can be put down to number of photons (or pseudo-particles) at anyone place at a certain time, is also the likelihood of finding a photon (or pseudo- particle) there (x) and then (t) what we need is something altogether different but mathematically similar - a wave equation for matter wave, the solutions to which, i.e. Ψ(x,t) - docsity.com 2 the matter waves - will be a valid description of how “small” things move – and the square of which Ψ(x,t)2 will give us the probability of finding the particle of the matter wave there (x) and then (t) Ψ(x,t) contains everything that is and can be know about the particle, to get the probability of finding the particle at some specific (x,t) we have to calculate Ψ(x,t)2 Born’s interpretation call the probability that particle will be found in the infinitesimal small interval dx about the point x P(x), probability density as it is per length unit, then Born’s interpretation is P(x) dx = Ψ(x,t)2 dx will be a number ≤ 1 (at time t) it is not possible to specify with certainty the position of a particle (x) - Heisenberg’s uncertainty principle – but it is possible to assign definitive values of probabilities for observing it at any place we care to calculate the square function for (at a given time) Ψ(x,t)2 is intensity of matter wave, a measurable quantity, while Ψ(x,t) is only a mathematical model for the matter wave, a non physical thing, can’t be measured docsity.com 5 now show that y = Ae-iω(t-x/v) is a solution to the classical wave equation first partial derivate of y with respect to x y v i x y     second derivate yv y v i x y 2 2 2 22 2 2     first partial derivate with respect to t yit y    second derivate yyit y 222 2 2     comparing the second derivates, difference is just 2 1 v otherwise they are identical so 2 2 22 2 1 t y vx y      which is the wave equation, so y(x,t) = Ae-iω(t-x/v) must be a solution to this equation docsity.com 6 complex wave functions / just like complex numbers Ψ = A + iB, A real part of function B imaginary part then Ψ* = A – iB , (i is replace everywhere by – i and one has the conjugate complex function) Ψ2 = Ψ* Ψ = Ψ Ψ* = A2 – i2 B2 = A2 + B2 is all real i2 = -1 the fundamental problem of quantum mechanics given the wave function at some instant, say t = 0, i.e. Ψ(x,0), find the wave function at some or all other times t - when there are forces acting on the particle Ψ(x,0) is the initial information on the particle, Newton’s mechanics analogue was initial position (x) and momentum (p) of a classical particle, now it is an infinite set of numbers a set of values, for all points x one value of Ψ(x,0) docsity.com 7 in Newton’s mechanics we obtain x(t) and p(t) by solving Newton’s second law am dt vmd FpddtF     , , an net force acting on the particle changed it’s momentum, change in position over kinematics Schrödinger’s equation (SE) propagates Ψ(x,0) forward in time, that’s what we want to know, given (within Heisenberg’s uncertainty) we know where a particle is and what its momentum there is, we want to calculate were will be at some time (t) and what will it’s momentum be at that time i.e. the initial Ψ(x,0) changes into Ψ(x,t) t ixU xm         )( 2 2 22 F = dx dU  is the force acting on the particle U(x) is the potential energy function of the Force 1. left hand side (LHS) of SE is first evaluated for Ψ(x,0), i.e. t = 0, as it is not dependent on time, i.e. we make partial derivations and add the influence of the potential energy function on Ψ(x,0) docsity.com 10 Ψ(x,t) = tiextx   )()()( if U(x) potential energy is function of x only (not of t) !!! )( )( tE dt td i   )()()( )( 2 2 22 xExxU dx xd m    with E = h f = 2π  f = ω  so ω =  E we can look at the e-iωt factor above which describes the time dependency if the potential energy does not depend on time – so that time dependency factor is t iE e   in equations above, E is the total energy, which we can normalize to be the kinetic energy plus the potential energy, (if we set rest energy E0= 0, as a reference form which energy is counted - which we can do arbitrarily) docsity.com 11 0)()}({ 2)( 22 2  xxUE m dx xd   rearranged for further use and called, time independent, steady- state, or stationary Schrödinger equation in one dimensions if we have an arbitrary potential energy function U(x) there are no explicit analytical solutions to this equation  must be “well behaved” just as Ψ has to