Lecture notes for Atomic and Molecular Physics, FYSA31, HT Joachim Schnadt

Transcription

1 Lecture notes for Atomic and Molecular Physics, FYSA31, HT 01 Joachim Schnadt September 1, 01

2 Chapter 1 Before we start 1.1 General information on the course The course has a course homepage at There you ll find a schedule, course plan, and in due course also the written hand-in exercises. The final mark for the course is composed of the average mark for the lab exercises and the mark you achieve in a written exam, scheduled for 6 October 01. Marks are given in percent, which translate into the Swedish scale as U: below 50%, G: 50-79%, VG: 80% and above. On an A-F scale the marks are: F: below 50%, E: 50-59%, D: 60-69%, C: 70-79%, B: 80-89%, A: 90% and above. Also the written hand-in exercises will be marked, but these marks do not play any role for the final course mark. However, you will have to achieve an average of at least 70% for the hand-in exercises in order to be allowed to enter the oral exam. The result of the oral exam will not be registered before you have completed all lab exercises and received marks for them. You will have to deliver two sets of hand-in exercises with due-dates 14 September (noon) and 8 September (noon). Late hand-in will generally not be accepted. If you either fail in handing in the problems or in achieving a sufficient mark for entering the oral exam, you will get the chance of solving a new set of hand-in problems (with a number of problems that scales inversely with how much you have handed-in to that point). Attendance of lectures and exercise sessions is not compulsory, but recommended. Exercise sessions are scheduled for 10 September and 4 September.

3 1. Teaching and learning Passively listening to lengthy lectures has been shown to be about the most ineffective way of learning that exists. Therefore I will restrict the amount of time used for traditional lecturing. Instead we will reserve plenty of time for solving exercises and for discussing. This requires good preparation from your side: you will get reading instructions and you are expected to be prepared at the next lecture. You will also get a smaller homework at each lecture, and one of you will be selected (randomly) to present a solution to the problem in the beginning of the next lecture. Please also note that not all material relevant to the exam will be presented in the lecture. All reading and all exercises, even if not discussed in the lectures, may have relevance in the exam. 1.3 What is allowed - and what isn t? You are not allowed to copy the solution (or parts of it) to hand-in exercises or lab reports from a fellow student or any other source (internet, books, etc.), offer your own solution to a hand-in exercise or lab report to a fellow student for copying. In contrast, you are allowed to collaborate with your colleagues in solving hand-in problems and preparing lab reports - as long as you hand in a version of the solution, which has been prepared by yourself. Please be aware of that offences against these rules may inflict serious punishments for you, i.e. you may have to appear before the Disciplinary Board of the University and you may be completely banned from University for a period of time of up to six months. All teachers, including the lab supervisors, are obliged to report any suspicion of cheating and fraud to the Disciplinary Board, and this practice will be enforced rigourously. 3

4 1.4 Reading I suggest that you have a look at a couple of different books on Atomic Physics and that you then obtain a copy of one that suits you. You might, for example, consider the following books: Christopher J. Foot, Atomic Physics, Oxford University Press, Oxford, 005. The Physics Library has a few copies. Fairly cheap. Wolfgang Demtröder, Atoms, Molecules, and Photons, Springer, Berlin, Heidelberg, 006. Available as e-book from the University Library. A lot of material. Anne P. Thorne, U. Litzén, and S. Johansson, Spectrophysics. The Physics Library has a number of copies. Available from Media-Tryck, Sölvegatan 14. Fairly cheap. Hermann Haken and Hans Christoph Wolf, The Physics of Atoms and Quanta, Springer, Berlin, Heidelberg, 003. The Physics Library has some copies. Available as e-book from the University Library. in Swedish: Staffan Andersson, Fredrik Bruhn, Jan Hedman, Leif Karlsson, Sten Lunell, Katharina Nilson, Johan Wall, Atom- och molekylfysik, Uppsala universitet. You will have the opportunity of ordering copies. You will also need some Quantum Mechanics knowledge. Your favoured Quantum Mechanics textbook will probably do nicely. You could use, for example, Sara McMurry, Quantum Mechanics. Sölvegatan 14. Available from Media-Tryck, J. J. Sakuari, Modern Quantum Mechanics, Revised Edition, Addison- Weasley, Reading, Massachusetts, What have you done previously? Already in FYSA1 you have seen nearly all the quantum mechanical concepts that we are going to talk about. These are: Particle-wave duality, de Broglie wavelengths: λ = h/p 4

5 Heisenberg s uncertainty relationship for position and momentum: x p h/ Energy quantisation, for example in the radiation of a black body with radiation density S ν, S νdν dω da = hv3 c dν dω e hν/kt 1 Wave functions: Ψ(x) or Ψ(x, t) (cf. Fig. 1.1) (NB: I often write the position vector as x with components x 1 = x, x = y, and x 3 = z, although I sometimes also might write it as r.) Wave functions are interpreted as the probability amplitude of finding the particle described by Ψ at a position x and at time t. Thus, the corresponding probability is Ψ(x, t), and it is normalised to 1: Ψ(x, t) d 3 x = 1 Orthogonality of wave functions: ϕ a and ϕ b are said to be orthogonal if ϕ a(x)ϕ b (x)d 3 x = δ ab Quantum mechanical operators and measurements, eigenvalue equations: Â = Ψ (x)âψ(x)d3 x, and ÂΨ(x) = aψ(x), i.e., Â = a The superposition principle: if ϕ a and ϕ b are wave functions, then also Ψ(x, t) = ϕ a (x, t) + ϕ b (x, t) is a wave function Complete mutually orthogonal sets Particle in a potential well Compatible and non-compatible variables Commutators [A, B] The canonical commutation relations: [x i, x j ] = [p i, p j ] = 0 and [x i, p j ] = i hδ ij The parity operator P : P Ψ p (x, t) = ±Ψ p ( x, t) 5

