1. Introduction to quantum mechanics, Summaries of Quantum Mechanics

Download 1. Introduction to quantum mechanics and more Summaries Quantum Mechanics in PDF only on Docsity! TFY4215/FY1006 — Lecturenotes 1 1 Lecture notes 1 1. Introduction to quantum mechanics Chapter 1 of the this course — Introduction to quantum mechanics — is covered by the present notes, “Lecture notes 1”. In this chapter, we consider some of the milestones in the historical development of quantum mechanics. Chapter 1 in Bransden & Joachain gives a more comprehensive review and should be consulted for more details. Quantum mechanics — to put it a bit too simply — is our theory of the submicroscopic world. This world is not so easily accessible for observations as macroscopic physics. This is why the discovery of most of the quantum phenomena, and their explanation, came as late as in the beginning of the last century. At the beginning of the quantum era — in the year 1900 — classical physics appeared to be an almost complete physical theory: • Newton’s mechanics had been established for more than two hundred years. • The wave nature of light was known from the beginning of the 19th century, based on diffraction experiments carried out by Young and Fresnel, among others. • The relation between electric and magnetic phenomena had been clarified, particularly through the work of Faraday and Maxwell, culminating with Maxwell’s equations in the 1860’s. Maxwell’s prediction of electromagnetic waves in 1865, and the experi- mental verification by Hertz in 1887, led to the understanding that also light is an electromagnetic wave phenomenon. • In addition, thermodynamics and statistical mechanics were developed during the 19th century. At the end of the century, these theories were by many people considered to give an almost complete description of the physical world. A few years later, it was clear that this was far from the truth: • Newton’s laws turned out to be a limiting case (for small velocities) of Einstein’s special theory of relativity. • Newton’s theory of graviatation was extended to (replaced by) Einstein’s general theory of relativity. • The discovery that the atom consists of a very small and compact nucleus surrounded by pointlike electrons, and certain properties of the interaction between atoms and radiation, led to the development of the theory of quantum mechanics. TFY4215/FY1006 — Lecturenotes 1 2 These theories and the development that followed later in the twentieth century have rev- olutionized our physical picture of the world, and have led to a number of advances in our understanding of atomic physics, nuclear physics, particle physics, astrophysics and cosmol- ogy, and also within chemistry, biophysics and biology and of course also within solid-state physics, including applications in technology. Quantum mechanics contains a number of elements which are rather revolutionary com- pared to classical mechanics and our classical way of thinking. Therefore we may state that the “birth” of this theory was rather “painful” and took a long time, starting in 1900 and cul- minating in 1925, when there was a sudden breakthrough. After that, it took only a couple of years to obtain a fully developed theory. The reason for the difficult birth lies in the fact that this theory not only disagrees with classical physics, but also is counter-intuitive; it does not agree with the physical intuition which we develop based on our every-day experiences with macroscopic phenomena. In what follows we shall review the main events in the development of quantum mechanics, and we shall see how the physics community gradually was forced to accept this revolutionary physical theory: 1.1 Planck’s radiation law (1900) The discovery of the radiation law is discussed in section 1.1 in Bransden & Joachain. (See also 1.2 in Hemmer.) As explained there, a small opening to a cavity kept at a temperature T will radiate as a black-body surface at this temperature: Planck found an empirical formula for the spectral emittance (emitted energy flux per unit area and unit frequancy) from a black surface, based on experimental results that had just been obtained, I(ν, T ) = 2πν2 c2 hν ehν/kBT − 1 . (T1.1) At equilibrium, and supposing that also the inside walls of the cavity are black, there must be a relation between this spectral emittance from the walls and the spectral energy density inside the cavity, u(ν, T ). It can be shown that I(ν, T ) = u(ν, T ) · c/4, so that u(ν, T ) = 4 c I(ν, T ) = 8πν2 c3 hν ehν/kBT − 1 . (T1.2) These two formulae are two equivalent ways of expressing Planck’s radiation law. What was really remarkable with Planck’s empirical formula, was that it was a very good fit to the experimental frequency distribution of the radiation, not only for one particular TFY4215/FY1006 — Lecturenotes 1 5 Efield = c|pfield|. The fact that electromagnetic radiation also contains a momentum |p| = E/c was verified experimentally by E.F. Nichols and G.F. Hull in 1903. If the electromagnetic pulse is to be regarded as a collection of light quanta (photons), then these quanta must also carry a momentum, in addition to the energy. (ii) Secondly, it follows from relativistic particle kinematics, E = mc2√ 1− v2/c2 , p = mv√ 1− v2/c2 =⇒ E p = c2 v , E = √ m2c4 + c2p2 , (T1.4) that E/p→ c when m→ 0. The reason that these hypotheses were met with such a large skepticism was that they seemed to be in contradiction with all the knowledge that