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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 HAϕn = En ϕn Aϕn − En ϕn Aϕn =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) xP ψ − xP V (X )ψ ∂ ∂ xψ + i V (x) xψ = −i V (x) ∂x ∂x = i V ′ (x) xψ = i xV ′ (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 xXP ψ = x xP ψ =x =x dp xp p pψ dp peipx/ ψ (p) ∂ ∂x dp xp pψ λ ϕn V (X )ϕn . 2 = x −i = −i x ∂ ψ (x) ∂x xP X ψ = = = dp xp p pX ψ dp dx′ px′ xp px′ 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′ xp px′ ψ (x′ ) ∂x ∂ = −i dx′ xx′ 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 xP ψ = −i ∂x xψ . 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 xP 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.) xψ = N√ ∞ −∞ dk e−k/k0 eikx . 2π dp e−p/p0 xp where p0 = k0 . To be true for any x requires √ 2πN ψ = √ dp p e−p/p0 . so that √ 2πN pψ = √ 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−2p/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−2p/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 1ψ (t) , 2ψ (t) , and 3ψ (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 + 2i2 + 3 ) 2 1 ω2 = √ (−1 + 3 ) 2 √ 1 ω3 = (1 − 2i2 + 3 ) 2 The components of ψ (t) are found via ψ (t) = e−iHt ψ (0) , giving 1ψ (t) =
√ √ 1 2 + e−i2(1− 2)t + e−i2(1+ 2)t 4 √ √ i 2ψ (t) = √ e−i2(1− 2)t − e−i2(1+ 2)t 22 √ √ 1 3ψ (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.
 Fall '08
 MichaelMoore
 mechanics, Work

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