Theoretical Physics III Quantum Mechanics Axel Groß 23 May 2005 Preface Theoretical Physics Master Quantum Mechanics Prof Dr Axel Groß Phone: 289–12355 Room No.: 3248 Email: agross@ph.tum.de http://www.ph.tum.de/lehrstuehle/T30g Contents Introduction – Wave Mechanics Fundamental Concepts of Quantum Mechanics Quantum Dynamics Angular Momentum Approximation Methods Symmetry in Quantum Mechanics Scattering Theory Relativistic Quantum Mechanics Suggested Reading: • J.J Sakurai, Modern Quantum Mechanics, Benjamin/Cummings 1985 • G Baym, Lectures on Quantum Mechanics, Benjamin/Cummings 1973 • F Schwabl, Quantum Mechanics, Springer 1990 III Preface Criteria for getting the Schein: • 50% of the points in the homework sets (at most two students can turn in the homework sets) • Passing the final exam These lecture notes are based on the class “Theoretical Physics – Quantum Mechanics” in the sommer semester 2002 at the Technical University Munich I am very grateful to Maximilian Lein who provided a LATEX version of the original notes which have been the basis for this text; furthermore, he created many of the figures Without his efforts this version of the lecture notes would not have been possible Mă unchen, September 2002 IV Axel Groß Contents Introduction - Wave mechanics 1.1 Postulates of Wave Mechanics 1.2 One-dimensional problems 1.2.1 Bound states 1.2.2 Transmission-Reflection Problems 1.2.3 Tunneling Through a Potential Barrier 1 2 Fundamental Concepts of Quantum Mechanics 2.1 Introduction 2.2 Kets, Bras, and Operators 2.2.1 Kets 2.2.2 Bra space and inner product 2.3 Operators 2.3.1 Multiplication of Operators 2.3.2 Outer Product 2.3.3 Base Kets and Matrix Representations 2.3.4 Eigenkets as Base Kets 2.3.5 Resolution of the Identity, Completeness Relation, or Closure 2.4 Spin 1/2 System 2.5 Measurements, Observables And The Uncertainty Relation 2.5.1 Compatible Observables 2.5.2 Uncertainty Relation 2.5.3 Change Of Basis 2.5.4 Diagonalization 2.6 Position, Momentum, And Translation 2.6.1 Digression On The Dirac Delta Function 2.6.2 Position and momentum eigenkets 2.6.3 Canonical Commutation Relations 2.7 Momentum-Space Wave Function 2.7.1 Gaussian Wave Packets 2.7.2 Generalization To Three Dimensions 9 10 10 11 12 12 12 13 13 13 14 15 15 18 20 22 22 23 23 25 27 28 29 Quantum Dynamics 3.1 Time Evolution and the Schră odinger Equation 3.1.1 Time Evolution Operator 3.1.2 Derivation of the Schră odinger Equation 3.1.3 Formal Solution for U (t, t0 ) 3.1.4 Schră odinger versus Heisenberg Picture 3.1.5 Base Kets and Transition Amplitudes 3.1.6 Summary 31 31 31 32 32 36 38 39 V Contents 3.2 Harmonic Oscillator 3.2.1 Heisenberg Picture 3.3 Schră odingers Wave Equation 3.4 Harmonic Oscillator using Wave Mechanics 3.4.1 Symmetry of the Wave Function 39 43 45 47 48 Angular Momentum 4.1 Rotations and Angular Momentum 4.2 Spin 12 Systems and Finite Rotations 4.3 Eigenvalues and Eigenstates of Angular Momentum 4.3.1 Matrix Elements of Angular Momentum Operators 4.3.2 Representations of the Rotation Operator 4.4 Orbital Angular Momentum 4.5 The Central Potential 4.5.1 Schră odinger Equation for Central Potential Problems 4.5.2 Examples for Spherically Symmetric Potentials 4.6 Addition of Angular Momentum 4.6.1 Orbital Angular Momentum and Spin 12 4.6.2 Two Spin 12 Particles 4.6.3 General Case 49 49 51 54 56 57 57 59 61 62 63 63 64 65 Approximation Methods 5.1 Time-Independent Perturbation Theory: Non-Degenerate Case 5.1.1 Harmonic Oscillator 5.2 Degenerate Perturbation Theory 5.2.1 Linear Stark Effect 5.2.2 Spin-Orbit Interaction and Fine Structure 5.2.3 van-der-Waals Interaction 5.3 Variational Methods 5.4 Time-Dependent Perturbation Theory 67 70 71 72 73 75 76 77 Symmetry in Quantum Mechanics 6.1 Identical Particles 6.2 Two-Electron System 6.3 The Helium Atom 6.3.1 Ground State 6.3.2 Excited States 83 83 85 87 87 88 Scattering Theory 7.1 Wave Packets 7.2 Cross Sections 7.3 Partial Waves 7.4 Born Approximation 89 89 91 91 92 Relativistic Quantum