Quantum Mechanics. P.J. Mulders

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1 Quantum Mechanics P.J. Mulders Department of Physics and Astronomy, Faculty of Sciences, Vrije Universiteit Amsterdam De Boelelaan 1081, 1081 HV Amsterdam, the Netherlands January 011 (vs.) Lectures given in the academic year

2 Introduction Voorwoord Het college Quantummechanica wordt dit najaar verzorgd door Prof. Piet Mulders geassisteerd door Drs. Wilco den Dunnen bij het werkcollege. Het college volgt in grote lijnen het boek Introduction to Quantum Mechanics, second edition van D.J. Griffiths (Pearson). In parallel zullen er aantekeningen verschijnen, die samenvattend van aard zijn. Het gehele vak beslaat 8 studiepunten en wordt gegeven in periodes, 3 en 4. Wekelijks worden uur hoorcollege gegeven, 1 uur wordt besteed aan een presentatie van een van de studenten en vragen, terwijl er uur werkcollege zijn. Daarnaast moeten er opgaven worden ingeleverd, die worden beoordeeld. We plannen om na blok 3 een deeltentamen af te nemen (schriftelijk op ). De opgaven en het deeltentamen vormen onderdeel van de toetsing. Het geheel wordt afgesloten met een (mondeling) tentamen (in principe in de week 1 t/m 6 maart 011). Dit tentamen, waarbij boek en uitgewerkte opgaven geraadpleegd mogen worden, gaat behalve over theoretische aspecten (afleidingen niet van buiten leren!) ook over opgaven die tijdens het werkcollege zijn behandeld. De hierboven genoemde regeling geldt voor studenten die actief deelnemen aan colleges en werkcolleges. Piet Mulders November 010

3 Introduction 3 Schedule (indicative) Material in book Material in notes Week Week 45.1, Week 46.,.3 3 Week 47.3,.4,.5 3 Week 48.5,.6, 3.1, 3. 4 Week , 3.4, 3.5, 3.6 Week Week 1 4. Week 4.3 Week Week 4 deeltentamen Week Week 6 5.1, 5. 1 Week 7 6.1, 9.1 Week 8 6., 6.3, 6.4, , 15 Week Week 10 (9) (11) (17), 19 Week 11 Literature 1. D.J. Griffiths, Introduction to Quantum Mechanics, Pearson 005. F. Mandl, Quantum Mechanics, Wiley C. Cohen-Tannoudji, B. Diu and F. Laloë, Quantum Mechanics I and II, Wiley J.J. Sakurai, Modern Quantum Mechanics, Addison-Wesley E. Merzbacher, Quantum Mechanics, Wiley B. Bransden and C. Joachain, Quantum Mechanics, Prentice hall 000

4 Introduction 4 Contents 1 Basic concepts of quantum mechanics Introduction Operators and their expectation values Commutators and uncertainty relations Time evolution 5.1 Stationary states General time evolution Spectrum and basic behavior of solutions Boundary conditions and matching conditions Three elementary properties of one-dimensional solutions One-dimensional problems The infinite square well Harmonic oscillator Free particle states The delta potential Bound states and scattering solutions for a square well potential Reflection and transmission through a barrier The Hilbert space of QM using Dirac notation Space of states = ket-space (Hilbert space) Scalar product and the (dual) bra-space Orthonormal basis Operators and matrices in QM Adjoint operator Hermitean operators Unitary operators Representations of states Coordinate-representation or back to wave functions Momentum-representation Angular momentum and spherical harmonics Spherical harmonics The radial Schrödinger equation in three dimensions Several observables and commutation relations Compatibility of observables Compatibility of operators and commutators Constants of motion The uncertainty relations The uncertainty relation for noncompatible observables The Heisenberg energy-time uncertainty relation

5 Introduction 5 9 The hydrogen atom Transformation to the center of mass Solving the eigenvalue equation Appendix: Generalized Laguerre polynomials A note on Bohr quantization Spin Definition Rotation invariance Spin states Why is l integer Matrix representations of spin operators Rotated spin states Combination of angular momenta Quantum number analysis Clebsch-Gordon coefficients Identical particles and relation with spin Permutation symmetry and the Pauli principle The states of the Helium atom Atomic multiplets Perturbation theory Basic treatment of time-independent perturbation theory Time-independent perturbation theory for degenerate states Examples of Time independent Perturbation Theory Fine structure in hydrogen: the mass correction Fine structure in hydrogen: the spin-orbit interaction Fine structure in hydrogen: the hyperfine splitting Refinements for Helium The fine structure of atoms Magnetic effects in atoms and the electron spin The Zeeman effect Spin-orbit interaction and magnetic fields Variational approach Basic treatment Application: ground state of Helium atom Application: ionization energies and electron affinities Time-dependent perturbation theory Examples of time-dependent interactions Application: Electron in rotating magnetic field Application: emission and absorption of radiation by atoms

6 Introduction 6 19 Scattering theory Differential cross sections Cross section in Born approximation Applications to various potentials

7 Basic concepts of quantum mechanics 1 1 Basic concepts of quantum mechanics 1.1 Introduction A classical system is determined by x(t) from which we get the velocity v(t) = ẋ(t), etc. The motion is obtained from Newton s equation, m d x dt = dv dx. It determines how we get from x(0) x(t) (given sufficient boundary conditions x(0) and ẋ(0)). Quantum mechanically, one works with a wave function ψ(x, t) satisfying the Schrödinger equation, i ψ t = m ψ +V(x,t)ψ(x). (1) x This determines how the system evolves from ψ(x, 0) ψ(x, t). The constant appearing in this equation is Planck s constant, h = Js. or more precise the reduced Planck s constant = h/π = Js = evs. This also determines the domain where quantum mechanics is needed to get a good description of nature, namely the domain where appropriate quantities are of the order of this constant. In this respect, note that energy time and/or to momentum distance or angular momentum have the same dimension as. Togoanyfurther, oneneeds to knowwhatto dowith ψ(x,t). It turns out that its physicalsignificance comes in by doing something with it. In particular ψ(x,t) dx is the probability to find the system at time t in an interval of size dx around the point x, or for a finite domain, P ab = b The function itself and the various ways to denote it, like a dx ψ(x,t). () ψ(x,t) = x ψ(t) = x,t ψ, (3) is referred to as probability amplitude (NL: waarschijnlijkheidsamplitudo) for state (NL: toestand) ψ (or ψ ) to be at place x at time t. The use of the notation ψ will be specified later. Where the system was before time t cannot be answered, one would have to do a measurement for that at the earlier time. But as soon as a measurement is done and the particle is found at position a, the state is no longer described by the wave function ψ but by ψ a, which would be a function peaking around point a. This sharp function will then spread according to the Schrödinger equation. Pretty weird! 1. Operators and their expectation values Suppose we know (we ll see later how) that we have a lot of identical particles, all being described by the same wave function ψ, of which we measure the position x. If these positions would be discrete, i.e. just a collection of possibilities x i we would measure how many (n i ) particles end up in each of the positions. That probability is P i = n i /N, where N = i n i is the total number of measurements. One has 1 = i x = i x = i P i, (4) x i P i, (5) x i P i, (6)

