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Unformatted text preview: TEORETISK FYSIK KTH ¨ TENTAMEN I KVANTMEKANIK FORDJUPNINGSKURS EXAMINATION IN ADVANCED QUANTUM MECHANICS Kvantmekanik f¨rdjupningskurs 5A1385/5A1329 f¨r F4 o o Tuesday June 5 2007, kl. 08.0013.00
Write on each page: Motivate in detail! Allowed material: Grading system: Examiner: 1. Spin A spin
1 2 1 2 Name, study program and year, problem number Insuﬃcient motivation leads reduction of points Summary of lectures, BETA, pocket calculator Max 3 points per problem Mats Wallin tel 5537 8475 particle particle is in the normalized spin state α β where α, β are constants. Determine the probabilities for the diﬀerent possible outcomes of a measurement of Sy and the expectation value Sy . 2. Particle in a harmonic oscillator potential Consider a particle with mass m in a simple harmonic oscillator potential. Calculate the ﬁrst order shift in the ground state energy to due to a weak perturbation H 1 = cp2 where p is the momentum operator and c is a constant. Compare the result with the exact solution. 3. Orbital angular momentum The wave function of a particle in a spherically symmetric potential is ψ (r) = (x + y + 3z )f (r) What are the possible values for L2 and for Lz that can be obtained in a measurement, and what are the corresponding probabilities? SEE NEXT PAGE! 1 4. Variational calculation Estimate the ground state energy of a simple harmonic oscillator using the variational trial wave function ψ (x) = e−λx where λ is to be varied. 5. Identical fermions Two noninteracting identical spin well potential:
1 2 fermions move in one dimension in an inﬁnite square 0 for 0 < x < L ∞ otherwise V (x) = What is the ground state energy and wave function in the following cases: (a) The two particles are constrained to a singlet spin state? (b) The two particles are constrained to a triplet spin state? 6. One dimensional scattering Use the LippmannSchwinger equation ψ ± = φ + 1 V ψ ± E − H0 ± iǫ to determine the transmission and reﬂection coeﬃcients for onedimensional scattering oﬀ an attractive delta function potential V (x) = −cδ (x) where c > 0 is a constant. GOOD LUCK! 2 TEORETISK FYSIK KTH Examination in Quantum Mechanics 070605 Solutions
α and assume without loss of generality that α is purely real, and that β β = β ′ + iβ ′′ . Normalization: 1. Set χ = χχ = (α, β ′ − iβ ′′ ) From this we get that αβ ′′  ≤ P (Sy = ± ) = Sy = ± χ 2 2
1 2 2 α β ′ + iβ ′′ = α2 + β ′2 + β ′′2 = 1 which is needed below. Probabilities: 1 = √ (1, ∓i) 2 α ′ β + iβ ′′
2 1 = √ (α ± β ′′ ∓ iβ ′ ) 2 2 = 1 1 1 = [(α ± β ′′ )2 + β ′2 ] = (α2 + β ′′2 + β ′2 ± 2αβ ′′ ) = ± αβ ′′ 2 2 2 where we used the normalization requirement in the last step. Note that the probabilities obey 0 ≤ P (Sy = ± ) ≤ 1, 2 as they should. Expectation value: P (Sy = + ) + P (Sy = − ) = 1 2 2 Sy = + P (Sy = + ) − P (Sy = − ) = αβ ′′ 2 2 2 2 Note that  Sy  ≤ /2. OK! 2. Shift in the ground state energy from ﬁrst order perturbation theory: ∆E = H 1 = c 0p2 0 = c 0 = −c i mω (a† − a) 2
2 0= mω mω 0(a† )2 − a† a − aa† + a2 0 = c 2 2 †2 † 2 where we used that 0(a ) 0 = 0a a0 = 0a 0 = 0 and [a, a† ] = 1 ⇒ 0aa† 0 = 01 + a† a0 = 1. Exact solution: p2 1 p2 1 + kx2 + cp2 = + kx2 ′ 2m 2 2m 2 ′ with m = m/(1 + 2mc). Exact new ground state energy: H= E= 1′1 ω= 2 2 k 1 = m′ 2 k 1 (1 + 2mc) ≈ m 2 k 1 mω c (1 + mc) = ω+ m 2 2 in agreement with the result from perturbation theory. 3 3. Rewrite the angular part of the state in terms of spherical harmonics: Y10 = ⇒ cos θ = 3 cos θ, Y1±1 = ∓ 4π 3 sin θe±iφ = ∓ 8π 3 sin θ(cos φ ± i sin φ) 8π 4π i √ (Y 1 + Y1−1 ) 321 4π 0 4π 1 √ (−Y11 + Y1−1 ), sin θ sin φ = Y1 , sin θ cos φ = 3 32 Use these results to rewrite the angular part of the wave function: ψ (θ, φ) = = 4π 3 = Normalize using Ylm Ylm ′
