PHYS 771, Quantum Mechanics, Final Exam, Fall 2011 Instructor: Dr. A. G. Petukhov. Solutions

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1 PHYS 771, Quantum Mechanics, Final Exam, Fall 11 Instructor: Dr. A. G. Petukhov Solutions 1. Apply WKB approximation to a particle moving in a potential 1 V x) = mω x x > otherwise Find eigenfunctions, eigenvalues and explain your results. pt) First of all, we relate energy E and the turning point a as a = E/mω. The WKB wave function for x > a reads: ψx) = A exp κx) x a ) κx)dx Using connection formulas Eq. 7.34), page 1) we obtain the wave function for < x < a: ψx) = A a ) cos kx)dx π/4 kx) x Now we can use the boundary condition ψ) =, which yields: a 1) ) kx)dx π 4 = π n + 1) 3) Using kx) = 1/ ) m[e 1/)mω x ] = mωa/ ) 1 x/a) we obtain a kx)dx = mωa 1 1 ξ dξ = mωa Using Eqs. 3) and 4) we find the eigenvalues: E n = ω n + 3 ) π 4 = πe ω 4) 5) As in the case of a quantum harmonic oscillator the WKB approximation replicates the exact eigenvalues. The classical analog of the system in question is a half-oscillator, which bounces from a perfectly elastic wall placed at the equilibrium position. The half-oscillator oscillates with natural frequency ω. The quantum of energy E in Eq. 5) is consistent with this value, E = ω. The infinite wall eliminates all even 1

2 eigenstates of the harmonic oscillator while the odd eigenvalues and eigenfunctions remain unchanged. The WKB eigenfunctions can be calculated straightforwardly using Eqs. 1), ) and 5). Introducing dimensionless coordinate ξ = mω/ ) 1/ x we obtain: in units of mω/ ) 1/, and ψ n ξ) = k n ξ) = 4n + 3 ξ, ξ < 4n + 3 kn ξ) cos[φ nξ) π/4], ξ < 4n + 3, where φ n ξ) = 1 ) ] [cos 1 ξ ξk n ξ) 4n Ψ 1 Ξ) Ξ FIG. 1: WKB wave function of a half-oscillator with n = 3 in the classically allowed region. Similarly κ n ξ) = ξ 4n + 3), ξ > 4n + 3 in units of mω/ ) 1/, and ψ n ξ) = κn ξ) exp[ S nξ)], ξ > 4n + 3, where S n ξ) = 1 [ )] ξ + κn ξ) ξκ n ξ) 4n + 3) ln 4n + 3 The wave function ψ 3 ξ) for the classically allowed region is shown in Fig. 1. expected, it oscillates properly but behaves incorrectly near the turning point ξ a = 15. As

3 . Compute the probability that the electron in a ground state of a hydrogen atom will be found at a distance from the nucleus greater than its energy would permit on the classical theory. 15 pt) The ground state wave function of a hydrogen atom reads: ψ 1 r) = 1 πa 3 exp r/a ) The classical turning point is determined from a condition: i.e. E = e a = e r c, r c = a Thus the probability in question can be calculated as an integral of ψ 1 over the non-classical volume Ω nc, which is an entire volume outside of the sphere with radius r c = a : P nc = ψ 1 r) dr = 1 Ω nc πa 3 a exp r/a )4πr dr = 13 exp 4) or P nc = Assume that a hydrogen atom is in the state corresponding to a maximum orbital angular momentum l = n 1). Find the following expectation values: a) r 5 pt) b) r 1 pt) c) 1r 5 pt) Express your answers in terms of the Bohr radius a and the principal quantum number n. We take advantage of the fact that the radial wave function for l = n 1 assumes a simple form: R n,n 1 r) = A n r n 1 exp γr) 6) 3

4 where γ = 1 na, 7) and A n is the normalization constant. To calculate A n γ) we write: or r R n,n 1dr = 1 Finally, A n Here we used: r r n R n,n 1r)dr = A n A n = γ)n+1 n)! r n exp γr)dr = A n)! n γ) = 1 n+1 r n exp αr)dr = n! α n+1 Now we can calculate expectation values of r p 8) r p = n, l, m r p n, l, m = π π The angular integral in Eq. 9) equals 1. Thus r p = A n Y m l θ, φ)y m θ, φ) sin θdθdφ l r R nlr p dr 9) r n r r p exp γr)dr = A n + p)! n 1) γ) n+p+1 Substituting A n and γ from Eqs. 8) and 7) we finally obtain: r p = n + p)! n ) p γ) p n)! = n + p)! ap n)! 11) Answer: a) b) c) r = a nn + 1/) r = a n n + 1)n + 1/) 1/r = 1/a n ) 4