8 Basic concepts of quantum mechanics which states that the probabilities add up to one, and weighing with x i or x i gives the average of x, x, etc., which can be extended to the average of any weight function w(x). The standard deviation (squared) is found by weighing with (x i x ). Since x is a number, one finds σ = x = i (x i x ) P i = x x. (7) For the quantum mechanical outcome of an x-measurements one has the continuous version of the above with all (real) possibilities for x, with probabilities ρ(x,t) = ψ(x,t). This implies the normalization and leads to specific expectation values (NL: verwachtingswaarden), 1 = dx ρ(x,t) = dx ψ (x,t)ψ(x,t), (8) x = dx xρ(x,t) = dx ψ (x,t)xψ(x,t), (9) x = dxx ρ(x,t) = dx ψ (x,t)x ψ(x,t), (10) which can be extended to any weight function and for which one has a standard deviation (NL: standaardafwijking) x of which the square is x = dx (x x ) ρ(x,t) = x x. (11) Thenotationontherighthandsidewithxbetweenthefunctionsψ andψ canbeconvenientlygeneralized to any operator Ô = dx ψ (x,t)ôψ(x,t), (1) where Ô is an operator acting on the function ψ. The left-handside is referred to as the expectation value of O. In the simple case above we would have the position operator ˆx that acts like ˆxψ(x,t) = xψ(x,t), but also the derivative d/dx is an example of an operation working on ψ and producing a new function (the derivative function). The normalization is in fact just the expectation value of the unit operator, ˆ1 = 1. Usually the hat on the operators will be omitted. Finally we note that an expectation value does depend on the state ψ, so the better notation would have been Ô ψ or ψ Ô ψ instead of just Ô. Just as one talks about the expectation value of ˆx, one now can talk about the expectation value of Ô (in state ψ). A first consistency check of quantum mechanics is the time independence of the normalization. This is straightforward, although it involves a bit of manipulation and doing partial integrations. One proves that d dx ψ (x,t)ψ(x,t) = 0, (13) dt for any normalizable function (Eq. 8) that satifies the Schrödinger equation (Eq. 1). A normalized continuous function must go to zero at x ±. For the probability in any finite interval one has a modified equation, dp ab = J(a,t) J(b,t), (14) dt where J(x,t) = i ( ψ ψ ) m x ψ x ψ is the probability current (NL: waarschijnlijkheidsstroom) or flux. (15)

9 Basic concepts of quantum mechanics 3 If we look at the time dependence of x, it is (in general) nonzero. We obtain d dt ˆx = i dx ψ (x,t) ψ(x,t) = ˆp x m x m, (16) with the momentum operator (NL: impulsoperator) ˆp given by ˆp x = i x. (17) The position and momentum are actually the basic quantities in quantum mechanical problems, as they are in classical mechanics. E.g. classically the energy can be written as a function of p and x. That same function is actually the righthand-side of the Schrödinger equation in Eq. 1, with i ψ(x,t) t = Ĥψ(x,t), (18) Ĥ = ˆp +V(ˆx,t) = +V(x,t). (19) m m x This is the energy operator or Hamiltonian, which is the most important operator because its role as time evolution operators. Other important operators are e.g. the angular momentum operators ˆl = ˆr ˆp, which in 3 dimensions play a key role. Explicitly, they are ( l x = i y z z ), (0) y ( l y = i z x x ), (1) z ( l z = i x y y ). () x 1.3 Commutators and uncertainty relations At this point, we just want to add a word of caution on the order of operators. Calculating  ˆBψ can be different from ˆBÂψ. Trying this for a simple wave function, for instance We have Applying the operatos once more, we get ψ(x) = N e αx / e iωt. ˆxψ(x) = xn e αx / e iωt ˆp x ψ(x) = i dψ dx = +i αxn e αx / e iωt ˆx ψ(x) = x N e αx / e iωt ˆp x ˆxψ(x) = i d ( xe αx / e iωt) = i ( 1 αx ) N e αx / e iωt, dx ˆxˆp x ψ(x) = i x d dx ( e αx / e iωt) = i αx N e αx / e iωt, ˆp xψ(x) = d ψ dx = ( α α x ) N e αx / e iωt,

10 Basic concepts of quantum mechanics 4 The various expectation values (familiarize yourselves with these type of integrals and see how you get them using Mathematica) are 1 = N e dx αx = N π ( α α = 1 N = π ˆx = N dx xe αx = 0, ˆp x = i N α ˆx = N ˆp x = N α dx xe αx = 0, dx x e αx = 1 α, dx ( 1 αx ) e αx = α. We note that for this wave function the expectation value of the position and momentum are zero, but the standard deviations are nonzero, x = 1/ α and p x = α/. This shows that in this particular case a wave function which is strongly peaked around zero (α is large) has a large standard deviation if one looks at the momentum. The product of expectation values is constant, however, x p x = /. In general, we will find for any wave function that ) 1/4, x p x. (3) This is an example (and the most famous one) of an uncertainty relation and as we will discover later, it is a property that is intimately connected with the fact that the operators ˆx and ˆp x do not commute, [ˆx, ˆp x ] = ˆxˆp x ˆp x ˆx = i. (4) This commutator is actually the most basic one in quantum mechanics, referred to as the canonical commutation relation. That it holds for any function is easily seen by calculating for an arbitrary function φ [ x, d ] φ(x) = x d dx dx φ d dφ dφ (xφ(x)) = x x φ(x) = φ(x). dx dx dx

