Lectures on Quantum Mechanics

Transcription

1 Giuseppe E. Santoro Lectures on Quantum Mechanics ICTP Diploma Course Printed on: May 7, 01

2 Preface These are just short notes intended as a guide to the Diploma Students taking their course in Advanced Quantum Mechanics.

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4 Contents 1 The basics of Quantum Mechanics A brief history 6 1. Gaussian wave-packets The Schrödinger equation: free-particle The Schrödinger equation The time-independent Schrödinger equation 14 One-dimensional problems 17.1 Infinite square well 17. Scattering by a step 18.3 Finite square well 19.4 Delta-function potential 0 3 Harmonic oscillator Commutators and Hermitean conjugates 1 3. Dimensionless variables 3.3 The algebraic solution 3 4 The hydrogen atom problem Orbital angular momentum 5 4. Separation of variables Dimensionless variables The Coulomb problem bound states 9 5 Theory of angular momentum Diagonalization of J and J z 3 5. Construction of spherical harmonics 33 6 Symmetries Space transformations Transformations on scalar functions 36

5 4 Contents 6.3 Symmetry properties of Hamiltonians Commuting observables Selection rules 41 7 Measurement in Quantum Mechanics Measurements in QM Expectation values of measurements Spin Stern-Gerlach experiment: the spin operators and states Experiments with Stern-Gerlach apparatus A single spin-1/ in a uniform field: Larmor precession Heisenberg representation 50 9 Spin-orbit and addition of angular momenta Relativistic effects 5 9. The origin of spin-orbit Construction of J-states out of p-states Generalization: addition of L and S Density matrices The density matrix for a pure state The density matrix for a mixed state Density matrices and statistical mechanics Density matrices by tracing out the universe Time-independent perturbation theory Two-state problem Non-degenerate case: general solution Degenerate case: first order Application: the fine-structure of Hydrogen 68 1 Time-dependent perturbation theory General case First-order solution Electromagnetic radiation on an atom 75

6 Contents 5 13 Many particle systems Many-particle basis states for fermions and bosons Two-electrons: the example of Helium. 81 Appendix A Second quantization: brief outline. 84 A.1 Why a quadratic Hamiltonian is easy. 88

7 1 The basics of Quantum Mechanics 1.1 A brief history The history of quantum mechanics starts right at the beginning of the The years from 1900 to 1930 might be really seen, as the title of a nice book by Gamow says, as the 30 years that shook physics. This story has to do with light (more generally, electromagnetic radiation) as well as electrons and atoms: while a full quantum treatment of light, however, necessarily requires a relativistic quantum theory (which we will not enter into), electrons and atoms can be very well described by a non-relativistic quantum theory, which will be the subject of our course. I urge you to read the first chapter of any of the many books in QM, for instance that of Gasiorowitz, for a full account of this very interesting story. I will be here very sketchy. The story itself, begins really with electromagnetic (e.m.) radiation, more precisely with the puzzle of the black-body radiation, an idealized furnace (oven) where the spectrum of the radiation inside was experimentally well known but escaping any classical physics explanation. It was Max Planck, in 1900, who postulated (very boldly, at that time), that everything could be explained if one assumed that the energy of an e.m. field of frequency ν was not arbitrarily and continuously changeable, but only multiples of the fundamental energy E = hν = ω, (1.1) h being the famous Planck s constant (ω = πν, and = h/π) would be admitted. This fixed all the difficulties that physicists had encountered in the black-body radiation problem. Constructing on this idea, soon after (in 1905) Einstein was able to explain in a very natural way the photoelectric effect: electrons are kicked out from the metal by individual collisions with a single quantum of radiation (a photon ): if the energy E = hν of the photon is not larger than the work function W (by definition, the energy you have to provide to extract an electron from a metal), then no electron will come out, no matter how intense you make the field (by shining very many photons on your metal). The final experimental confirmation of the particle-like nature of the e.m. radiation came with Compton (in 193), who performed a true scattering experiment of X-rays from electrons, where all the results could be rationalized by applying energy and momentum conservation to the collision event; the momentum that appeared to be associated to the X-ray had to be: where k = π/λ is the wave-number. p = E c = hν c = h λ = h π = k, (1.) π λ In the atomic world, the first bricks of a quantum theory came with Bohr, in 1913 (at that time a post-doc of Rutherford), who solved the mystery of the atomic spectra seen in emission and adsorption. Rutherford had put forward a planetary model for the atoms, with the nucleus occupying only a small central volume and the electrons orbiting around it. However, why those orbiting electrons would not emit radiation (being accelerated) as the electrons in a synchrotron do, was a mastery. Bohr elaborated a model for quantifying the allowed orbits, and their energies

8 Gaussian wave-packets 7 E n (which were, once again, discrete) in such a planetary system, with results which were in surprising agreement with spectral observations. In particular, the various radiation lines observed could be accounted for as differences between the quantized energy values E n that Bohr had calculated, as: hν n,m = E n E m. (1.3) In 195 de Broglie realized that the orbits of Bohr s atom appeared to possess a nice wave-like feature: an orbit of energy E n (with n = 1,, ) had a radius r n and a circumference πr n which appear to contain exactly n wavelengths λ = h/p (p being the momentum of the electron in the orbit): πr n = nλ. (1.4) This mysterious wave-like nature of electrons was revealed later on (before 1930) in diffraction experiments, by Davisson and Gerner, and, independently, by G. Thompson (son of the Thompson who discovered the electron!). Notice that electrons of energy E = 160 ev have momenta p (in c.g.s), and therefore the resulting λ 10 8 cm. de Broglie was, however, too slow in elaborating a full mathematical treatment of his electron waves. In 196, Schrödinger came out with a beautiful theory that put de Broglie insight on a firm ground, and explained in a very natural way the results of Bohr. 1. Gaussian wave-packets Question: how are we supposed to explain the motion of a material particle with a wave? A single plane-wave e ikx cannot certainly be representing anything localized, because it is periodic in space, with a period λ = π/k. Therefore, in order to look for something localized, which might be exchanged for a point-particle, at least from reasonably far away, we must superimpose many plane waves and create a wave-packet. We do that, by taking a superposition of the form: ψ(x) = + dk g(k)e ikx. (1.5) π g(k) can in principle be any weight function representing how much of the wave e ikx is included in the superposition, for each k. In order to carry out the calculations explicitly, we will assume that g(k) has the simple Gaussian form: g(k) = Ce α(k k0), (1.6) where C is a normalization constant which we choose so that dk g(k) = 1. The reason why we impose that the square of g(k) is normalized, and not g(k) itself, will be clear in a while. Now, the simple Gaussian integral that you certainly have encountered many times is: dk e ak = 1. (1.7) π a Based on this, it is simple to show that: C = ( ) 1/4 α. π By defining σ k = 1/(4α), it is also quite simple to show that σ k is the width of the function g(k), i.e., its second moment is: (k k 0 ) = dk (k k 0 ) g(k) = C Prove, as an exercise, the latter identity. dk (k k 0 ) e (k k 0) σ k = σk. (1.8)