Mechanics 8.1 Relativistic Spin Zero Particles 8.2 Klein’s Paradox 8.3 Dirac Equation 93 93 96 98 VI 67 Introduction - Wave mechanics We will start by recalling some fundamental concepts of quantum wave mechanics based on the correspondence principle 1.1 Postulates of Wave Mechanics The state of a system is described by its wave function Ψ(x, t) The probability density is defined as ρ(x, t) ≡ |Ψ(x, t)| (1.1) |Ψ(x, t)| d3 x describes the probability to find the particle at time t in the volume element d3 x at x Physical observables correspond to operators that act on the wave function For example, the momentum p and the energy E are represented by the following derivatives p → i E →i ∇, (1.2) ∂ ∂t (1.3) Starting from the Hamilton function H of classical mechanics, the time-dependent Schră odinger equation is given by E= p2 + V (x) → i Ψ(x, t) = 2m ∂t − 2m ∇2 + V (x) Ψ(x, t) , (1.4) i.e i ∂ Ψ(x, t) = H Ψ(x, t) ∂t (1.5) with the Hamiltonian H= − 2m ∇2 + V (x) (1.6) Energy eigenstates are given by the time-independent Schră odinger equation (H E) (x, t) = (1.7) 1 Introduction - Wave mechanics Figure 1.1: Square-well potential 1.2 One-dimensional problems For the sake of simplicity, we consider piecewise continuous potentials Assume that the potential has a step at a The time-independent Schră odinger equation can be reformulated as 2m d2 Ψ = − E − V (x) Ψ(x) dx2 (1.8) The second derivative Ψ makes at most a finite jump at a So both, Ψ and Ψ are continuous at a 1.2.1 Bound states The square well potential (see Fig 1.1) is given by V (x) = −V0 θ(a − |x|) = |x| > a −V0 |x| ≤ a V0 > real number (1.9) Bound states exist if −V0 ≤ E < The time-independent Schră odinger equation becomes 2mE |x| > a (1.10) Ψ =κ Ψ κ≡ Ψ = −q Ψ q≡ 2m(E + V0 ) |x| ≤ a (1.11) Basic solutions are Ψ= c+ e+κx + c− e−κx c1 e+iqx + c2 e−iqx |x| > a |x| ≤ a (1.12) e+κx is not normalizable for x > a, analogously e−κx for x < a Furthermore, since V (x) is an even potential, the solutions can be characterized according to their symmetry If we have even symmetry, the solution will be Ψ(x) = A cos qx |x| ≤ a e−κ|x| |x| > a (1.13) 1.2 One-dimensional problems For odd symmetry, we get Ψ(x) = B sin qx |x| ≤ a ±e−κ|x| x >< a (1.14) Assume that Ψ has even symmetry and that it is continuous A cos qa = e−κa (1.15) Ψ has to be continuous, too From that, we get Aq sin qa = κe−κa κ ⇒ tan aq = q (1.16) This is a transcendental equation that cannot be solved analytically Now assume odd symmetry B sin qa = e−κa κ ⇒ − cot qa = q Bq cos qa = −κe−κa (1.17) Again a transcendental equation that can only be solved graphically 2m(−E) κ κa = = a = q qa qa a κ = for qa = ⇒ 2mV0 q 2mV0 a2 − q a2 qa (1.18) The lowest energy state is always even Whenever tan qa = κ/q or − cot qa = κ/q, we have a solution There is at least one crossing point The number of states is given by √ 2a 2mV0 (1.19) NS = π with [α] nearest integer greater than α Even and odd states alternate Figure 1.2: Graphical solution of the square well problem Now let us assume that the potential walls are infinitely high, i.e V0 → ∞ Ψn = (−1)k cos(k + 1/2) πx a sin kπ a x n = 2k + n = 2k (1.20) − V0 (1.21) Eigenenergies En = 2m nπ 2a Introduction - Wave mechanics j in j trans j refl V0 x x=0 Figure 1.3: Potential step 1.2.2 Transmission-Reflection Problems Now we treat an one-dimensional potential step (1.22) V (x) = V0 θ(x) For x < 0, the potential is 0, for x > 0, the potential is V0 The Schră odinger equation for x < and x > is given by 2mE d2 Ψ = − Ψ = −k Ψ dx2 d2 Ψ 2m(E − V0 ) =− Ψ = −k Ψ 2 dx x0 (1.24) Let E > V0 Suppose a particle is incident from the left ΨI (x) = e+ikx + re−ikx +ik x (1.25) (1.26) ΨII (x) = te r and t are the reflection and transmission amplitudes Continuity of Ψ and Ψ at x = implies that 1+r =t k−k r= k+k ik(1 − r) = ik t (1.27) 2k t= k+k (1.28) Let us discuss the physical meaning of this Consider the probability flux jI (x) = = = m m Ψ∗I d ΨI dx (e−ikx + r∗ eikx )ik(eikx − re−ikx ) m ik − |r| −re−i2kx + r∗ ei2kx =real number k (1 − |r| ) ≡ jin − jref = m t∗ e−ik x (ik )teik x m k |t| ≡ jtrans = m (1.29) jII (x) = (1.30) Symmetry in Quantum Mechanics with   χ↑↑     √1 χ + χ ↑↓ ↓↑ χ(ms1 , ms2 ) =  χ ↓↓     √1 χ↑↓ − χ↓↑ triplet triplet (6.19) triplet singlet The first three states correspond to the symmetric triplet, the last one to the antisymmetric singlet Note that x1 , ms1 ; x2 , ms2 |P12 |α = x2 , ms2 ; x1 , ms1 |α ; hence we get for a fermion x1 , ms1 ; x2 , ms2 |α = − x2 , ms2 ; x1 , ms1 |α (6.20) Due to the fact that we were able to separate the spin and the position dependence of the wave function, the permutation operator has to decompose into a space and a spin permutation operator: space spin P12 = P12 P12 (6.21) In this case, the spin permutation operator is given by + S1 · S2 |α → P12 |α ⇒ Φ(x1 , x2 ) → Φ(x2 , x1 ) spin = P12 χ(ms1 , ms2 ) → χ(ms2 , ms1 ) (6.22) The combined wave function has to be antisymmetrical, so if the spatial part is symmetrical, the spin state has to be antisymmetrical and vice versa |Φ(x1 , x2 )| d3 x1 d3 x2 is the probability for finding electron around x1 and electron around x2 Assume that there is no mutual interaction, i e H ≡ H(p1 , p2 , x1 , x2 ) = p21 p2 + + Vext (x1 ) + Vext (x2 ) 2me 2me (6.23) In such a case, the eigenfunctions are separable: Φ(x1 , x2 ) = √ ϕA (x1 )ϕB (x2 ) ± ϕA (x2 )ϕB (x1 ) (6.24) The + sign is for the spin singlet, the − sign for the spin triplet Thus we get for the probability |Φ(x1 , x2 )| d3 x1 d3 x2 = 2 2 |ϕA (x1 )| |ϕB (x2 )| + |ϕA (x2 )| |ϕB (x1 )| ±2 ϕA (x1 )ϕB (x2 )ϕ∗A (x2 )ϕ∗B (x1 ) d3 x1 d3 x2 (6.25) exchange density This implies that in the triplet state |ϕ(x1 , x2 = x1 )| = 0, whereas for a singlet state we get an enhanced probability to find both particles at the same point in space: 2 |ϕ(x1 , x2 = x1 )| = |ϕA (x1 )| |ϕB (x2 )| this 86 does violate the Pauli exclusion principle, because the electron have opposite spin 6.3 The Helium Atom 6.3 The Helium Atom The simplest system where the problem of identity plays an important role is the helium atom Still the two-particle Schră odinger equation with the basic Hamiltonian H= p21 p2 2e2 2e2 e2 + − − + 2me 2me r1 r2 r12 (6.26) where r1 = |x1 |, r2 = |x2 |, and r12 = |x1 − x2 |, cannot be solved analytically Hence this system also serves as an example for the application of approximation methods The total spin is a constant of motion, because there is no external magnetic field so that the Hamiltonian commutes with the total spin operator Hence, the spin state is either a singlet or a triplet Since there is no analytical solution for this problem, we will first apply perturbation theory with the perturbation e /r12 The ground state configuration of the unperturbed system is 1s , i e both electrons are in n = 1, l = The spatial part of the wave function is then symmetric so that only the spin singlet function is allowed 6.3.1 Ground State The unperturbed wave function is the product of the single-electron wave functions and the proper spinor: Ψ100 (x1 )Ψ100 (x2 )χsinglet = Z −Z/a0 (r1 +r2 ) e χsinglet πa30 (6.27) The ground state energy of H0 is E0 = −2 × e2 = −8 Ryd = −108.8 eV 2a0 (6.28) This is considerably less than the experimental value of E0exp = −78.8 eV Let’s apply first-order perturbation theory: e2 r12 (1) ∆1s2 = 1s2 Z −2Z/a0 (r1 +r2 ) e2 d x1 d3 x2 e π a60 r12 = (6.29) Let r< and r> be the smaller and larger radius of r1 and r1 and γ is angle between x1 and x2 : = r12 r12 + r22 − 2r1 r2 cos γ ∞ ∞ = l−0 l r< P (cos γ) l+1 l r> l = l−0 l r< 4π Y ∗ (ϑ , ϕ1 ) Ylm (ϑ2 , ϕ2 ) l+1 2l + lm r> m=−l (6.30) After some algebra we get for the first-order shift (1) ∆1s2 = (1) ⇒ E0 e2 2a0 = 8− e2 ≈ −74.8 eV ≈ 95% · E0exp 2a0 (6.31) 87 Symmetry in Quantum Mechanics J I J E_{100} + E_{nlm} para 1P1 1s2p ortho 3P2,1,0 para 1S0 1s2s ortho 3S1 para 1S2 1s2 Figure 6.2: The Helium spectrum Now we apply the variational method Assume that the nucleus is screened by the charge cloud of the other electron Thus our