11 Time evolution 5 Time evolution.1 Stationary states We return to our important Schrödinger equation, i ψ(x,t) t = Ĥ(ˆx, ˆp x)ψ(x,t) = m d ψ(x,t) dx +V(x)ψ(x,t). In the case that the potential in the Hamiltonian is independent of the time, one can easily see that the Schrödinger equation allows separable solutions of the form ψ(x,t) = φ(x)f(t). Inserting this ansatz in Eq. 1, one finds with ψ(x,t)/ t = φ(x)df/dt and ψ(x,t)/ x = f(t)dφ/dx, i df dt f(t) = E = m d φ dx +V(x)φ(x) φ(x), (5) where E must be t-independent but also x-independent, thus constant. We obtain two equations m the time-independent Schrödinger equation and a time dependence d φ +V(x)φ(x) = Eφ(x), (6) dx i df dt = Ef(t) = f(t) e iet/. (7) Thus we get (in general many) stationary states (NL: stationaire toestanden), ψ En (x,t) = φ n (x)e ient/, (8) where φ n (x) are eigenfunctions (NL: eigenfuncties) of the (time-independent) Hamiltonian with eigenvalues E n, Ĥ(x,p)φ n (x) = E n φ n (x). (9) It is easy to see that if φ n is normalized, then Ĥ φ n = E n, Ĥ φn = E n, and thus H φn = 0. (30) Thus, for an eigenstate φ n of Ĥ the standard deviation is zero and the state has a precise energy. To find real energies the Hamiltonian must satisfy certain conditions, to be precise it must be hermitean (to be discussed later). Actually, when measuring an energy, it must be one of the eigenvalues of Ĥ and the system is immediately after that measurement in a state φ n (x). The set of eigenvalues of Ĥ is known as its energy spectrum.. General time evolution The eigenvalues of the Hamiltonian form the energy spectrum of a quantum theory. One can show that normalizable solutions φ n (x) of the time-independent Hamiltonian can be choosen real, while the eigenvalues of the Hamiltonian E n must be real. Furthermore the eigenfunctions satisfy dx φ m (x)φ n(x) = δ mn, (31)

12 Time evolution 6 where δ mn is known as the Kronecker delta, which is 1 if m = n and δ mn = 0 if m n. Such functions are called orthogonal (NL: orthogonaal). The stationary states as in Eq. 8 have a well-determined energy and a known time dependence which is a simple phase. The probability ρ(x,t) is for a stationary state independent of the time. This is not true in general. But starting with a given function at t = 0, one can expand it in the set of eigenfunctions, which turn out to form a complete set of functions. This means that we can write ψ(x,0) = c n φ n (x) with c n = dx φ n (x)ψ(x,0). (3) n Since the Schrödinger equation is a linear equation and we know the time-dependence of ψ En (x,t) we immediately deduce that ψ(x,t) = c n φ n (x)e ient/. (33) Such solutions in general are complex and the probability is not time-independent. They show oscillations. If the initial state ψ(x,0) is a linear combination of φ 1 (x) and φ (x) the oscillations will be of the form with sine or cosine with argument cos((e 1 E )t/ ), hence have a frequency f = (E 1 E )/h or an angular frequency ω = πf = (E 1 E )/..3 Spectrum and basic behavior of solutions This is best illustrated by considering the one-dimensional hamiltonian H = ˆp x/m+v(ˆx), stationary solutions of the form ψ(x,t) = φ(x) exp( iet/ ), are obtained by solving [ d ] m dx +V(x) φ(x) = Eφ(x), rewritten as d dx φ = m (E V(x)) φ(x). (34) }{{} k (x) V(x) E ~k (x) ~κ (x) x φ (x) oscillatory exponential x We distinguish two situations: (i) k (x) 0 in the region where E V(x). In that case the change of slope is opposite to the sign of

13 Time evolution 7 the wave function, which implies that the solution always bends towards the axis. Looking at the case that k (x) = k = m(e V 0 )/ (constant potential V 0 ) one has: or φ(x) = A sinkx+b coskx, (35) φ(x) = A e ikx +B e ikx. (36) (ii) k (x) = q (x) 0 in the region where E V(x). In that case the change of slope has the same sign as the wave function, which implies that the solution always bends away from the axis. Looking at the case that k (x) = q = m(v 0 E)/ (constant potential V 0 ) one has: φ(x) = A sinhqx+b coshqx, (37) or φ(x) = A e qx +B e qx. (38).4 Boundary conditions and matching conditions In order to find proper solutions of the Schrödinger equation, which is a second order (linear) differential equation, one needs d φ/dx, hence φ(x) and dφ/dx must be continuous. The second condition implies dφ dx dφ a+ǫ dx = a ǫ a+ǫ a ǫ dx d φ dx = m a+ǫ a ǫ dx (V(x) E)φ(x) = 0 Thus continuity not necessarily requires a continuous potential, but it is necessary that the potential remains finite. In points where the potential becomes infinite one must be careful, e.g. by studying it as a limiting case. Cases in which the potential makes a jump are studied by imposing matching conditions from both sides, namely equating the wave function and its derivative. Because one in quantum mechanics often takes some freedom in the normalization (to be imposed later), one often chooses to equate the ratio of derivative and wave function, the socalled logarithmic derivative, lim ǫ 0 (dφ/dx) φ(x) (dφ/dx) = lim ǫ 0 a ǫ φ(x). (39) a+ǫ The behavior at infinity (let s simply assume the potential to be zero) depends on the energy, e.g. if E = q /m < 0 one must have lim φ(x) = x Ce qx 0, and lim φ(x) = x C e qx 0. (40) If the energy E = k /m > 0 one could have in case of an wave from the left with incoming flux k/m (allowing to be general for a reflected wave) the boundary condition lim φ(x) = x eikx +A R e ikx. (41) In a particular physics problem, one might look for a solution with only a transmitted wave lim φ(x) = A T e ikx. (4) x with a transmitted flux A T k/m and thus a transmission probability (transmitted/incoming flux) T = A T. With the above ingredients several one-dimensional problems can be solved. Some examples