6 Figure 1.1: Spectral distribution of the radiation density S (λ) of a blackbody (Figure taken from teledetection/report1/node16.html). The one-dimensional oscillator and ladder operators â,â Vibrational spectra During the course we will re-encounter many of these concepts, although they sometimes might look somewhat different from what you are used to. 1.6 The Dirac notation Table 1.1 shows a couple of quantum mechanical concepts both in the way you have written them previously and in the notation introduced by Paul Dirac ( ), one of the founders of quantum mechanics. The notation is called Dirac notation or bra-ket notation. The main difference between wave function and Dirac notation, which indeed also is a conceptual difference, is that the wave function notation always specifies coordinates, while the Dirac notation is coordinate-free. If coordinates have to be specified, e.g. when you want to describe the outcome of a position measurement, you specify that you want to use the position basis written as { x } (the curly brackets indicate that the position basis is made up from a set of vec- 6

7 Concept Notation that you have Dirac notation used before Wave function Ψ(x) = x Ψ Quantum mechanical state Scalar product Ψ ϕ a (x, t)ϕ b (x, t)d 3 x = ϕ a ϕ b Normalisation Orthogonality ψ(x, t) d 3 x = 1 ϕ a (x, t)ϕ b (x, t)d 3 x = δ ab = ϕ ϕ = ϕ a ϕ b = δ ab Operator  A Eigenvalue equation ÂΨ(x) = aψ(x) = x A Ψ = a x Ψ A Ψ = a Ψ Expectation value  = Ψ (x, t)âψ(x, t)d3 x = A = Ψ A Ψ Table 1.1: The Table shows some quantum mechanical concepts in both conventional and Dirac notation. tors x ). Depending on your problem you also might specify other bases, such as the momentum basis { p }. The Dirac notation also makes obvious that there exists a difference between the concept of a wave function and that of a quantum mechanical state: The quantum mechanical state contains all information on the considered system. The wave function is the probability amplitude for a position measurement on the considered system, which implies that the probability for a position measurement is given by Ψ(x, t). Thus the wave function 7

8 contains a lot of useful information, actually very often the information one is interested in - but it does not contain all information on the considered state. As an example we may take the hydrogen atom. In Dirac notation, the quantum mechanical state of a hydrogen atom in the state with quantum numbers n, l, m l, m s the knowledge of which fully characterises the state of the hydrogen atom is written nlm l m s. In wave function notation, the probably of finding the same hydrogen atom within a volume element d 3 x at x is written ψ n,l,ml,m s (x) d 3 x. The connection between both notations is provided by x nlm l m s = ψ n,l,ml,m s (x). Task: Think about and discuss the differences between the two concepts of a quantum mechanical state and a wave function. Which essential physical concept lies behind the difference? 8

9 Chapter The hydrogen atom in classical and semi-classical physics We start out with the most basic problem of Atomic Physics, the description of the hydrogen atom. The starting point is the classical description of the hydrogen atom. Already in the 19 th century it became clear that the experimental observations did not agree with any classical solution, however, and therefore Niels Bohr introduced a new, phenomenological model, now known as the Bohr Model, which could explain some of the observations. The Bohr Model was one of the highlights of early quantum mechanics, but was soon replaced by a better, truly quantum mechanical model. The description of the quantum mechanical model is the goal of the first part of the course, and on the way towards the quantum mechanical solution we will introduce all necessary quantum mechanical concepts..0.1 Statement of the problem The hydrogen atom is the smallest of all atoms. It contains one electron and a proton, which are held together by Coulomb attraction. Hence, we are dealing with a two-body problem characterised by a force acting along a line between the proton and electron and a potential, which depends on the distance r e r p only: V = V ( r e r p ). Classical physics show that such a two-body problem can be reduced to a one-body central field problem. 9

10 e - r ep r e p + r p R O Figure.1: Illustration of the vectors involved in the description of the hydrogen atom. R is the vector to the centre-of-mass of the hydrogen atom..0. The classical approach Figure.1 shows the notation that we are going to use in the following treatment. Newton s second law gives us the following: where r ep = r e r p. Likewise, m e r e = e V ( r e r p ) = dv dr ep r e r p r ep, m p r p = e V (r ep ) = dv dr ep r p r e r ep. The vectors r e and r p can be exchanged against the vectors R and r ep the centre-of-mass and between the proton and electron, respectively: to r e = me+mp r M e = m e r M e + mp r M e + mp r M p mp r M p = = m er e +m p r p M + m p M r ep = R + m p M r ep, (.1) and, in the same way, r p = R m e M r ep. (.) 10

11 M is the mass of the total system of the proton and electron. Adding the two equations gives that is m e r e + m p r p = m e R + m e m p r M ep + m p R m e m p r M ep = M R = = dv dr ep r ep r ep + dv dr ep r ep r ep = 0, M R = 0, (.3) which means that the centre-of-mass moves without acceleration, quite as expected for a hydrogen atom in no outer field. Taking the difference of equations (.1) and (.) one sees that 1 (m { e r e m p r p ) = 1 me R + m e m p r M ep m p R + m e m p r } M ep = m e m p r ( ) M ep µ r ep = 1 dv r e r p dr ep r ep dv dr ep ˆr ep, where ˆr ep denotes the unit vector along r ep and µ is the reduced mass of the system. In other words, we have shown that the separation of proton and electron, r ep, obeys a Newtonian law µ r ep = dv dr ep ˆr ep, (.4) and that the two-body problem can be reduced to a one-body problem. V is the central potential, which in the case of the hydrogen atom is the Coulomb potential e V =. (.5) 4πε 0 r ep Since the force is purely radial, i.e. F (r ep ) = f(r ep )ˆr ep = dv dr ep ˆr ep, we see that f(r ep ) = dv dr ep. (.6) Angular momentum The torque r ep F is zero. This means that the angular momentum l = r ep p is a constant of the motion. It follows that the motion is strictly planar: r ep l = 0, and we may choose a coordinate system such that x = r ep cos ϕ, y = r ep sin ϕ. (.7) 11