in order to give sensible results for probabilities, i.e. finite everywhere including +-  , single valued for any x, continuous, and “smooth” – which is dx d must also be continuous and single valued (the Serway book says here: wherever U(x) has a finite value, other books say all the time) – all of them are mathematical conditions, so called boundary conditions if we can separate the variables, we also get Ψ(x,t)2 = 2 , meaning all probabilities we calculate from Ψ(x,t) will not depend on time, are static or stationary docsity.com 12 expansion to three dimensions straightforward 0),,()},,({ 2 ),,( ),,( ),,( 2 2 2 2 2 2 2        zyxzyxUE m dz zyx dy zyx dx zyx      consequence at least 3 quantum numbers, taking account of the spin of the electron it will be 4 for electrons confined to be in an atom let’s look at a free particle in the plane wave approximation, also called a harmonic wave free non-relativistic particle means no force on it F = 0 = dt dU  , no force means no potential energy U(x), and no dependence of the potential energy on t, as particle is free, all energy is kinetic E = ½m v2 one dimensional time independent SE simplifies to docsity.com 15 we may have as well calculated 2)(x to find the probability of finding the particle for any x we want 2)(x = *)(x )(x = 1202)(  AeAAeAe kxiikx analyzing the graph we see that the probability of finding the particle in any one segment of equal length Δx or dx is absolutely the same as it is a constant, so the particle has equal probabilities to be at any place, there is no most likely place docsity.com 16 so let’s assume we have a free particle moving to the right, expressed by wave function, see what happens if we put it into Schrödinger equation Ψ(x,t) = )]sin()[cos)( tkxitkxAAe tkxi   where A is a constant, let’s differentiate partially for x and t and put our derivates into the time dependent (one dimensional) Schrödinger equation       iAei t tkxi )(     2)(2 2 2 )( kAeik x tkxi  as it is a free particle, it is not under the influence of a force, so it has constant (time and position independent) net potential energy U(x) = U0, which may be zero or any other value (remember potential energy levels can be set arbitrarily) t ixU xm         )( 2 2 22 plugging our derivates in  )()(2 0 2 2 iiUk m   which we can divide by Ψ!!! docsity.com 17 and we get   0 22 2 U m k as we know PEKEEtotal  so what is m k 2 22 with k2 = (2π / λ)2 and p2 = (h / λ)2 = m2v2 m k 2 22 = ½ m v2 is kinetic energy of the free particle moving to the right QED, formalism makes sense as long as there is no net force, a particle does not change momentum, and moves in a straight line at constant speed, uniform linear motion – just the same for macroscopic particles is stated in Newton’s first law, Newton’s first law is contained in, i.e. it is actually a solution of Newton’s second law, just as harmonic (plane) wave is a solution of, i.e. is (contained in), Schrödinger’s law free particle solution can also be written as docsity.com 20 in all cases P is just the area under a curve this sets a strict condition to Ψ if it is not only to be a function that happens to solve the Schrödinger equation, but also to represent the pilot/guiding/matter wave of a real particle the area under the curve has to be finite so that it can normalized to be 1 or 100 %, so Ψ(x,t) has to go to zero for x1 and x2 otherwise Ψ(x,t)2 would not go to zero and the area under the curve would not be finite docsity.com 21 example: Bohr radius in hydrogen atom boundary conditions must be fulfilled for Ψ(x,t) to represent a real particle docsity.com 22 well behaved functions Ψ and  must be “well behaved” in order to give sensible results for probabilities, i.e. finite everywhere, single valued for any x (and t), continuous, i.e. having x (and t) values everywhere (unless V(x) is infinite) and “smooth” – which is dx d must also be continuous wherever U(x) has a finite value in addition to being a solution of the Schrödinger equation so boundary conditions and requirements of normalization will make it possible for us to decide which solution of Schrödinger equation represent real particles and which are a purely mathematical construct docsity.com 25 (analogously: applying the definition of ω = 2π f and E = h f (Plank-Einstein equation) gives physical meaning to Δω Δt ≈ 1) so we don’t violate with the mathematical model for the