trial function is 3 Zeff e−Zeff /a0 (r1 +r2 ) πa30 p2 p2 ¯ = ˜ H 0| + |˜ − 2m 2m Zeff = 2· − 2ZZeff + x1 , x2 |˜ = Ze2 Ze2 ˜ e2 ˜ ˜ 0| + |0 + ˜0| |0 r1 r2 r12 e2 Zeff a0 (6.32) where Zeff is the optimization parameter This equation is minimized when Zeff = 27/16 ≈ 1.6875 Hence, E0 = −77.5 eV, which is already rather close to the measured value of E0exp = −78.8 eV 6.3.2 Excited States Now we will consider the excited states 1s nl The energy is E = E100 + Enlm + ∆E To first order, we get ∆E = e2 r12 =I ±J (6.33) Here, we take to + sign for the singlet, the − sign for the triplet I is the direct integral and the exchange integral, respectively, which are given by e2 r12 d x1 d3 x2 I= |Ψ100 (x1 )| |Ψnlm (x2 )| J= Ψ100 (x1 )Ψnlm (x2 ) re12 Ψ∗100 (x2 )Ψ∗nlm (x1 )d3 x1 d3 x2 (6.34) He in the singlet state is called parahelium, while He in the triplet state is known as orthohelium The resulting energy spectrum is shown in Fig 6.2 Note that the energies are spin-dependent although there is no spin term in the Hamiltonian 88 Scattering Theory The interaction potential between two particles cannot be measured directly, only its consequences in scattering Therefore it is very important to understand the outcome of scattering processes By comparing measured and calculated scattering distributions, the interaction potentials between the scattering particles might be derived In this short chapter we will only cover the basic concepts of scattering theory 7.1 Wave Packets We now consider the scattering of particles by a fixed “short range” potential This means the particle is free except for the time of interaction As a consequence, Coulomb scattering is excluded which requires a special treatment Assume that an incident particle with average momentum p = k at t = t0 is described by a wave packet Ψ(r, t) = d3 k ik·r e ak (2π)3 (7.1) We expand Ψ(r, t) in exact eigenstates of the potential problem Ψ(r, t) = d3 k Ψk (r)Ak (2π)3 (7.2) with 2m with Ek = 2 k 2m ∇2 + Ek Ψk (r) = V (r)Ψk (r) (7.3) > At later times t > t0 , we have d3 k Ψk (r)Ak e−iEk (t−t0 )/ (2π)3 Ψ(r, t) = (7.4) We construct the energy eigenstates Ψk by using the Green’s functions approach, 2m ∇2 + Ek G(r, k) = (r) (7.5) Insertion into the Schră odinger equation leads to Ψk (r) = Φ0 (r) + G(r − r , k)V (r ) Ψk (r ) d3 r (7.6) where (r) is a solution of the homogenous Schră odinger equation, i e for a free particle without any potential acting on it Proof Operate with 2m ∇ + Ek on both sides of equation (7.6) 89 Scattering Theory We interpret this result as follows: the total wave Ψk (r) consists of the incident wave Φ0 (r) plus the scattered wave The second term V (r)Ψk (r) acts as a source for the scattered wave The propagator G(r − r , k) gives us the amplitude of the scattered wave at r due to a unit source at r It is given explicitly by solving the equation ∇2 + Ek eip·r d3 p = δ(r) (2π)3 Ek − p2 2m G(r, k) = d3 p eip·r m =− 2 (2π) Ek − 2π ir 2m p 2m (7.7) ∞ peipr dp p2 − k −∞ (7.8) The integral is evaluated by the theory of complex functions G(r, k) = − m keikr m eikr · 2πi · = − 2 π ir 2k 2π r (7.9) Let r be far from the potential k |r − r | ≈ kr − kr · rˆ = kr − k · r , (7.10) where k = kˆ r is the wave vector in the far field The asymptotic solutions are the of the form eikr fk (Ωr ) (7.11) r where Ωr specifies the direction of r fk (Ωr ) is called the scattering amplitude Ψk (r) = eik·r + fk (Ωr ) = − m 2π e−ik ·r V (r )Ψk (r ) d3 r (7.12) Insert (7.11) in (7.4) and use ak = Ak : Ψ(r, t) = Ψ0 (r, t) + d3 k ei ak (2π) kr−Ek (t−t0 )/ fk (Ωr ) r (7.13) where Ψ0 (r) = d3 k ak ei(k·r−Ek (t−t0 )/ (2π)3 ) (7.14) is the wave packet at time t if there were no potential present Now let us take a closer look at the second term of (7.13) Assume that ak is strongly peaked about k0 and that fk (Ωr ) slowly varies around k0 Then we can ˆ replace eikr fk (Ωr ) by eik·k0 r fk0 (Ωr ) and obtain fk0 (Ωr ) ˆ r, t) Ψ0 (k (7.15) r This equation says that the total wave function after scattering