Download Solution to Harmonic Oscillator Problem Set #3 in Physics 324 and more Assignments Quantum Mechanics in PDF only on Docsity! 1 Physics 324 Solution to Problem Set # 3 Due 10/27/03 1. A harmonic oscillator consists of a 1 g mass on a spring. Its oscillation frequency is 2 Hz and its amplitude of oscillation is 1 cm. Approximately, what is the magnitude of the energy quantum number associated with this classical oscillator? The energy of the classical oscillator is KE+PE = kA2/2 where A = 1 cm is the amplitude of oscillation and k = mω2 is the spring constant and ω the angular oscillation frequency. Here, k = 1 g · (2π · 2 Hz)2 = 158 g/s2; E = 158 g/s2 · (1 cm)2/2 = 79 erg The energy of a quantum mechanical oscillator is given by: En = (n + 1/2)h̄ω ≈ nh̄ω for large n. Therefore: En = nh̄ω = 79 erg => n = 79 erg 2π · 2 Hz · 1.06× 10−27 erg − s = ≈ 5.9 × 1027 The classical oscillator corresponds to a superposition of quantum mechanical energy eigenstates whose quantum numbers, n, are of order 6 × 1027. 2. From Eqns. 2.48 and 2.54 in your textbook, we find that the normalized wavefunction for the lowest energy state of a harmonic oscillator potential can be written as: ψ0(x, t) = (mω πh̄ )1/4 e−mωx 2/2h̄ e−iωt/2 (a) Using Eqns. 2.51 and 2.54 in your textbook, find an expression for the normalized wavefunction of the first excited state of the harmonic oscillator, ψ1(x, t). From Eqn. 2.51, we have ψ1(x) = A1a+e−mωx 2/2h̄ = (iA1ω √ 2m)xe−mωx 2/2h̄ and from Eqn. 2.54 we find that A1 = −i(mω/πh̄)1/4/ √ h̄ω. Combining these, we have: ψ1(x) = iω √ 2m · −i√ h̄ω (mω πh̄ )1/4 · x e−mωx 2/2h̄ = (4m3ω3 πh̄3 )1/4 x e−mωx 2/2h̄ which agrees with the result of Eqn. 2.69. Therefore, with E1 = 3h̄ω/2, we have: ψ1(x, t) = (4m3ω3 πh̄3 )1/4 x e−mωx 2/2h̄ e−3iωt/2 2 (b) What are the expectation values of position and momentum, 〈x〉 and 〈p〉, for the two states ψ0(x, t) and ψ1(x, t) ? (You may use symmetry arguments to reduce the work required.) Like the infinite square well potential, |ψn(x)|2 is symmetric about the center of the potential (x = 0 for the harmonic oscillator potential). Therefore, 〈x〉 = 0 for every energy eigenstate, and because the expectation values of energy eigenstates are constant in time, 〈p〉 = m d〈x〉/dt = 0. (c) Compute 〈x2〉 and 〈p2〉 for state ψ0(x, t). 〈x2〉 = ∫ ∞ −∞ |ψ(x)|2 x2 dx = 2 (mω πh̄ )1/2 ∫ ∞ 0 x2 e−mωx 2/h̄ dx Using ∫ ∞ 0 y2e−y 2 dy = √ π/4 and y = x √ mω/h̄, we have: 〈x2〉 = 2 (mω πh̄ )1/2 ( h̄ mω )3/2 · √ π 4 = h̄ 2mω (Note, the result has units of length2, as it must.) For 〈p2〉, the shortcut is to use the result that expectation values behave as classical variables: 〈p2〉 2m + 〈V (x)〉 = E0 = h̄ω 2 => 〈p2〉 2m + mω2 2 〈x2〉 = h̄ω 2 and now use our result that 〈x2〉 = h̄/(2mω) : 〈p2〉 2m + mω2 2 h̄ 2mω = h̄ω 2 => 〈p2〉 = 2mh̄ω 4 = mh̄ω 2 Alternatively, we can compute 〈p2〉 directly: 〈p2〉 = ∫ ∞ −∞ ψ∗0(x) ( − ih̄ d dx )2 ψ0(x) dx = −h̄2 √ mω πh̄ ∫ ∞ −∞ e−mωx 2/2h̄ d 2 dx2 e−mωx 2/2h̄ dx Letting z = ( √ mω/h̄)x and d2(e−z 2/2)/dz2 = (z2 − 1)e−z2/2, we have: 〈p2〉 = −h̄2 √ mω πh̄ √ mω h̄ ∫ ∞ −∞ (z2 − 1) e−z 2 dz = 2h̄mω√ π ∫ ∞ 0 (1 − z2) e−z 2 dz = 2h̄mω√ π [√π 2 − √ π 4 ] = mh̄ω 2
