HW3_Solutions - PHYS851 Quantum Mechanics I Fall 2008 HOMEWORK ASSIGNMENT 3 Fundamentals of Quantum Mechanics 2.8 We are given H= and H |ϕn = En

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Unformatted text preview: PHYS851 Quantum Mechanics I, Fall 2008 HOMEWORK ASSIGNMENT 3: Fundamentals of Quantum Mechanics 2.8 We are given H= and H |ϕn = En |ϕn 1 a.) We have [X, H ] = 2M [X, P 2 ] because [X, V (X )] = 0 Now [X, P 2 ] = XP 2 − P 2 X = XP 2 − P XP + P XP − P 2 X = [X, P ]P + P [X, P ] = 2i P . So that i XH − HX = P m hitting from the left with ϕn | and from the right with |ϕn′ gives 12 P + V (X ) 2m ϕn |XH |ϕn′ − ϕn |HX |ϕn′ = Using H |ϕn = En |ϕn then gives −(En − En′ ) ϕn |X |ϕn′ = which gives the desired result ϕn |X |ϕn′ = − i ϕn |P |ϕn′ m i ϕn |P |ϕn′ m i ϕn |P |ϕn′ m(En − En′ ) b.) Taking the square modulus of the previous result gives | ϕn |X |ϕn′ |2 = 2 m2 (En − En′ )2 2 | ϕn |P |ϕn′ |2 Multiplying both sides by (En − En′ )2 and summing over n′ gives (En − En′ )2 | ϕn |X |ϕn′ |2 = = = where we have used n |ϕn n′ m2 2 n′ | ϕn |P |ϕn′ |2 ϕn |P |ϕn′ ϕn′ |P |ϕn m2 2 n′ m2 ϕn |P 2 |ϕn ϕn | = I to obtain the final result. 1 2.9 We are given the eigenvalue equation H |ϕn = En |ϕn . a.) [A, H ] = AH − HA ϕn |[A, H ]|ϕn = ϕn |AH |ϕn − ϕn |HA|ϕn = En ϕn |A|ϕn − En ϕn |A|ϕn =0 b.) Let H = α.) [V (X ), P ] = [V (X ), P ]. For an arbitrary state |ψ , we have x|[V (X ), P ]|ψ = V (x) x|P |ψ − x|P V (X )|ψ ∂ ∂ x|ψ + i V (x) x|ψ = −i V (x) ∂x ∂x = i V ′ (x) x|ψ = i x|V ′ (X )|ψ 1 2 2m P + V (X ). [H, P ] = 21 [P 2 , P ] + m So to be true for any |ψ and x requires [H, P ] = i V ′ (X ). [H, X ] = 1 2 2m [P , X ] + [V (X ), X ] = 1 2 2m [P , X ]. [P 2 , X ] = P 2 X − XP 2 = P [P, X ] + [P, X ]P = −2i P = P 2 X − P XP + P XP − XP 2 = −P [X, P ] − [X, P ]P Which gives [H, X ] = −i m P . For [H, XP ] we find [H, XP ] = HXP − XP H = HXP − XHP + XHP − XP H = [H, X ]P + X [H, P ] = −i m P 2 + i X V ′ (X ). β .) ϕn |P |ϕn = i m ϕn |[H, X ]|ϕn = 0, according to part a.) γ .) We have P2 |ϕn . Ek ≡ ϕn | 2m Using the previous result [H, XP ] = −i m P 2 + i X V ′ (X ), we find P2 1 i = XV ′ (X ) + [H, XP ]. 2m 2 2 2 Thus ϕn | P2 |ϕn 2m = = 1 i ϕn |XV ′ (X )|ϕn + ϕn |[H, XP ]|ϕn 2 2 1 ϕn |XV ′ (X )|ϕn . 2 When V (X ) = V0 X λ , we have V ′ (X ) = λV0 X λ−1 , so that XV ′ (X ) = λV0 X λ = λV (X ). The relation to the kinetic energy is thus Ek = 2.10 x|XP |ψ = x x|P |ψ =x =x dp x|p p p|ψ dp peipx/ ψ (p) ∂ ∂x dp x|p p|ψ λ ϕn |V (X )|ϕn . 2 = x −i = −i x ∂ ψ (x) ∂x x|P X |ψ = = = dp x|p p p|X |ψ dp dx′ px′ x|p p|x′ x′ |ψ 1 2π i 2π dp dx′ px′ eip(x−x )/ ψ (x′ ) ∂ ∂x dp dx′ x′ eip(x−x )/ ψ (x′ ) ′ ′ =− = −i ∂ dp dx′ x′ x|p p|x′ ψ (x′ ) ∂x ∂ = −i dx′ x|x′ x′ ψ (x′ ) ∂x ∂ = −i dx′ δ(x − x′ )x′ ψ (x′ ) ∂x ∂ = −i xψ (x) ∂x (1) ∂ ∂x Yes, they could have been found just using P → −i ∂ states, by using x|P |ψ = −i ∂x x|ψ . and |ψ → ψ (x), or more correctly 3 3.1 a.) N eip0 x/ ψ (x) = √ . x2 + a2 Now dx |ψ (x)|2 = 1, so that N2 ∞ −∞ dx x2 1 =1 + a2 N2 So we find N = b.) P = = = = c.) P = = ψ |P |ψ ∞ −∞ π =1 a a/π . √ a/ 3 a 1 √ dx 2 π −a/ 3 x + a2 √ √ 1 arctan(1/ 3) − arctan(−1/ 3) π 1 3 2 √ dx |ψ (x)| 3 −a/ √ a/ 3 dx ψ |x x|P |ψ dx ψ ∗ (x) ∂ ψ (x) ∂x e−ip0 x/ ∂ eip0 x/ √ dx √ x2 + a2 ∂x x2 + a2 = −i = −i = −i = p0 = p0 a π a π ip0 eip0 x/ xeip0 x/ e−ip0 x/ √ −2 dx √ x2 + a2 x2 + a2 (x + a2 )3/2 x a dx 2 dx |ψ (x)|2 + i π (x + a2 )2 as the second term is zero due to odd parity. 4 3.3 ψ (x, 0) = N a.) x|ψ = N√ ∞ −∞ dk e−|k|/k0 eikx . 2π dp e−|p|/p0 x|p where p0 = k0 . To be true for any x requires √ 2πN |ψ = √ dp |p e−|p|/p0 . so that √ 2πN p|ψ = √ e−|p|/p0 p1 The probability is then P = −p 1 dp |ψ (p)|2 p1 −p 1 p1 0 = = = = 2πN 2 4πN 2 4πN 2 dp e−2|p|/p0 dp e−2p/p0 e−2p1 /p0 − 1 − p0 2 2πN 2 p0 1 − e−2p1 /p0 Taking p1 → ∞ and setting P → 1 then gives N= 2πp0 b.) If the measurement is performed at time t the probability is the same, as the momentum distribution is constant in time for a free particle. c.) ψ (x, 0) = N =N ∞ −∞ ∞ 0 dk e−|k|/k0 eikx dk ek(ix−1/k0 ) + N 0 −∞ dk ek(ix+1/k0 ) 1 1 =N− + ix − 1/k0 ix + 1/k0 2N/k0 = 2 x2 + 1/k0 now X = 0 by odd parity, and X2 = = x2 4N 2 dx 2 2 2 k0 (x + 1/k0 )2 ∞ x2 4N 2 dx (2k0 ) 2 (1 + x2 )2 k0 0 5 = π 4N 2 2 (2k0 ) 4 k0 = 2πN 2 /k0 2 = 1/k0 so that ∆X = X 2 = 1/k0 . P2 . 2πN 2 2 k0 0 2 p3 0 k0 4 2 k2 0 2 ∞ −∞ ∞ Now p = 0 by symmetry, so ∆p = P2 = = = = so we have ∆P = √ P 2 = k0 / 2. dp p2 e−2|p|/p0 dp p2 e−2p/p0 (2) √ This gives ∆X ∆P = / 2, which is larger than the minimum allowed by the Heisenberg uncertainty principle. 6 1. Consider the system with three physical states {|1 is: 1 −2i H= 1 Find the eigenvalues {ω1 , ω2 , ω3 } and eigenvectors {|ω1 , |ω2 , |ω3 } of H . Assume that the initial state of the system is |ψ (0) = |1 . Find the three components 1|ψ (t) , 2|ψ (t) , and 3|ψ (t) . All your answers must be written using proper Dirac notation. Answer: The eigenvalues are solutions to det |H − ω I | = 0 Taking the determinate in Mathematica gives 4ω + 4ω 2 − ω 3 = 0 which factorizes as ω (ω 2 − 4ω − 4) = 0 which has as its solutions ω1 = 2(1 − ω2 = 0 ω3 = 2(1 + √ 2) √ 2) , |2 , |3 }. In this basis, the Hamiltonian matrix 2i 1 2 −2i 2i 1 the corresponding eigenvectors are √ 1 |ω1 = (|1 + 2i|2 + |3 ) 2 1 |ω2 = √ (−|1 + |3 ) 2 √ 1 |ω3 = (|1 − 2i|2 + |3 ) 2 The components of |ψ (t) are found via |ψ (t) = e−iHt |ψ (0)| , giving 1|ψ (t) = √ √ 1 2 + e−i2(1− 2)t + e−i2(1+ 2)t 4 √ √ i 2|ψ (t) = √ e−i2(1− 2)t − e−i2(1+ 2)t 22 √ √ 1 3|ψ (t) = −2 + e−i2(1− 2)t + e−i2(1+ 2)t 4 The mathematic script I used to work this problem is here: 7 8 ...
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This note was uploaded on 10/25/2010 for the course PHYSICS PHYS 851 taught by Professor Michaelmoore during the Fall '08 term at Michigan State University.

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