12 Using these cylindrical coordinates one finds for the length of the angular momentum l = µr ep ϕ, (.8) which, like l, is a constant of the motion. It turns out that also the z- component of the angular momentum is a constant of the motion. l z = µ (xÿ yẍ) (.9) Remark: Here we find both parallels and differences to the quantum mechanical treatment, as we will see later during the course. Classically, l, l, and l z are constants of the motion. Quantum mechanically, in contrast, l is no constant of the motion, while both l and l z are. The reason is that quantum mechanics does not allow the full determination of the angular momentum. Task: Show that the torque r ep F vanishes. Show also that the angular momentum l = r ep p is a constant of the motion. Show that the motion of the electron around the nucleus (or rather the motion of the nucleus and electron around the centre-of-mass) is planar. Finally, show also that l and l z are constants of the motion. Phyiscally, why must l z be a constant of the motion? Centrifugal barrier The goal of a classical treatment would be to obtain the (classical) trajectory of the electron around the nucleus r ep (t). The starting point for finding r ep (t) would be the energy E = E kin + V = 1 µ (ẍ + ÿ ) + V (r ep ) = 1 µ ( r ep + r ep ϕ ) + V (r ep ) = = 1 µ r ep + l µr ep + V (r ep ) = 1 µ r ep + V eff (r ep ), (.10) 1

13 l µ r ep Energy V eff - A / r ep r ep Figure.: Illustration of the shape of the effective potential. l where V eff (r ep ) contains the centrifugal barrier. For a 1/r-potential the µr ep potential curves are drawn in Figure.. Remark: We will also encounter the centrifugal barrier in the full quantum mechanical treatment of the hydrogen atom. Remark: The energy, which we consider above, is not a constant of the motion, i.e., we have neglected an important contribution. Which one? If we had the full energy, we could, in principle, determine the classical trajectory by calculating the time dependence t(r ep ) and then inverting it to give r ep (t). Even in the classical theory this becomes very demanding, due to the occurring integrals. One gets quite a bit further by instead explicitly considering the forces which act on the electron and neglecting radiation. With the potential given by Eq. (.5), the relevant forces are the Coulomb force and the balancing centrifugal force: e 4πε 0 r ep = µv r ep = µω r ep, 13

14 where ω is the angular frequency ω = v/r ep. Solving for ω gives ω = e /4πε 0 µr 3 ep = 1 T, (.11) with T the time period for one revolution. This corresponds exactly to Kepler s third law, with the gravitational potential replaced by a Coulomb potential: Kepler s third law states that the orbital period T is proportional to the cube a 3 (= r 3 ep) of the semi-major axis of its orbit. From the balance of forces we also find an expression for the energy: E = 1 µv + V (r ep ) = 1 µv e /4πε 0 r ep = e /4πε 0 r ep. (.1) This is this far classical physics takes us under the neglect of radiation. In order to get a step further we now need to introduce the quantum hypothesis of Planck. It was Niels Bohr in Copenhagen ( ), who did this first in his semi-classical model of the atom. If you are interested you can find his original work on the internet (N. Bohr, Philosoph. Mag. Ser. 6 (1913) 6, 1, accessible from db=all)..0.3 The Bohr Model (1913) In contradiction to classical theory, Bohr assumed that there are only certain allowed orbits for the electron, and that radiation due to electron acceleration is not allowed. The only way radiation can be emitted or absorbed is by electrons passing from one orbit to another. The lost or gained energy is assumed to be a multiple of Planck s energy quantum hω = hν: E = k hω, i.e., ω = E k h. (.13) Without any loss of generality we here assume that E > 0. From Eq. (.1) an expression for the difference of the energies of an electron in different orbits is found ( E = e /4πε 0 1 r ep 1 r ep ), i.e., ( ) ω = 1 e /4πε 0 1 k h r ep 1 e /4πε 0 r ep k h 14 r ep. r ep

15 Here it has been assumed that r ep r ep, which is true only for two orbits with a large distance of the electron from the nucleus. This is no further limitation, since the physics for large orbits carry over to smaller orbits. We can now equate the classical result in Eq. (.11) with the present value: ω = e /4πε 0 = (e /4πε 0) ( r ep ) µr 3 ep 4k h, i.e., r ep ( r ep ) = 4 e /4πε 0 h m e m e µ k r ep = 4a 0 m e µ k r ep, = r ep = a 0 r ep me µ k. a 0 is the Bohr radius (a 0 = 0.59 Å= m). Let k = 1 and assume that there exists a quantum number n (a quantum number is always an integer number) so that r ep = an x m x e µ, r ep = a(n + 1) x m x e µ. This ansatz comes about from requiring that the orbit radius should scale with a label (the quantum number n) and that it needs to have the dimension of a length therefore the length parameter a. For n 1 and using a Taylor expansion of (n + 1) x one finds { ( ) x a me µ r ep = (a(n + 1) x an x ) m x e µ n x + x ( m e µ ) x 1 = ax(n + 1) ( ) x 1 x 1 m e µ (n + 1) x 1 n x ( m e µ = a 0 r me = x ep a µ 0 an = a = n1 x a0 x axn ( ) x 1 x 1 m e µ ( ) x me 4 µ ( ) 1 x me. µ ) x } = By construction, a should be independent of n, and this implies x = and a = a 0, which gives r ep = a 0 n m e µ, (.14) 15 =