5 4. Consider the Schrödinger equation SE) for a three-dimensional isotropic harmonic oscillator, V = 1/)mω x + y + z ) = 1/)mω r. a) Solve SE by separation of variables in Cartesian coordinates 15 pt) b) Find the eigenvalues and their degree of degeneracy 1 pt) c) Solve SE by separation of variables in spherical coordinates and assuming the eigenfucntions to be of the form ψr, θ, ϕ) = r l exp mωr / ) fr)yl m θ, φ), where fr) is an associated Laguerre polynomial of the variable mωr / with half-integral indices. Obtain the eigenvalues and establish correspondence with 4b) pt). a) We start with 3D SE: / x + / y + / z ) ψx, y, z) + V x, y, z)ψx, y, z) = Eψx, y, z) m 1) In Cartesian coordinates the potential is separable, i.e. V x, y, z) = V x x) + V y y) + V z z). Therefore we can represent the wave function as a product: ψx, y, z) = ϕ x x)ϕ y y)ϕ z z) 13) Using the standard separation of variables procedure we obtain three independent 1D eigenvalue problems: d ϕ x x) + 1 m dx mω x ϕ x x) = E x ϕ x x) 14a) d ϕ y y) + 1 m dy mω y ϕ y y) = E y ϕ y y) 14b) m d ϕ z z) dz + 1 mω z ϕ z z) = E z ϕ z z) 14c) and E = E x + E y + E z 15) The solutions of Eqs. 14a)-14c) are well known: 5

6 i. Eigenfunctions: ϕ ni x i ) = A ni e ξ i / H ni ξ i ), 16) where A n = 1/ n n!) 1/ mω/π ) 1/4, ξ i = mω/ ) 1/ x i, and are Hermite polynomials; H n ξ) = 1) n e ξ dn dξ n e ξ ) ii. Eigenvalues: E ni = ωn i + 1/), 17) where n i =, 1,,... b) Using Eqs. 15) and 17) we obtain: E = ω n x + n y + n z + 3 ) = ω n + 3 ), 18) where n = n x +n y +n z, and n i =, 1,... To find the degeneracy of the level with given n we have to answer the question: How many ways are there to distribute n identical items among three boxes?. We first note that the number of ways to distribute m identical items among two boxes is m + 1. Then, if we imagine that the number of items in the first box is n m then the number of ways to distribute the remaining m items between second and third boxes is m + 1. We have to sum up all possible cases, i.e. m =, 1,..., n. Thus: Degree of Degeneracy = n m= c) In spherical coordinates we use V r) = 1/)mω r and m + 1) = 1 n + 1)n + ) 19) ψr) = 1 r ur)y m l θ, ϕ), ) This leads to the familiar 1D SE for ur) with the effective potential containing a centrifugal term: [ d u dr + k λ r ] ll + 1) u = 1) r where k = me/ and λ = mω/. Due to the centrifugal term ur) r l+1 for small r. Using this fact we are looking for a solution in the form: ur) = r l+1 exp λr /)fr), ) 6

7 in analogy with the linear harmonic oscillator. Introducing instead of r a dimensional variable ρ = λr, we obtain: ρ d f df + l + 3/ ρ) dρ dρ 1 l + 3/ ε)f =, 3) where ε = k /λ = E/ ω. Substitution of the power series fρ) = n c n ρ n into Eq. 3) yields the following recurrent relation: c n+1 = n ε + l + 3/) nn + l + 5/) c n. 4) This recursion leads to fρ) expρ) for ρ, i.e. to diverging ur), unless: ε l + 3/) = n r, 5) where n r =, 1,,.... If Eq. 5) holds the series truncates and fρ) becomes a Laguerre polynomial. Thus Eq. 5) defines the eigenvalues: E = ω n r + l + 3 ) = ω n + 3 ) 6) Again, the energy depends on a single composite quantum number and must be degenerate. It follows from Eq. 6) that for a given n the quantum number l must assume the values l = n, n,..., if n is even and l = n, n,..., 1 if n is odd. For each l the degeneracy is l + 1. Thus we have to find the sum: Degree of Degeneracy = n l=,... l + 1) = 1 n + 1)n + ) 7) Exactly the same formula can be obtained for odd n. This result coincides with that of 4b. 7