9 8 The basics of Quantum Mechanics Still based on the Gaussian integral in Eq. (1.7), by completing the square, it is a simple matter to show the following generalization: dk π e ak e kj = 1 a e J a. (1.9) There is no restriction on J being a real quantity, in this integral. So, if we substitute J = ix, where x is real, the equality still holds, and reads: dk π e ak e ikx = 1 a e x a. (1.10) This very useful identity, expressed in words by saying that the Fourier transform of a Gaussian in k-space is a Gaussian in x-space, with an inversely proportional width, will be used now to calculated ψ(x). Indeed, a straightforward calculation leads to: + ψ(x) = Ce ik0x dk e αk e ikx = π ( ) 1/4 1 e ik0x e x 4α, (1.11) πα i.e., ψ(x) is a plane-wave e ik0x with a Gaussian envelope. If we consider ψ(x), it will be a simple Gaussian with a width σ x = α. Notice that the widths of the distributions in k- and x-space are related by a simple inverse proportionality, i.e.: σ x σ k = α 1 α = 1. (1.1) We will see in a while that this is a realization of a minimum Heisenberg-uncertainty-principle wave-packet (we will see that Heisenberg uncertainty principle requires x p /). Let us now consider the time dependence of this wave-packet. For a single plane wave e ikx the time-dependence has the form: ψ k (x, t) = e i(kx ω kt), where ω k is the angular frequency of the wave. Such a form guarantees that the wave moves with a phase velocity v k = ω k k, since, indeed, ψ k (x, t) = e ik(x v kt), i.e., the wave amplitude depends only on x v k t. Now consider the full time evolution of ψ(x), which naturally reads: ψ(x, t) = + dk g(k)e i(kx ωkt) = π + dk g(k)ψ k (x, t). (1.13) π The crucial question is the choice of the k-dependence of ω k. For the case of an electromagnetic wave in vacuum, we have ω k = kc, and the phase velocity of each plane-wave component is identical to the speed of light: v k = c. As a consequence, it is very simple to prove that, in such a case, the wave would move rigidly, undistorted, with the speed of light, i.e., ψ(x, t) = ψ(x ct, 0). One refers to such a case as non-dispersive. On the contrary, whenever each k-component moves with a different phase-velocity v k, the resulting wave has a dispersive behaviour: we will soon show that this implies a rigid motion with a group velocity v g = dω k /dk plus an inevitable spreading of the envelope. Clearly, to possibly describe the motion of a non-relativistic particle, we have to chose this second route. In order to carry on the calculation, we will need to Taylor expand ω k (which is in general a non-linear function of k) around the peak wave-vector k 0. To fully justify this expansion,

10 The Schrödinger equation: free-particle 9 let us assume that α is large, so that g(k) is quite narrow around its peak at k = k 0. The Taylor expansion of ω k around k = k 0, up to second order, reads: where ω k = ω k0 + v g (k k 0 ) + β(k k 0 ) +, (1.14) v g = dω k dk, (1.15) k0 and β = (1/)d ω k /dk k0. By substituting these expressions in ψ(x, t) we easily get: ψ(x, t) = Ce i(k0x ω k 0 t) + dk e (α+iβt)k e ikx = π C (α + iβt) e i(k0x ω k 0 t) e (x vg t) 4(α+iβt), (1.16) where we have used that fact that the Fourier transform in Eq. (1.10) holds even when a is complex, as long as with a positive real part (so that the integrand does not blow up), which is the case here, since α > 0. Consider now ψ(x, t). It reads: ψ(x, t) = 1 α α(x vg t) π α + β t e (α +β t ). (1.17) Notice that, since β 0, there is a spreading of the wave-packet, i.e., its width inevitably increases: this is completely independent of the sign of β. Now, the velocity with which the center of the wave-packet moves should be related to the classical momentum in the usual non-relativistic way: v g = dω k dk = p 0 k0 m = k 0 m, (1.18) where we have used de Broglie s identification of the momentum with k 0. It is then clear that the only way this identity holds is that: which is the standard non-relativistic kinetic energy of a particle. E k = ω k = k m = p m, (1.19) Exercise 1.1 Consider the problem of spreading of a Gaussian wave-packet for a free non-relativistic particle. Calculate how much the width of the wave-packet changes in t = 1 second in the following two cases: a) the particle is an electron (m = kg) and the initial wave-packet has dimensions x = 10 5 m, or b) x = m; c) the particle has mass m = 10 3 kg and initial size x = 10 3 m. It is convenient to calculate, in all cases, the fractional change of the width given by: x t x x. 1.3 The Schrödinger equation: free-particle Let us now verify that a wave-packet of the form given in Eq. (1.16) or more generally one obtained from Eq. (1.13), as long as ω k has the non-relativistic form in Eq. (1.19) satisfies the following Schrödinger equation ψ(x, t) i t = ψ(x, t), (1.0) m where = / x + / y + / z is the Laplacian (in three-dimension, in one-dimension only the x-component survives). Verifying this statement is very simple. We just prove that it is true