16 where n, as said before, is an integer number. Approximating m e µ (or defining a 0 using the reduced mass of the hydrogen atom rather than the electron rest mass) one gets the more commonly seen For the energy one finds r ep = a 0 n. (.15) E = e /4πε 0 a 0 1 n µ m e m e µ : E = e /4πε 0 a 0 1 n. (.16) This result for the energy of the hydrogen atom turns out to be correct even in a proper quantum mechanical treatment! The result applies, approximately, also for atoms composed of a heavier nuclei and an electron. The Rydberg constant and Rydberg formula One can express the energy in wave numbers instead of the here used energy. Wave numbers ( ν = 1/λ = ν/c = E/hc) are a commonly used quantity in some kinds of spectroscopy. In wave numbers the difference of energies becomes: ν = e /4πε 0 a 0 hc µ m e ( 1 n 1 n ) R µ m e ( 1 n 1 n ) R ( 1 n 1 n ). (.17) The Rydberg constant of indefinite mass is R = m 1. The formula is called Rydberg formula. Good to remember: With n = 1 and n = 1 you find hcr = 13.6 ev, which is the ground state ionisation energy for an atom with one electron and an indefinitely heavy nucleus, but of course the value also applies to the hydrogen atom due to the small difference between m e and µ. For real nuclei with atomic number Z and mass m z the Rydberg constant becomes ( µ m Z R Z = R = R R 1 m ) e. (.18) m e m Z + m e m Z In addition, we need to take into account that real nuclei have a charge +Ze, so that the Coulomb interaction with the electron becomes Ze /4πε 0 r ep. 16

17 The energy of a hydrogen-like atom (i.e., nucleus plus one electron) is then (remember that a 0 contains the factor e /4πϵ 0!) Standing waves E = e /4πε 0 Z a 0 n = hcr Z Z n. (.19) The derivation of the distance between electron and nucleus in the Bohr model is more straightforward if one applies the wave/matter duality (which was not known to Bohr at the time he developed his model). Any stationary state of the electron must be described by a standing wave. This condition can be written as πr ep = nλ electron, (.0) with n an integer number. From de Broglies wave/matter duality we find the wavelength associated with the electron as which gives for the speed of the electron λ electron = h p = h µv, (.1) v = n h. πµr ep From the balance of centrifugal and central forces we have and hence r ep = e /4πε 0 µv = e 4πε 0 (πµr ep ) µn h, r ep = n h 1 1 e /4πε 0 µ 4π = a 0n m e µ, which is the same result as we found before in Equation (.14)..0.4 Experimental observations One of the primary driving forces for the introduction of new atomic models in the 19th and 0th century and among these models in particular the Bohr Model was the observation that atoms emit and absorb light only at certain, discrete energies. This was first shown by Gustav Kirchhoff and Robert Bunsen in 1859, and they probably used setups similar to those shown in Figure.3 and.4 (well, without the computer and diode). Also the 17

18 Figure.3: Typical optical emission setup. The source is hot, so that the source atoms are heated up and emit light (Figure from Demtröder, Atoms, Molecules and Photons, Springer, Berlin, Heidelberg, 006). Figure.4: Typical optical absorption setup. The light source needs to be continuous (Figure from Demtröder, Atoms, Molecules and Photons, Springer, Berlin, Heidelberg, 006). emission series of the hydrogen atom, for which an empirical formula was developed by Johann Jakob Balmer in the 1880s, will have been measured by similar setups. Both the emission and absorption series of the hydrogen atom are shown in Figure.5. Balmer s formula was later generalised by Janne Rydberg to formula (.17), which also describes the other emission and absorption series of the hydrogen atom. The Balmer series corresponds to transitions to/from the Bohr orbit with label n =. Its main part is found in the visible regime. Other series are the Lyman (from/to n = 1), Paschen (from/to n = 3), Brackett (from/to n = 4), and Pfund series (from/to n = 5). The wavelength, energy, and wavenumber regimes of these series are specified in Table.1. 18

19 Figure.5: The Balmer series in emission and absorption (Figure from 19

20 Energy (ev) Wavelength (nm) Wavenumber (cm 1 ) Lyman (n=1) n = n = Balmer (n=) n = n = Paschen (n=3) n = n = Table.1: Energy, wavelength, and wavenumber regimes of the Lyman, Balmer, and Paschen series..0.5 Moseley s Law In the 1910s Henry Gwyn Jeffreys Moseley measured the x-ray spectra of many elements by bombarding the materials by high-voltage electrons. This leads to the excitation of inner electrons. The excited atoms will eventually decay under the emission of x-rays with characteristic energies. Moseley found a relationship between the frequency of the emitted x-rays and the atomic number: ν Z. (.) This relationship is today called Moseley s Law and is in qualitative agreement with Rydbergs formula { } Z ν = cr n Z. n The formula and measured energies do not comply too well, since the formula is formulated for atoms with one electron. The agreement can be improved by the introduction of shell-dependent screening factors σ n : { (Z σn ) ν = cr (Z σ n ) }. n n 0

21 Figure.6: Moseley s recording of the frequencies of the characteristic x-ray emitted by atoms with atomic number Z (from Phil. Mag. 7, 703 (1914)). 1

22 The term screening describes the effect of the other electrons, which do not participate in the optical transition, onto the nuclear charge seen by the electron, which jumps from one orbit to another. As is seen from the formula, the charge is reduced to an effective charge (Z σ n )e. The term shell describes the orbits in the Bohr Model and are labelled as K-shell for n = 1, L-shell for n =, M-shell for n = 3, and so on..0.6 Shortcomings of the Bohr model The Bohr model correctly predicts the energies observed in the emission and absorption series of atomic hydrogen. Nevertheless, it has a number of severe shortcomings, which made it necessary to introduce a fully quantum mechanical model. Some of the shortcomings are: The Bohr model does not comply with wave/matter duality: it does not obey Heisenberg s uncertainty relation x p h. Both position and momentum are fully classical and can be determined to any precision simultaneously. In the Bohr model the orbital angular momentum for hydrogen in its ground state is l = n h (show this by yourself!). From experiment it is know that it actually is zero. The Bohr model fails in explaining the intensities of the spectral lines. The spectra of heavier atoms can only be explained if ad-hoc assumptions are made (see Moseley s law). The fine and hyperfine structure in the spectra of hydrogen and other atoms cannot be explained. An explanation of this structure actually goes beyond non-relativistic quantum mechanics, as well, but can be treated when making some assumptions. In this course we will treat the non-relativistic quantum mechanical theory of the hydrogen atom and other atoms. Before we do that, we introduce the concept of the spin by means of a consideration of the Stern-Gerlach experiment. This experiment provides a very nice and pretty straightforward illustration of fundamental quantum effects, and therefore we already introduce spin now, although we will not need it in the first quantum mechanical treatment of the hydrogen atom. We will also use it to repeat and introduce some quantum mechanical concepts and tools.