is given by the wave packet which one would have in the absence of any scattering plus a scattered part ˆ r, t) corresponds to the value The scattered term has a pictorial interpretation Ψ0 (k of the wave function at point r if all the potential did was to bend the trajectory of the particle from the forward direction towards r The probability amplitude that this bending has occured is just given by the factor fk0 (Ωr )/r Ψ(r, t) = Ψ0 (r, t) + 90 7.2 Cross Sections 7.2 Cross Sections Experimental results are usually expressed in terms of cross sections The differential cross section is defined as dσ # of particles scattered into unit solid angle per unit of time ≡ dΩ # of incident particles crossing a unit area per unit of time (7.16) With respect to the wave function (7.11), we note that the number of incident particles crossing a unit area per unit of time is just proportional to the probability flux due to the first term of Eq (7.11) Equivalently, the number of particles scattered into an unit solid angle per unit of time is given by the probability flux according to the second term of Eq (7.11) Thus we have r2 |jscatt | dσ = = |fk0 (Ωr )| dΩ |jincid | (7.17) The total cross section given by dσ/dΩ integrated over all angles is the total probability of a particle being scattered with respect to the total probability that crossed a unit area in front of the target Thus |fk0 (Ωr )| dΩ σ= (7.18) Note that cross sections have the unit of an area 7.3 Partial Waves Consider a spherical symmetric potential Then it is convenient to expand the plane waves in angular momentum eigenfunctions Assuming that k is parallel to the zˆ-axis, the expansion is given by ∞ eik·r = il (2l + 1)Pl (cos ϑ)jl (kr) l=0 ∞ = il (2l + 1)Pl (cos ϑ) hl (kr) + h∗l (kr) (7.19) l=0 where jl is a Bessel function and hl are the Hankel functions Only m = terms enter because of k zˆ: Pl (cos ϑ) = 4π Yl0 (ϑ, ϕ) 2l + (7.20) Consequently, fk (Ωr ) depends only on ϑ Asymptotically, the energy eigenstates of the potential problem can be written as Ψk (r) = ∞ il (2l + 1)Pl (cos ϑ) h∗l (kr) + ei2δl hl (kr) l=0 = eik·r + ∞ il (2l + 1)Pl (cos ϑ)(ei2δl − 1)hl (kr) (7.21) l=0 91 Scattering Theory where we have used the representation (7.19) for eik·r The factors δl appearing in the exponential term are called phase shifts Thus, the scattering process is completely described in terms of the phase shifts ei2δl is called the partial wave scattering amplitude Total cross sections can be expressed as σ= 4π k2 ∞ ∞ (2l + 1) sin2 δl = l=0 σl (7.22) l=0 One can show that if < δl < π/2, the potential is attractive, i e V < Here, the frequency of the wave functions increases If δl < 0, the potential is predominantly repulsive 7.4 Born Approximation We assume that the potential has only a small effect on all partial waves Approximate Ψk (r) by eik·r in (7.12) Then we have fk (Ωr ) = − m 2π ei(k−k )·r V (r ) d3 r = − m V (k − k) , 2π (7.23) where V (k) corresponds to the Fourier transform of V (r) This approximation is known as the Born approximation As an example, we will apply this approximation to the so-called Yukawa potential.: 4πa e−κr =⇒ V (q) = r q + κ2 dσ a2 ⇒ = κ2 dΩ 4Ek sin2 ϑ2 + 2m V (r) = a (7.24) For a = e2 and κ → this results reduces to the classical Rutherford cross section for Coulomb scattering However, this is just accidental since the Born approximation is not valid for the long-range Coulomb potential 92 Relativistic Quantum Mechanics So far, we did not consider any relativistic effects The kinetic energy of a classical particles is given by E= p2 2m (8.1) Quantum mechanically, the wave function of a particle ψ(r, t) = ei(pÃrEt)/ (8.2) obeys the Schră odinger equation i (r, t) = (−i ∇) ψ(r, t) ∂t 2m (8.3) Thus, the Schră odinger equation can be obtained from the energy momentum relation (8.1) by making the following correspondences: E → i ∂ ∂t p → −i ∇ Furthermore, the introduction of electromagnetic fields is described by a pair of ∂ ∂ a scalar and a vector