14 Time evolution 8 of potential wells and barriers will be discussed below. Given a particular potential we note three domains: (i) No solutions exist when E V min. In that case one has everywhere k (x) 0 and there is no way that the exponentially behaving solutions at ± can be matched without a region where the solution bends back towards the axis, i.e. a region where E V(x). (ii) In the regin V min < E < V asym (where V asym is the lowest asymptotic value of the potential) one has a discrete energy spectrum corresponding to a discrete set of solutions: starting with a vanishing exponential at, say, x =, one will find in general a solution that at x = + behaves as a linear combination of two exponential functions (as in Eq. 38). Only for very specific energies the coefficient of the growing exponential e qx will be zero and one finds a normalizable solution. These localized solutions, found for discrete energies, are referred to as bound states. (iii) In the region E V asym one has a continuous spectrum of which the solutions at infinity oscillate or equivalently are complex-valued waves (with definite momentum). One always can find such a solution. This is even true if the asymptotic values of the potential at x = ± are not equal. Finally, considering an infinitely high potential (e.g. for x > 0) as the limit of a large potential V 0 for x > 0 one must match on to the solution e qx with q = mv 0 /, leading to (dφ/dx) mv0 lim = lim x 0 φ(x) V 0 =, (43) which requires φ(0) = 0 and a finite derivative as only nontrivial possibility..5 Three elementary properties of one-dimensional solutions Property 1: In one dimension any attractive potential has always at least one bound state. Property : For consecutive (in energy) solutions one has the node theorem, which states that the states can be ordered according to the number of nodes (zeros) in the wave function. The lowest energy solution has no node, the next has one node, etc. Property 3: Bound state solutions of the one-dimensional Schrödinger equation are nondegenerate. Proof: suppose that φ 1 and φ are two solutions with the same energy. Construct known as the Wronskian. It is easy to see that W(φ 1,φ ) = φ 1 (x) dφ dx φ (x) dφ 1 dx, (44) d dx W(φ 1,φ ) = 0. Hence one has W(φ 1,φ ) = constant, where the constant because of the asymptotic vanishing of the wave functions must be zero. Thus (dφ 1 /dx) = (dφ /dx) d φ 1 φ dx lnφ 1 = d dx lnφ d ( ) ( ) dx ln φ1 φ1 = 0 ln = constant φ 1 φ, φ and hence (when normalized) the functions are identical. φ

15 One-dimensional problems 9 3 One-dimensional problems 3.1 The infinite square well We start by considering the potential V(x) = 0 for x < a/ and V(x) = for x a/. As discussed in the previous section, the wave function is zero for x > a/. For E 0 one would have solutions of the form φ(x) = Ae qx +Be qx, with q = m E /, which does not have a solution (one finds A = B = 0 from the conditions φ(a/) = φ( a/) = 0). For E > 0 one has solutions, which in general are of the form φ(x) = A sin(kx)+b cos(kx), with k = me/, which leads to either A = 0 or B = 0. Check that the solutions separate into (including normalisation) the even solutions ( nπ ) φ n (x) = a cos a x with E n = n π ma (odd n values), (45) and the odd solutions φ n (x) = a sin ( nπ a x ) with E n = n π ma (even n values). (46) The number of nodes is given by n 1. It is easy to check that the solutions are orthogonal, 3. Harmonic oscillator dx φ n(x)φ m (x) = δ mn. (47) The harmonic oscillator potential is encountered in very many applications. It is a natural approximation for a system in equilibrium. Around the minimum of any potential function one can write V(x) = V(x min )+ 1 k(x x min) +..., and after a redefinition of the energy (shift) and the coordinates (choose x min = 0), one has V(x) = 1 kx 1 mω x. The latter involves the angular frequency ω = k/m, which appears in the classical solutions. From mẍ = F = dv/dx = kx one finds as classical solution The energy x(t) = A cos(ωt+ϕ), p(t) = mẋ(t) = mωa sin(ωt+ϕ) E = p m +V(x) = mω A, is time-independent. We note that it can be written as E = (mωx+ip)(mωx ip) m = A (t)a + (t), m

16 One-dimensional problems 10 where A ± (t) = (mωx±ip) = mωae ±i(ωt+ϕ). For the time-independent Schrödinger equation for the harmonic oscillator in one dimension the energy expression is taken as the Hamiltonian with x and p operators. It is written as [ ] p H φ(x) = m +V(x) φ(x) = [ m d dx + 1 mω x ] φ(x) = Eφ(x). (48) To handle it mathematically, it is conventient to make this equation dimensionless by introducing ξ = α x, leading to [ α d ] m dξ + mω α ξ φ(ξ) = Eφ(ξ). To fix α one can equate the multiplicative factors in both terms, α /m = mω /α which gives α = mω/, leading to [ ω 1 d dξ + 1 ] ξ φ(ξ) = Eφ(ξ). (49) Rather than solving this as a differential equation (done in the appendix below), we use an algebraic approach to solve this problem. For this we write the Hamiltonian as a product of two (conjugate) operators, but one must be careful in this case, since the operators x and p (or equivalently ξ and d/dξ do not commute. To be precise we have the basic canonical commutation relation [x,p] xp px = i (see Eq. 4). This also means that the classical quantities A and A + mentioned above, do not commute after replacing the position and momentum by operators. Introducing a ± = 1 ( ξ d ) 1 = (mωx ip), (50) dξ mω one then finds that these operators satisfy [a,a + ] = 1 (51) and one must be careful to rewrite H. The correct result is ( H = ω a + a + 1 ). (5) Finding the eigenvaluesofthis is donein afew steps. It is sufficient tostudy the eigenvaluesofn a + a, which means that we are looking for (normalized) eigenfunctions φ n satisfying Nφ n = a + a φ n nφ n. (53) It is easy to see that the eigenvalue n satisfies n = dx φ n Nφ n = dx φ n (a +a φ n ) = dx (a φ n ) (a φ n ) 0. Here one must realize that while dx φ (xφ) = dx (xφ) φ and dx φ (pφ) = dx (pφ) φ (x and p are hermiteanoperators)thisisnottrueforthea ± operators,forwhichonehas dxφ (a ± φ) = dx(a φ) φ. Returning to the eigenvalue n, we thus know it is real and non-negative. The beauty of a ± is that they can be used to construct new eigenfunctions, N(a + φ n ) = a + a a + φ n = a + (a + a +1)φ n = (n+1)a + φ n, (54) N(a φ n ) = a + a a φ n = (a a + 1)a φ n = (n 1)a + φ n. (55)

