Modern Physics. Luis A. Anchordoqui. Department of Physics and Astronomy Lehman College, City University of New York. Lesson X November 12, 2015

Transcription

1 Modern Physics Luis A. Anchordoqui Department of Physics and Astronomy Lehman College, City University of New York Lesson X November 2, 205 L. A. Anchordoqui (CUNY) Modern Physics / 25

2 Table of Contents Schrödinger Equation Particle in a central potential 2 Stern-Gerlach experiment 3 Klein-Gordon Equation L. A. Anchordoqui (CUNY) Modern Physics / 25

3 L. A. Anchordoqui (CUNY) Modern Physics / 25

4 Schrödinger Equation Particle in a central potential Prescription to obtain 3D Schrödinger equation for free particle: substitute into classical energy momentum relation differential operators E = p 2 2m () E iħ t and p iħ (2) resulting operator equation acts on complex wave function ψ( x, t) ħ2 2m 2 ψ = iħ t ψ (3) Interpret ρ = ψ 2 as probability density ψ 2 d 3 x gives probability of finding particle in volume element d 3 x L. A. Anchordoqui (CUNY) Modern Physics / 25

5 Schrödinger Equation Particle in a central potential Continuity equation We are often concerned with moving particles e.g. collision of particles Must calculate density flux of particle beam j From conservation of probability rate of decrease of number of particles in a given volume is equal to total flux of particles out of that volume t V ρ dv = j ˆn ds = S (last equality is Gauss theorem) V j dv (4) Probability and flux densities are related by continuity equation ρ t + j = 0 (5) L. A. Anchordoqui (CUNY) Modern Physics / 25

6 Schrödinger Equation Particle in a central potential Flux To determine flux... First form ρ/ t by substracting wave equation multiplied by iψ from the complex conjugate equation multiplied by iψ ρ t ħ 2m (ψ 2 ψ ψ 2 ψ ) = 0 (6) Comparing this with continuity equation probability flux density j = iħ 2m (ψ ψ ψ ψ ) (7) Example free particle of energy E and momentum p has ρ = N 2 and j = N 2 p/m ψ = Ne i p x iet (8) L. A. Anchordoqui (CUNY) Modern Physics / 25

7 Schrödinger Equation Particle in a central potential Time-independent Schrödinger equation for central potential Potential depends only on distance from origin hamiltonian is spherically symmetric V( r) = V( r ) = V(r) (9) Instead of using cartesian coordinates x = {x, y, z} use spherical coordinates x = {r, ϑ, ϕ} defined by x = r sin ϑ cos ϕ y = r sin ϑ sin ϕ z = r cos ϑ r = x 2 + y 2 + z 2 ϑ = arctan (z/ ) x 2 + y 2 (0) ϕ = arctan(y/x) Express the Laplacian 2 in spherical coordinates 2 = ( r 2 r 2 ) + ( r r r 2 sin ϑ ) + sin ϑ ϑ θ r 2 sin 2 ϑ 2 ϕ 2 () L. A. Anchordoqui (CUNY) Modern Physics / 25

8 Schrödinger Equation Particle in a central potential To look for solutions... Use separation of variable methods ψ(r, ϑ, ϕ) = R(r)Y(ϑ, ϕ) [ ( ħ2 Y d 2m r 2 r 2 dr ) + R ( dr dr r 2 sin ϑ Y ) + sin ϑ ϑ ϑ Divide by RY/r 2 and rearrange terms [ ( ħ2 d r 2 dr 2m R dr dr )] + r 2 (V E) = ħ2 2mY [ sin ϑ R r 2 sin 2 ϑ 2 ] Y ϕ 2 + V(r)RY = ERY ( sin ϑ Y ) + ϑ ϑ sin 2 ϑ Each side must be independently equal to a constant κ = 2m ħ2 l(l + ) Obtain two equations sin ϑ d R dr ( r 2 dr ) 2mr2 dr ( sin ϑ Y ) + ϑ ϑ sin 2 ϑ What is the meaning of operator in angular equation? 2 ] Y ϕ 2 ħ 2 (V E) = l(l + ) (2) 2 Y = l(l + )Y (3) ϕ2 L. A. Anchordoqui (CUNY) Modern Physics / 25

9 Schrödinger Equation Particle in a central potential Angular momentum operator ˆ L = ˆ r ˆ p = iħ ˆ r ˆ (4) in cartesian coordinates ( ˆL x = ŷ ˆp z ˆp y ẑ = iħ y z ) y z ( ˆL y = ẑ ˆp x ˆp z ˆx = iħ z x ) z x ( ˆL z = ˆx ˆp y ˆp x ŷ = iħ x y ) x y (5) commutation relations [ ˆL i, ˆL j ] = iħ ε ijk ˆL k and [ ˆL 2, ˆL x ] = [ ˆL 2, ˆL y ] = [ ˆL 2, ˆL z ] = 0 (6) ˆL 2 = ˆL 2 x + ˆL 2 y + ˆL 2 z L. A. Anchordoqui (CUNY) Modern Physics / 25

