Quantum Physics Lecture 8

Transcription

1 Quantum Physics Lecture 8 Applications of Steady state Schroedinger Equation Box of more than one dimension Harmonic oscillator Particle meeting a potential step

2 Waves/particles in a box of >1 dimension b Consider 2-D box Potential U = 0 between x = 0 and x = a And between y = 0 and y = b U infinite elsewhere Wavefunction Ψ expressed as Ψ( x.y) = f ( x) g( y) Steady state Schrödinger Equation inside box, where U = 0 2 2m 2 Ψ( x.y) x 2 ( ) + 2 Ψ x.y y 2 ( ) = f ( x) g ( y) Ψ x.y = EΨ x.y y U = 0 0 a x - Separation of variables ( ) ( Recall HΨ = EΨ ) Put in - noting partial differentials!

3 Waves/particles in a 2-D box (cont.) 2m g y 2 ( ) 2 f ( x) x 2 + f x ( ) 2 g( y) y 2 = E f x ( ) g( y) Thus Only one term is x dependent, and it equals a constant ( ) x 2 f x ( ) 2 f x = C So 2 f ( x) x 2 2 2m 2 f ( x) x 2 f ( x) + 2 g ( y) y 2 g ( y) = Cf ( x) Which we have seen before ( ) = Asin nπ x Solutions, using boundary conditions, are f x with a Similarly, for y dependence g( y) = Bsin mπ y b Hence the energy levels in the box are E n,m = 2 2m With TWO quantum numbers n,m needed to specify the state n 2 π 2 C = n2 π 2 a 2 = E + m2 π 2 a 2 b 2

4 Waves/particles in a 2-D box (cont.) Ψ is specified by the quantum numbers n & m There are as many states as there are possible n,m combinations (N.B. n & m are positive) Two distinct wave functions are DEGENERATE if they have the same energy. e.g. the states 1,3 and 3,1 are degenerate if a = b Examples: 2,3 and 3,2 wavefunctions If a/b is irrational there are no degeneracies Readily extended to 3-D. Useful, especially when filling box with more than one particle. C.f. black body cavity

5 Harmonic Oscillator Examples: mass on spring, diatomic molecule Hooke s Law: restoring force F = -kx kx = m d 2 x dt 2 d 2 x dt 2 + k m x = 0 x = Acosωt Where ω = k m Potential energy U = (1/2)kx 2 (potential well, parabolic) Expect (1) quantised energy, (2) E min 0 (3) particle outside the classical limits A

6 Apply SSSE to Harmonic Oscillator d 2 ψ dx 2 + 2m 2 E 1 2 kx 2 ( )ψ = 0 or d 2 ψ dx 2 + write y = 2mE 2 km km 2 x 2 ψ = 0 12 x and dy 2 = d 2 ψ dy 2 + α y 2 ( )ψ = 0 α = 2mE 2 km km dx 2 = 2E km m hν k = 2E ω Book typo!

7 Results for Harmonic Oscillator Solution requires: α = 2n +1 for n = 0,1,2, 3... d 2 ψ dy 2 E n E= hν 2 hν 2 ( 2n +1 ) = ( n + 1 n = ω 2 α = ω 2 2n + 1 ω 2)hν E o = 1 2 hν E 0 = 1 2 ω ( ) = n a.k.a. zero point energy + ( α y2 )ψ = 0 Recall Planck s assumption & blackbody formula (Lecture 6): Oscillator energies assumed E n = nω Later, found additional detail & statistics, C v etc. but Right ideas on quantisation: fortunate guesswork!

8 Wavefunctions of Harmonic Oscillator Comparing with infinite square well case, ψ has approximately same shape except: (1) width of well is changing (2) ψ extends beyond classical limit (3) amplitude increases at well edge Look at (2) and (3) via the probability density..

9 Probability Density of Harmonic Oscillator (1) Finite probability of particle outside the classical limits (2) the quantum picture only approaches the classical picture at large n values (classical - probability maximum at extremities of oscillation - slower) - Correspondence Principle Footnote: compare square well [En n2] and harmonic oscillator [En (n + 1/2)] and Bohr H atom [En -1/n2] for energies. Differing shapes of the potential wells.

10 Particle meeting step potential U=U o U=U o 2 types: step-up step-down U=0 U=0 Particle direction left-to-right ; potential flat apart from step region; Particle energy E and step size U o ; 3 distinct situations: up E < U o up E > U o down E > U o Find elements of propagating and decaying wavefunctions; Solutions may not be standing waves (as previously) cannot draw ( complex ψ) but can always draw ψ 2 (real).