17 One-dimensional problems 11 Thus a ± φ n = c ± φ n±1. From Eq. 3. we see that c = n. Similarly one easily sees that c + = n+1. Thus if φ n is a solution of the Hamiltonian, a + φ n = n+1φ n+1 and a φ n = nφ n 1 (56) are also eigenfunctions of the Hamiltonian with energies increased or lowered by ω. But this poses a problem with Eq. 3.. We may end up with negative eigenvalues, unless there is a solution such that a φ = 0. This solution exist, namely φ 0. Thus n = 0 is necessarily one of the eigenvalues of N. All other solutions then can be constructed from this. Summarizing, we have find that ( Hφ n = n+ 1 ) ωφ n, (57) with n = 0,1,,... and eigenfunctions satisfying φ n = (a +) n n! φ 0, (58) where φ 0 satisfies a φ 0 = 0 of which the solution is easily found, ( a φ 0 = ξ + d ) φ 0 (ξ) = 0 = φ 0 (ξ) e ξ /. (59) dξ Note actually that the functional form of the solutions is not needed to find the spectrum (eigenvalues of Hamiltonian) or to find expectation values of x, p, or combinations of them. The operators x and p can simply be expressed in a ± of which the action on the φ n s is known. Appendix: Hermite polynomials The problem of the one-dimensional harmonic oscillator in essence reduces to the differential equation for which the solutions are given by y +g 0 (x)y = 0 with g 0 (x) = n+1 x (60) where H n are polynomials of degree n. They are normalized as y(x) = e x / H n (x). (61) dx e x [H n (x)] = n n! π, (6) and satisfy the differential equation [ d dx x d ] dx +n H n (x) = 0. (63) Some useful properties are H n (x) = ( ) n dn x e dx n e x (Rodrigues formula), (64) xh n (x) = 1 H n+1(x)+nh n 1 (x), (65) d dx H n(x) = nh n 1 (x). (66)

18 One-dimensional problems 1 4 Some explicit polynomials are H 0 (x) = 1, H 1 (x) = x, H (x) = 4x, H 3 (x) = 8x 3 1x Free particle states Plots of the H n (x) or HermiteH[n,x] functions for n = 0, 1,, and 3. The next example is the case that V(x) = 0. In this case one only has positive energy solutions that do not vanish asymptotically, but oscillate. We will see that there is a continuous spectrum. Introducing E k ω k k /m or k = me, the Schrödinger equation takes the form d dx φ(x) = k φ(x), (67) with as in Eq. 35 or 36 simple solutions, for the case V = 0 generally referred to as plane waves, φ k (x) = Ae ikx. (68) The stationary time-dependent solutions are ) ψ k (x,t) = A exp (ikx k m t = Ae ikx iωkt) = Ae ik(x vphaset), (69) where v phase = ω k /k = k/m is the phase velocity. We note that the density of this solution is constant, ρ = A. Thefunctionscorrespondtodefinitemomentap = k (theyareeigenfunctionsofthemomentum operator). Using the expression for the current (Eq. 15) one gets J(x,t) = A k/m = ρp/m = ρv, where v is the normal classical velocity, v = p/m = k/m, not to be confused with the phase velocity. As discussed before, we now also know the evolution of any system, described by an arbitrary wave function ψ(x,0) = φ(x) at time t = 0. We have to expand this solution into plane waves, dk φ(x) = π φ(k)e ikx with φ(k) = dx φ(x)e ikx, (70) where the second equation tells us how to find the coefficient. This is just the continuous versionof Eq. 3 with n dk/π and c n φ(k). The procedure is know as Fourier transformation and its inverse. The use of π versus π in the integrations is just a matter of convention. The full time-dependent solution becomes the wave packet dk ψ(x,t) = π φ(k)e ikx iωkt. (71) One might ask what the velocity is of the corresponding density ψ(x,t). This can be studied, generalizing the case of ω k = k /m to an arbitrary situation where ω(k) is the dispersion relation. Suppose

19 One-dimensional problems 13 the wave packet is strongly peaked around k k 0, in that case p k 0. By expanding the frequencies around k 0 one finds ω(k) ω(k 0 )+ dω dk (k k 0 ) ω 0 +v g (k k 0 ). (7) k0 Here the group velocity v g = dω/dk = de/dp has been introduced, which is also the classical velocity from basic classical mechanics. We get ψ(x, t) dk π e i(ω0 vgk0)t φ(k)e ikx i(ω0+vg(k k0))t dk π φ(k)e ik(x vgt) e i(ω0 vgk0)t φ(x v g t) (73) which up to a phase that doesn t affect ψ is the same wave function as at t = 0, but shifted as if moving with the group velocity v g. 3.4 The delta potential First, let s discuss the properties of the delta-function. The Dirac delta-function is in fact a distribution (mapping functions into numbers), defined via dx f(x)δ(x) = f(0), (74) or dx f(x)δ(x a) = f(a). It can be considered as the limit of a peaked function, e.g. δ(x) = lim ǫ 0 φ ǫ (x), (75) where φ ǫ (x) = 0 if x > ǫ/ and φ ǫ (x) = ǫ if x ǫ/. Other examples are Some properties are δ(x) = lim Λ sinλx δ(x) = lim Λ πx = lim Λ Λ π e Λ x, (76) Λ Λ dk dk π eikx = π eikx. (77) δ( x) = δ(x), δ(ax) = 1 a δ(x), xδ(x) = 0, dx f(x)δ(x a) = f(a), d θ(x) = δ(x), dx where θ(x) is the Heaviside function, θ(x) = 0 for x < 0 and θ(x) = 1 for x 0. Note that Mathematica can work with this function, e.g. to define a square well potential with depth V 0 using the expression V(x) = V 0 θ( 1 a+x)θ(1 a x).

20 One-dimensional problems 14 Bound state(s) of the delta potential Given a one-dimensional potential of the form The Schrödinger equation becomes φ (x) a V(x) = δ(x). (78) ma δ(x)φ(x) = me φ(x). (79) By integrating from (0 ǫ) tot (0+ǫ) one shows that for a solution of the Schrödinger equation limφ (x) limφ (x) = φ(0). (80) x 0 x 0 a For this potential there (always) exist one bound state solution. For E = q /m we get φ(x) = e qx voor x < 0, φ(x) = e qx voor x > 0. The condition in Eq. 80 gives q = 1/a. Thus we find a bound state with energy E = /ma. Scattering states of the delta potential For positive energies we have plane wave solutions except for x = 0 and we can calculate the transmission and reflection coefficient (A T en A R ) for a wave coming from the left (from x ). The wave function corresponding to this problem is ( taking E = k /m) φ(x) = e ikx +A R e ikx for x < 0, φ(x) = A T e ikx for x > 0. The boundary conditions at x = 0 become (using continuity and for derivatives Eq. 80), 1+A R = A T, ikaa T (ika ikaa R ) = A T. From this one obtains A R = 1 1 ika, A T = ika 1 ika. One can easily show that the flux is conserved. Calculating the flux for plane waves e ±ikx one has J(x) = ± k/m, we find that flux for x < 0 is k(1 A R )/m and for x > 0 is k A T /m. Indeed the results satisfy 1 A R = A T. 3.5 Bound states and scattering solutions for a square well potential We consider the potential V(x) = V 0 for x < a and V(x) = 0 for x a.