had been gained about the wave nature of light. How could light be both a wave and a stream of particles? The natural answer to this question was to discard Einstein’s photon hypothesis, and this was the ruling opinion for several years after Millikan’s work. This easy way out of the problem was finally blocked in 1923, through the discovery of the Compton effect. A small exercise: From Einstein’s photon hypothesis and Planck’s radiation law (T1.2) it follows that the number of photons per unit volume and frequency in an oven at temperature T is n(ν, T ) = u(ν, T )/hν. Show that the number of photons per unit volume, n(T ) = ∫∞ 0 n(ν, T )dν, is n(T ) = 2.029 · 107 T3 m−3 K−3, and that the average energy of these photons is E = u(T ) n(T ) = 2.701 kB T, given that ∫∞ 0 x2(ex − 1)−1dx = 2.4041. Another exercise: Show that a photon with wavelength λ has the energy Eγ = 1.24 · 10−6 eVm λ . (T1.5) [Hint: At the end of these notes you will find Planck’s constant in units of both eVs and Js=Nms. Presently, you will find that is is most practical to use h = 4.136 · 10−15 eVs.] The work function W (that is, the binding energy for the most loosely bound electrons) for cesium (Cs) and gold (Au) are respectively 1.9 eV and 4.8 eV. Decide whether photoelectric emission is possible when visible light (0.4 µm < λ < 0.7 µm) is incident on these two metals. [Hint: Show that the photon energies for visible light lie in the range 1.77 eV <≈ hν <≈ 3.1 eV.] Another small exercise: A 25-watt light bulb emits around 10% of its effect in the visible region. Estimate roughly the number of photons emitted per second in this λ-region. With night adaptation the eye can see a light source which is so weak that around 5 photons pass the pupil per second. Assume a pupil area of ∼0.5 cm2. How far away can the bulb be seen? [Hint: Assume that there is no absorption between the bulb and the eye. Since we want only an estimate, you can use a photon energy in the middle of the visible spectrum, λ = 0.55µm. Answer: ∼ 2500 km.] TFY4215/FY1006 — Lecturenotes 1 6 1.3 Compton’s exsperiment (1922–23) The figure shows a sketch of Compton’s exsperiment. Monochromatic X-rays were scattered on a piece of graphite. The scattered radiation was observed at an angle θ which could be varied. The wavelength of this secondary radiation was observed, using so-called Bragg scattering on a calcite crystal. The result was a sensation: Compton observed secondary radiation with a wavelength λ′ larger than the primary wavelength λ, and the difference λ′ − λ (the Compton shift) depended on the scattering angle θ. This phenomenon could not be understood in terms of classical electromagnetic theory: If the primary radiation is considered as an electromagnetic wave with wavelength λ and frquency ν = c/λ, the oscillating electric field in the wave will cause the graphite electrons to oscillate with the same frequency ν. According to Maxwell’s theory, these oscillating electrons will radiate an electromagnetic wave with the same frequency and hence the same wavelength as that of the primary wave. (This is analogous to the way an electric antenna works.) However, this was not what Compton observed. It should be noticed that the measurement of λ′, by scattering the secondary radiation on the calcite crystal, is based on the wave nature of the radiation. Compton’s result, however, can only be understood by accepting Einstein’s hypothesis: Let us consider the process as a collision between a single particle-like light quantum (photon) with energy Eγ = hν = hc/λ and momentum pγ and a single electron which is approximately free, being at rest before the collision. Thus, before the collision the electron has the momentum p = 0 and the energy E = mec 2. Since the secondary photon leaves the collision with a momentum p′γ in the direction θ (the so-called scattering angle), we see that the electron necessarily must receive a momentum p′ = pγ − p′γ. TFY4215/FY1006 — Lecturenotes 1 7 Thus the electron is sent away (“recoils”), with the momentum p′ and the energy E ′ =√ m2 ec 4 + c2(p′)2. This increase of the electron energy of course comes at the expence of the photon energy; the secondary photon must be emitted with a lower energy and a larger wavelength than the primary one, which is precisely what Compton observed. This increase of the wavelength can be calculated, using the relativistic conservation laws for energy and momentum: hc λ +mec 2 = hc λ′ + E ′ [(E ′)2 = (cp′)2 +m2 ec 4] pγ + 0 = p′γ + p′. Using these formulae, it is (in principle) a simple matter to show that 1 λ′ = λ+ h mec (1− cos θ), (T1.6) which is known as the Compton relation. We notice that the Compton shift, λ′ − λ = h mec (1− cos θ) ≡ λC(1− cos θ), (T1.7) is independent of the primary wavelength, but depends on the scattering angle θ. 2 After doing this calculation and checking that his experiments agreed with the resulting formula Compton could draw the following dramatic conclusion: “The present theory depends essentially upon the assumption that each electron which is effective in the scattering scatters a complete quantum (photon). It involves also the 1It is easy to derive the Compton relation. We start by solving the above equations for E′ and p′, and then calculate (E′/c)2 − (p′)2, which is equal to (mec) 2: E′2 c2 − (p′)2 = [ mec+ h( 1 λ − 1 λ′ ) ]2 − (pγ − p′γ)2 = m2 ec 2. By writing out the squares (where pγ ·p′γ = pγpγ′ cos θ) and using that pγ = h/λ and p′γ = h/λ′ you will find that 2mehc( 1 λ − 1 λ′ ) + 2h2 λλ′ (cos θ − 1) = 0. The Compton relation follows by multiplying this equation by λλ′/(2mehc). 