10 Schrödinger Equation Particle in a central potential We can always know: length of angular momentum plus one of its components E.g. choosing the z-component resentation of the angular momentum, with fixed L z and L 2, but complete uncer L. A. Anchordoqui (CUNY) Modern Physics / 25

11 Schrödinger Equation Particle in a central potential Angular momentum vector in spherical coordinates ( ˆL x = iħ sin ϕ ) + cot ϕ cos ϕ θ ϕ ( ˆL y = iħ cos ϕ ) cot ϑ sin ϕ ϑ ϕ ˆL z = ħ ϕ Form of ˆL 2 should be familiar ˆL 2 = ħ 2 [ sin ϑ ϑ ( sin ϑ ) + ϑ sin 2 ϑ 2 ] ϕ 2 (7) (8) Eigenvalue equations for ˆL 2 and ˆL z operators: ˆL 2 Y(ϑ, ϕ) = ħ 2 l(l + )Y(ϑ, ϕ) and ˆL z Y(ϑ, ϕ) = ħmy(ϑ, ϕ) L. A. Anchordoqui (CUNY) Modern Physics / 25

12 Schrödinger Equation Particle in a central potential Solution of angular equation sin ϑ (sin ϑ Ym ϑ l (ϑ, ϕ) ϑ ) + 2 Yl m (ϑ, ϕ) sin 2 ϑ ϕ 2 = l(l + )Yl m (ϑ, ϕ) Use separation of variables Y(ϑ, ϕ) = Θ(ϑ)Φ(ϕ) By multiplying both sides of the equation by sin 2 ϑ/y(ϑ, ϕ) [ sin ϑ d ( sin ϑ dθ )] + l(l + ) sin 2 ϑ = d 2 Φ Θ(ϑ) dϑ dϑ Φ(ϕ) dϕ 2 (9) 2 equations in different variables introduce constant m 2 : d 2 Φ dϕ 2 = m2 Φ(ϕ) (20) sin ϑ d ( sin ϑ dθ ) = [m 2 l(l + ) sin 2 ϑ]θ(ϑ) dϑ dϑ (2) L. A. Anchordoqui (CUNY) Modern Physics / 25

13 Schrödinger Equation Particle in a central potential Solution of angular equation First equation is easily solved to give Φ(ϕ) = e imϕ Imposing periodicity Φ(ϕ + 2π) = Φ(ϕ) m = 0, ±, ±2, Solutions to the second equation Θ(ϑ) = APl m (cos ϑ) P m l associated Legendre polynomials Normalized angular eigenfunctions Y m l (ϑ, ϕ) = Spherical harmonics are orthogonal: π 2π 0 0 (2l + ) (l m)! 4π (l + m)! Pm l (cos ϑ)e imϕ (22) Yl m (ϑ, ϕ) Yl m sin ϑdϑdϕ = δ ll δ mm, (23) L. A. Anchordoqui (CUNY) Modern Physics / 25

