MATH325 - QUANTUM MECHANICS - SOLUTION SHEET 11

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1 MATH35 - QUANTUM MECHANICS - SOLUTION SHEET. The Hamiltonian for a particle of mass m moving in three dimensions under the influence of a three-dimensional harmonic oscillator potential is Ĥ = h m + mω r, where r = r = x + y + z and the radial part of the Laplacian operator is rad = r + r r. (i Given that the normalised ground state wave function is ψ (r = Ae β r, where A is real, determine β and the ground state energy E. Calculate also the normalisation constant A. (ii The potential is perturbed by the addition of a term λr 5 where λ is small. Use first order perturbation theory to obtain an approximation to the perturbed ground state energy in the form E + λk where K is a constant which you should find. [Standard results from perturbation theory may be assumed without proof.] [ The integrals you need are all of the form I n You can find them from the recursion relation r n e β r dr. I n = n β I n, with I = β, I = β ] (i To find β and E we plug the ground-state wave-function into Schrödinger s equation. Ĥψ (r = E ψ (r h m ψ (r + mω r ψ (r = E ψ (r h m ( r + r r Ae β r + mω r Ae β r = E Ae β r h m 3β e β r h m β4 r e β r + mω r e β r = E e β r h m 3β h m β4 r + mω r = E

2 Equating coefficients (to make this equation true for all r gives β 4 = m ω mω h β = h and E = 3 h m β = 3 hω. We now know β and E. To find A we impose 4πr ψ dr = 4π A r e β r dr = 4π A I =. From the recursion relation so A π 3 β 3 = I = β I = 4β 3 A = β 3 π 3 4 = ( 3 mω 4 hπ choosing A to be real and positive (which we are free to do. (ii The new Hamiltonian is Ĥ = h m + mω r }{{} Ĥ + λ r 5 }{{} λĥ We were told in the lectures that the new energy of the nth energy level is E n = E n + φ n λĥ φ n + O(λ, ( where E n is the nth eigenvalue and φ n is the nth eigenstate of the original unperturbed Hamiltonian Ĥ. ( Standard results from perturbation theory may be assumed without proof means that you are expected to know and understand equation (*, but you don t have to prove it in an exam. In this problem Ĥ is just r 5, and the eigenstate we are interested in is ψ from part (i. So (* becomes The recursion relation gives E E + ψ λr 5 ψ = E + r dr 4πr ψ λr 5 r = E + λ 4π A dr r 7 e β r = E + λ 4π A I 7 K = 4π A I 7. I 7 = 6 β 4 β β I = 3 β 8

3 and we know A = β 3 /π 3 from part (i, so K = β 5 = ( 5/ h. π mω The ground-state energy of the perturbed system is E = 3 hω + Kλ + O(λ.. The Hamiltonian for a harmonic oscillator is d H = h m dx + mω x. In order to estimate the ground state energy using the variational method consider trial wave function 5 ψ(x = 6a 5 (a x ( x a ψ(x = otherwise. (i Show that the trial wave function ψ is normalised. (ii Show that, in the state ψ, H = 5 h 4 ma + 4 mω a. (iii Use the variational method to derive an estimate for the ground state energy by minimising H with respect to a. (iv Compare your result with the exact answer. (i ψ dx = 5 a (a 4 a x + x 4 dx = 5 [ a 4 x 6a 5 a 6a 5 3 a x 3 + ] a 5 x5 = a (ii For kinetic energy we can either calculate h m ψ dx or h m ψ ψ dx. Both must give the same answer, but as explained in the lectures, the ψ integral is often easier. T = m (ˆpψ (ˆpψ dx = h m ψ dx = h 5 m 6a 5 a a(x dx = 5 h 4 ma V = ψ V (x dx = 5mω (a 4 x a x 4 + x 6 dx = 3a 5 a 4 mω a H = T + V = 5 h 4 ma + 4 mω a. a

4 (iii The minimum of H happens at d H = 5 h da ma mω a = a 4 = 35 h m ω 5 H min = hω = hω. 4 (iv The exact eigenvalue is hω, so our estimate is about % too high. (You should know that the energy levels of a -d harmonic oscillator are E n = (n+ hω, with n =,,,. 3. A particle of mass m moves in three dimensions subject to a potential V (r = α, r 3 where α is a positive constant, and r = r = (x + y + z. (i Show that the wave function ψ(r = Ae ar (where a > is normalised if A = a3. π (ii Show that the expectation values of the kinetic energy, T = m p = h m and the potential energy V of the particle in the state ψ are given by T = h a m, and V = α a 3. (iii Use the variational method to show that the ground state energy of a particle in the given potential is not more than E = 7 α 4 m 3 π 8 h 6. You may assume the following: (a the non-zero stationary value for α corresponds to a minimum; (b the radial part of the Laplacian in spherical polars is (c r + r r ; r n e βr dr = n! β n+ (n =,,..., r e βr dr = β 3 (β >.. The important thing to remember when doing integrations in polar co-ordinates is that the volume element in polar co-ordinates is dv = r dr sin θ dθ dφ, which simplifies to

5 4πr dr for spherically symmetric integrands. (Imagine dividing space up into spherical shells - a shell of radius r and thickness dr will have the volume 4πr dr because 4πr is the surface area of a sphere. To integrate a function of r, θ, φ over all space the correct limits are π π r dr sin θ dθ dφ f(r, θ, φ. To integrate a function of r alone, you can either use the above formula, or simply (i To normalise we want ψ ψ =, i.e. 4πr dr f(r. = 4πr ψ dr = 4π A r e ar dr = 4π A! (a 3 A = a3 π. (ii Because the wave function only depends on r we just need the radial part of, given as hint (b, when we calculate T ψ = radψ = T = h m ( r + r The integrand for V is easier V = ( Ae ar = A r a a r dr 4πr ψ ( ψ = h m A 4π = h m 4a3 e ar dr (a r are ar ( a! a! (a 3 (a 4πr ψ V (r dr = 4π A α r e ar dr = αa 3 π. = h a m. (iii The integrals for T and V depended on this being a 3-d problem. But from now on, everything else is the same as in a -dimensional problem. This has a minimum at H = T + V = h a m αa 3 h a m 3 αa π = a = 9πα m 8 h 4. The minimum value of H with this trial function is H min = 7π 8 α 4 m 3 h 6 E.