11 10 The basics of Quantum Mechanics for a single plane-wave ψ k (x, t) = e i(k x ω kt), and then, by linearity of the Schrödinger equation, it will be true also for any combination of plane-waves, as in Eq. (1.5). Verifying that ψ k (x, t) solves Eq. (1.0) is very simple and is left as an exercise (done in class). Moreover, if we define the momentum operator as: ˆp = i, (1.1) it is simple to show that the plane-waves are eigenstates of the momentum operator (remember de Broglie): ˆp e ik x = ( k) e ik x. (1.) It is also simple to show that the term appearing in Eq. (1.0) can be also written as: m = ˆp m, (1.3) i.e., the same form as the kinetic energy of a free non-relativistic particle. 1.4 The Schrödinger equation In general, for a particle which is subject to a potential V (x), the Schrödinger equation (SE) reads: ( ) ψ(x, t) ˆp i = t Ĥψ(x, t) = m + V (x) ψ(x, t). (1.4) The operator Ĥ appearing in this equation is the Hamiltonian of the system, which has the standard non-relativistic form, except that it is an operator, acting of functions (wave-functions). We must think of the problem in terms of a space of functions (the ψ s) on which we act with linear operators (ˆp, ˆx, Ĥ). The SE is first-order in time. So, given any initial wave-function ψ(x, t = 0), we can in principle calculate ψ(x, t) at any later time t > 0 in a perfectly deterministic way. There is a formal way of expressing this solution, in terms of the so-called evolution operator. Let us consider for simplicity the case in which the Hamiltonian is time-independent. (If the Hamiltonian is time-dependent, as in the presence of external varying fields, the evolution operator can be still defined, but its expression is more complicated.) In general, the exponential of an operator Ô is defined as: def eô = 1 + Ô +!Ô 1 + = n=0 1 n!ôn, (1.5) where 1 is the identity operator, i.e., multiplying by 1. One should be careful, because Some of the usual rules of ordinary exponential functions do not hold for exponential of operators, for instance: eô1 eô eô1+ô, (1.6) unless the two operators commute (i.e., Ô 1 Ô = ÔÔ1). However, some other rules do apply here too. For instance: d dt eôt = ÔeÔt = eôt Ô. (1.7) The proof of this equality (done in class) is very simple, starting from the Taylor expansion of the exponential.

12 The Schrödinger equation 11 Consider now the following operator: Û(t) = e iĥt. (1.8) It is known as evolution operator, for a reason that will be clear in a moment. First of all, one immediately has Û(t = 0) = 1, and: As a consequence, if we define the time-evolved state ψ(x, t) = i d iĥt iĥt e = Ĥ e. (1.9) dt Û(t)ψ(x, t = 0) = e iĥt ψ(x, t = 0), (1.30) it is a very simple matter to prove (done in class) that it satisfies correctly the initial-time condition at t = 0 and the SE. We ask now how to define, given the fact that our basic objects are wave-functions, the probability density P (x, t) for finding the particle at some position x at time t. Obvious requirements for P (x, t) are: a) P (x, t) 0 ( ) d b) dx P (x, t) = 0 dt c) dx P (x, t) = 1. (1.31) The first requirement is common to all probability densities (real and non-negative functions); the second tells us that the total probability is conserved in time. As a consequence of this conservation, we will be able to rescale P in such a way to impose the standard normalization of the total probability given in c) above. How is P (x, t) related to ψ(x, t)? For one thing, ψ(x, t) is inevitably a complex wave-function! Even if we start from a real ψ(x, t = 0) the SE-evolution will make ψ(x, t) complex. A possible guess would be to take P = ψ(x, t), but we will show that any such guess will violate the conservation requirement in b), except for the following one suggested by Max Born (in strict analogy with the energy density of the electromagnetic field, in terms of the field itself): P (x, t) = ψ(x, t) = ψ (x, t)ψ(x, t). (1.3) To show this in a nice way, it is better to introduce a few formal tools, which will be anyway useful in the following. Notice that this form immediately implies interference between waves, because ψ 1 (x, t) + ψ (x, t) ψ 1 (x, t) + ψ (x, t) A few formal tools First of all, we introduce in the space of wave-functions ψ a structure of a linear space, which is rather obvious: if c 1 and c are arbitrary constants, and ψ 1, ψ wave-functions, then c 1 ψ 1 + c ψ is also a possible wave-function. Moreover, we introduce in this space a scalar products, by analogy with the usual euclidean scalar product of ordinary vectors: ψ 1 ψ def = dx ψ1(x)ψ (x) = ψ ψ 1. (1.33) In this way, the norm ψ of a function is defined as: ψ ψ = dx ψ(x) = ψ. (1.34) The mathematicians call this norm (there are many other possible scalar products and norms that you can introduce) the L -norm. In this way, the linear space of all possible normalizable wavefunctions (i.e., with ψ < ), forms a space known as Hilbert space, which we will denote by H

13 1 The basics of Quantum Mechanics (to distinguish it from the Hamiltonian H or Ĥ). Notice immediately that our beloved plane-waves ψ k (x) = e ik x are immediately rather bizarre, being ψ k =. We will keep using them because they are useful as states on which we expand physical normalizable wave-functions (remember the wave-packet discussion, and the Fourier expansion); we will also soon learn how to regularize them properly on a finite-volume, through a procedure very familiar in condensed-matter (periodic boundary conditions). Of all the linear operators that one can imagine to act on the wave-functions, of particular importance are the so-called Hermitean operators, which are defined as those operators such that, for any ψ 1 and ψ, satisfy: ψ 1 Ôψ = Ôψ 1 ψ, (1.35) i.e., the operator Ô is Hermitean if it can move back-and-forth on the two sides of the scalar product. Many (indeed all!) of the operators encountered so far are Hermitean. To start with, the operator ˆx, which simply multiplies by x a ψ, is Hermitean because x is real, and therefore (try writing both terms explicitly to fully convince yourself): ψ 1 ˆxψ = ˆxψ 1 ψ. (1.36) Similarly, any real potential V (x) corresponds to an Hermitean operator. Perhaps surprisingly (at least at a first sight, because of the i), the momentum operator is Hermitean: the very simple proof (done in class) uses integration by parts and the extra minus sign in it makes for the necessary complex conjugate of i: ψ 1 ˆpψ = ˆpψ 1 ψ. (1.37) Powers of an Hermitean operators are Hermitean (just move one operator at a time from the right to the left of the scalar product). Therefore: Also, the sum of two Hermitean operators is Hermitean: ψ 1 ˆp ψ = ˆpψ 1 ˆpψ = ˆp ψ 1 ψ. (1.38) ψ 1 (Ôa + Ôb)ψ = (Ôa + Ôb)ψ 1 ψ, (1.39) as the linearity of the scalar product immediately shows. Finally, the Hamiltonian Ĥ of a particle in a potential is Hermitean because it is the sum of two Hermitean operators, ˆp and V (x). Exercise 1. Of the following operators acting on functions, tell which ones are linear operators, and, out of these, which ones are Hermitean operators: 1) O 1ψ(x) = x 3 ψ(x), ) O ψ(x) = xdψ/dx, 3) O 3ψ(x) = λψ (x), 4) O 4ψ(x) = e ψ(x), 5) O 5ψ(x) = dψ/dx + a, 6) O 6ψ(x) = R x dx (x ψ(x )). With these definitions and with the use of the SE and of its conjugate: it is very simple to show that: i ψ (x, t) t = (Ĥψ(x, t) ), (1.40) i d dx ψ (x, t)ψ(x, t) = ψ Ĥψ Ĥψ ψ = 0, (1.41) dt where the last equality follows from the fact that Ĥ is Hermitean.