2The quantity h/(mec) ≡ λC is characteristic for Compton scattering on electrons, and is known as the Compton wavelength (for electrons). Its numerical value is λC ≡ h mec = 2.426 310 238(16)× 10−12 m. TFY4215/FY1006 — Lecturenotes 1 10 (iii) Planck’s quantization condition for the energy transfer between matter and radiation: The energy is radiated or absorbed in quanta hν = ∆E, where ν = c/λ is the frequency of the radiation and ∆E is the energy loss or gain for matter. B. Bohr’s model. Bohr built his model on the following ideas: 1. The idea of stationary states: (T1.10) Bohr assumes that the atom can exist in a discrete set of states of motion, each with a well-defined energy. This implies, firstly, a quantization principle: The energy can only take a discrete set of values. This of course contradicts one of the basic principles of classical physics: In a one-electron atom, e.g., the electron is moving in a 1/r2 force field, and just like for a satellite moving in the field of the earth, a continuum of energies should then be allowed. Secondly, Bohr’s assumption breaks with classical physics on another point. Due to the 1/r2 force, the electron is being accelerated the whole time. According to Maxwell’s equations, it should then radiate energy continuously, and thus gradually loose its energy, spiralling towards the nucleus. Bohr’s assumption implies that no radiation is emitted when the atom is in one of its stationary states, so that the energy is kept constant in such a state. (This is part of the meaning of the word “stationary”.) 2. The idea of the quantum leap (or jump): (T1.11) With the assumption above, Bohr could interprete the discrete sets of spectral lines as the result of the atoms making sudden quantum leaps, jumping between discrete energy levels, and emitting an energy quantum hν = ∆E, where ∆E is the difference between the energies before and after the jump, in agreement with Planck’s assumption. For hydrogen, with the observed lines (T1.9), these energy quanta would be hν = hc λ = hcR∞ ( 1 n2 − 1 m2 ) , n = 1, 2, 3, · · · , m = n+ 1, n+ 2, n+ 3, · · · . Bohr’s assupmtion thus corresponds to the following experimental values for the discrete energy levels of hydrogen: En = −hcR∞ 1 n2 , n = 1, 2, · · · . Here the numerical value of the constant hcR∞ is approximately 13.6 eV. TFY4215/FY1006 — Lecturenotes 1 11 The two ideas above, of stationary states and quantum jumps, have turned out to survive; they are central aspects of quantum mechanics. However, in order to explain why the atomic energies are discrete, and in order to be able to predict the size of these energies, Bohr introduced two additional assumptions, which turned out to be wrong. 3. The electron in a hydrogen atom is moving in circular orbits, respecting Newton’s laws. In the hydrogen atom, the potential energy of the electron in the field of the proton is V (r) = −e e 4πε0r ≡ −k r , with k ≡ e2 4πε0 . The force on the electron then is F = −∇V (r) = − e2 4πε0r2 êr where the minus sign means that the force is directed opposite to the unit vector êr, that is, towards the proton. For a circular motion, it then follows from Newton’s second law that the centripetal acceleration is a = v2 r = |F| me = e2 4πε0mer2 ≡ k mer2 . It is then straghtforward to show that the velocity v, the angular momentum L = mvr and the kinetic and total energies K and E are proportional to respectively r−1/2, r1/2 and r−1 : 5 v(r) = √ k mer ∝ r−1/2, L = mevr = √ kmer ∝ r1/2, K(r) = 1 2 mev 2 = 1 2 k r ∝ 1/r, E(r) = K(r) + V (r) ( = 1 2 V (r) = −K(r) ) = −1 2 k r ∝ 1/r. This classical calculation allows all values (> 0) for the orbital radius r and (< 0) for the energy E. In order to obtain the all-important energy quantization, Bohr therefore needed one more (new and revolutionary) assumption. After some trial and error arrived at the following one: 4. The angular momentum of the electron must be discretized, and can only be a multiple of h/2π ≡ h̄, that is, 6 L = mvr = nh̄, n = 1, 2, · · · . According to the relations above, such a quantized angular momentum corresponds to quan- tized values also for the radius and the energy: L = √ kmer = nh̄ =⇒ r = rn = n2h̄2 kme = n2 4πε0h̄ 2 e2me =⇒ 5Note that with another definition of the coefficient k, these calculations also hold for a satellite moving in a circular orbit around the earth. 6In modern quantum mechanics, the quantity h/2π ≡ h̄ (“h-bar”) occurs much more frequently than Planck’s constant h itself. TFY4215/FY1006 — Lecturenotes 1 12 En = − k 2rn = −k 2me 2h̄2 1 n2 = −1 2 ( e2 4πε0h̄c )2 mec 2 1 n2 , n = 1, 2, · · · . In the formula for the energies, the quantity mec 2 = 510 998.910(13) eV ≈ 0.511 MeV (electron rest energy) has the dimension of an energy. This means that the quantity in parantheses must be a dimensionless constant. This constant plays a very important role in quantum physics, and is called the fine-structure constant: α ≡ e2 4πε0h̄c = 1 137.035 999 679(94) ( fine-structure constant ) . (T1.12) Inserting numbers, we now find that the lowest energy according to Bohr’s theoretical model is E1 ≈ −13.6 eV, corresponding precisely to the lowest experimental energy (the “ground-state” energy). Also the smallest radius (of the ground state) does in fact play an important role in quantum mechanics (as we shall see later) and is called the Bohr radius: a0 ≡ 4πε0h̄ 2 e2me = 0.529 177 208 59(36)× 10−10m (Bohr radius). (T1.13) What was very sensational with Bohr’s model was that he was able to “explain” the 1/n2 dependence of the experimental hydrogen energies without introducing any new parameter — Planck’s constant did the job not only for light, as Planck and Einstein had found, but also for the atom. When Einstein was told about this new theory, he exclaimed: “Then this is one of the greatest discoveries”. (Since then we have learnt that Planck’s constant always enters in quantum mechanics, whether we consider particles or radiation, or both for that matter.) Now we must hasten to add that in spite of this success, assumptions number 3 and 4 in Bohr’s theory have turned out to be wrong. It is correct that the angular momentum is quantized, and it even turns out that h̄ plays an important role in this quantization. However, the ground state turns out to have zero angular momentum (not h̄ as in Bohr’s model). What is even more important is that the notion of classical orbits for the elctrons has to be discarded. Thus, to explain the stationary states of e.g. the hydrogen atom, we are today using a quantum-mechanical theory where the wave nature of the particles plays the central role. The classical-mechanical way of thinking (which is so tempting for all of us), where the particle has a well-defined orbit r = r(t), is wrong when it comes to phenomena on the atomic or sub- atomic scale. TFY4215/FY1006 — Lecturenotes 1 15 averaged over time. This wave function is a solution of the wave equation for surface waves in water. This wave equation for surface waves is approximately linear for small wave heights. Therefore, if we let both 1 and 2 be open, the resulting wave height (wave function) at the point P will be given approximately by the sum of h1 and h2 (cf the superposition principle): h12 ≈ h1 + h2 = A(θ)[cos(kr1 − ωt) + cos(kr2 − ωt)]. As you know, we then get an interference pattern behind the dam, with constructive inter- ference (h12 = 2h1 and I12 = 4I1) where the path difference between the two waves is an integer number of wavelengths, r2 − r1 = d sin θ = nλ, and destructive interference (h12 = 0 and I12 = 0) where d sin θ = (n+ 1 2 )λ. (See the figure.) 1.6.b A double-slit experiment with monochromatic light (Young, 1802) With b << λ and 1 or 2 open we again get slowly varying intensity distribution Ii(θ) (see the dashed line in the figure). With both 1 and 2 open, we get the same behaviour of I12 as for the water waves (solid line in the figure). Based on such an experiment, Young concluded in 1802 that light must be a wave phenomenon. This was a decisive blow against Newton’s TFY4215/FY1006 — Lecturenotes 1 16 corpuscular theory of light, which had been dominating for a century. We should note that Young was not able to decide what kind of wave this was. It took sixty years before it was understood by Maxwell that the interference and diffraction effects observed by Young and Fresnel (and others) could be understood in terms of interference of electromagnetic waves, and that the wave functions which then interfere are classical electromagnetic fields. Long after Young (well into the last century in fact) it was discovered that this kind of experiment can also be used to demonstrate the non-classical particle nature of light: If the intensity of the incident light is gradually decreased, the intensity distribution I12(θ) on the screen of course also decreases in the same way, but the form of this distribution is the same, as long as we are able to observe it. At some point it becomes invisible to our eyes, but it can still be observed, e.g. by the use of a photographic plate and sufficiently long exposure time. (This technique is used in astronomy to observe very distant and faint light sources.) A single photon can then darken a point on the plate. Such a system works even if the intensity is so low that we register only one photon per day (e.g.). This way, the particle nature of light becomes very evident. The fact that a single photon is registered at some definite point on the plate (instead of being “smeared out” in any manner), in a way does not agree with the classical wave theory, which predicts an intensity distribution I12(θ). This is an unavoidable consequence of the particle nature of light, which clearly is a quantum (as opposed to classical) phenomenon. Thus wave theory can not predict what happens with a single photon (or with two, or with three,...). We need to collect a large number of photons before we begin to see that the interference pattern emerges. Thus we can conclude that: The classical interference pattern I12(θ) is valid in the sense that it gives the probability distribution of the photons on the screen (suitably normalized). The probability of observing a sin- gle photon at an angle θ then is P (θ) ∝ I12(θ). (T1.15) With this probability interpretation of the classical wave theory, there is no contradiction between the wave properties and the particle properties of light. Both properties are present. The wave property determines the interference (both here and in scattering on a grating or on a crystal). The particle property shows up in the detection, and also in e.g. Compton scattering. This probability theory, as already stated, does not allow us to predict what happens with a single photon. And there is no other theory that can help us with this. Thus the photons, which do not differ in any way, experience unpredictable and different fates behind the double-split screen. This element of arbitrariness or unpredictability is a central feature of quantum physics, which most of us have some difficulties in accepting. 