potential (Φ, A) i ∂t → i ∂t − eΦ, −i ∇ → −i ∇ − ec A 8.1 Relativistic Spin Zero Particles For the relativistic case, we will also use the energy-momentum relation E = p2 c2 + m2 c4 (8.4) in order to derive a relativistic wave equation If we apply our correspondences, we obtain the Klein-Gordon equation, i ∂ c ∂t 2 Ψ(r, t) = i ∇ Ψ(r, t) + m2 c2 Ψ(r, t) (8.5) or if we write it another way, we get ∂2 m2 c2 − ∇ + Ψ(r, t) = c2 ∂t2 (8.6) This looks like a classical wave equation known, e.g., from electrodynamics, with an extra term (m2 c2 / ) The Klein-Gordon equation is a second-order differential equation in time This has some important and unexpected consequences: 93 Relativistic Quantum Mechanics Both Ψ(r, t) and ∂Ψ ∂t must be known as initial conditions Hence we have to specify more initial information about the particle As it turns out, this extra information corresponds to the charge of the particle The wave function Ψ(r, t) = ei(p·r−Et)/ with either sign of E = ± p2 + m2 c2 solves the Klein-Gordon equation Thus we have negative eigenenergy solutions for free particles In fact, the + sign corresponds to particles whereas the − corresponds to antiparticles! There is another strange consequence of the Klein-Gordon equation Let Ψ be a solution of the Klein-Gordon equation Then Ψ|Ψ = constant Thus, |Ψ∗ (r, t)Ψ(r, t)| cannot be interpreted as a probability density anymore However, the Klein-Gordon equation (8.5) is invariant under complex conjugation Thus Ψ∗ is also a solution of (8.5) Therefore we can write Ψ∗ m2 c2 ∂2 m2 c2 ∂2 2 − ∇ + Ψ − Ψ − ∇ + Ψ∗ = 2 c2 ∂t2 c2 ∂t2 (8.7) This gives us a continuity equation: ∂ρ + ∇ · j(r, t) = ∂t (8.8) with the current defined as usual j(r, t) = 2im Ψ∗ ∇Ψ − Ψ ∇Ψ∗ , (8.9) but an modified density defined by ρ(r, t) = i ∂Ψ∗ ∗ ∂Ψ − Ψ Ψ 2mc2 ∂t ∂t (8.10) We consider a free particle at rest, i e p = The wave function corresponding to the positive energy solution of the Klein-Gordon equation is Ψ(r, t) = e−imc t/ (8.11) Now we will transform this system to a frame moving with velocity −v Thus the momentum and the energy are p = γ(v)mv Ep = γ(v)mc2 (8.12) where γ(v) is given by γ(v) = 1− v2 c2 (8.13) In the new frame, the wave function is given by Ψ(r , t ) = ei(p·r −Ep t )/ (8.14) −mc2 t = p · r − Ep t (8.15) The quantity 94 8.1 Relativistic Spin Zero Particles is a Lorentz scalar, and thus it is the same in all frames Using (8.9) and (8.10), we obtain Ep mc2 p c2 j(r , t ) = = ρ(r , t )p = vρ(r , t ) m Ep (8.16) ρ(r , t ) = (8.17) Thus, the current is just the density times the relativistic velocity of the particle pc2 /Ep Let us now consider the negative energy solutions Starting again with a particle at rest, we have Ψ(r, t) = e+imc t/ ⇒ ρ(r, t) = −1 (8.18) The particle density is negative! We identify the negative particle density with the positive density of antiparticles Now we the same step-by-step considerations that we did before We transform to a frame moving with velocity −v We then get Ψ(r , t ) = eimc t/ = e−i(p·r −Ep t )/ (8.19) Thus, the antiparticle of energy Ep and momentum p corresponds to a particle of energy −Ep and momentum −p! The flux and the density also change analogously ρ(r , t ) = − Ep mc2 j(r , t ) = − p c2 = ρ(r , t )p m Ep (8.20) What happens to a particle in an electromagnetic field? In the presence of an electromagnetic field, the Klein-Gordon equation becomes ∂ − eΦ(r, t) i c2 ∂t Ψ(r, t) = e −i ∇ − A(r, t) c + m2 c2 Ψ(r, t) (8.21) If we take the complex conjugate of this equation, we obtain ∂ i + eΦ(r, t) c ∂t Ψ∗ (r, t) = e −i ∇ + A(r, t) c + m2 c2 Ψ∗ (r, t) (8.22) This says that if Ψ is a solution of the Klein-Gordon equation, then Ψ∗ is a solution to the Klein-Gordon equation with the opposite sign of the charge and the same mass Comparing the solutions for positive and negative energies, we note that in fact the complex conjugate of a positive energy solution is a negative energy solution Hence it follows that particles and antiparticle must have opposite charges! In