14 385 Cuando ℓ = 0, tenemos m = 0. Ahora PY = #φ) y la= constante normalización # de 0,0 ( + 9 cos2 θ) sen3 θe3 i φ 5,3 (θ,# Particle in a central potential 32Equation uación representamos las superficies rschr = o dinger Yl,m (θ, φ)π# para estos armóes c0,0 = / 4π. Por tanto, TABLE I: Associated Legendre polynomials. éricos Cuya representación 0 2 Siendo las correspondientes gráficas Y0,0 (θ, φ) =. 4π 0 P00 = Diferenciales Ecuaciones II 0 P = cos # 2 P20 =es(3 cos2 # La gráfica correspondiente 3 P30 (5 cos3 )/ cos #)/2 # P3 P sin # Ecuaciones Diferenciales II P2 = 3 cos # sin # P22 = 3 sin2 # Métodos y desarrollo autofunciones = 3(5 cos2de#separación )/2 de sinvariables # P32 = 5encos # sin2 # We can then write the eigenvalue equations for these two operators: andl 0 (#, ') L 2 Y (#, ') = }2 l(ll +=)Y 3 [Capítulo 3 sin3 P33 = 5 # ℓ= = L z Y (#, ') = }my (#, ') and (42) used Resolución la they ecuación where we already the fact de that shareradial common eigenfunctions. Then, by its very nature we can label these eigenfunctions by l and m, i.e. Yl,m (#,según '). k sea nulo o no. Distinguimos dos casos 3.4] La ecuación de Helmholtz en coordenadas esféricas ± 2 2 Y such (#, ')that l = 0,, 2, and 0 that radial Si k = 0Show la EXERCISE z are integers Y0ecuación (#, ') the allowed values for!0l and! m!!!y,± (θ, φ)! Y!the ') Y(#, φ)!,0 (θ, mz = l,, l, l. [Hint: This result can2 be inferred from commutation relationship.] r R + 2rℓR= 2 ℓ(ℓ + )R = 0 Si ℓ = podemos tener tres casos: m =, 0,. Debemos evaluar las funciones We Pnow go back to the Schro dinger equation in spherical coordinates de Legendre,0 y P,. Acudiendo a la fórmula (3.7) obtenemos l = 2 Diferenciales II Ecuaciones 47 and we consider the angular and radial d (ξ 2 ) = 2ξ, dξ! 2 d P, (ξ) = ξ 2 (ξ 2 ) = 2 ξ 2Para. ℓ = 2 es fácil obtener d ξ2!2!!y2,2 (θ, φ)! Y2±2 (#, ')!!!! 2!!!Y2, (θ,! 2 Y (θ, φ) 2,0 Y2±φ)(#, ') Y20 (#, ') P,0 (ξ) =! Por ello, Y,0 (θ, φ) = Y, (θ, φ) = " 3 sen θ ei φ, 8π " Y, (θ, φ) = Y2,0 (θ, φ) = " " 3 8π Por último, como ejercicio dejamos el cálculo de L. A. Anchordoqui (CUNY) " 5 ( + 3 cos2 θ), 6π " 5 5 iφ Y2, (θ, φ) = sen θ cos θ e, Y2, (θ, φ) = sen θ cos θ e i φ, 8π 8π " " iφ 5 5 sen Yθ2,2e(θ,. 2 2iφ φ) = sen θ e, Y2, 2 (θ, φ) = sen2 θ e 2 i φ. 32π 32π 3 cos θ, 4π Modern Physics " / 25

15 Schrödinger Equation Particle in a central potential Solution of radial equation ( d r 2 dr(r) ) 2mr2 dr dr ħ 2 (V E) = l(l + )R(r) (24) to simplify solution u(r) = rr(r) ħ2 d 2 [ u 2m dr 2 + V + ħ 2m define an effective potential ] l(l + ) r 2 u(r) = Eu(r) (25) V (r) = V(r) + ħ2 l(l + ) 2m r 2 (26) (25) is very similar to the one-dimensional Schrödinger equation Wave function need 3 quantum numbers (n, l, m) ψ n,l,m (r, ϑ, ϕ) = R n,l (r)y m l (ϑ, ϕ) (27) L. A. Anchordoqui (CUNY) Modern Physics / 25

16 Stern-Gerlach experiment 24 Chapter 7 The Hydrogen Atom in Wave Mechanics Stern-Gerlach apparatus z axis + m l Oven Slit Magnet Screen Beam FIGURE of atoms 7.6 Schematic passes diagram through of a the region Stern-Gerlach where there experiment. is nonuniform A beam of B-field atoms passes through a region where there is a nonuniform magnetic field. Atoms with their Atoms magnetic withdipole theirmoments magnetic in opposite dipoledirections moments experience in opposite forces in directions opposite directions. experience forces in opposite directions experience a net upward force and are deflected upward, m l = (µ L,z =+µ B ) are deflected downward. The at undeflected. After passing through the field, the beam strikes a scr visible image. When the field is off, we expect to see one i L. A. Anchordoqui (CUNY) Modern Physics / 25

17 m passes through a region where there is a nonuniform magnetic field. Atoms with their If magnetic dipole moments in opposite directions experience forces in opposite directions. sc Results of Stern-Gerlach experiment If (a) th experience a net upward force and are deflec va m l = (µ L,z =+µ B ) are deflected downw it undeflected. an After passing through the field, the beamth visible image. When the field is off, we expec si center of the screen, because there is no deflec hy expect three images of the slit on the screen on m l = 0), one above the center (m l =+), and l If the atom were in the ground state (l = 0), m screen whether the field was off or on (recallw t If we had prepared the beam in a state with l = (a) (b) Image of slit with field turned off (left) the field on. The number of images that appea T With the field on two images of slitvalues, appear FIGURE which7.7 is equalthe to 2l results +. With of the the pos m it follows that 2l + hasthevalues,3,5,7 Small divisions in the scale represent 0.05 Stern-Gerlach mm experiment. (a) The ba an oddimage number of the of images slit withon the the field screen. turnedhowe nu the experiment off. (b) Withthefieldon, with hydrogentwoimages in the l = s be six images of theonslit theappear. screen! The Even small moredivi- sionsininthe thel scale = 0state,wewouldfindnot the left confusi (s hydrogen represent FIGURE 7.6 Schematic Stern-Gerlach diagram experiment of the Stern-Gerlach experiment. A beam of atoms L. A. Anchordoqui (CUNY) Modern Physics / 25