′ x + y + 3z = sin θ cos φ + sin θ sin φ + 3 cos θ = r 1 i √ (−Y11 + Y1−1 ) + √ (Y11 + Y1−1 ) + 3Y10 2 2 −1 + i 1 1 + i −1 √ Y1 + √ Y1 + 3Y10 2 2 = δl,l′ δm,m′ : 4π 3 ψ ψ = 4π 4π (1 + 1 + 32 ) = 11 3 3 −1 + i 1 1 + i −1 1 √ Y1 + √ Y1 + 3Y10 ⇒ ψ (θ, φ) = √ 11 2 2 We therefore get l = 1 with certainty, and m = +1 with probability P = 1/11, m = −1 with probability P = 1/11, and m = 0 with probability P = 9/11. 4. Normalization:
∞ −∞ ∞ ψ 2 dx = 2c2
∞ e−2αx dx =
0 c2 = 1 ⇒ c2 = α α
2 ∞ −∞ Kinetic energy: T= Since
dψ 2 dx 1 2m ψ ∗ p2 ψdx =
−∞ 1 2m ∞ (pψ )∗ (pψ )dx =
−∞ 2m dψ dx 2 dx is an even function we have
2 ∞ 0 T= m dψ dx
∞ 2 α2 dx = m ∞ 0 −αe
∞ −αx 2 22 α3 2 1 α dx = = m 2α 2m Potential energy: 1 V = mω 2 2 Total energy: E= T + V = Optimize: x2 ψ 2 dx = αmω 2
−∞ 0 x2 e−2αx dx = αmω 2 α2 mω 2 + 2m 4α2
2 mω 2 2 = (2α)3 4α2 2 mω 2 mω dE = − = 0 ⇒ α2 = √ dα2 2m 4α4 2 Minimum energy: √ mω 2 2 ω mω √+ E= =√ 2m 2 4 mω 2
2 4 5. Single particle energy eigenstates: ψn (x) = 2 nπx n2 2 π 2 sin , En = , n = 1, 2, 3, . . . L L 2mL2 (a) The singlet spin state, S = 0, is antisymmetric with respect to spin exchange. To make the overall state antisymmetric with respect to exchange the spatial part of the state must be symmetric. The lowest energy is therefore obtained by having both particles in the single particle ground states. The ground state is non degenerate: 1 ψ0 (x1 , x2 , S, m) = ψ1 (x1 )ψ1 (x2 )S = 0, m = 0 = ψ0 (x1 )ψ0 (x2 ) √ ( ↑  ↓ −  ↓  ↑ ) 2 The ground state energy is E0 = 2
22 π π = 2 2mL mL2 22 (b) The triplet spin state S = 1 is symmetric with respect to exchange, so the spatial part must be antisymmetric with respect to exchange to make the state overall antisymmetric. This state is three fold degenerate: 1 ψ0 (x1 , x2 , S, m) = √ (ψ1 (x1 )ψ2 (x2 ) − ψ2 (x1 )ψ1 (x2 )) S = 1, m = 0, ±1 2 Ground state energy: E0 = (1 + 22 ) π 5 2π2 = 2mL2 2mL2
22 6. In the position representation the LippmannSchwinger equation becomes
∞ ψ ± (x) = φ(x) +
−∞ dx′ x
∞ −∞ 1 x′ V (x′ )ψ ± (x′ ) = E − H0 ± iǫ = φ(x) + 2m
2 dx′ G± (x, x′ )V (x′ )ψ ± (x′ )
2 k2 2m Calculate the Green functions, using E = G± (x, x′ ) = = 2m
2 2m
2 x 1 x′ = E − H0 ± iǫ dk ′ dk
′ dk ′′ xk ′ k ′  eik x dk √ 2π
′′
′ = 2m
2 ∞ 1 eik (x−x ) dk ′ 2 2π −∞ k − k ′2 ± iǫ where ǫ > 0 means a diﬀerent constant on the last line. Evaluate using residue integra′ ′ δ (k ′ − k ′′ ) e−ik x √ = 2 k2 2 k ′2 − 2m ± iǫ 2π 2m 1 k ′′ k ′′ x′ = E − H0 ± iǫ
′′ ′ = tion. There are two poles in the complex plane: k ′ = ±k again the last ǫ is a diﬀerent constant) which gives
′ ′ 1± iǫ k2 → ±k ± iǫ (where 1 eik (x−x ) G (x, x ) = × 2πiRes 2 2π k − k ′2 ± iǫ
± ′ 5 We need to distinguish two cases to make the semicircle contour integral vanish: 1. For x − x′ > 0 we must close the integral above the real axis and thus take k ′ = k + ǫ eik (x−x ) G (x, x ) = i d 2 (k − k ′2 + iǫ) dk′
+ ′
′ ′ k′ =+k+iǫ =− i +ik(x−x′ ) e 2k 2. For x − x′ < 0 we must close the integral below the real axis and thus take k ′ = −k − ǫ, and in this case we get a minus sign in front since we circulate the pole in the clockwise direction: ′ ′ eik (x−x ) i ′ − ′ G (x, x ) = −i d 2 = − e−ik(x−x ) ′2 − iǫ) 2k (k − k dk′ ′
k =−k−iǫ With φ(x) = eikx (unnormalized) and V (x) = −cδ (x) = − d m δ (x) we now get 2 ψ ± (x) = eikx − dG± (x, 0)ψ ± (0) Determine the constant ψ ± (0): ψ ± (0) = 1 + From this we ﬁnd ψ ± (x) = eikx + This gives for x > 0: ψ + (x) = eikx + 1 id ± ψ (0) ⇒ ψ ± (0) = id 2k 1 − 2k id/2k ±ikx e 1 − id/2k 2 id/2k ikx id/2k eikx e = 1+ 1 − id/2k 1 − id/2k 1 1 − id/2k
2 and the transmission coeﬃcient is T = 1+ id/2k 1 − id/2k
2 = = 1 1 = 1 + (d/2k )2 1 + (mc/ 2 k )2 Similarly we ﬁnd for x < 0: ψ − (x) = eikx + and the reﬂection coeﬃcient is id/2k R= 1 − id/2k (Test: (d/2k )2 1 + =1 T +R= 1 + (d/2k )2 1 + (d/2k )2 OK!)
2 id/2k −ikx e 1 − id/2k = (d/2k )2 (mc/ 2 k )2 = 1 + (d/2k )2 1 + (mc/ 2 k )2 Note that both T and R are independent of the sign of the delta function potential. 6 ...
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 BelmaŞimsek

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