14 The Schrödinger equation 13 Before leaving this formal section, let us introduce the concept of expectation value of an operator. By analogy with classical probability, a natural choice for the average position at time t < x > (t) is: < x > (t) = dx xp (x, t) = dx ψ (x, t)xψ(x, t) = ψ(t) ˆxψ(t). Similarly, and more generally, the mean value (or expectation value) of an operator Ô on a state ψ(t) is defined as: < O > (t) def = ψ(t) Ôψ(t) = dx ψ (x, t)ôψ(x, t). (1.4) Notice that Ô might involve derivatives, which should be applied only to the state to its immediate right, and therefore, generally speaking, expectations values of operators cannot be expressed as integrals where the probability P (x, t) appears! Exercise 1.3 Prove that the expectation value of an Hermitean operator is a real quantity (done in class). Exercise 1.4 Write down the expectation value of the momentum operator, and observe explicitly that: 1) it is real even though there is an imaginary unit i in its expression; ) it cannot be expressed in terms of P (x, t). A final word on the notation. The Dirac s notation is strictly related to what we have written above, except for adding an extra : ψ 1 Ôψ = ψ 1 Ô ψ, (1.43) where the right hand side is in the Dirac s notation, with the operator sitting between a bra ψ 1 and the ket ψ. Exercise 1.5 Define the uncertainty in x and p on a state ψ with the usual rules: ( x) = ψ x ψ ψ x ψ, ( p) = ψ p ψ ψ p ψ. Next define the following quantity: Z D(α) = dx (x x )ψ(x) + iα(p p )ψ(x) 0, which is non-negative, by inspection, for every value of α. Expand the modulus square appearing in D(α) to show that: D(α) = ( x) + α ( p) α 0. From the fact that this quadratic function of α is always non-negative, deduce something on the discriminant of the parabola, and show that Heisenberg s uncertainty principle follows: x p.

15 14 The basics of Quantum Mechanics 1.4. The continuity equation and the current To convince ourself that ψ(x, t) is the fundamental quantity of Quantum Mechanics, and not P (x, t), let us write down the partial differential equation that P obeys. By using the SE and its conjugate it is immediate to show that: t P (x, t) = t (ψ (x, t)ψ(x, t)) = 1 i ( ) ψ (Ĥψ) (Ĥψ) ψ. (1.44) Now, you can verify directly that the potential V (x) disappears completely from this expression, and only the kinetic term survives: i ( P (x, t) = ψ (x, t)( ψ(x, t)) ( ψ(x, t)) ψ(x, t) ). (1.45) t m It takes now a small calculus exercise, based on identities like (ψ ψ) = ψ ψ + ψ ψ, to show that the right hand side of the latter equation is nothing but the divergence of a current J(x, t): t P (x, t) = J(x, t) J(x, t) = i m ( ψ (x, t)( ψ(x, t)) ( ψ(x, t)) ψ(x, t) = 1 m (ψ (x, t)(ˆpψ(x, t)) + (ˆpψ(x, t)) ψ(x, t)). (1.46) In this form, the equation for P expresses the differential form of the conservation of probability, and is nothing but the familiar continuity equation which you encounter in electrodynamics. ) 1.5 The time-independent Schrödinger equation Suppose the system is isolated, so that its Hamiltonian is time independent, Ĥ. (If the system is subject to external fields, for instance an external electromagnetic field, then the Hamiltonian is explicitly time-dependent, Ĥ(t), and the evolution operator is not the simple exponential we have considered previously.). We already know that the formal solution of the SE, given an initial-time ψ(x, t = 0), can be written as: ψ(x, t) = Û(t)ψ(x, t = 0) = e iĥt ψ(x, t = 0). (1.47) Our question, now, is if and when such a wave-function can factorized into a space and time function: ψ(x, t) = f(t)φ(x). (1.48) It is now difficult to show (done in class) that the only possibility this can happen is that φ(x) = ψ(x, t = 0) is one of the eigenstates of Ĥ Ĥφ n (x) = E n φ n (x), (1.49) and, as a consequence, the time-dependent solution factorizes with f(t) = e ient/, i.e.: ψ(x, t) = e ient/ φ n (x). (1.50) Notice that, for these solutions, P (x) = ψ(x, t) = φ n (x) is time-independent, and, therefore, the current must have zero divergence: J n (x) = 0. A general mathematical theorem guarantees that the full set of eigenstates φ n of Ĥ provides an orthonormal basis set for the Hilbert space: we will see several examples of this statement, with

16 The time-independent Schrödinger equation 15 n being a discrete set of labels, or even a continuous one. We will use a notation, for simplicity, where n is assumed to be discrete. The fact that the energy eigenstates form an orthonormal basis means that: φ n φ m = δ n,m, and that every function ψ 0 (x) can be expanded in terms of the φ n : ψ 0 (x) = n c n φ n (x). Indeed, the expansion coefficients c n are simply given (by using the orthonormality of the basis) as: c n = φ n ψ 0. Armed with the knowledge of the basis set {φ n }, we can readily solve any time-dependent Schrödinger problem with assigned initial condition ψ 0 (x): ψ(x, t) i = Ĥψ(x, t) t ψ(x, t = 0) = ψ 0 (x). (1.51) Indeed, all we need to do is to expand the initial condition ψ 0 (x) on the eigenstates φ n (x), ψ 0 (x) = n c nφ n (x), and write down the state at time t as: ψ(x, t) = n c n e ient/ φ n (x). (1.5) Free-particles and plane-wave normalization. The simplest case it that of a free particle. Here the energy eigenvalues are simply E k = k /(m), associated to wave-functions φ k (x) = e ik x. Notice that here k stands for the index n of the previous section, and is in general a continuous multidimensional label. The spectrum of Ĥ, that is the set of all its possible eigenvalues, is here continuous, and extends in the region E [0, + ). A few observations: 1) while k univocally specifies φ k, the same energy E k = E is obtained by, in general, infinitely many k, all those with k = k E = me/ (a sphere in 3d, a circle in d, and points in 1d). ) All the φ k are not normalizable in the infinite volume case. A trick often used in condensed matter is to consider the system as confined to a cube of side L, hence of volume Ω = L 3 in 3d (Ω = L d, in general dimension d), with periodic boundary conditions (PBC), i.e., we postulate that: φ(x + Le i ) = φ(x) where e i is any of the three coordinate versors. This requirement (done in class) restricts the possible allowed k-vectors to a discrete (although still infinite) set: k = π L (n 1, n, n 3 ), (1.53) with n i Z (i.e., positive and negative integers, or zero). The correctly normalized free-particle eigenstates with PBC in the cube will then be: φ k (x) = 1 Ω e ik x, (1.54) with the k given by Eq Notice that now the φ k form a nice orthonormal basis for periodic functions in the cube of volume Ω, since: φ k φ k Ω = 1 dx e i(k k) x = δ k,k. Ω Ω