7 This has been the source of much debate through the years. Einstein, for example, who was one of the fathers of quantum theory, could never reconcile himself with this feature of 7The unpredictability feature of quantum physics is also evident if we observe an ensemble of α-radioactive nuclei, e.g. of the isotope 226Ra. These unstable nuclei, which decay by emitting an α particle (helium nucleus), 226 88Ra =⇒ 222 86Rn + 4 2He, TFY4215/FY1006 — Lecturenotes 1 17 quantum mechanics. He is often cited for the following statement: “God does not throw dice”. There have been many attempt to modify the theory to remove the unpredictability. (Keywords here are “hidden variables” and Bell’s inequalities.) So far, it seems that this element of quantum mechanics can not be removed. It is something we have to live with. 1.6.c A double-slit experiment with electrons The double-slit experiment with electrons played absolutely no role in the development of quantum mechanics, but is included here because it shows the wave nature of particles very clearly. 8 In order to get an interference pattern for visible light one needs to work with a slit width comparable with the wavelength, that is, <∼ 1 micrometer. If one wants to make a similar experiment with particles, much smaller slit widths are required than for light, and similarly for the distance d between the slits. This is because the de Broglie wavelength for particles are very small. Thus, for non-relativistic electrons we find from de Broglie’s formula a wavelength λ = h p = h√ 2meEkin = h mec √ mec2 2Ekin = λC √ 511000 eV 2Ekin , that is, λ = h p = 12.264 Å √ 1 eV Ekin ( de Broglie wavelength for non-rel. electrons ) . (T1.16) Even with a kinetic energy as small as 1 eV, the de Broglie wavelength thus is as small as 12 Å. So slow electrons are very difficult to handle experimentally. In practice one must therefore work with faster electrons, with wavelengths which are even smaller. This is why the first double-slit experiment with electrons that I have heard about was done as late as in 1961 (C. Jönsson). are completely identical. Still they do not decay simultaneously. We can measure the average lifetime, which is 1620 years, and we can observe the distribution of the lifetimes. But we can not predict how long a single nucleus will survive. Starting out with a large number, we can state that the expected number (surviving) after 1620 years is reduced by a factor 1/e. After another 1620 years the number is reduced by another factor 1/e. In fact at any point in time, the laverage lifetime of those who have survived is 1620 years. Thus, there is no difference between those who died early and those who survive for a long time. This of course is not very easy to understand intuitively. 8As mentioned earlier, the wave nature of electrons was demonstrated in experiments carried out in 1926–27 by Davisson & Germer and by G.P. Thomson. TFY4215/FY1006 — Lecturenotes 1 20 Does each single electron pass through one of the slits or through both? Here it is difficult to give a clear answer. What can be stated with certainty, is that both slits have to be open if we want to observe the interference pattern I12(θ) ∝ |Ψ12|2. It is also a fact that nobody has observed “half” electrons (or “half” photons). Thus there is no indication that the electron can be divided into two parts, one part travelling through slit 1 and the other through slit 2. We can of course modify our experiment so that we monitor whatever passes through slit 1 and 2. Experiments (with one electron at a time) then show that the electron either passes through 1 or 2, not through both at the same time. However, with an experiment which allows such a monitoring, it turns out that there is no interference pattern; we get the same smooth distribution as with only one open slit. Thus we have to conclude that in an experiment where the interference pattern is observed, we are not able to answer the question about which slit the electron passed. This is only one of many examples where questions which seem to be perfectly reasonable according to classical physics and our “macroscopic” way of thinking, can not be answered in the submicroscopic and quantum-mechanical world. A small exercise: a. Calculate the de Broglie wavelength λ for electrons with a kinetic energy of 200 eV. [Answer: 0.8762 Å.] b. Suppose that electrons in an old-fashioned TV tube have a kinetic energy 20 keV, that is a factor 100 higher than in the previous question. What is then the (non-relativistic result for) the wavelength in this case? [Answer: A factor 10 smaller than in the previous case.] c. Use the relativistic formula p = √ E2/c2 −m2 ec 2 (where E = Ekin+mec 2) to show that the ratio between the relativistic (i.e. correct) value of the momentum p and the non-relativistic one (p = √ 2meEkin) is √ 1 + Ekin/2mec2. How big is the relative error of the wavelength in question b? [Answer: The correct value for the wavelength is about 1 percent smaller than the value found in question b.] 1.7 Wave equations. Motivation for Schrödinger’s equa- tion Surface waves in water are desribed by a certain wave equation. Radio waves, light waves and other electromagnetic waves are in general described by the so-called classical wave