Physics 3810, Spring 2009 - Exam 2 Solutions: Hilbert Spaces, States, Uncertainty, Exams of Quantum Mechanics

Download Physics 3810, Spring 2009 - Exam 2 Solutions: Hilbert Spaces, States, Uncertainty and more Exams Quantum Mechanics in PDF only on Docsity! Phys 3810, Spring 2009 Exam #2 1. Give short definitions of the terms (as used by Griffiths) a) Hilbert space of functions. This is the set of functions for which the modulus squared can be integrated over the entire applicable space to give a finite result, i.e. they can be normalized. (It is a vector space, following the rules for vector addition, with an inner product defined by an integral.) b) Determinate state. Pertaining to a specific observable, these are states where measurement of that observable always gives the same result. They are also eigenstates of the observable's operator. c) Degenerate states. States which are linearly independent but which have the same eigenvalue (for a particular oper- ator/observable). d) Compatible observables. Compatible observables are those for which we can from simultaneous eigenstates, that is, states which are eigenvectors of both operators. For compatible observables, the corresponding operators must commute. 2. Give a brief but correct summary of the energy–time uncertainty relation ∆t∆E ≥ h̄ 2 . It isn't what most people say. If we consider a particular observable, say, Q and consider the length of the time it takes for Q to change by one standard deviation (and call it ∆t) then ∆t and the energy uncertainty ∆E are related by ∆E∆t ≥ h̄ 2 It's not about the amount of time for which God lets you violate conservation of energy. Sheesh. 3. (An easy one) An operator  (representing observable A) has two normalized eigen- states ψ1 and ψ2, with eigenvalues a1 and a2, respectively. Operator B̂ has two normalized eigenstates φ1 and φ2 with eigenvalues b1 and b2 respectively. The eigenstates are related by ψ1 = (2φ1 − φ2)/ √ 5 ψ2 = (φ1 + 2φ2)/ √ 5 a) Observable A is measured and the value a2 obtained. What is the state of the system (immediately) after this measurement? If A is measured and the value a2 is the result, the system is then immediately put into state ψ2. 1 b) If B is now measured, what are the possible results and what are their probabilities? Since the system is now in state ψ2, the possible results of a measurement of B are b1 and b2. Squaring the coefficients of φ1 and φ2 in the expression for ψ2, these values would occur with probabilities b1 : 1 5 and b2 : 4 5 respcitively. 4. Evaluate the commutator of the kinetic energy and square of the coordinate operator, [ p2 2m , x2 ] . Evaluate and explain all the steps. We could use [x, p] = ih̄, but here I'll derive it from scratch; consider [p2, x2]. Since p2 = −h̄2 d2 dx2 , the action of the operator p2x2 on a function f(x) gives p2(x2f(x)) = −h̄2 d 2 dx2 (x2f(x)) = −h̄2 d dx (2xf(x) + x2f ′(x)) = −h̄2(2f(x) + 2xf ′(x) + 2xf ′(x) + x2f ′′(x)) = −h̄2 ( 2f(x) + 4x(i/h̄)p f(x) + x2(−1/h̄2)p2 ) f(x) = (−2h̄2 − 4xih̄p + x2p2)f(x) Then [p2, x2]f(x) = (p2x2 − x2p2)f(x) = (−2h̄2 − 4ixh̄p)f(x) Include the 1/2m factor, [ p2 2m , x2 ] = − h̄ 2 m − 2ixh̄ m p 5. In the space of states consisting of |1〉, |2〉, |3〉, express as a matrix the operator H = a|1〉 〈1| + a|2〉 〈2| + c|3〉 〈3| + |1〉 〈2| + |2〉 〈1| + +|2〉 〈3| + |3〉 〈2| where |1〉 =   1 0 0   , |2〉 =   0 1 0   , etc. Outline how you would find the eigenvalues and eigenvectors of H. (You don’t need to actually find them.) The diagonal elements ofH are (a, a, c) and ``connecting'' the states |1〉 and |2〉 is the coefficient , likewise for |2〉 and |3〉. We get: H =   a  0  a  0  c   2 Since a is inversely proportional to the mass, we have 〈r〉1 = ( 1 207 ) 3 2 (0.529 × 10−10 m) = 3.83 × 10−13 m d) Find the wave length of the first Lyman transition in muonium. The wave length is inversely proportional toR which in turn is proportional tom. So λ is inversely proportional to m. As the wavelength of the first Lyman transition is 1 λ = R(1 − 1 4 ) =⇒ λ = 1.2 × 10−7 m the corresponding wavelength for muonium is λ(µ) = ( 1 207 ) (1.21 × 10−7 m) = 5.9 × 10−10 m 5 Useful Equations Math ∫ ∞ 0 xne−x/a = n! an+1 ∫ ∞ 0 x2ne−x 2/a2 dx = √ π (2n)! n! (a 2 )2n+1 ∫ ∞ 0 x2n+1e−x 2/a2 dx = n! 2 a2n+2 ∫ b a f dg dx dx = − ∫ b a df dx g dx+ fg ∣ ∣ ∣ ∣ ∣ b a δ(x) = 1 2π ∫ ∞ −∞ eikx dx Numbers h̄ = 1.05457 × 10−34 J · s me = 9.10938 × 10−31 kg mp = 1.67262 × 10−27 kg e = 1.60218 × 10−19 C c = 2.99792 × 108 m s Physics ih̄ ∂Ψ ∂t = − h̄ 2 2m ∂2Ψ ∂x2 + VΨ Pab = ∫ b a |Ψ(x, t)|2 dx p→ h̄ i d dx ∫ ∞ −∞ |Ψ(x, t)|2 dx = 1 〈x〉 = ∫ ∞ −∞ x|Ψ(x, t)|2 dx 〈p〉 = ∫ ∞ −∞ Ψ∗ ( h̄ i ∂ ∂x ) Ψ dx σ = √ 〈j2〉 − 〈j〉2 σxσp ≥ h̄ 2 − h̄ 2 2m d2ψ dx2 + V ψ = Eψ φ(t) = e−iEt/h̄ Ψ(x, t) = ∞ ∑ n=1 cnψn(x)e −iEnt/h̄ = ∞ ∑ n=1 Ψn(x, t) ∞ Square Well: En = n2π2h̄2 2ma2 ψn(x) = √ 2 a sin (nπ a x ) ∫ ψm(x) ∗ψn(x) dx = δmn cn = ∫ ψn(x) ∗ f(x) dx ∞ ∑ n=1 |cn|2 = 1 〈H〉 = ∞ ∑ n=1 |cn|2En Harmonic Oscillator: V (x) = 1 2 mω2x2 1 2m [p2 + (mωx)2]ψ = Eψ a± ≡ 1√ 2h̄mω (∓ip+mωx) [A,B] = AB −BA [x, p] = ih̄ H(a+ψ) = (E + h̄ω)(a+ψ) H(a−ψ) = (E − h̄ω)(a+ψ) a−ψ0 = 0 ψ0(x) = (mω πh̄ )1/4 e− mω 2h̄ x2 ψ1(x) = (mω πh̄ )1/4 √ 2mω h̄ xe− mω 2h̄ x2 Free particle: Ψk(x) = Ae i(kx− h̄k2 2m )t vphase = ω k vgroup = dω dk 6 Ψ(x, t) = 1√ 2π ∫ ∞ −∞ φ(k)ei(kx− h̄k2 2m t) dk φ(k) = 1√ 2π ∫ ∞ −∞ Ψ(x, 0)e−ikx dx Delta Fn Potl: ψ(x) = √ mα h̄ e−mα|x|/h̄ 2 E = −mα 2 2h̄2 fp(x) = 1√ 2πh̄ exp(ipx/h̄) [Â, B̂] ≡ ÂB̂ = B̂ Φ(p, t) = 1√ 2πh̄ ∫ ∞ −∞ e−ipx/h̄Ψ(x, t) dx σ2Aσ 2 B ≥ ( 1 2i 〈[Â, B̂]〉 )2 σxσp ≥ h̄ 2 ∆E∆t ≥ h̄ 2 − h̄ 2 2m [ 1 r2 ∂ ∂r ( r2 ∂ψ ∂r ) + 1 r2 sin θ ∂ ∂θ ( sin θ ∂ψ ∂θ ) + 1 r2 sin2 θ ∂2ψ ∂φ2 ] + V (r)ψ = Eψ ψ(r, θ, φ) = R(r)Θ(θ)Φ(φ) d2Φ dφ2 = −m2Φ sin θ d dθ ( sin θ dΘ dθ ) +[`(`+1) sin2 θ−m2]Θ = 0 Y 00 = √ 1 4π Y 01 = √ 3 4π cos θ Y ±11 = ∓ √ 3 8π sin θe±iφ Y 02 = √ 5 16π (3 cos2 θ − 1) Y ±12 = ∓ √ 15 8π sin θ cos θe±iφ etc. a = 4π0h̄ 2 me2 = 0.529×10−10 m En = − [ m 2h̄2 ( e2 4π0 )2 ] 1 n2 ≡ E1 n2 for n = 1, 2, 3, . . . where E1 = −13.6 eV. R10(r) = 2a −3/2e−r/a R20(r) = 1√ 2 a−3/2 ( 1 − 1 2 r a ) e−r/2a R21(r) 1√ 24 a−3/2 r a e−r/2a λf = c Eγ = hf 1 λ = R ( 1 n2f − 1 n2i ) where R = m 4πch̄3 ( c2 4π0 )2 = 1.097×107 m−1 7