21 One-dimensional problems 15 One does not find a finite solution for E < V 0. One would have exponential behavior in all x-ranges and since k (x) < 0 everywhere, one must have a solution which everywhere either has a positive or a negative derivative. For V 0 < E < V 0, finite solutions are of the form φ(x) = C e qx for x a, (81) φ(x) = A sin(kx)+b cos(kx) for x a, (8) φ(x) = Ce qx for x a (83) (see Eq. 40), where E = q /m and E+V 0 = k /m. The matching conditions for wave functions and derivatives give C e qa = A sin(ka)+b cos(ka), A sin(ka)+b cos(ka) = Ce qa, (q/k)c e qa = A cos(ka)+b sin(ka), A cos(ka) B sin(ka) = (q/k)ce qa. It is easy to convince one self that there are two classes of solutions, even solutions with A = 0 and C = C, odd solutions with B = 0 and C = C. For the solutions one obtains from the matching of logarithmic derivatives in the point x = a k tan(ka) = q (even), (84) k cot(ka) = q (odd). (85) Introducing the dimensionless variable ξ = ka and writing ξ 0 = mv 0 a /, one has qa = ξ0 ξ and E = (ξ0 ξ )/ma. The variable ξ runs from 0 ξ ξ 0. The conditions become ξ tan(ξ) = 0 ξ ξ (even), (86) ξ tan(ξ) = ξ 0 ξ (odd). (87) One sees that there always is an even bound state in the region 0 ξ minimum{ξ 0,π/}, and then depending on the depth of the potential (i.e. as long as ξ ξ 0 ) a first odd bound state (with one node) between π/ ξ π, a second even bound state (with two nodes) between π ξ 3π/, etc. Next one might look for solutions with positive energy, in particular of the kind with boundary conditions as in Eqs. 41 and 4, φ(x) = e ikx +A R e ikx for x a, (88) φ(x) = A sin(kx)+b cos(kx) for x a, (89) φ(x) = A T e ikx for x a, (90) where E = k /m and E +V 0 = K /m. The matching conditions become e ika +A R e ika = A sin(ka)+b cos(ka), A sin(ka)+b cos(ka) = A T e ika, i k k K e ika ia R K eika = A cos(ka)+b sin(ka), A cos(ka) B sin(ka) = ia T k K eika.

22 One-dimensional problems 16 Eliminating A and B one obtains for the (complex) amplitudes of the reflected and transmitted waves, A R = e ika i(k k ) sin(ka) kk cos(ka) i(k +K ) sin(ka), (91) A T = e ika kk kk cos(ka) i(k +K ) sin(ka). (9) In contrast to negative energies, one has always solutions for positive energy. The interpretation of the specific ansatzofthe wavefunction isanincomingwavefromthe left, φ i (x) = e ikx (with fluxj i = k/m), a reflected wave, φ r (x) = A R e ikx (with flux j r = A R k/m), and a transmitted wave, φ r (x) = A T e ikx (with flux j t = A T k/m). The probabilities for reflection and transmission are R = A R = = T = A T = = (K k ) sin (Ka) 4k K cos (Ka)+(k +K ) sin (Ka) (K k ) sin (Ka) 4k K +(K k ) sin (Ka) 4k K 4k K cos (Ka)+(k +K ) sin (Ka) (93) 4k K 4k K +(K k ) sin (Ka), (94) satisfying 1 R = T (flux from the left = flux to the right). Note that there exists particular energies, for which the wave vector K satisfies Ka = nπ, in which case R = 0 and T = 1, i.e. the potential is invisible for those particular waves. 3.6 Reflection and transmission through a barrier We consider the situation of a positive square potential, also called a barrier potential, V(x) = +V 0 for x < a and V(x) = 0 for x a. In this case we only have scattering solutions for E > 0. The case E V 0 is even completely similar to the scattering solutions in the previous section. We have the same expression for k = me/, but now K = m(e V 0 )/. The quantities k and K are the wave numbers in the different regions, respectively. In terms of k and K the expressions for reflection and transmission remain the same as in Eqs. 93 and 94. One again has the situation that there are energies for which Ka = nπ and the barrier is invisible. The case0 E V 0 appearsat first sightdifferent becausethe wavefunctions in the region a x a become exponentional, e.g. linear combinations of cosh(qx) and sinh(qx) with q = m(v 0 E)/. But this in fact is nothing else than using complex wave numbers, K iq (note that sign doesn t matter), leading instead of Eqs. 91 and 9 to A R = e ika i(k +q ) sinh(qa) kq cosh(qa) i(k q ) sin(qa), (95) A T = e ika kq kq cosh(qa) i(k q ) sinh(qa). (96) leading to the reflection and transmission probabilities through a barrier, R = A R = T = A T = (k +q ) sinh (qa) 4k q +(k +q ) sinh (qa) (97) 4k q 4k q +(k +q ) sinh (qa). (98)

23 One-dimensional problems 17 In this situation one always has T < 1. In the case that qa 1, which is the case if E V 0 and mv 0 a / 1, the transmission coefficient reduces to the tiny probability T 16k q (k +q ) e 4qa 16 E V 0 e 4qa. In fact, only the exponential is the important part in this probability. For a barrier with a variable potential, the result becomes the famous WKB formula for tunneling, which is the product of the consecutive exponential factors for small barriers, ( x ( ) ) 1/ m T exp dx x 1 [V(x) E] 1/, (99) where x 1 and x are the points where V(x 1 ) = V(x ) = E with V(x) E in the region x 1 x x. Hitting a barrier Considering the situation of a potential V(x) = +V 0 for x < 0 and V(x) = 0 for x 0. For the solutions with 0 E = k /m V 0, one can find solutions of the form φ(x) = Ce qx for x 0, (100) φ(x) = e ikx A R e ikx for x 0, (101) where q = m(v 0 E) and we have already eliminated one overall constant by choosing the coefficient before e ikx, the incoming wave (from the right), to be unity. By equating the logarithmic derivatives from right and left, one finds the solution for A R, which can be written as A R = q +ik q ik = eiδ with tanδ(e) = k q = E V 0 E. (10) We see as expected from flux conservation that A R = 1 (there is no transmission) and can be expressed in terms of a phase shift δ(e). When V 0, the phase shift disappears and the wave function on the right-hand side becomes a sin(kx) with φ(0) = 0. For energies E > V 0 one will have both reflection and transmission.