23 Chapter 3 Quantum mechanics 3.1 The Stern-Gerlach experiment The Stern-Gerlach experiment was conceived in 191 by Otto Stern and Walther Gerlach and carried out by Stern and Gerlach in Frankfurt in 19. A beam of (neutral) Ag atoms is transmitted through an inhomogenous magnetic field and further onto a photographic plate (see Fig. 3.1). Ag has altogether 47 electrons, out of which 46 form a spherical shell with no resulting angular momentum. The total angular momentum of the neutral Ag atom is thus given by the spin of the 47th electron, or, in other words, the neutral Ag atom behaves like a single electrons when it comes to its interaction with a magnetic field 1. The magnetic moment of the atom is proportional to the spin, µ S, (3.1) and the force onto the atom is then, entirely as in classical physics, given by the gradient of the scalar product of the magnetic moment and the magnetic field, F z = z (µb) = µ B z z z. (3.) Classically, one would expect that, assuming that the spin angular momentum corresponds to a spinning motion of the electron, all values of µ z are 1 The existence of an angular momentum intrisic to the electron has to be postulated in non-relativistic quantum mechanics it is an experimental fact, which finds one of its clearest evidences in the Stern-Gerlach experiment, although the concept was unknown at the time of the experiment and introduced first a couple of years later. 3

24 Figure 3.1: Stern-Gerlach experiment (Figure taken from 4

25 assumed between µ and µ. Hence, the photographic plate should exhibit the recording of a beam of atoms elongated along z. What instead is observed are two distinct components that correspond to µ z S z = ± h, (3.3) where h is the Planck constant h = ev s. The Stern-Gerlach experiment thus shows that, if the magnetic isotropy of space is removed by application of a magnetic field in a particular direction, the component of the spin in this direction is quantised, i.e., it can assume only particular values. The Stern-Gerlach experiment represents one of the most direct proofs of quantisation and has had a profound influence on the development of quantum mechanics Sequential Stern-Gerlach experiments Of course it is possible to apply magnetic fields in other directions of space than in the (arbitrarily chosen) z-direction. Then one can also let the silver atom beam transmit sequential arrangements of Stern-Gerlach setups, such as shown schematically in Fig. 3.. In each of these sequential Stern-Gerlach experiments the beam traverses first a magnetic field in the z-direction, which leads to a split-up of the atomic beam into a beam with electrons with S z = + h/ and one with S z = h/. Only the former is allowed to enter the second Stern-Gerlach setup with a magnetic field in either the z- or x-direction. As expected, if the beam is transmitted through a second magnetic field in the z-direction only one beam with S z = + h/ exits the setup. Also as expected, the application of a magnetic field in the x-direction leads to a splitting-up of the beam into two beams with S x = + h/ and S x = h/, respectively. If one now again stops one of the beams, namely that with S x = h/, and lets the other go through yet another Stern-Gerlach setup with a magnetic field in the z-direction, something strange happens: classically, one would expect only one beam to exit this setup with S z = + h/. The reason is that the first beam stopper already removed all electrons with S z = h/. What is observed instead is two beams: one with S z = + h/ and one with S z = h/! It is very important to see that this behaviour is entirely unclassical. It was only the advent of the quantum mechanical theory which could render this result plausible. Quantum mechanically, S x and S z cannot be determined simultaneously, S x and S y are non-compatible observables. The determination of S x in the second Stern-Gerlach experiment destroys all information obtained in the first Stern-Gerlach setup with a magnetic field in the z-direction. A very peculiar behaviour indeed! 5

26 z x Figure 3.: Sequential Stern-Gerlach experiments (Figure taken from J. J. Sakurai, Modern Quantum Mechanics). We have to leave the classical description behind, and instead go over to an alternative formulation, which makes use of the concept of states. It is said that the Ag atoms are in spin states, which are represented by vectors in a two-dimensional vector space. We write these as S z, + = 1 S x, + 1 S x,, (3.4) S z, = 1 S x, S x,, (3.5) S x, + = 1 S z, S z,, (3.6) S z, = 1 S z, S z,, (3.7) i.e., all states are written as superpositions of other states. The outcome of the sequential Stern-Gerlach experiments can now be explained: after the first experiment only the beam with atoms in the S z, + is allowed to enter the second setup. This beam is a superposition of the S x, + and S x, states, and thus after the second setup with a magnetic field in the x-direction two beams are found. Only the S x, + beam is allowed to go on but the atoms of this beam with S x, + are from (3.6) seen to be either in the S z, + or S z, state! Now, of course there is a third component of the spin, the S y component. It is incorporated in the above system by complex addition: 6