pulse/wave bundle/wave packet Heisenberg’s uncertainty principle and this describes a real particle, in addition, the area under a pulse will of course be finite, so we can normalize our wave function Expectation values and Operators the solutions to the Schrödinger equation contain everything that can be known (i.e. which the uncertainty principle allows us to know) about the movement of an entity that is a wave-particle with mass so lets extract the (arithmetic) mean position – which is also called the expectation value, (your book states here incorrectly the average position, an average does not refer to a distribution/population but the arithmetic mean does),    n i ii xfn x 1 1 where f is the dimensionless frequency of occurrence of one particular value of x (forget about p 215 lower half and p 216 top paragraph, I am pretty sure that is incorrect as I did not find a similar Modern Physics treatment in Beiser and Tipler docsity.com 26 short maths into if the “sample of x values” is large the mean of these values may be taken as an estimate of the distribution/population mean the sum of all discrepancies form the mean is zero    n i i xx 1 0)( the variance of the mean      n i ii xxfn s 1 22 )( 1 1 for large n, one can approximate n with n-1 and use the variance of the population      n i ii n i ii xfn xf n x 1 2 1 2 )( 1 1 )( 1 )var(  as a measure of variance of the sample standard deviation (σ) is the square root of the variance and another measure of the amount of scatter in the data if σ = 0 then var(x) = 0, there is no spread in the data and the distribution is called sharp docsity.com 27 the uncertainly principle now tells us that particle positions (x) can only been know with probabilities, i.e. its distribution is never sharp and always fuzzy back to the expectation value, <x>     dxtxtxxx ),(*),( where Ψ(x,t) has to be normalized definition the arithmetic mean of x that would be expected from measurements of the positions of a large number of particles with the same wave function! don’t confuse with probability of finding a particle in an infinitesimal interval around x – it’s completely different things, so P = 0 may be compatible with a finite expectation value <x> e.g. for an infinite square well and even quantum number wave functions: P(L/2) = 0, but <x> = L/2 because Ψ2 and also 2 are symmetric about that point to calculate we have the definition of the expectation value <x>     dxxxxx )(*)(  we need normalized wave functions, and they are docsity.com 30 so f(x) can be potential energy U(x) for example however no function p = p(x) exist by virtue of the uncertainty principle, Δpx Δx ≥ 2  if both of these entities vary in a fuzzy way there simply can’t be a relation between the two of them (there is simply no classical path in quantum mechanics) p = mv but p ≠ p(x) in quantum mechanics there is the same problem with expectation value of E, a there is an uncertainty principle as well ΔE Δt ≥ 2  only if we are considering a stationary state, i.e. when there is no time dependency and no Δt, no such uncertainty, we will have sharp values for energy so what we need here are operators operator is a mathematical concept telling us what to do with the operand that follows it e.g. )( xi   (x2 t) means that one has to take the partial x derivate of the function (x2 t) and multiply it with i  so )( xi   (x2 t) = xt i 2  what is )( xi   (cos x ) = - )( i  sin x docsity.com 31 entities for which we have operators are called observables as they have physical meaning and can be observed (although subject to the uncertainty principle) now )( xi   is actually the momentum operator [p] that gives us the expectation value of the momentum <p>        dxtx xi txp ),(*),(  note that the order of factors is important, there is only one way of doing it correctly similarly          dxtx xixi txp )},({*),(2  first one operator is applied to its operand yielding the operand for the second operator (which will again stand to the right of the operator) for example: calculate the expectation value <p> for the ground state wave function in the infinite square well, docsity.com 32 we know it is a stationary state (standing wave) so it is time independent, we know the particle is trapped in the well, so it is never outside, so we can restrict the integral to the well the (normalized and time independent) wave function for that state