24 The Hilbert space of QM using Dirac notation 18 4 The Hilbert space of QM using Dirac notation 4.1 Space of states = ket-space (Hilbert space) Quantum mechanical states are denoted u and form a linear vector space over the complex numbers (C), the Hilbert space H = { u }, u 1 H u H c 1 u 1 +c u H (103) c 1,c C Given a basis { u 1,..., u 1 } for an N-dimensional Hilbert-space (N can be infinite!) consisting of a collection linearly independent kets, we can express every ket in this basis (completeness), u H u = 4. Scalar product and the (dual) bra-space N c n u n. (104) For elements u, v H we can construct the complex number u v C, for which u v = v u, If u = c 1 u 1 +c u then v u = c 1 v u 1 +c v u. Note that this implies u u 0. n=1 u v = v u = c 1 v u 1 +c v u = c 1 u 1 v +c u v. Beside the ket-space we can also introduce the dual bra-space, H = { u }, which is anti-linear meaning that u = c 1 u 1 +c u u = c 1 u 1 +c u. (105) The scalar product is constructed from a bra-vector and a ket-vector ( bra(c)ket ). 4.3 Orthonormal basis A state u is normalized when u u = 1. Two states u and v are orthogonal when u v = 0. In a linear vector space an orthonormal basis can be constructed, in which every state can be expanded, Basis { u 1, u,...} with u m u n = δ mn. If u = n c n u n, then c n = u n u (proof) and we can write u = n u n u n u = }{{} c n c 1 c.. (106) Note that u u = 1 n c n = 1, (107) hence the name probability amplitude for c n.

25 The Hilbert space of QM using Dirac notation 19 The collection { u 1, u,...} forms an orthonormal basis for the bra-space. In most one dimensional examples that we considered so far, the Hilbert space was the space of square integrable functions on the real axis, with the scalar product φ 1 φ It is straightforward to check all properties of a scalar product. 4.4 Operators and matrices in QM dx φ 1 (x)φ (x). (108) An operator A acts in the Hilbert-space H, i.e. v = A u = Au H. If u = c 1 u 1 +c u then A u = c 1 A u 1 +c A u (A is linear). The matrix element of A in states u and v is given by u A v. If u = v we call this expectation value of A, if u v we call this transition matrix element. The unit operator acts as I u = u and can with the help of a complete orthornormal basis { u n } be written as I = u n u n, (109) n directly following from Eq. 106 and known as completeness relation. If u = n c n u n = n u n u n u then we can write for A u A u = A u n u n u n = ( ) u m u m A u n u n u = u m A mn c n }{{}}{{} m,n m n A mn c n A 11 A 1... c 1 = A 1 A... c and the matrix element of A is given by (110) u A u = m u m A u n u n u m,n u u }{{}}{{}}{{} c m A mn c n A 11 A 1... = (c 1 c...) A 1 A..... c 1 c. (111) 4.5 Adjoint operator The adjoint operator A is defined by giving its matrix elements in terms of those of the operator A, u A v v A u. (11) We note that de bra-state Au = u A. This is proven in the following way: for every v is Au v = v Au = v A u = u A v. In matrix language one thus has that A = A T.

26 The Hilbert space of QM using Dirac notation Hermitean operators Definition: An operator A is hermitean when u Au = Au u. By applying this to a state c 1 u + c v with arbitrary coefficients one sees that this is equivalent with i.e. A is self-adjoint. u Av = Au v u A v = u A v A = A, The consequences of A = A for transition matrix elements and expectation values are u A v = u A v = v A u (113) u A u = u A u real expectation values (114) For the eigenvalues (a n ) and eigenstates ( n ), of a hermitean operator, A n = a n n we have n A n = a n are the (real) eigenvalues. Eigenstates corresponding with nondegenerate eigenvalues are orthogonal, A n = a n n A m = a m m m n = 0. a m a n If eigenvalues are degenerate, we can construct orthogonal eigenstates (possibly by using other, commuting, operators). Eigenstates can be normalized, n n = 1. Thus, eigenstates form an orthonormal basis, m n = δ mn. Using this basis A is diagonal, A = a n a n n = 0 a... (115) n..... The expectation value of a hermitean operator can be written as u A u = u n a n n u = a n c n. (116) }{{}}{{} n c n n c n This coincides with the interpretation of c n as the probability to find the state n and obtain the result a n in a measurement. 4.7 Unitary operators Definition: An operator U is unitary when U 1 = U, or UU = U U = I. It is easy to prove that a unitary operator conserves scalar products, Uv Uw = v w (117) With a unitary matrix we can transform an orthonormal basis { u n } in another such basis {U u n }.

27 Representations of states 1 5 Representations of states 5.1 Coordinate-representation or back to wave functions With the Dirac notation at hand, we can formalize some issues on wave mechanics. We have used ψ and ψ(r) more or less interchangeable. To formalize we simply write down the eigenvalue equation for the hermitean operator operator ˆr, denoting the eigenstates as r and the eigenvalues as r. Thus We have the following properties: ˆr r = r r, (118) (i) Orthogonality: r r = δ 3 (r r ) (See section 3.4 for the definition) (ii) The value ψ(r) C is simply the coefficient in the expansion ψ = (iii) The identity: I = d 3 r r r (iv) The operator expansion: ˆr = d 3 r r r r d 3 r r r ψ. }{{} ψ(r) We of course should check that the above is consistent and agrees with the wave formulation of quantum mechanics. The most important check is to calculate the behavior of the function ˆrψ. Using property (ii) and the definition 118 in the form r ˆr = r r (hermiticity) we get ˆrψ(r) = r ˆr ψ = r r ψ = rψ(r), so indeed the operator does what we already saw before. We can check other consistency requirements, Normalization of ψ(r): ψ ψ = d 3 r ψ r r ψ = d 3 r ψ(r) = 1 }{{} ψ (r) }{{} ψ(r) Scalar product of two states: ψ φ = d 3 r ψ r r φ = d 3 r ψ (r)φ(r) }{{} ψ (r) }{{} φ(r) Expectation value: ψ ˆr ψ = d 3 r ψ r r r ψ = d 3 r r ψ(r) Knowing that V(ˆr)ψ(r) = V(r)ψ(r) and ˆpψ(r) = i ψ(r) we have among others V(ˆr) = d 3 r r V(r) r, (119) ˆp = d 3 r r ( i ) r. (10) 5. Momentum-representation Consider the hermitean operator ˆp and denote the eigenstates as p and the eigenvalues as p. Thus ˆp p = p p. (11) We already have seen that in the coordinate representation for p : ˆpφ p (r) = pφ p (r) φ p (r) = ( ) i ρ exp }{{} p r r p (1)