the presence of an electromagnetic field, we still have the continuity equation (8.8), but with modified currents and density given by j(r, t) = e e Ψ∗ −i ∇ − A Ψ + Ψ i ∇ − A Ψ∗ , 2m c c (8.23) and ρ(r, t) = Ψ∗ 2mc2 i ∂ − eΦ Ψ + Ψ ∂t −i ∂ − eΦ Ψ∗ ∂t (8.24) 95 Relativistic Quantum Mechanics V = eΦ p x x=0 Figure 8.1: Potential step 8.2 Klein’s Paradox In this section we will see that the negative energy of the antiparticles implies that a particle-antiparticle pair could be created without violating energy conservation! In order to derive this phenomenon, we will now take a look at an electrostatic potential barrier of height V = eΦ, basically a potential step – but this time, we look at it relativistically As in the nonrelativistic case, we will determine the stationary solutions of the problem For x < 0, there is no potential The solution to this problem is a superposition of an incident and a reflected wave, Ψ(x) = aeipx/ + be−ipx/ x 0, the Klein-Gordon equation in the presence of an electrostatic field according to Eq (8.21) is given by (Ep − V )2 Ψ(x) = − c2 ∂ Ψ(x) + m2 c4 Ψ(x) ∂x2 (8.26) The time dependence of the stationary solutions is given by e−iEp t/ The stationary wave function is of the form Ψ(x) = deikx , (8.27) where the wave vector k is related to the energy via 2 c k + m2 c4 = (Ep − V )2 We specify the boundary conditions by assuming that Ψ and Thus, b= p− k a p+ k d= 2p a p+ k (8.28) ∂Ψ ∂x are continuous (8.29) Now, there are three cases Case Ep > V + mc2 Then the energy is sufficient to surmount the barrier, and we get k= (Ep − V )2 − m2 c4 c which is analogous to the non-relativistic case 96 (8.30) 8.2 Klein’s Paradox Case Ep − mc2 < V < Ep + mc2 Then k = iκ is imaginary with κ= m2 c4 − (Ep − V )2 c (8.31) In this case, the wave is reflected at the barrier; the density in the barrier is given by ρ(x) = Ep − V |d| e−2κx mc2 (8.32) So ρ(x) > for Ep > V , but ρ(x) < for Ep < V This means that we reflect positively charged particles from the wall but find negatively charged particles within the wall Case V > Ep + mc2 Now we expect that is is even more impossible to cross the barrier since the barrier height has been further increased However, from (8.28) it follows that k is real again so that we have a particle current propagating to the right The group velocity of the waves for x > is given by vg = ∂Ep ∂( k) (8.33) Using Eq (8.28), we get ∂ ∂k ∂ (Ep − V )2 ∂k c2 k ⇒ vg = Ep − V 2 c k = ⇒ 2 c k = (Ep − V ) ∂E ∂k (8.34) Here, the denominator is negative, i e k < for a wave packet travelling to the right (x > 0) This also implies that the reflection coefficient b/a > 1, i e more wave is reflected than is incident! The charge density for x > according to Eq (8.24) is ρ= Ep − V |d| < mc2 (8.35) The current is also negative for x > These results can be interpreted by assuming that the incident particle induces the creation of pairs of particles and antiparticles if the potential is larger than mc2 The created antiparticles experience an attractive potential −eΦ because of their opposite charge The created particles travel to the left Together with the totally reflected incident wave the outgoing current to the left is larger than the incident one Still the total outgoing current equals the incident current due to charge conservation Let us now consider energy conservation The energy of the created particles is given by E= Ep √ c2 k + m2 c4 − V x < 0, particle x > 0, antiparticle (8.36) For the antiparticle at the right, we have taken into account that the electrostatic potential has the opposite sign for the opposite charge If we take the positive square 97 Relativistic Quantum Mechanics root of both sides of Eq (8.28), which means that the square root at the right hand side becomes V − Ep because of V > Ep + mc2 , we find that c2 k Ep + + m2 c4 − V = (8.37) This means that it takes zero energy to create a particle-antiparticle pair since the energy of the antiparticle is not only merely less than mc2 but negative