equation, which can be deduced from Maxwell’s equations. It is then natural to pose the following question (as Schrödinger did in 1925): What is the wave equation that has de Brogle’s harmonic waves as solutions? TFY4215/FY1006 — Lecturenotes 1 21 1.7.a Free particle, with sharp momentum p = êxpx + êypy + êxpz and energy E = p2/2m For a free particle, with well-defined momentum p and energy E = p2/2m — e.g. the elctrons incident on the two-slit screen — we shall see that it is easy to find a suitable wave equation. We saw that de Broglie’s hypothesis, k = p/h̄ and ω = E/h̄, (de Broglie’s hypothesis) (T1.18) corresponds to a plane, harmonic wave with the form Ψ(r, t) ∝ ei(k·r−ωt) = ei(p·r−Et)/h̄ = ei(pxx+pyy+pzz−Et)/h̄. Like Schrödinger we can ask ourselves: What is the simplest differential equation that is satisfied by this plane wave Ψ? To find the answer, we shall introduce a few mathematical operators which turn out to play central roles in quantum mechanics, namely p̂x ≡ h̄ i ∂ ∂x , p̂y ≡ h̄ i ∂ ∂y and p̂z ≡ h̄ i ∂ ∂z . (T1.19) A small exercise: A) When an operator acts on a function, the result usually is a new func- tion. As an example, you should let the operator ∂/∂x act on the functions exp(−x2), sin kx and cos kx. B) In some cases you will find that an operator acting on some function results in the same function multiplied by a constant. This function is then called an eigenfunction of the operator in question, and the constant is called an eigen- value of the operator. As an example, show that exp(ikx)[= cos kx+ i sin kx] is an eigenfunction of the operator p̂x = (h̄/i)∂/∂x, and find the eigenvalue. C) Show that the function e−x 2/2 is an eigenfuntioon of the operator ĥ = −1 2 d2/dx2 + 1 2 x2 and determine the eigenvalue. Applying the operator p̂x on the function Ψ, we see that p̂xΨ = h̄ i ∂ ∂x ei(pxx+pyy+pzz−Et)/h̄ = h̄ i ipx h̄ Ψ = pxΨ. (T1.20) This equation is a so-called eigenvalue equation, on the form Operator ·Ψ = eigenvalue ·Ψ. (T1.21) It states that the function Ψ is an eigenfunction of the operator p̂x with an eigenvalue equal to the momentum component px. In the same manner, we find that p̂yΨ = pyΨ, p̂zΨ = pzΨ and p̂Ψ = pΨ, where 11 p̂ ≡ êx h̄ i ∂ ∂x + êy h̄ i ∂ ∂y + êz h̄ i ∂ ∂z ≡ h̄ i ∇ and p = êxpx + êypy + êzpz. (T1.22) 11Here, êx etc are unit vectors, while the “hat”̂over p means that p̂ is an operator. TFY4215/FY1006 — Lecturenotes 1 22 These eigenvalue equations play a very important role in quantum mechanics, and you should bear them in mind: de Brogle’s abstarct plane harmonic wave Ψ, which “describes” a particle (or particles) with sharply defined momentum p, with com- ponents px, py, pz, is an eigenfunction of the operators p̂, p̂x, p̂y, p̂z. The eignvalues are the sharp momentum values p, px, py, pz. (T1.23) We call these operators momentum operators. We can also introduce an energy oper- ator, Ĥ ≡ p̂2 x + p̂2 y + p̂2 z 2m = p̂2 2m = − h̄2 2m ( ∂2 ∂x2 + ∂2 ∂y2 + ∂2 ∂z2 ) = − h̄2 2m ∇2, which if possible plays an even more central role in the theory. Since p̂2 xΨ = p̂x(pxΨ) = p2 xΨ, we obviously have ĤΨ = p2 x + p2 y + p2 z 2m Ψ = EΨ. Thus Ψ is an eigenfunction of the operator Ĥ with the (kinetic) energy as eigenvalue. During this course, we shall get very well acqainted with the energy operator, which is in general called the Hamilton operator, or simply the Hamiltonian. (Therefore the symbol Ĥ.) We should also notice that Ψ [proportional to exp(−iEt/h̄)] is an eigenfunction of the operator ∂/∂t: ∂ ∂t Ψ = −iE h̄ Ψ. Including an extra factor ih̄, we see that ih̄ ∂ ∂t Ψ = EΨ. Thus, comparing with the relation ĤΨ = EΨ, we discover (as Schrödinger did) that de Broglie’s plane wave Ψ satisfies the following partial differential equation: ih̄ ∂Ψ ∂t = − h̄2 2m ∇2Ψ. ( Schrödinger’s equation for a free particle ) (T1.24) This is the Schrödinger equation for a free particle, the simplest equation satisfied by de Broglie’s plane waves. It has turned out that this simplest equation is also the one that works (in non-relativistic quantum mechanics). An important point is to note how the imaginary unit i entered in the ”derivation” above. We note that the appearence of the factor i is is an unavoidable consequence of tha fact that the non-relativistic energy-momentum relation E = p2/2m is linear in E and quadratic in p. This is what leads to a complex wave equation and complex de Brogle waves. The real and imaginary parts of de Broglie’s wave Ψ do not satisfy this equation separately. Thus we are forced to work with a complex wave function. TFY4215/FY1006 — Lecturenotes 1 25 satisfies the time-dependent Schrödinger equation. [Check this.] To find these eigenfunctions (ψ(r)) is a purely mathematical task, which Schrödinger solved readily, and which we too shall solve not as fast later in this course. It turns out that the equation above has eigenfunction solutions for all energies E > 0. For negative erergies, on the other hand (corresponding to bound states of the electron and the proton), Schrödinger found that this equation has energy eigenfunction solutions only for a discrete set of energy eigenvalues, and this set turned out to be precisely the set of ”experimental” energies mentioned above! We shall study these solutions in great detail later in the course, so let us here try only to convey a rough idea about how the quantum-mechanical description of the hydrogen atom works. For the lowest energy, E1 = −13.6 eV, Schrödinger found one eigenfunction of the energy