is  L x L  sin 2 as there are no i(s) in it the conjugate complex of that functions * is also L x L  sin 2 so     L dx L x LxiL x L p 0 )sin( 2 ))(sin( 2   simplifies to 0)cos()sin( 2 0   L dx L x L x LLi p  this is of course because sin x = 0 at the nodes!! so the expectation value <p> is zero, what does it mean, simply the particle is just as likely moving to the right as it is moving to the left, the arithmetic mean must, thus, give zero generally operators are written in sharp straight brackets, i.e. [p] or with a “caret”, i.e. p̂ as there are many more observables, entities with physical meaning that are allowed to be known by the uncertainty principle, there are many more operators that give us expectation values of these observables docsity.com 35 the “pretty compact” version of the Schrödinger equation Eigenvalues and Eigenfunctions for simplicity we deal here only with time independent wave functions, if something is in a steady state the uncertainty principle ΔE Δt ≥ 2  does not apply, there is all the time in the world, so the energy has settled into a stationary state an exact value, it is only when it jumps between stationary states that there is a Δt again, and with it an uncertainty of energy that shows up in a widths of a spectral line “eigen” is German and means self, so what is meant here is combinations of real numbers (values) and functions that are equivalent to the action of an operator on these functions. (if you know about systems of linear equations and matrix representation, you have the very same things, combinations of vectors with values that are “self” solutions to the problem, was invented in Göttingen by Jordan and Hilbert, who told Born und Heisenberg about it, …) mathematical definition [G]Ψn = gnΨn where e.g. operator 2 2 dx d has eigen function xe2 what is the eigenvalue to this functions and operator xxxx ee dx d e dx d dx d e dx d 2222 2 2 42)(  docsity.com 36 as the eigen function was just xe2 the (generally real) number 4 is for that function exactly equivalent to the operator 2 2 dx d back to physics eigenfuctions are here again solutions to the Schrödinger equation, we deal only with time independent form if we are looking at stationary states such as in the case of a particle in a box we get sharp values for certain operators such as the total energy operator, so there is no expectation value for energy as there is no arithmetic mean of measurements on many identical particles, if we are dealing with an eigenvalue/eigenvector problem the eigenvalue is just one value, e.g. a definitive energy for every eigenfunction, , eigenfunction and eigenvectors are refereeing to a set of quantum numbers that are integers from particle in an infinite square well, you know, energy comes only in discrete values, En, these are the eigenvalues to the eigenfunctions n so time independent Schrödinger equation can be written most compactly [H] n = En n for correct description of atoms we will have a second set of eigenvalues and eigenfuctions, because angular momentum is in nature also quantized not only energy, so there will be another quantum number actually there will be two more sets of eigenfunctions and eigenvalues as a state of an electron in an atom is described by 4 quantum numbers docsity.com 37 Model: Particle in a box with infinitely large potential barriers, infinite square well infinite barriers, the particle is always confined, never outside exercise: deriving form of the wave function under the boundary conditions it’s a stationary state, so we use time independent Schrödinger equation 0)()}({ 2)( 22 2  xxUE m dx xd   so 0)( x outside the box inside box U(x) = 0 )( )( 0 )(2)( 2 2 2 22 2 xk dx xdxmE dx xd    solutions of this ordinary partial equation are sin kx and cos kx so most general solution is kxBkxAx cossin)(  inside the box 0 < x <L docsity.com 40 for 2 3 )(x the particle is never at 1/3 and 2/3 L how does the particle get over these points ??? well it’s particle-wave duality not just a particle of which we have an intuitive idea how it is supposed to move – something we can’t grasp with or brain having evolved over time looking only at classical phenomena – and of course, there is no path the wave-particle could follow so far we only looked at 2)(xn to make calculations of actual probabilities, we need to normalize the wave functions dx L xn AdxxP LL )(sin1)( 0 22 0 2    there is a trigonometric identity: 2 sin2 Θ = 1 – cos2Θ so we get dx L xn Adx L xn A LL ) 2 cos(1 2 1 )(sin1 0 2 0 22    now cos L x2 integrates to sin L x2 which is zero at x = 0 and x = L docsity.com 41   L dx 0 1 is just L – 0 = L so 1 = 2 2LA from which we get the normalization factor A = L 2 our wave functions ready to be evaluated for probability densities in % are, thus, )sin( 2 )( L xn L xn   with n = 1, 2, 3, … one more thing on the infinite square well the lowest energy state is given by n = 1 what would happen is n = 0 ? 