18 Stern-Gerlach experiment Uhlenbeck-Goudsmit-Pauli hypothesis Magnetic moment connected via intrinsic angular momentum µ S = e 2m e g e S (28) For intrinsic spin only matrix representation is possible Spin up and down are defined by ( ) ( ) 0 spin up spin down. (29) 0 Ŝ z spin operator is defined by Ŝ z = ħ 2 ( 0 0 ) (30) L. A. Anchordoqui (CUNY) Modern Physics / 25

19 Stern-Gerlach experiment Ŝ z acts on up and down states by ordinary matrix multiplication Ŝ z = ħ ( ) ( ) 0 = ħ ( ) = ħ (3) Ŝ z = ħ 2 ( 0 0 ) ( 0 ) = ħ 2 ( 0 ) = ħ (32) 2 As for orbital angular momentum [Ŝ i, Ŝ j ] = iħ ɛ ijk Ŝ k Ŝ x = ħ 2 ( 0 0 ) and Ŝ y = ħ 2 ( 0 i i 0 ) (33) Only 4 hermitian 2-by-2 matrices indentity + Pauli matrices ( ) ( ) ( ) 0 0 i 0 σ = σ 0 2 = σ i 0 3 = 0 (34) L. A. Anchordoqui (CUNY) Modern Physics / 25

20 he spin angular moexhibits space quanfigure shows the two tations of the spin spin. 2 particle, such S z Stern-Gerlach experiment Spin up h 2 m s = 2 0 S = 3 2 h h 2 m s = 2 Spin down L. A. Anchordoqui (CUNY) Modern Physics / 25

21 Klein-Gordon Equation Relativistic wave equation Schrodinger equation violates Lorentz invariance and is not suitable for particle moving relativistically Making the operator substitution starting from relativistic energy momentum relation 2 ψ t ψ = m 2 ψ (ħ = c = ) (35) Introducing the covariant form p µ i µ µ = ( ) t, and µ = ( ) t, (36) we can form invariant (D Alembertian) operator 2 µ µ µ µ ψ + m 2 ψ ( 2 + m 2 )ψ = 0 (37) Recall ψ( x, t) is scalar complex-valued wave function L. A. Anchordoqui (CUNY) Modern Physics / 25

22 Klein-Gordon Equation Negative probability density? Multiplying KG by iψ minus complex conjugate equation by iψ t [i(ψ t ψ ψ t ψ )] }{{} ρ +. [ i(ψ ψ ψ ψ )] = 0 (38) }{{} j Consider motion free particle of energy E and momentum p from (38) ρ = 2 E N 2 and j = 2 p N 2 ψ = N e i( p. x Et) (39) Probability density ρ is timelike component of 4-vector ρ E = ±( p 2 + m 2 ) /2 (40) In addition to acceptable E > 0 solutions we have negative energy solutions which have associated negative probability density! L. A. Anchordoqui (CUNY) Modern Physics / 25

23 Klein-Gordon Equation Antimatter Pauli and Weisskopf inserted charge e in continuity equation j µ = i e (ψ µ ψ ψ µ ψ ) (4) interpreting j µ as electromagnetic charge-current density j 0 represents a charge density not a probability density and so the fact that it can be negative is no longer objectable Stückelberg and Feynman negative energy solution describes a particle which propagates backwards in time or positive energy antiparticle propagating forward in time L. A. Anchordoqui (CUNY) Modern Physics / 25

24 Klein-Gordon Equation L. A. Anchordoqui (CUNY) Modern Physics to be continued / 25

25 Klein-Gordon Equation Thor s spinless positron Consider spin-0 particle with (E, p, e) generally referred to as the spinless electron Electromagnetic 4-vector current is Taking antiparticle e + of same (E, p) j µ (e ) = 2e N 2 (E, p) (42) j µ (e + ) = +2e N 2 (E, p) = 2e N 2 ( E, p) (43) exactly same current of the original particle with E, p As far as system is concerned emission of antiparticle with energy E is same as absorption of particle of energy E Negative-energy particle solutions going backward in time describe positive-energy antiparticle solutions going forward in time L. A. Anchordoqui (CUNY) Modern Physics / 25