operator Ĥ, on the form ψ1(r) = C1e −r/a0 with a0 ≡ 4πε0h̄ 2 e2me = Bohr radius. (T1.28) As we have seen above, this means that the wave function Ψ1(r, t) = C1e −r/a0e−iE1t/h̄ (T1.29) is a solution of Schrödinger’s time-dependent equation. Thus the radius a0 of Bohr’s inner- most circular orbit plays a role also in Schrödinger’s solution. We note that this is a very different role. Here, r is the distance from the proton to the electron. We note that this wave function is spherically symmetric (depends on r but not on angles). The time dependence is given by the factor exp(−iE1t/h̄), which is a complex number running around the complex unit circle, with absolute value | exp(−iE1t/h̄)| = 1. The spatial factor e−r/a has its maximal value when the distance r is equal to zero. It is important to note that Schrödinger’s time- independent equation has no eigenfunction with a lower energy than E1. For the next lowest energy, E2 ≈ −3.4 eV, Schrödinger found four infependent eigenfunctions. Three of these depend both on r and on the angles (θ, φ). Thus, Schrödinger found that his time-dependet equation has a special set of solutions, on the form Ψ(r, t) = ψ(r)e−iEt/h̄, which implies that these are also eigenfuntions of the energy operator Ĥ. It was of course very promising that the energy eigenvalues are precisely the discrete energies which are observed experimentally for the stationary hydrogen states. These are simple mathematical facts, which may be somewhat unfamiliar for us presently, but which are easy to digest when we get some more experience. It was much more challenging in 1925 (and still is) to come to grips with the physical interpretation of these mathematical facts. We understand that the strange wave-function solution Ψ1(r, t) of Schrödinger’s time-dependent wave equation must have something to do with the stationary ground state of the hydrogen atom. But what is the connection between this strange function and the physical state? This was not clear to Schrödinger either. However, after a while (in 1926) Max Born came up with a constructive suggestion. He proposed that the absolute square |Ψ1(r, t)|2 could be interpreted as a probability density for the position of the electron. (Remember the corresponding interpretation in our discussion of the double-slit experiment.) Since the phase factor exp(−iE1t/h̄) is a number on the unit circle in the complex plane, the absolute value of this exponential factor is equal to 1, as stated above. (For a real number a we TFY4215/FY1006 — Lecturenotes 1 26 have in general that | exp(ia)|2 = exp(−ia) exp(ia) = 1.) Thus the phase factor is of no importance for the probability density, which becomes time independent: |Ψ1(r, t)|2 = |C1|2e−2r/a0 . This is of course satisfactory when we are trying to describe a stationary state. Inspired by this we shall from now on call all solutions of the Schrödinger equation on the form Ψ(r, t) = ψ(r)e−iEt/h̄ stationary solutions. (T1.30) What does it mean to state that |Ψ1|2 is the probability density of the position of the electron? The answer is, in analogy with the definition of for example mass density, that the probability density multiplied with a volume element d3r, |Ψ1(r, t)|2d3r, gives the probability of finding the electron within this volume element. The probability of finding the electron somewhere is of course equal to 1. Therefore we must require that our wave function satisfies the following normalization condition ∫ |Ψ(r, t)|2d3r = 1 (normalization condition), (T1.31) where the integral is over the whole space. It can be shown that this condition is satisfied with C1 = (πa3 0)−1/2. To understand this a little better, let us imagine that we prepare a large number of hydrogen atoms in the state Ψ1 (the ground state), and for each of them measure the position of the electron. In analogy with the double-slit experiment, we then cannot predict the outcome of a single measurement. However, the distribution of the measured positions will for a large number of measurements agree well with the theoretical probability distribution |Ψ1(r, t)|2 = |C1|2e−2r/a0 . The figures shown below illustrate simulated results of three such series of measurements, with 10, 100 og 1000 “measured” positions. These results were obtained using a Matlab program involving a random number generator. The program is available on the home page. By running this program you will get a better impression of the 3D character of the distribution. (The program generates the points successively, while the distribution is made to rotate around the z-axis.) The measured positions are seen to be most densely spaced near the origin (that is, near the proton), which indeed is where the probability density |Ψ1|2 is largest. TFY4215/FY1006 — Lecturenotes 1 z Zz 10 points 100 points 27 TFY4215/FY1006 — Lecturenotes 1 30 square integrable, as we say. However, this can be repaired by constructing a wave group. Since the time-dependent Schrödinger equation is both linear and homogeneous, the su- perposition principle:is valid: A sum of two solutions of the Schrödinger equation is also a solution. ( superposition prinsiple ) . (T1.35) This holds also if the sum is repaleced by an integral: Ψ(x, t) = ∫ φ(k)ei(kx−ωt)dk, (k = p/h̄, ω = h̄k2/2m). (T1.36) Here, φ(k) is a smooth distribution of wave numbers around a central value k0. (According to de Broglie, these wave numbers correspond to momenta smoothly distributed around a central value p0 = h̄k0.) From