17 16 The basics of Quantum Mechanics The fact that φ k forms a basis set, simply a restatement of the fact that any periodic function ψ(x + Le i ) = ψ(x) can be expanded in Fourier series: ψ(x) = k φ k (x) φ k ψ = 1 e ik x ψ k, (1.55) Ω where the Fourier coefficient ψ k is simply given by: ψ k = Ω φ k ψ = dx e ik x ψ(x). (1.56) It is now quite simple to take the thermodynamic limit L. Consider, indeed, the wavefunctions: 1 φ k = eik x. (π) d Their overlap, restricted to the volume Ω, tends towards the Dirac s delta function: φ k φ k Ω = Ω (π) d φ k φ k Ω = Ω (π) d δ Ω k,k δ(k k ). To prove this, you should start from the natural discretization of the integral dk, which is: Ω ( ) d π Ω dk, L k k and show that, for any function f(k): ( ) d π f(k) = f(k Ω ) L (π) d δ k,k k Ω dk f(k )δ(k k ). As a consequence, you can take the limit of the Fourier sum in Eq. 1.55, obtaining: ψ(x) = 1 e ik x Ω dk ψ k eik x Ω (π) d ψ k. (1.57) k The Ω limit of the Fourier coefficient ψ k in Eq gives: ψ k = dx e ik x ψ(x), (1.58) which we recognize to be the Fourier transform ψ k of the function ψ(x). Eq simply defines the inverse Fourier transform. Notice that, when L, k becomes a continuous vector.

18 One-dimensional problems Let us begin with some general remarks on the time-independent SE. First of all, the spectrum of the Hamiltonian (i.e., the set of all its eigenvalues) is constrained to be not below the minimum of the potential (if one exists): E n V min. The proof of this is simple (done in class). Quickly, one can show that if Hψ n (x) = E n ψ n (x) for an eigenstate, then by taking the scalar product with ψ n (assumed to be normalized) we get: ( E n = dx ψn ) ( m ψ n + V (x) ψ n (x) ψ = dx ) n m + V (x) ψ n(x), (.1) from which it is immediate to conclude that E n cannot be less then V min, because kinetic energy is in general positive, and the wave-function is never really localized on the minimum of V. A second general consideration regards bound versus extended states. A bound state is associated to a discrete eigenvalue of H, and to a normalizable wave-function, usually quite well localized in space. On the contrary, an extended state, exemplified by the familiar plane-wave e ik x, is not normalizable in the infinite volume, and associated to an eigenvalue belonging to a continuous part of the spectrum. We have already seen this for the free particle. We will see more examples of both shortly. A third general remark has to do with the continuity conditions of a solution of the SE at the boundary between regions were the potential has some discontinuity. Generally speaking, a solution of the SE must be continuous, with a continuous first derivative (indeed, ψ should admit a second derivative!). For simplicity, assume we are in one-dimension. Suppose that x 0 is a point were the potential has possibly a finite (jump) discontinuity V. By integrating the SE over the interval [x 0 ɛ, x 0 + ɛ], writing: m x0+ɛ x 0 ɛ dx ψ (x) = m [ψ (x 0 + ɛ) ψ(x 0 ɛ)] = x0+ɛ x 0 ɛ dx [E V (x)]ψ(x), and taking the limit ɛ 0 we can quickly prove that the first derivative ψ (x) is continuous at x 0, unless the potential V (x) has an infinite jump V at x 0, or some delta-function contribution..1 Infinite square well The first problem we have solved in class is that of an infinite square-well in the region [0, L]. In other words, the potential is infinite for x < 0 and x > L, and V = 0 in [0, L]. Due to the fact that ψ(x) must be 0 for x < 0 and x > L and it must be continuous everywhere (no exception to this rule), we concluded that the appropriate boundary conditions for ψ(x) are: ψ(0) = ψ(l) = 0. We solved this problem by taking combinations of e ikx and e ikx (in the free-particle region [0, L]) of the form:

19 18 One-dimensional problems ψ k (x) = A k sin kx. The boundary condition ψ k (0) = 0 is automatically satisfied. As for the condition ψ k (L) = 0, it implies sin kl = 0, or: k = nπ with n = 1,, (.) L Negative n s do not bring extra solutions, because sin kx = sin kx. The corresponding energy eigenvalue is E k = k m = π n ml. (.3) Each solution is non degenerate, i.e., only one eigenstate is associated to each eigenvalue. Notice that the spectrum is now purely discrete, as an effect of the confining potential. For L we recover a dense continuous spectrum in [0, + ). We discussed some properties of the box wave-functions. First of all, we showed that ψ k ˆp ψ k = 0 (because you can always take the ψ k to be real). Next, we calculated ψ k ˆp ψ k in terms of energy eigenvalues by using the fact that ˆp = mĥ. We commented on the Heisenberg uncertainty principle, showing that ( x) ( p) grows with the quantum number n. We also commented on the increasing number of zeroes of the eigenstates, and the corresponding kinetic energy increase, since you can always re-express the kinetic energy as an integral of ψ. Exercise.1 Solve the same infinite square well problem by choosing the origin in the center of the region, which is therefore [ L/. + L/], showing that now you have cos kx and sin kx solutions, alternating (the ground state is even, the first excited state is odd, and so on). Plot the first eigenstates. Calculate the eigenstate normalization factors, and show the orthogonality of different eigenstates. Obviously, the eigenfunctions of the Hamiltonian form a orthonormal basis in the chosen interval: they are the basis element of the Fourier series. For instance, with the symmetric choice [ L/, +L/] the Fourier series reads: ψ(x) = ( nπ ) a n cos L x + ( nπ ) b n sin L x. (.4) n=1,3, n=,4, Exercise. Write the expressions for the Fourier coefficients a n and b n, taking proper care of the correct normalization of the basis functions cos kx and sin kx.. Scattering by a step Next, we did a scattering problem: a potential step of V 0 > 0 for x > 0. There is no loss of generality in doing so: if V 0 < 0 you can reset the zero of energy and view it as a step on the left of the origin. Since the minimum of the potential is 0, we have E 0. We should distinguish two cases: i) E > V 0 and ii) E V 0. For the case i), E > V 0 we have that in both x < 0 and x > 0 the problem is a free-particle problem, but with energy, respectively, E and E V 0 (because you can bring the constant potential on the right-hand side of the SE). We therefore consider candidate solutions of the form: { e ψ k (x) = ikx + re ikx for x 0 te iqx for x 0, (.5) with positive k and q, so that the ψ k describe a wave incoming from the left of the origin, partially reflected (reflection amplitude r), and transmitted to the right of the origin (transmission amplitude t). The overall normalization factor is not important (this is why we put the coefficient of e ikx to 1) because the ψ k is intrinsically not-normalizable, as any extended (unbound) state is. In order