27 S y, + = 1 S z, + + i S z,, (3.8) S y, = 1 S z, + i S z,. (3.9) 3. The vector space of quantum mechanics 3..1 Requirements In the preceding section we have seen that the quantum mechanical states can be described as vectors in a vector space. So far we had restricted ourselves to a two-dimensional space with basis { S z, +, S z, } (alternatively: { S x, +, S x, } or { S y, +, S y, }, but it is convention to use the { S z, +, S z, } basis). The concept can be generalised to include all necessary physical quantities. What is needed is the following: a complex vector space with a norm (i.e., a length of the vectors), which requires a scalar (inner) product, with infinite dimensions (the number of dimensions defines the number of possible measurement outcomes, and, for example, a position measurements thus requires an infinite number of dimensions), which is complete (mathematically: every Cauchy sequence converges, which in turn implies that each superposition of vectors from that space also converges to a vector in that space, even though the difference between the vectors might be arbitrarily small). 3.. States: kets The requirements are exactly the requirements which fulfilled by a Hilbert space, named after the German mathematician David Hilbert ( ). Thus the quantum mechanical state space is a Hilbert space, with physical states represented by ket vectors α. Of course, the following states are also vectors in the quantum mechanical state space: α + β = γ, c α = α c, where c is a complex number. Actually, c α represents the same state as α alone. 7

28 3..3 Operators Now that we have defined the physical states, we also have to tell how to arrive at a measurement of a physical quantity such as S z above. First of all, a physical quantity which can be observed is called an observable. Observables are represented by operators on the vector space. Operators act on states: A α As in mathematics, operators may have eigenvectors or eigenstates, as they are called in quantum mechanics. If the eigenstates of an observable A are denoted by a, a, a,..., then A a = a a, A a = a a,..., where the a, a,... are numbers. The introduction of operators in this way makes it possible to properly describe what happens in the sequential Stern-Gerlach experiments above. The z-component of the spin angular momentum is an observable and thus represented by an operator S z. It has eigenstates S z, + and S z, with eigenvalues h/ and h/ (you already see that the eigenvalues will represent the outcomes of a measurement!): S z S z, + = h S z, +, (3.10) S z S z, = h S z,. (3.11) Similar definitions apply to the operators S x and S y. In the sequential Stern-Gerlach experiments the first Stern-Gerlach setup is a measurement of the spin in the z-direction, which corresponds to an application of S z to the states of the silver atom beam, which can be in either state S z, + or S z, ; the outcome of the measurement is thus either h/ or h/ and leaves the atoms in the corresponding states S z, ± : S z S z, ± = ± h S z, ±. (3.1) For the second and third experiment in Fig. 3. the second setup traversed by the S z, + beam corresponds to a measurement of the spin in the x- direction. According to (3.4), an application of S x to S z, + yields ± h/ 8

29 and leaves the atoms in states S x, ±. In the third experiment of Fig. 3. only S x, + is allowed to traverse the final magnetic field in z-direction. This corresponds again to a measurement of S z. An application of the operator S z to (3.6) yields immediately the outcome of the experiment: the z-component of the spin can either be h/ or h/, and after measurement the atoms are either in state S z, + or S z,. Other examples for observables are the position and momentum operators x and p: x x = x x, (3.13) p p = p p. (3.14) Observe the difference between operators x and p and their eigenvalues x and p, which are numbers! Finally, two operators A and B are equal if their application to an arbitrary state yields the same results: A = B if (and only if) A α = B α α. (3.15) 3..4 Dimensionality of the vector space As already indicated above, the dimensionality of the quantum mechanical vector space of interest i determined by the number of possible outcomes of a measurement (we will typically restrict ourselves to considering only a relevant subspace of the total Hilbert space of quantum mechanics, which comprises dimensions for all possible outcomes of measurements of all physical quantities). In the Stern-Gerlach experiment one really has only two different outcomes of the measurement, which in a bit more colloquial language are called spin-up and spin-down. You might argue that you could measure the z-component of the spin in the x-, y-, and z-directions, with a total of six dimensions. However, equations (3.4) to (3.9) show that the outcome of a measurement of S x and S y always can be described in terms of the z-component of the spin (of course the use of the z-component is a convention); hence two dimensions are enough. Any spin state can be written as superposition of the two possible eigenstates of S z : Ψ = c + S z, + + c S z,. More generally, if an observables has eigenstates { a } (observe the use of curly brackets, which again indicates a set) then a general state can be written as 9

30 Ψ = a c a a. (3.16) 3..5 States: bras Now from Mathematics it is known that each vector space has a dual space, which also is a vector space a mirror image, so to say. Of course also the quantum mechanical Hilbert space has a dual space. The vectors of this dual space are written as a, and they are called bra s. The bra which corresponds to a ket c a is c a, where c is the complex conjugate of c. The scalar product (inner product) of a bra a and a ket b is written as It has the following property: a b. (3.17) a b = b a. (3.18) This implies that a a is a real number (how do you see that?). The scalar product of a bra with its corresponding ket is positive or zero: a a 0. (3.19) This property is important since the scalar product defines the norm and thus the length of a vector. Since the square of the length of a quantum mechanical vector a a is coupled to its probability interpretation (cf. Table 1.1) this value has to be zero or larger and cannot be negative. It is also practical (but not necessary) to require that a a = 1. (3.0) For an non-normalised state a this length normalisation is of course achieved by a = 1 a a a. (3.1) Finally, two states a and b are said to be orthogonal if a b = b a = 0. 30

31 3..6 More on operators Operators in bra space and Hermitian operators Since we have operators on ket space, we also need operators on bra space. In the Dirac notation operators are defined on kets from the left and on bras from the right side (the arrows merely indicate, into which direction the operators act in the equations): b = X a, b = a Y. The operator in dual (bra) space that corresponds to the operator X in ket space is X, the Hermitian adjoint of X. Thus but X a does not correspond to a X, X a corresponds to a X. Operators have matrix representations (we will come back to this point soon); the Hermitian adjoint of an operator is then obtained by taking the complex conjugate of the transpose of the operator s matrix in a matrix representation: Hermitian adjoint: a ij a ji. (3.) The operator is then said to be Hermitian (or self-adjoint) if the Hermitian adjoint is the same as the matrix, which corresponds to the operator. An example of Hermitian operators is given by the Pauli matrices: a ij ( a ji ( ) ( 0 i i 0 ) ( 0 i i 0 ) ( ) ( ) ) The eigenvalues of a Hermitian operator A are real, and, furthermore the eigenkets of A corresponding to different eigenvalues are orthogonal. This is easily shown: Assume the A is Hermitian. Then A a = a a a A = a A = a a 31