11 Apply SSSE to step-up potential (E < U o ) Classically, particle is reflected! d 2 ψ dx 2 + 2m 2 E U ( )ψ = 0 For region I, solution is complex exponentials: ψ I = Aexp( ik 1 x) + Bexp( ik 1 x) where k = 2mE 1 For region II, solutions are real exponentials: ψ II = C exp k 2 x ( ) + Dexp( k 2 x) where k 2 = 2m ( U E o ) Also, require C = 0, to keep ψ finite at large +ve x: ψ II = Dexp( k 2 x)

12 Apply boundary matching ψ II = ψ I at x = 0 Dexp(0) = Aexp(0) + Bexp(0) D = A + B dψ II /dx = dψ I /dx at x = 0 -k 2 Dexp(0) = ik 1 Aexp(0) - ik 1 Bexp(0) (ik 2 /k 1 )D = A - B Convenient to write A and B in terms of D: ψ I = D ik 2 k1 exp ik 1x ( ) + D 2 1 ik 2 k1 meaning? incident particle(s) reflected particle(s) ( ) ψ II = Dexp k 2 x particle(s) probability decays into wall! ( ) exp ik 1x

13 Particle reflection Reflection Coefficient R = B*B/A*A R = 1 i k 2 k i k * 2 k1 * k 1 i 2 k i k 2 k1 = 1 + i k 2 1 i k 2 k 1 k 1 1 i k i k 2 k 1 k 1 = 1 As expected, in agreement with classical picture! Exercise: use [exp(iθ) = cosθ + isinθ] to show ψ I = Dcos k 1 x ( ) D k 2 k1 sin k 1x ( ) can be generalised as sin(k 1 x + φ) i.e. standing wave! (combination of equal incident and reflected waves)

14 Plotting ψ and ψ 2 ψ ψ 2 Because in this case ψ is a pure standing wave, it can still be plotted; note how ψ matches onto ψ (red) As U o increases, k 2 increases (less penetration of wall) and ψ moves closer to simple sin(k 1 x). Plot of ψ 2 always possible Signature of perfect reflection (R = 1): minimum value of ψ 2 = 0

15 Apply SSSE to step-up potential (E > U o ) Classically: kinetic energy decrease and particle not reflected! d 2 ψ dx 2 + 2m 2 E U ( )ψ = 0 For region I, solution is complex exponentials as before: ψ I = Aexp( ik 1 x) + Bexp( ik 1 x) where k 1 = 2mE For region II, solution is also complex exponentials: ( ) ψ II = C exp( ik 2 x) + Dexp( ik 2 x) k 2 = 2m E U o where Require D = 0, no negative-going wave for x > 0, as all particles are incident in +ve x direction! Not so for x < 0, there is some reflection! ψ II = Cexp( ik 2 x)

16 Apply boundary matching ψ II = ψ I at x = 0 Cexp(0) = Aexp(0) + Bexp(0) C = A + B dψ II /dx = dψ I /dx at x = 0 ik 2 Cexp(0) = ik 1 Aexp(0) - ik 1 Bexp(0) (k 2 /k 1 )C = A - B Convenient to write B and C in terms of A: ψ I = A exp( ik 1 x) + A k k 1 2 ( ) k 1 + k 2 exp ik 1 x meaning? incident particle(s) reflected particle(s) ( ) particle(s) with reduced energy! ψ II = A 2k 1 exp ik k 1 + k 2 x 2

17 Particle reflection Recall Reflection Coefficient R = B*B/A*A R = k k 1 2 k 1 + k 2 2 Recall the equations for k 1 and k 2 k 2 k = 2m ( E U 0 ) 1 R = 1 1 U 0 E U 0 E 2 = 1 k 2 k1 1 + k 2 k1 2mE = 1 U 0 E 2 for E > U o and R = 1 for E < U o

18 Particle meeting step-down potential compare: step-up with step-down Reverse situations, simply exchange k 1 and k 2 : ψ I = A exp ik 2 x ψ II = Cexp( ik 1 x) ( ) + Bexp( ik 2 x) Proceed to determine matching etc. Significant point is that some reflection occurs here too; Origin of reflectivity is change in potential U For quantum lemmings, some reflect from cliff edge.

19 Step-up, where E < U o - total reflection - but can exist in wall! Step-up, where E > U o - decreased kinetic energy - and partial reflection! General conclusions Step-down, where E > U o - increased kinetic energy - also partial reflection! solve, match, R and plot