Spring 2009 Physics Exam 3: Problems on Spin and Quantum Mechanics, Exams of Quantum Mechanics

Download Spring 2009 Physics Exam 3: Problems on Spin and Quantum Mechanics and more Exams Quantum Mechanics in PDF only on Docsity! Phys 3810, Spring 2009 Exam #3 1. If we operate with L+ on Y 2 2 (θ, φ), what do we expect to get? Show that we get this result. Since L+ is the raising operator, and there is no state ``above'' Y 2 2 , we expect it to give zero! To show this explicitly, L+Y 2 2 (θ, φ) = h̄e +iφ ( ∂ ∂θ + i cot θ ∂ ∂φ ) √ 15 32π sin2 θ e+2iφ = h̄ √ 15 32π e+iφ (2 sin θ cos θ + i cos θ sin θ(2i)) e+2iφ = 0 In the second step we used cot θ = cos θ sin θ to cancel a factor of sin θ; in the third step, the stuff in the paranthesis cancels. 2. What did the Stern–Gerlach experiment demonstrate? Through exerting different forces on electrons in different spin orientations with a non-uniform magnetic field, the Stern-Gerlach expt showed that when we make a measurement of the electron's spin in a given direction, there are only two results, thus showing the quantization of spin magnetic moment (and by extension, spin angular momentum). 3. Suppose a spin-1 2 particle is in the state χ = 1√ 6 ( 2 1 − i ) What are the probabilities of getting +h̄/2 and −h̄/2 if you measure Sx? What is the expectation value of Sx for this state? Decompose the state into eigenstates of Sx. Using the eigenvectors for Sx, this gives 1√ 6 ( 2 1 − i ) = a√ 2 ( 1 1 ) + b√ 2 ( 1 −1 ) and solve for a and b. Get: 2√ 3 = a + b 1 − i√ 3 = a− b which (adding and subtracting) gives 2a = 3 − i√ 3 =⇒ a = 3 − i√ 12 2b = 1 + i√ 3 =⇒ b = 1 + i√ 2 1 so we've shown χ = 3 − i√ 12 χ (x) + + 1 + i√ 12 χ (x) − so the probability to measure +h̄/2 for Sx is Px,+ = 1 12 (3 − i)(3 + i) = 10 12 = 5 6 and the probability to measure −h̄/2 for Sx is Px,− = 1 12 (1 + i)(1 − i) = 2 12 = 1 6 One can use these probabilities to get the expectation value of S x: 〈Sx〉 = 5 6 (+h̄/2) + 1 6 (−h̄/2) = 4 12 h̄ = h̄ 3 One can also compute it directly: 〈Sx〉 = χ†Sxχ = 1 6 ( 2 1 + i ) h̄ 2 ( 0 1 1 0 )( 2 1 − i ) and this gives: 〈Sx〉 = h̄ 12 ( 2 1 + i ) ( 2 1 − i ) = h̄ 12 (2 − 2i + 2 + 2i) = 4h̄ 12 = h̄ 3 4. Construct the S2, Sz and S+ matrices for a spin–2 particle. S2 = 6h̄2       1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1       Sz = h̄       2 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 −1 0 0 0 0 0 −2       The action of S+ on the states is, for example S+|2 − 2〉 = h̄ √ 2 · 3 − (−2)(−1)|2 − 1〉 = 2h̄|2 − 1〉 Proceeding in this way, we get S+|2 − 2〉 = 2h̄|2 − 1〉 S+|2 − 1〉 = √ 6h̄|2 0〉 S+|2 0〉 = √ 6h̄|2 1〉 S+|2 1〉 = 2h̄|2 2〉 S+|2 2〉 = 0 S+ = h̄       0 2 0 0 0 0 0 √ 6 0 0 0 0 0 √ 6 0 0 0 0 0 2 0 0 0 0 0       2 in this exercise. The value 1 was chosen just because the solution for u(r) is something like a sine function near the origin and the slope of sin(x) is 1 at the origin, so the solution will be a function of (convenient) order 1, but still not normalized. 5 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− h̄k 2 2m )t vphase = ω k vgroup = dω dk Ψ(x, t) = 1√ 2π ∫ ∞ −∞ φ(k)ei(kx− h̄k 2 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 = B̂Â 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 6 − 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 Y ±22 = √ 15 32π sin2 θ e±2iφetc. u(r) ≡ rR(r) − h̄ 2 2m d2u dr2 + [ V + h̄2 2m l(l + 1) r2 ] u = Eu 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 L = r × p [Lx, Ly] = ih̄Lz [Ly, Lz] = ih̄Lx [Lz, Lx] = ih̄Ly Lz = h̄ i ∂ ∂φ L± = ±h̄e±iφ ( ∂ ∂θ ± i cot θ ∂ ∂φ ) L2 = −h̄2 [ 1 sin θ ∂ ∂θ ( sin θ ∂ ∂θ ) + 1 sin2 θ ∂2 ∂φ2 ] L2fml = h̄ 2l(l + 1)fml Lzf m l = h̄mf m l [Sx, Sy] = ih̄Sz [Sy, Sz] = ih̄Sx [Sz, Sx] = ih̄Sy S2|s m〉 = h̄2s(s+1)|s m〉 Sz|s m〉 = h̄m|s m〉 S±|s m〉 = h̄ √ s(s+ 1) −m(m± 1) |s m±1〉 χ = ( a b ) = aχ+ + bχ− where χ+ = ( 1 0 ) and χ− = ( 0 1 ) S 2 = 3 4 h̄2 ( 1 0 0 1 ) Sz = h̄ 2 ( 1 0 0 −1 ) 7