28 Representations of states (This defines ρ). We then have the following properties: (i) Orthogonality: p p = ρ(π ) 3 δ 3 (p p ) d 3 p (ii) Expansion of state: ψ = (π ) 3 ρ p p ψ }{{} ψ(p) d 3 p (iii) Identity: I = (π ) 3 ρ p p d 3 p (iv) Operator expansion: ˆp = (π ) 3 ρ p p p Switching of representation is achieved via, d 3 p ψ(r) = r ψ = (π ) 3 ρ r p p ψ = d 3 ( ) p i ρ exp (π ) 3 ρ p r ψ(p), (13) ψ(p) = p ψ = d 3 r p r r ψ = d 3 r ρ exp ( i ) p r ψ(r), (14) as we have seen before being Fourier transforming. Choices for the normalization of plane waves are ρ = 1 or ρ = (π ) 3 (non-relativistic) or ρ = E (relativistic). For instance consistency of Eq. 1 and (iv) can be checked, d 3 p r ˆp ψ = (π ) 3 ρ r p p p ψ = d 3 ( ) p i (π ) 3 ρ p ρ exp p r ψ(p) d 3 ( ) p i = i ρ exp (π ) 3 ρ p r ψ(p) = i ψ(r) (15) and finally we note that the expectation value of the momentum squared can be calculated from ψ(r) or from ψ(k), p = d 3 r ψ (r) ( ) d 3 k ψ(r) = (π) 3 ρ k ψ(k). (16)

29 Angular momentum and spherical harmonics 3 6 Angular momentum and spherical harmonics In this section we study the (three) angular momentum operators ˆl = ˆr ˆp = i r, looking for eigenvalues and eigenstates. The angular momentum operators are best studied in polar coordinates, x = r sinθ cosϕ, (17) y = r sinθ sinϕ, (18) z = r cosθ, (19) from which one gets r x/r y/r z/r x θ ϕ = x cotθ y cotθ r sinθ y x 0 The ˆl operators are given by ( ˆl x = i y z z ) y ( ˆl y = i z x x ) = i z ( ˆl z = i x y y ) x and the square ˆl becomes ˆl = l x +l y +l z = [ 1 sinθ ( = i y z. ) +cotθ cosϕ ϕ sinϕ θ ( cosϕ +cotθ sinϕ θ ϕ, (130) ), (131) = i ϕ, (13) ( sinθ ) + 1 θ θ sin θ ] ϕ. (133) From the expressions in polar coordinates, one immediately sees that the operators only acts on the angular dependence. One has ˆl i f(r) = 0 for i = x,y,z and thus also ˆl f(r) = 0. Being a simple differential operator (with respect to azimuthal angle about one of the axes) one has ˆl i (fg) = f (ˆl i g)+ (ˆl i f)g. 6.1 Spherical harmonics We first study the action of the angular momentum operator on the Cartesian combinations x/r, y/r and z/r (only angular dependence). One finds ( x ) ( y ( y ) ( x ( z ) ˆl z = i, ˆlz = i, ˆlz = 0, r r) r r) r which shows that the l operators acting on polynomials of the form ( x ) n1 ( y ) n ( z n3 r r r) do not change the total degree n 1 +n +n 3 l. They only change the degrees of the coordinates in the expressions. For a particular degree l, there are l+1 functions. This is easy to see for l = 0 and l = 1. For l = one must take some care and realize that (x +y +z )/r = 1, i.e. there is one function less than the six that one might have expected at first hand. The symmetry of ˆl in x, y and z immediately

30 Angular momentum and spherical harmonics 4 implies that polynomials of a particular total degree l are eigenfunctions of ˆl with the same eigenvalue λ. Usingpolarcoordinatesoneeasilyseesthatfortheeigenfunctionsofl z onlytheϕdependencematters, since The eigenfunctions are of the form f m (ϕ) e imϕ, where the actual eigenvalue is m and in order that the eigenfunction is univalued m must be integer. For fixed degree l of the polynomials m can at most be equal to l, in which case the θ-dependence is sin l θ. It is easy to calculate the ˆl eigenvalue for this function, for which one finds l(l+1). The rest is a matter of normalisation and convention and can be found in many books. In particular, the eigenfunctions of l and l z, referred to as the spherical harmonics, are given by ˆl Yl m Yl m (θ,ϕ), (134) ˆl z Yl m (θ,ϕ) = m Yl m (θ,ϕ), (135) with the value l = 0,1,,... and for given l (called orbital angular momentum) l+1 possibilities for the value of m (the magnetic quantum number), m = l, l+1,...,l. Given one of the operators, l or l z, there are degenerate eigenfunctions, but with the eigenvalues of both operators one has a unique labeling (we will come back to this). Note that these functions are not eigenfunctions of l x and l y. Using kets to denote the states one uses l,m rather than Y l m. From the polynomial structure, one immediately sees that the behavior of the spherical harmonics under space inversion (r r) is determined by l. This behavior under space inversion, known as the parity, of the Y m l s is ( )l. The explicit result for l = 0 is Explicit results for l = 1 are Y 0 0 (θ,ϕ) = 1 4π. (136) 3 Y1 1 (θ,ϕ) = x+iy 8π r 3 Y1 0 (θ,ϕ) = 4π Y 1 1 (θ,ϕ) = 3 4π z r = 3 x iy = 8π r 3 = 8π sinθeiϕ, (137) cos θ, (138) 3 8π sinθe iϕ. (139) The l = spherical harmonics are the(five!) quadratic polynomials of degree two, Y ± (x y )±ixy 15 (θ,ϕ) = 3π r = 3π sin θe ±iϕ, (140) 15 Y ±1 z(x±iy) 15 (θ,ϕ) = 8π r = 8π sinθ cos θe±iϕ. (141) 5 Y 0 3z r 5 ( (θ,ϕ) = 16π r = 3cos θ 1 ), (14) 16π where the picture of Y 0 is produced using Mathematica, SphericalPlot3D[Abs[SphericalHarmonicY[,0,theta,phi]], {theta,0,pi},{phi,0,*pi}]. The spherical harmonics form a complete set of functions on the sphere, satisfying the orthonormality