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Modern Physics II: Quantum Mechanics HW Set 9 - Uncertainty Principle & Separable Systems, Assignments of Physics

The solutions to problem 1 and problem 2 from homework set 9 of modern physics ii: quantum mechanics. Problem 1 deals with the calculation of expectation values, uncertainty products, and commutation relations for spin states. Problem 2 discusses the separability of the schrödinger equation for a two-dimensional harmonic oscillator system.

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Pre 2010

Uploaded on 08/30/2009

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Download Modern Physics II: Quantum Mechanics HW Set 9 - Uncertainty Principle & Separable Systems and more Assignments Physics in PDF only on Docsity! PHY 373 Modern Physics II: Quantum Mechanics, Homework Set 9 Solutions Matthias Ihl 11/28/2007 Note: I will post updated versions of the homework solutions on my home- page: http://zippy.ph.utexas.edu/~msihl/teaching.html 1 Problem 1 The correctly normalized spin state is χ = 1√ 13 ( 3 2i ) . (1) The spin operators are defined as Sx = ~ 2 ( 0 1 1 0 ) , (2) Sy = ~ 2 ( 0 −i i 0 ) , (3) Sz = ~ 2 ( 1 0 0 −1 ) . (4) (a) The expectation values can be easily calculated 〈Sx〉 = 〈χ|Sx|χ〉 = 0, (5) 〈Sy〉 = − 6 13 ~, (6) 〈Sz〉 = 5 26 ~. (7) 1 (b) Note that S2x = S 2 y = S 2 z = ~2 4 ( 1 0 0 1 ) . Therefore, 〈S2x〉 = 〈S2y〉 = 〈S2z 〉 = ~ 2 4 . We have (∆Si) 2 = 〈S2i 〉 − 〈Si〉2. Thus, (∆Sx) 2 = 〈S2x〉 = ~ 2 4 , (8) (∆Sy) 2 = 〈S2y〉 − 〈Sy〉2 = ~ 2 4 − 36~ 2 169 = ( 5~ 26 )2 , (9) (∆Sz) 2 = 〈S2z 〉 − 〈Sz〉2 = ~ 2 4 − 25~ 2 626 = ( 6~ 13 )2 . (10) (c) The uncertainty products are given by (∆Si)(∆Sj) ≥ ∣ ∣ ∣ 1 2i 〈[Si, Sj]〉 ∣ ∣ ∣ . (11) Observe that [Si, Sj] = i~ǫijkSk, so that ∣ ∣ ∣ 1 2i 〈[Si, Sj]〉 ∣ ∣ ∣ = ∣ ∣ ∣ ~ 2 〈Sk〉 ∣ ∣ ∣ . It is easy to check that (∆Sx)(∆Sy) = 5 52 ~ 2 ≥ ~ 2 |〈Sz〉| = ~ 2 5~ 26 , (∆Sy)(∆Sz) = 15 169 ~ 2 ≥ ~ 2 |〈Sx〉| = 0, (∆Sx)(∆Sz) = 3 13 ~ 2 ≥ ~ 2 |〈Sy〉| = 3 13 ~ 2. 2 Problem 2 (a) If we make an ansatz ψ(x, y) = X(x)Y (y), the problem is completely separable, i.e., we can write the Schrödinger equation − ~ 2 2m ( ∂2 ∂x2 + ∂2 ∂y2 ) ψ(x, y) + 1 2 mω2(x2 + y2)ψ(x, y) = Eψ(x, y), (12) as − ~ 2 2m 1 X ∂2X ∂x2 + 1 2 mω2x2 = Ex, (13) − ~ 2 2m 1 y ∂2Y ∂y2 + 1 2 mω2y2 = Ey, (14)
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