wave theory we then remember that the group velocity of this wave group Ψ(x, t) is vg = dω dk ∣∣∣∣∣ k0 = h̄k0 m = p0 m ≡ v0. (T1.37) This way we can construct a wave group that is normalizable, so that∫ ∞ −∞ |Ψ(x, t)|2dx = 1. (For simplicity we are here working in one dimension.) Thus the wave group is moving with precisely the velocity (v0 = p0/m) wich we should require for this wave group. (v) Let us return to the ground state of the hydrogen atom. Even if this state is stationary, with a probability distribution which is time independent (”does not move”), it is important to relize that the electron is not at rest. This is because the kinetic energy is far from being equal to zero. This is understood as follows: We know that that the total energy is well defined (sharp), E1 = −13.6 eV, while the expectation value of the potential energy is 〈V 〉 = −27.2 eV. Since 〈K + V 〉 = 〈E 〉 = E1, the expectation value of the kinetic energy is 〈K 〉 = E1 − 〈V 〉 = 13.6 eV. A kinetic energy of this size in fact corresponds to a velocity of the order of αc = c/137. [Show this.] Some control questions 1. Which physical observable corresponds to the operator p̂x = h̄ i ∂ ∂x ? 2. Which operator K̂ corresponds to the kinetic energy K = 1 2 mv2 = p2/2m ? 3. Show that the de Broglie wave Ψ3 = exp[i(px− p2t/2m)/h̄] is an eigenfunction of the momentum operators p̂x, p̂y and p̂z and of the kinetic-energy operator K̂, and determine the respective eigenvalues. 4. Same for Ψ4 = exp[i(−px− p2t/2m)/h̄]. 5. Which physical momenta do the two de Broglie waves above correspond to? 6. Show that cos kx is not an eigenfunction of p̂x. TFY4215/FY1006 — Lecturenotes 1 31 7. Vis at Ψ3 oppfyller den tidsavhengige Schrödingerligningen (T1.24) for en fri partikkel. 8. Show that the real part of Ψ3 does not satisfy (T1.24). 9. According to the superposition prinsiple (T1.35), the wave function Ψ9 = Ψ3 + Ψ4 is a solution of (T1.24), and therefore is an acceptable wave function for a free particle. Is Ψ9 an eigenfunction of the operator K̂ ? Is Ψ9 an eigenfunction of p̂x ? Does Ψ9 describe a physical state with well-defined momentum? 10. At thermal equilibrium, the atoms in a one-atomic gas will (according to the so- called equipartition principle) have an average energy of 1 2 kBT per degree of freedom, that is, 3 2 kBT per atom. When neutrons are slowed down in a so-called moderator in a nuclear reactor, they end up as ”thermal” neutrons, with an average kinetic energy 3 2 kBT . Suppose that T = 300 K and find the average kinetic energy in electron volts (eV). Calculate the de Broglie wave length of the neutron corresponding to this kinetic energy K. Do you think that the wave nature of these neutrons would be revealed by scattering them on a crystal? [Answer: λ = 1.46 Å.] 11. When an atom is scattered by the atoms in a crystal, the scattering involves the Coulomb force between the electron and the atomic charges (electrons and nuclei). We then say that the electron experiences electromagnetic interactions. What kinds of interactions do you know about? Can you figure out what kind(s) of interaction are acting when neutrons are scattered on the crystal? [Remember that the neutrons are electrically neutral.] 12. Show that the function ψ = Ceikx is an eigenfunction of the momentum operator p̂ (see (T1.22)), and determine the eigenvalue. 13. Write down (and memorize) den time-dependent Schrödinger equation and the time- independent Schrödinger equation for a particle with mass m moving in a three-dimensional potential V (r). 14. What is the form of these equations when the potential is one-dimensional, V = V (x)? [Hint: The classical expression for the energy then is E = p2 x/2m+ V (x).] Physical constants15 Light velocity in vacuum c 2.997 924 58·108 m s−1 Planck’s constant h 6.626 068 96(33)·10−34 J s =4.135 667 333·10−15 eVs Planck’s constant /(2π) (“h-strek”) h̄ 1.054 571 628(53)·10−34 J s =6.582 118 99(16)·10−16 eVs Proton charge e 1.602 176 487(40)·10−19 C Permittivity of vacuum ε0 = 1/(µ0c 2) 8.854 187 817...·10−12 F m−1 Permeability of vacuum µ0 4π · 10−7 N A−2 (eksakt) Electron mass me 0.510 998 910(13) MeV/c2 =9.109 382 15(45)·10−31 kg 15Uncertainties in the tabulated values are given in the following way: 1.2345(13) means 1.2345±0.0013. 1 eV=1.602 176 487(40)·10−19 J. Tabulated values from 2010. TFY4215/FY1006 — Lecturenotes 1 32 Proton mass mp 938.272 013(23) MeV/c2 =1.672 621 637(83)·10−27 kg Neutron mass mn 939.565 346(23) MeV/c2 Fine-structure constant α = e2/(4πε0h̄c) 1/137.035 999 679(94) Classical electron radius re = e2/(4πε0mec 2) 2.817 940 2894(58)·10−15 m Electron Compton wavelength λe = h/(mec) 2.426 310 2389(16)·10−12 m Electron Compton wavelength /(2π) −λe = h̄/(mec) = re/α 3.861 592 6459(53)·10−13 m Bohr radius a0 = 4πε0h̄ 2/(mee 2) 0.529 177 208 59(36)·10−10 m = re/α 2 Rydberg energy 1 2 α2mec 2 = h̄2/(2mea 2 0) 13.605 691 93(34) eV Gravitational constant GN 6.674 28(67)·10−11 m3 kg−1 s−2 Avogadro’s number NA 6.022 141 79(30)·1023 mol−1 Boltzmann’s constant kB 1.380 6504(24)·10−23 J K−1 =8.617 343(15)·10−5 eV K−1 Stefan–Boltzmann’s constant σ = 2π5k4 B/(15h3c2) 5.670 400(40)·10−8 W m−2 K−4 The greek alphabet In quantum mechanics (and in physical literature in general) we need many more symbols than there are letters in the alphabet. Therefore it is customary to include letters from e.g. the greek alphabet. Below you will find upper-case and lower-case letters and waht they are called and pronounced in Norwegian.