0 )( 2 2  dx xd  is to be solved ! solution is wave function docsity.com 42 BAxboxinside )_( this wave function has to be zero at x and L, this requires both A and B to be zero !!!_0)_( everywhereBAxboxinside  is we look at the probability of finding this particle   L particlesuchnoistheredxtx 0 2 ____%00),( so n = 0 and E = 0 are not possible !!! returning to wave function for particle in infinitely deep box / infinite square well L xn Axn  sin)(  for each of the quantum numbers, 1, 2, 3, … there is a specific wave function describing everything that is permitted to be known by the uncertainty principle one last thing on the infinite square well with assuming impenetrable walls of infinite height, we actually violated one of the boundary conditions for physical meaningful wave functions: docsity.com 45 speed corresponding to the diminished total energy kinetic energy KE = E-PE= E-U > 0 but if total E is smaller than the height of the potential energy walls of the well, i.e. E-PE = E-U < 0, there is no kinetic energy left to roam freely, so classical particle can’t be outside the wall and moving, it is trapped forever in 0 < x < L In quantum mechanics, because of the condition x tx   ),( or dx d x x     )( must be continuous, i.e. slopes must be continuous, a particle leaks out into the potential walls !!! This is because Ψ is never zero outside the well, so the probability of finding the particle there Ψ2 is not zero either, so the particle is actually there !!! so lets look at the parts of the wave function that penetrated into potential walls solutions to the (time independent) Schrödinger equation x I Cex  )( for x < 0, section I, where C is a constant we can use for fit to the second segment docsity.com 46 x III Dex  )( for x > L, section III, where D is a constant we can use for fit to the second segment and 2 )(2  EUm   is a positive constant, as U is taken to be zero in the well and some positive value outside the well so we have an exponential decay, that’s pretty fast, from of the constant α, we can see that the heavier the particle is and/or the larger the difference U – E, (i.e. the larger – KE of the bound state) the faster )(x decays in the walls, if the walls are infinitely wide, the wave function decays to zero general solution for region II is x mE Bx mE AxII  2 cos 2 sin)(  as U(x) = 0 in the well and k =  mE2 as usual but as sin (0) = 0 we have to set A = 0 and can only use the second part with “cos” functions that “cos” function’ has to match with the functions for section I and III at x = 0 and L and its first derivate with respect to x has to match as well (smoothness condition of wave functions that describes real particles) for x = 0 and L docsity.com 47 this can only be achieved for certain energy levels En which are all smaller that their counterparts from the infinite square well of the same widths 2 2222 __inf 82 mL hn m k E wellsquareiniten   example say we have n = 1 and max 2 1 )( IIII L   , i.e. II at the wall is only half the maximal value of this function at the center so we can say cos kL = 1/2 kL = 60° = π/3 k = π/3L k =  12mE from above, resolved for E1 2 22 2 22 __ 1 18 1 18 mL h mL E wellsquarefinite   for that particular scenario exactly 2.25 times smaller due to the particular height and widths of the square potential well, i.e. U and L, that results in the value of the wave function at L just being half the maximum value (which we have in the center of the well) on can also see form the graph that the wavelength that fit into a finite square well (with leakage into the barriers) are somewhat docsity.com 50 deposition, surface diffusion, interdiffusion are random events, smaller band gap semiconductor (alloy) usually larger lattice constant, (one way of self- assembly, resulting in “cake with raisins”) epitaxially grown quantum dots compressively