20 Finite square well 19 to be a solution with energy E, we must have (verify) E = k /m and E V 0 = q /m. So, for any given E, k > 0 and q > 0 are univocally defined: k = me, and q = m(e V 0 ). The continuity of ψ k (always true) and of ψ k (because the potential has just a finite jump), lead to the following two equations: t = 1 + r tq = k(1 r), (.6) whose solution gives: t = k k + q and r = k q k + q. (.7) The other case, ii) E < V 0, is solved in a similar way, but with the choice of an evanescent wave te qx for x 0, with q = m(v 0 E). The calculations are very similar to case i), and simply require substituting q iq, obtaining: t = k k + iq and r = k iq k + iq. (.8) We noted that r is a pure phase factor, and that, for V 0 +, we have q +, r 1 and t 0, and the wave-function becomes i sin kx for x 0, and vanishes for x 0, i.e., it is still continuous in 0, but its derivative becomes discontinuous in the origin (an infinite V 0 does not preserve continuity of ψ. We discussed the degeneracy of solutions, the physical meaning of the reflection and transmission amplitudes r and t, the expression for the current (which must be constant everywhere)..3 Finite square well Next we considered the case of a finite, symmetrical, potential well in [ a, a]: the potential is set to 0 in [ a, a], and is equal to V 0 for x a. Once again we must have E 0, and we can have two cases: i) E < V 0 (search of bound states), and ii) E > V 0 (scattering states, with transmission and reflection coefficients). In case i), E < V 0 we write the wave-function as: ae qx for x a ψ k (x) = be ikx + ce ikx for x a de qx for x a, (.9) with k = me, and q = m(v 0 E). Next we discussed how to use parity, taking d = a and c = b for even solutions, and d = a and c = b for odd solutions. In both cases, once continuity of ψ and ψ are imposed at point x = a, they are automatically guaranteed at the other point x = +a. We arrived at the (graphical) solution for the bound-state eigenvalues, proving that there is at least one bound state independently of a and V 0, while the total number of bound states present (both even and odd) depends on the depth of the well. Exercise.3 Solve the finite symmetric square well for the scattering states with E > V 0, by setting up, as usual in scattering problems, ψ k (x) = e iqx + re iqx for x a, ψ k = ae ikx + be ikx for x a, and ψ k (x) = te iqx for x +a, with k = me, and q = p m(e V 0). Calculate the current everywhere, and deduce a relationship between t and r. Show that there are peculiar energy values (corresponding to resonances for the infinite square well) at which the transmission coefficient becomes 1 (resonant tunneling).

21 0 One-dimensional problems Exercise.4 By exploiting the results of the symmetric finite potential well (in particular, the odd solutions of it) solve the case of a square well with an infinite wall in the origin, i.e., such that the potential is + for x < 0, 0 for [0, a], and V 0 for x > a..4 Delta-function potential Finally, we considered a rather singular potential, an attractive delta-function in the origin: V (x) = V 0 δ(x) with V 0 > 0. We discussed how the boundary condition for the derivate ψ should be imposed, and found the bound state of the problem. Exercise.5 Discuss scattering solutions with E 0 for both the delta-function potential, in both attractive and repulsive cases.

22 3 Harmonic oscillator The problem we consider now is very important, because it arises in many contexts as an approximation to the real, more complicated, potential (close to a minimum). The Hamiltonian reads: Ĥ = P m + 1 mω X = H x + H y + H z, (3.1) where we have explicitly indicated the separation into three one-dimensional Hamiltonian each referring to a given Cartesian component (α = x, y, z): H α = P α m + 1 mω X α. (3.) Obviously, these three Hamiltonians commute among each other. We now illustrate how to solve the 3D problem by the method of separation of variables. It is easy to prove (done in class) that if we solve the one-dimensional problems: H α ψ (α) n (X α ) = E n (α) ψ n (α) (X α ), (3.3) then Ψ n,m,l (X, Y, Z) = ψ n (x) (X)ψ m (y) (Y )ψ (z) l (Z) is an eigenstates of Ĥ with energy E n,m,l = E n (x) + E m (y) + E (z) l. So, it is enough to solve a one-dimensional harmonic oscillator (indeed, one might generalize a bit the 3D Hamiltonian and consider a different ω α for each component). Omitting all indices, we write the Hamiltonian as: 3.1 Commutators and Hermitean conjugates H = P m + 1 mω X. (3.4) The concept of commutator is central to QM. The order in which two operators are applied to a wave-function matters. Even in the finite-dimensional case, when the operators are simple matrices, this is so. For instance, consider two matrices A, and B defined as follows: A = ( ) , B = ( ) (3.5) Then, it is easy to show that: ( ) 0 1 A B = 1 0, B A = ( ) (3.6) We define the commutator of A and B as: [A, B] def = A B B A. (3.7) Consider, for instance A = ˆX and B = ˆP. When acting on a generic wave-function ψ(x):