32 Multiply the first equation by a from the left and the second by a from the right, which gives: Taking the difference yields a A a = a a a, a A a = a a a. 0 = (a a ) a a. If a = a, then a a = 1, and thus a = a = a and thus a is real. Hence the eigenvalues of a Hermitian operator A are real. If, in contrast, a a, then a a = a a 0 and hence a a = 0 or, in other words, a and a are orthogonal. The eigenvectors, which corresponds to different eigenvalues of the same Hermitian operator A are orthogonal! Operator multiplication and outer product Two operators might act onto a state one after the other (this corresponds to two subsequent measurements). In general, operators do not commute, i.e., if X and Y are two operators, then in general XY a Y X a. You can rationalise this fact by considering a real problem. You are familiar with the Heisenberg position-momentum uncertainty principle, which states that x p h/. This implies that first measuring the momentum and then the position of a quantum mechanical particle does not necessarily give the same measurement result as first measuring the position and then the momentum - x and p do not commute. You should also realise that the order of operators (in ket space), which corresponds to the order of measurements, is from right to left. The rightmost operator acts first onto the state. We have already considered operators in bra space and stated that X a corresponds to a X. From mathematics we know that the Hermitian adjoint of an operator product is given by (XY ) = Y X. 3

33 Of course this also makes sense physically, since operators in bra space act to the left. One can build operators from a product of states. The outer product of two states a and b is defined as X = b a ; forming such an outer product results in the formation of an operator, which here is called X. The Hermitian adjoint of this operator is X = a b. Finally, we have already encountered matrix elements (cf. Table 1.1) of the form b X a (you will soon see why these constructs are called matrix elements). For these the following relationship is valid: b X a = a X b. Of course, if X is Hermitian, i.e., if X is an observable, then b X a = a X b Bases of state space In section 3..4 we already discussed briefly that the state space of interest is spanned by the eigenvectors of the observable of interest. For example, in the Stern-Gerlach experiment the spin state space is two-dimensional and spanned by S z, + and S z,, which are the eigenvectors of the observable S z. After an S z measurement the silver atom is either in state S z, + or in state S z,. S z, + cannot be described in terms of S z, and vice versa; S z, + and S z, are linearly independent, which implies that they are orthogonal: S z, + S z, = 0. S z, + and S z, form a basis of our spin state space. More generally, if we call the basis vector of the state space of interest a, a,..., and if we normalise them, a N = a a a a N = a a a..., then a, a,... form a complete orthonormal basis of state space. Thus an arbitrary normalised state α in state space can be written as a sum of basis vectors, which here are assumed to be already normalised (i.e., we skip the index N): 33

34 α = a c a a. (3.3) The coefficients c a are retrieved by multiplying equation (3.3) with a from the left: α can thus also be written as a α = c a. α = a a α a = a a a α. This in turn implies the very important relationship that a a = 1 (3.4) a (3.4) is called the closure or completeness relation (SV: fullständighetsrelationen). The relation expresses the completeness of the basis { a }. The meaning of the coefficients c a = a α is that they give the probability amplitude for finding the system described by α to be in state a, i.e., c a is the corresponding probability. We also can assign a meaning to the outer product a a : it projects the state it acts on (it is an operator!) onto the state a. a a is said to be a projector: a a α = a a a a a α = a a δ a a c a = c a a Matrix representations Of course one always can choose different bases for the state space of interest. For example, one basis for the spin space of the Stern-Gerlach experiment is the { S z, +, S z, } basis. Other choices would be, e.g., { S x, +, S x, } or { S y, +, S z, }. For a particular basis { a (n) }, with n = 1... N, operators and vectors can be written in matrix notation : a (1) a X (1) a (1) a X ()... a (1) a X (N) a () a X (1) a X ˆ= () a X ()... a () a X (N), (3.5) a (N) a X (1) a (N) a X (N) You should read the sign ˆ= as corresponds to. 34

35 α ˆ= a (1) α a () α... a (N) α, (3.6) α ˆ= ( a (1) α, a () α,..., a (N) α ). (3.7) Equation (3.5) shows clearly why the construct b X a is called a matrix element. Observe that in equation (3.5) to (3.7) above there is no equality between what is to the left and the right of the ˆ= sign this is a correspondence. If you want to write an equation that corresponds to equation (3.5) then you should apply the closure relation two times: X = a a a a X a a. (3.8) Exercise: Matrix notation vs Dirac notation 1. Write the operator β α in matrix notation using the basis { a (n) }.. Consider, again, the electronic spin. If we define + S z, + and S z, and knowing that S z ± = ± h ± write the matrix/vector notations for +,, S z, S + h +, and S h The position operator and the position basis of state space The action of the position operator is defined by x x = x x, (3.9) where x is an operator (actually, it is an observable) and x a scalar the outcome of the position measurement represented by x. Typically, the possible values of x are infinitesimally closely spaced. Just think of a free particle anywhere in space. Since x is an observable its eigenvectors must span a subspace of the quantum mechanical state space. This is the position subspace. Since the eigenvalues are infinitesimally closely spaced, there must exist an indefinite number of them, and hence the x cannot form a discrete basis, but the basis must be continuous and the number of dimensions of the position subspace 35

