Download Homework Set 7 Solutions - Quantum Mechanics | PHY 373 and more Assignments Physics in PDF only on Docsity! PHY 373 Modern Physics II: Quantum Mechanics, Homework Set 7 Solutions Matthias Ihl 04/18/2007 Note: I will post updated versions of the homework solutions on my home- page: http://zippy.ph.utexas.edu/~msihl/teaching.html We will frequently work in God-given units c = ~ = 1. The casual reader may also want to set 1 = 2 = π = −1. 1 Problem 1 The average energy of an harmonic oscillator can be calculated using some tools of thermodynamics: 〈H〉 = TrHe −βH Tre−βH , (1) where Z := Tre−βH is called the partition function and is a weighted sum of all the accessible states of a quantum system (weighted by a Boltzmann distribution). This can also be written as 〈H〉 = −∂β logZ. (2) The partition function in this case is actually a geometric series yielding Z = Tre−βH = ∞ ∑ n=0 e− ~ω kT (n+ 1 2 ) = e− ~ω 2kT ∞ ∑ n=0 e− ~ω kT n = e− ~ω 2kT 1 1 − e− ~ωkT . (3) With this result we can compute the average value of the energy: 〈H〉 = ~ω eβ~ω − 1 + ~ω 2 . (4) 1 The last term is the zero-point energy of the harmonic oscillator. All this generalizes straightforwardly to more than one harmonic oscillators, yielding 〈E〉 = ∑ k ( ~ωk eβ~ωk − 1 + ~ωk 2 ) . (5) The second contribution from the zero-point or vacuum energies diverges and has to be subtracted in order to get a finite result (this is a standard procedure in Quantum Field Theory (QFT)). 2 Problem 2 This problem is similar to Problem 2 of homework 6 since we assume that we do not have spin-orbit coupling, which would change the eigenstates etc. of the system. Here the eigenkets are given by |l,ml; s = 1, ms〉 := |l,ml〉 ⊗ |s = 1, ms〉, (6) where for spin 1 we have Sz|s = 1, ms〉 = ~ms|s = 1, ms〉, (7) with ms = −1, 0,+1. Now the spectrum of the Hamiltonian is simply H|l,ml; s = 1, ms〉 = El,ms |l,ml; s = 1, ms〉, (8) where El,ms = ~ 2l(l + 1) 2mR2 − κ~Bzms, (9) where the l-dependent part is (2l+ 1) times degenerate since −l ≤ ml ≤ +l. 3 Problem 3 The state ψ(θ, φ) = 〈θ, φ|ψ〉 = √ 3 4π sin φ sin θ (10) can be rewritten as ψ(θ, φ) = √ 3 4π 1 2i (eiφ − e−iφ) sin θ = 1√ 2 ( Y m=1l=1 − Y m=−1l=1 ) . (11)