strained and possess random distribution of atoms → ordinarily strained QDs docsity.com 51 over time, atomic ordering as observed by transmission electron microscopy Figure 1: (In,Ga)Sb agglomerates in GaSb matrix; (a) [001] plan-view HRTEM image which was recorded at 500 ºC and after a thermal treatment in the electron microscope at temperatures of the order of magnitude of the growth temperature for several hours, ref. 14, suggesting that the transformed structure of this QD rather than its original sphalerite prototype structure is thermodynamically stable; (b) <110> cross section Z-contrast STEM image, showing a QD with atomic ordering in every forth ± (002) plane; power spectra as inserts; c) [001] atomic resolution Z-contrast STEM images of structurally transformed In(As,Sb) QDs in InAs matrix. (from one of my papers, see web pages, if interested) PSU’s new more than $ 1,000,000 microscope, would itself not be possible without modern physics and Schrödinger’s equation, because design of electromagnetic lenses is quite involved, ray optics does not do the job, it’s too crude an approximation 5 nm c 5 nm a b docsity.com 52 Quantum states of Harmonic Oscillator, a very useful approximation say a particle is in potential well subject to a linear restoring force xKF   with force constant K corresponding potential energy is U(x) = ½ K x2 we had something like it as a mass on a spring, a very long pendulum with a small elongation, …. anything that is limited to small excursions (x) around a stable equilibrium position near the stable equilibrium position, say x = a, the potential energy can be approximated by a parabola: U(x) = U(a) + ½K(x-a)2 under the condition that the curvature of that parable must match that of U(x) at the point x = a, this condition is fulfilled if K = 2 2 dx Ud │a and U(a) is potential energy in equilibrium position a, which we can of course define as the zero level from which all potential energies are measured, analogously we can use coordinate shift and define a = 0 on the x axis with these two conventions we have U(x) = U(a) + ½K(x-a)2 = 0 + ½K(x-0)2 = ½K x2 docsity.com 55 that leakage into the barriers is shown below as well for 2 n so the particle is actually at these positions with certain probabilities for large quantum numbers classical physics (dashed lines) and quantum physics (curves) give corresponding probabilities of finding the particle docsity.com 56 Transitions between states and selection rules energy levels revealed when system makes transitions, either to a higher energy state as a result of excitation (absorption of energy) or to a lower energy state as a result of relaxation (de-excitation, emission of energy , if it is an electron this is usually electromagnetic radiation) form classical physics: if a charge q is accelerated, it radiated electromagnetic radiation, remember that’s how X-rays are produced, if a charge oscillates, the radiation is of the same frequency as the oscillation if we have charged particle (charge q), we define charge density nnn q  * this quantity is time independent, stationary state, i.e. does not radiate, quantum mechanical explanation of Bohr’s postulate, let’s say n is the ground state with this wave function n goes a certain (eigen-value) energy En , as long as the charged particle is in this energy state it does not radiate, it does neither lose nor gain energy docsity.com 57 say it gained just the right amount of energy to go to an excited state, this means eigen-value (energy) and wave function eigen- function change let’s now consider how the particle returns to the ground state only if a transition form one wave function (m) to another wave function (n) is made, the energy changes ΔE = Em –En from one definitive value (excited stationary state, e.g. m) to the other definitive value (relaxed stationary state, e.g. n), Em > En as wave function for a particle that can make a transition, we need time dependent wave function Ψ(x,t), as it is two different states m and n, we have a superposition Ψm,n(x,t) = a Ψm(x,t) + b Ψn(x,t) initially say a = 1, b = 0, electron in excited state, m while in transition a < 1, b <1, electron is oscillating between states finally a = 0, b = 1, electron in relaxed state, n we can calculate frequency of this oscillation expectation value that a particle can be in a transition is     dxtxtxxx nm ),(*),( docsity.com