36 is infinite. Therefore, the sum in the closure relation must be replaced by an integral: a a = 1 a An arbitrary state α can then be expanded as dx x x = 1. (3.30) or, in three dimensions: α = dx x x α, (3.31) α = d 3 x x x α. (3.3) If you compare this to the discrete expansion α = a a a α = a c a a, then you see that x α must be a probability amplitude, as are the coefficients c a. From before you know that probability amplitudes related to a position measurement are given by the wave functions, and we can identify Scalar products of position states For discrete bases we found for the basis states that ψ(x ) = x α. (3.33) a a = δ a a, where δ a a is the Kronecker delta. For a continuous basis such as the position basis things are bit more complicated: x x = δ(x x ), (3.34) which looks very similar to the expression for the discrete basis. However, δ(x x ) is the delta function, a generalised function with the following two properties: dx dδ(x) dx f(x) = dx δ(x x 0 )f(x) = f(x 0 ), (3.35) dx δ(x) df(x) dx 36 = df(x) dx. (3.36) x=0

37 The latter property further implies that dδ( x) dx = dδ(x) dx, x dδ( x) = δ(x). dx If the delta-function has a vector argument, this implies that it represents the product of the delta functions of the vector components: δ(x x 0 ) = δ(x x 0 )δ(y y 0 )δ(z z 0 ). (3.37) Comparison of relations for a discrete basis and for a continuous basis The following list compares the expressions for a discrete basis and a continuous basis (here the position basis has been chosen, but it could be any continuous basis, e.g., the momentum basis): Orthonormality: a a = δ a a x x = δ(x x ), Closure relation: a a a = 1 dx x x = 1, Expansion of an arbitrary state: α = a a a α α = dx x x α, Normalisation: a a α = 1 dx x α = 1, Scalar product of arbitrary states: β α = a β a a α β α = dx β x x α, Matrix elements of a matrix A with eigenstates { a (i) } : a (j) A a (i) = a (i) δ ij x x x = x δ(x x ). Question: Can you think of any observables (other than the position), which typically would have a continuous spectrum of eigenvalues? Task: Using equation (3.33), rewrite the closure relation for a discrete basis in wave function notation. Give a physical interpretation of the outcome. 37

38 3.3 The spin vectors and spin operators Explicit construction Already in section we wrote down expressions for the states S z, ±, S x, ±, and S y, ±. While we at that time postulated how they have to look like exactly, we will now construct them explicitly. The same we will do for the spin operators S x, S y, and S z. We start by again considering the third of the sequential Stern-Gerlach experiments in Figure 3.. Since the outcome of this experiment clearly shows that the silver atoms coming out of the Stern-Gerlach analyser with the magnetic field in the x-direction can be in both the S z, + and S z, states, and since the probability for both of these possibilities is 1/ we can write or, equivalently, S z, + S x, + = S z, S x, + = 1, S z, + S x, + = S z, S x, + = 1. (3.38) From equation (3.38) we get the normalisation factors for expressing the states S x, + and S x, in terms of the states S z, ±. By convention we furthermore set the complex phase factor of the S z, + state in the expressions for S x, + to 1, while the complex phase factor of the S z, is yet to be determined: S x, + = 1 S z, e iδ 1 S z,. (3.39) Since we know that, necessarily, S x, S x, + =0 (why?) we can easily obtain an expression for S x,. Suppose that S x, = A S z, + + B S z,, where A and B are complex constants. Again, by convention, A is set to 1, and only B has to be determined from equation (3.39). Since and thus S x, = 1 S z, + + B S z,, 38

39 S x, S x, + = 1 S z, + S z, eiδ 1 S z, + S z, + B S z, S z, + + B e iδ 1 S z, S z,, and since furthermore S z, + S z, + = 1, S z, S z, = 1, and S z, S z, + = S z, + S z, = 0, we find B = e iδ 1 and thus S x, = 1 S z, + 1 e iδ 1 S z,. (3.40) In the same way expressions are obtained for S y, ± : S x, ± = 1 S z, + ± 1 e iδ S z,. (3.41) δ 1 and δ can now be determined from the scalar product of S y, ± with S x, ±, which by symmetry can be determined from equation (3.38): S y, ± S x, ± = 1. Putting in the expressions for S x, ± and S y, ±, equations (3.40) and (3.41), one obtains 1 1 ± e i(δ 1 δ ) 1 =, i.e., δ δ 1 = ±π/. We choose δ 1 = 0 and δ = π/ and thus arrive at S x, ± = 1 S z, + ± 1 S z,, (3.4) S y, ± = 1 S z, + ± i S z,. (3.43) This means that the expressions for the components of the spin in the x and y directions now have been fully expressed in terms of the z-component of the spin. We still have to write the observables S x, S y, and S z in terms of the z- component of the spin, since such a formulation will tell us exactly how these observables act on the different spin states. This is straightforwardly done by using the fact that any observable can be written in terms of a sum of projectors onto the eigenstates of the observable. This is seen by applying the closure relation (3.4) two times to an observable A: 39

40 A = a a a a A a a = a a a a, (3.44) since a A a = a δ a a. The eigenvalues of S z (and likewise of S x and S y )are known from experiment to be ± h/. Since the eigenstates of S z are S z, + and S z, one readily obtains S z = h { S z, + S z, + S z, S z, }. (3.45) The expressions for S x and S y are obtained in the same way starting from S x,y = h { S x,y, + S x,y, + S x,y, S x,y, }, and then putting in the expressions for S x, ± and S y, ±, equations (3.4) and (3.43), respectively. This gives S x = h { S z, + S z, + S z, S z, + }, (3.46) S y = h { i S z, + S z, + i S z, S z, + }. (3.47) 3.3. The commutators and anticommutators of the spin operators Definition of commutators and anticommutators We will see shortly that the commutators of observables play an immensely important role in quantum mechanics. The commutator [A, B] of two operators A and B is defined as Likewise, the anticommutator {A, B} is defined as Exercise It is now your task to show that [A, B] AB BA. (3.48) {A, B} AB + BA. (3.49) where [S i, S j ] = iε ijk hs k, (3.50) 40