23 Harmonic oscillator However: ˆX ˆP ψ(x) = ( i )X d dx ψ(x). ˆP ˆXψ(X) = ( i ) d ( Xψ(X) = ( i ) ψ(x) + X d ) dx dx ψ(x) =. Therefore, on any ψ(x) we have that: [ ˆX, ˆP ] = i. (3.8) Exercise 3.1 Prove that the following equalities hold: [AB, C] = A[B, C] + [A, C]B, [A, BC] = [A, B]C + B[A, C]. With these equalities, it is simple to calculated the following commutators: Exercise 3. Show that [ ˆX, ˆP ] = i ˆP, [ ˆX, ˆP 3 ] = 3i ˆP, and, in general, [ ˆX, ˆP n ] = ni ˆP n 1, or, for functions f( ˆP ) which can be expanded in Taylor series: [ ˆX, f( ˆP )] = i f ( ˆP ), where f denotes the derivative. Similarly, show that [ ˆX n, ˆP ] = ni ˆX n 1, and [f( ˆX), ˆP ] = i f ( ˆX). An important mathematical concept related to operators is that of the Hermitean conjugate of an operator A, denoted with A. The definition is simple. For any two vectors ψ 1 and ψ in the Hilbert space, one has to have: ψ 1 Aψ = A ψ 1 ψ. (3.9) In words, you can bring an operator acting on the ket in such a way that it acts on the bra, if you take the Hermitean conjugate. Hermitean operators, evidently, are the Hermitean conjugate of themselves, i.e., A = A. Exercise 3.3 If A and B are Hermitean operators, show that: (A + ib) = A ib = A ib. 3. Dimensionless variables It is always a good practice, both in analytical work as well as in numerical implementations on the computer, to work with dimensionless variables, so that parameters like the mass m, or the frequency ω, or universal constants like, drop out of the problem. In the present case, for instance, it is a good idea to work with a length l such that the potential energy of the spring and the typical confinement energy for a particle in a box of the same length are equal: ml = 1 mω l. This provides a typical oscillator length, defined by:

24 l = mω. The algebraic solution 3 Define now a dimensionless length ˆx = ˆX/l and a dimensionless momentum ˆp = ˆP l/. It is simple to verify that their commutator is [ˆx, ˆp] = i, i.e., effectively as if = 1. It is now simple to verify that the Hamiltonian reads: Ĥ = ωĥ with ĥ = 1 (ˆp + ˆx ), (3.10) where we have defined a dimensionless Hamiltonian ĥ = Ĥ/( ω). It is now clear that this procedure is totally equivalent to setting m = 1, ω = 1, = 1, as often stated in books or articles. You should always be aware, however, of the exact meaning of this statement, and be ready to reintroduce the appropriate parameters and universal constants if asked to produce explicit numbers. Summarizing, we measure lengths in units of l, masses in units of m, energies in units of ω, momenta in units of /l, times in units of /( ω) = 1/ω, etc. 3.3 The algebraic solution Define now: a = 1 (ˆx + iˆp). (3.11) a is quite clearly not Hermitean (see exercise 3.3). Its Hermitean conjugate is: a = 1 (ˆx iˆp). (3.1) In x-representation, both are quite simple operators: a = 1 ( x + d ) and a = 1 ( x d ) dx dx The commutator of a and a is immediately calculated: [a, a ] = 1.. (3.13) It is simple to invert the definition of a and a to re-express ˆx and ˆp in terms of a and a, obtaining: ˆx = 1 ( a + a ) and ˆp = 1 ( a a ). (3.14) i As done in class (do it as an exercise), one can then show that: ( ĥ = a a + 1 ). (3.15) Let us start finding a state ψ 0 which is annihilated by a, i.e. such that: a ψ 0 = 0. Such a state has a simple expression in x-representation, ψ 0 (x) = x ψ 0, since it obeys the equation: ( 1 x + d ) ψ 0 (x) = 0, dx which can be rewritten as: and is solved by: where C a normalization constant. d dx ψ 0(x) = xψ 0 (x), ψ 0 (x) = Ce x /,

25 4 Harmonic oscillator Exercise 3.4 Calculate the normalization constant C appearing in the state ψ 0. Evidently, ψ 0 is an eigenstate of ĥ with eigenvalue 1/: ĥ ψ 0 = 1 ψ 0. It is also easy to show that ψ 0 must be the ground state of ĥ: Exercise 3.5 Given any state ψ, the expectation value of ĥ satisfies: ψ a a + 1 «ψ = 1 + a ψ 1. Construct now the state ψ 1 = a ψ 0. Using the commutator [a, a ] = 1 it is simple to show that ψ 1 is an eigenstate with energy 3/. In real space, the representation of ψ 1 is simple: ψ 1 (x) = x ψ 1 = 1 ( x d ) ψ 0 (x). dx Calculate and plot this wave-function. More generally, we are now going to use an important property of the number operator ˆn = a a. (3.16) Exercise 3.6 Prove (for instance by recursion) that: ˆn a n ψ0 = n a n ψ0, or, in words, that the state with n operators a applied to ψ 0 is an eigenstate of the number operator ˆn = a a, with eigenvalue n, i.e., ˆn counts the number of times a appears. Based on this, and observing that ĥ = ˆn+1/, we realize that the eigenstates of ĥ are constructed as: ( ψ n = C n a ) ( n ψ0 = ĥ ψ n = n + 1 ) ψ n, where C n is a normalization constant. It is easy to show that a and a allow us to move up and down in the ladder of states ψ n. Exercise 3.7 Prove that the states ψ n = (a ) n n! ψ 0, are correctly normalized. Deduce also that: a ψ n = n + 1 ψ n+1 and a ψ n = n ψ n 1. Notation: You often find n in place of ψ n and 0 in place of ψ 0. The state 0 is called the vacuum of a.

26 4 The hydrogen atom problem The hydrogen atom problem consists of a proton (more generally, a nucleus of charge +Ze), with an electron (charge e) orbiting around it. The Hamiltonian is: H = p 1 + p Ze m 1 m r r 1, (4.1) where m 1 indicates the mass of the nucleus, m that of the electron, and (r 1, p 1 ) and (r, p ) denote coordinate and momenta for nucleus and electron, respectively. This is a typical two-body problem, and, as in classical mechanics, one solves it by separating variables into center-of-mass and relative motion. To this end, one introduces the center of mass R = (m 1 r 1 +m r )/M and the total momentum P = p 1 + p. It is a simple matter to show that these are canonically conjugate variables, i.e., their commutator is [R α, P β ] = i δ α,β. Regarding the relative motion, an obvious coordinate is the relative coordinate r = r r 1. However, the relative momentum is less trivial: the naive guess p = p p 1 is wrong, since it is not canonically conjugate to r, and does not commute with R either, as it should. The way to go is to pose p = β p β 1 p 1 and determine β 1 and β requiring canonical conjugation. The solution is: p = m 1 M p m M p 1. (4.) Therefore, the problem admits now separation of variables, since the total Hamiltonian is: H = P M + p µ Ze. (4.3) r The total wave-function can be written as Ψ(R, r) = Φ(R)ψ(r). The center-of-mass problem is just a free-particle problem, with wave-function Φ(R) = e ik R and energy E k = k /M (there is a translation symmetry for the whole atom): we will neglect it from now on. The relative particle wave-function ψ(r) obeys the 3D problem: ( p µ Ze r ) ψ(r) = Eψ(r). (4.4) This would be still a quite intricate problem (we would need a computer if the potential was not spherically symmetric): luckily enough, we can still separate variables in spherical coordinates. This require introduction of a very important operator: the angular momentum. 4.1 Orbital angular momentum The classical mechanics angular momentum is defined as: L = r p.