because of the large potential V 8.3 Dirac Equation Now we consider relativistic spin 12 particles In order to describe the properties of the spin under Lorentz transformations, a new dynamical variable α has to be introduced This α behaves as a vector under spatial rotations Hence the commutation relations with S are given by αi , Sj = αi Sj − Sj αi = iεijk αk (8.38) where S is the spin operator Let β be the parity operator in the spin space Note that the spin representation of rotations is double valued, hence β can be either +1 or −1 For convenience we choose β = The standard representations of α and β for spin are α= σ σ β= 0 −1 (8.39) where the elements stand for × matrices; here, the σ’s are the Pauli matrices This means that a relativistic spin 21 particle has four internal states – twice as many as in the nonrelativistic case In equivalence to the Klein-Gordon equation, these additional states are associated with negative energy eigensolutions, i.e with antiparticles From Eq (8.39) now follows that αβ + βα = 0, (βαi )2 = −1 (8.40) Let γ µ = (β, βα) and (E/c, p) be four-vectors Then the scalar product of these two four vectors, βE − βα · p c (8.41) has to transform as a scalar In order to derive the meaning of this quantity, we square it: βE − βα · p c = β2 E2 E + (βα · p)2 −(ββα + βαβ ) · p c c =1 = E c ⇒ βE − βcα · p = mc2 98 −p2 =0 − p2 = m2 c2 (8.42) (8.43) 8.3 Dirac Equation ∂ and p → i ∇ in (8.43) and multiply it by Now, we use the correspondence E → i ∂t β From that we immediately get the Dirac equation for relativistic spin 21 particles, i ∂Ψ = −i cα∇ + βmc2 Ψ ∂t (8.44) From the Dirac equation, the Hamiltonian can be derived to be H = cα · p + βmc2 (8.45) In the presence of an electromagnetic field, the Dirac equation becomes by invoking the usual correspondences i ∂ − eΦ Ψ = cα · ∂t i e ∇ − A + βmc2 ψ c (8.46) We will now show that the gyromagnetic ratio of the electron, g = 2, falls out of the Dirac equation To show that we consider the nonrelativistic limit of the Dirac equation First we write the wave function as ψ= φ χ (8.47) , where φ and χ are each two-component tensors Then the Dirac equation takes the form of coupled differential equations, e ∇ − A · σχ + (eΦ + mc2 )φ i c e ∇ − A · σφ + (eΦ − mc2 )χ i c ∂φ =c ∂t ∂χ i =c ∂t i (8.48) (8.49) In the non-relativistic limit, the energy of the particle is can be approximated to leading order by E ≈ mc2 Hence we have χ ∝ e−iEt/ = e−imc t/ (8.50) and thus i ∂χ = mc2 χ + ∂t Inserting this into (8.49) yields with mc2 mc2 χ = c i (8.51) eΦ e ∇ − A · σφ − mc2 χ , c (8.52) e ∇ − A · σφ c (8.53) or χ= 2mc i This means that χ is smaller than φ by a factor of ∼ v/c Inserting (8.53) in (8.48) and using (a · σ)(b · σ) = (a · b) + iσ · (a × b) (8.54) 99 Relativistic Quantum Mechanics and (∇ × A + A × ∇)φ = Bφ , (8.55) where B is the magnetic field, we obtain i ∂φ = ∂t 2m e ∇− A i c φ− e σ · Bφ + (eΦ + mc2 )φ 2mc (8.56) This corresponds exactly to the nonrelativistic equation for a spin 12 particle in an electromagnetic field, the so-called Pauli equation It immediately follows that the magnetic moment associated with the spin of the particle is e /2mc which means that the gyromagnetic moment is g = 2, i.e we not have to put it in This is one of the great triumphs of the Dirac equation Note that the Dirac equation does not include the influence of vacuum fluctuations which lead to the small α/2π corrections to g, where α is the fine structure constant 100 ... Mechanics Fundamental Concepts of Quantum Mechanics Quantum Dynamics Angular Momentum Approximation Methods Symmetry in Quantum Mechanics Scattering Theory Relativistic Quantum Mechanics Suggested... J.J Sakurai, Modern Quantum Mechanics, Benjamin/Cummings 1985 • G Baym, Lectures on Quantum Mechanics, Benjamin/Cummings 1973 • F Schwabl, Quantum Mechanics, Springer 1990 III Preface Criteria...Preface Theoretical Physics Master Quantum Mechanics Prof Dr Axel Groß Phone: 289–12355 Room No.: 3248 Email: agross@ph.tum.de