27 6 The hydrogen atom problem In quantum mechanics, the corresponding operator is obtained by assuming that r and p are operators. The three Cartesian components of L are immediate: L x = yp z zp y L y = zp x xp z L z = xp y yp x. (4.5) Commutators between the different components of L are easy to establish, using the usual rules for commutators, such as [AB, C] = A[B, C] + [A, C]B. The result (worked out in class) is: [L x, L y ] = i L z, and similar cyclic relations, which can be all summarized as: [L α, L β ] = i ɛ αβγ L γ. (4.6) where ɛ αβγ is the totally antisymmetric tensor appearing also in the vector product of two ordinary vectors: (a b) γ = ɛ αβγ a α b β (sum over repeated indices are always assumed). By changing variables from Cartesian to spherical coordinates, x = r sin θ cos φ y = r sin θ sin φ z = r cos θ, (4.7) it is relatively simple to show that all components of L do not depend on r, but only on the angles θ and φ: ( L x = i + sin φ ) + cos φ cot θ θ φ ( L y = i cos φ ) + sin φ cot θ θ φ L z = i φ (4.8) Exercise 4.1 Verify that the three components of angular momentum are given by Eq To this end, it might be useful to show that: A y sin θ cos φ cos θ cos φ 1 0 sin φ sin θ sin θ sin φ cos θ sin φ + cos φ sin θ cos θ sin θ 0 z r 1 r θ 1 r φ 1 A. (4.9) Another important operator is: L = L x + L y + L z, (4.10) which can be expressed in spherical coordinates as: L = ( θ + 1 tan θ θ + 1 sin θ ) φ. (4.11)

28 Separation of variables 7 Exercise 4. Verify the expression in Eq for L. It is simple to show (done in class) that [L, L α ] = 0. Therefore, we can find simultaneous eigenfunctions of L and, say, L z (we will prove this result, in general, later on). We will indeed show, in the next lecture, that we can construct functions of θ and φ, known as spherical harmonics, which are common eigenfunctions of L and L z : with l integers and m = l, + l. L Y l,m (θ, φ) = l(l + 1)Y l,m (θ, φ) L z Y l,m (θ, φ) = my l,m (θ, φ). (4.1) The reason why all this is useful is that p (i.e., the Laplacian) can be easily expressed in terms of L and derivatives with respect to r. Classically, consider the plane containing r and p, and convince yourself that L = r p = r (p p ) = r p (r p). Quantum mechanically, things are slightly more complicated because r and p do not always commute. You can still prove (done in class) that: L = r p (r p) + i r p. (4.13) This equality can be shown either directly, by components, or by doing the following exercise: Exercise 4.3 Using the fact that L i = ɛ ijk x jp k, with the convention of summing on repeated indices, construct L = L il i, and use the fact that ɛ ijk ɛ ilm = (δ jl δ km δ jmδ kl ) to prove the result for L. Moreover, if you calculate you can immediately conclude that: r = x r x + y r y + z r z = 1 r r r, r p = i r r. (4.14) Therefore, using Eqs and 4.14 we can easily rewrite the Laplacian as: p = r r ( r ) r r r r r + 1 r L = r r r + 1 r L. (4.15) 4. Separation of variables This expression for the Laplacian clearly separates the r-variable from the θ and φ variables, which are contained only in L. As a result of that, you can factorize the wave-function ψ(r) as ψ k,l,m (r) = R k,l (r)y l,m (θ, φ), where k is an extra quantum number (still not specified) which labels the many possible solutions of the equation for R, which reads:

29 8 The hydrogen atom problem ( µ r + ) ( ) l(l + 1) R k,l + r r µr Ze R k,l = E k,l R k,l. (4.16) r Notice that this is a one-dimensional differential equation, which, however, has not the standard form of a SE. The normalization condition is also quite different from a 1D normalization, since it reads: + dr r R l,l (r) = 1. It is therefore tempting to define a new function 0 u k,l (r) = rr k,l (r), which, to start with, will have a standard normalization + 0 dr u k,l (r) = 1. Moreover, you can immediately verify that the non-standard derivative terms for R become completely standard when written in terms of derivatives of u: ( r + r ) R k,l = 1 u k,l r r r. When written in terms of u k,l the SE becomes therefore a standard 1D SE in the half-line [0, ): ( u k,l ) l(l + 1) µ r + µr Ze u k,l = E k,l u k,l, (4.17) r with a standard normalization condition: + 0 dr u k,l (r) = 1. (4.18) 4.3 Dimensionless variables We now get rid of all the physical constants (, e, µ, Z) by switching to dimensionless variables. The procedure is quite similar to that for the harmonic oscillator. Let us define a length a such that potential energy and kinetic energy of confinement are of the same order: Solving this equation we get: Z e a = µa. a = a B = µe Z, (4.19) which we recognize to be the effective Bohr radius of the problem. The corresponding energy is: E H = Z e a B = µe4 Z. (4.0) This quantity is, for Z = 1, the Hartree and is equal to E H Z (7.)eV. Measuring lengths in units of a B, energies in units of E H and masses in units of µ defines a system of units called atomic units, often abbreviated as a.u.. If ρ = r/a B is the dimensionless length and Ẽ = E/E H the dimensionless energy, defining ũ k,l (ρ) = a B u k,l (r = a B ρ), we get the dimensionless SE in the form: 1 ( ũ k,l l(l + 1) ρ + ρ 1 ) ũ k,l = ρ Ẽk,lũ k,l, (4.1) with the normalization condition: + 0 dρ ũ k,l (ρ) = 1. (4.)