Perturbation Theory Examples - Lecture Notes | CEM 987, Study notes of Chemistry

Download Perturbation Theory Examples - Lecture Notes | CEM 987 and more Study notes Chemistry in PDF only on Docsity! Perturbation Theory Examples A. Charged particle in a one dimensional box Consider a particle of mass m and charge q moving in the potential 0V for x & x= ∞ < > a a≤ and V f . The wavefunction is zero where the potential is infinite and so we need only consider the motion between 0 & a. The Schrodinger equation is with 0 0or x= ≤ 0 0 0 0 N N NH EΦ = Φˆ 2 2 0 22 d m dx = −Ĥ , 0 2N N xsin a a π Φ = and 22 0 2N NE m a π =     with . If we introduce a constant electric field, 1N ,= 2 3, ˆF Fx= = − along the x axis the charged particle experiences an additional potential energy V q and so the perturbed problem is Fx N ( )0 N NĤ qFx E− Φ = Φ . The first order correction to the energy of the Nth state is ( )1 0 0 0 0 2N N N N N qFaE qFx qF x= Φ − Φ = − Φ Φ = − where we have used the fact that the average value of x is a/2 for all states. Note that each state is shifted by the same amount and whether the energy increases or decrease depends on the signs of q & F. The second order correction to the energy is given by ( ) ( ) 2 20 0 02 2 0 0 0 2 2 1 N K N N K N K NN K qFx xqF E E E E N K≠ ≠ Φ − Φ Φ Φ = = − −∑ ∑ 0 K where 22 0 1 2 E m a π =     . The required matrix element is 0 0 2 0 0 2 2a N K N x K x ax x sin sin dx sin N sin K d a a a ππ π ξ ξ ξ π Φ Φ = =∫ ∫ ξ Exercise: Evaluate 0 sin N sin K d π ξ ξ ξ∫ ξ J. F. Harrison Michigan State University 9/19/2006 1 B. Harmonic Oscillator with a cubic perturbation Suppose we are interested in estimating the eigenvalues and eigenvectors associated with the Hamiltonian 2 2 2 3 0 3 2 1ˆ ˆ 2 2 dH kx x H m dx xλ λ= − + + = + The first order correction to the energy of the state 0nΦ is then zero by parity arguments. (1) 0 3 0 0n n nE xλ= Φ Φ = The second order correction is 2 2 20 0 0 3 0 3 (2) 0 0 ˆ ( ) ( ) n p n p n n pp n p n p n V x n x p E h n p h n pE E λ λ ν ν ≠ ≠ ≠ = = = − −−∑ ∑ ∑ Φ Φ Φ Φ So we need to evaluate matrix elements of 3x between various harmonic oscillator states. Using recursion relationships between the Hermite polynomials many texts show that , 1 , 1 1 ( ) 2 2j i j i ij j ji x j xδ δ α α− + + = + = and this result allows us to evaluate matrix elements of lx for any positive integer value of using the resolution of the identity. Since the set of eigenfunctions Φ is complete we may write l 0n 0 0 0 0 1̂ n n n n n n ∞ ∞ = = = Φ Φ =∑ ∑ So 2 2 0 ( )ij n i x j x i xx j i x n n x j ∞ = = = =∑ or in a more explicit matrix form 2 0 ( ) ( ) ( )ij in nj n x x x ∞ = =∑ Using the above result for ( )ijx the sum can be evaluated and we obtain J. F. Harrison Michigan State University 9/19/2006 2 ( )22 2 2 2 1 ( ) ( ) 2 2 2 qFd k E m d k ξ ξ ξ ξ    − + Φ = + Φ        This is the harmonic oscillator equation, so, as we have seen above 21/ 2( ) ( )n n nN H e αξξ α ξ −Φ = and ( ) 21( )2n 2E h n qF kν= + − Note that all of the levels have been lowered by the same amount and the wavefunctions have all been shifted along the x axis so that they are centered at qFx k = . If q and F are both positive the equilibrium point is shifted in the x+ direction as expected. The explicit dependence of the first two wavefunctions on the electric field is ( ) ( ) ( ) ( ) ( ) 1/ 4 0 1/ 43 1 2 2 /2 /2 ( ) 4( ) qFx k qFx k x qFx x k e e α α α π α π − − − − Φ = Φ = − Now let’s use perturbation theory to solve the problem. The perturbation is and the first order correction to the energy is zero by parity. The second order correction is then qFx− ( )22 22 pn( ) n p n p n xp qFx n ( qF )E hv( n p ) h ( n p )ν ∞ ∞ ≠ ≠ − = = − −∑ ∑ From above the only terms that will appear in the summation are 1p n= ± so we have 2 22 2 1 1 1 1 ( ) n ,n n.n n ( x ) ( x )( qF )E hν − +   = +  −  2 2 1 1 1 2 2n,n n ,n n n( x ) & ( x ) α α− + + = = so ( )22 2 ( ) n qF E k = − J. F. Harrison Michigan State University 9/19/2006 5 Note that because the first order correction to the energy vanishes, the third order correction has the form (3) 0 0 np pt tn n np ntp n t n x x x E E E≠ ≠ =∑∑ which is identically zero. Indeed all higher corrections must be zero and the exact energy of the oscillator in the field is ( )21 2 2n qF E h ( n ) k ν= + − as found in the exact solution. In this problem the second order correction is the total correction. The correction to the wavefunction is given by ( ) ( ) ( ) ( ) 0 0 0 1 1, 1 , 1(1) 0 0 1 111 1 2 p n nnp n n n n n n n p n x x xqF qF qF n n h n p h hν ν α ν − +− + − + ≠  − − − = = + = −  − −   ∑ Φ Φ Φ Φ Φ + Φ Exercise: We have seen that the exact solution for the ground state of the harmonic oscillator in a constant electric field is ( ) ( )1/ 40 2 /2 ( ) qFx kx e αα π −− Φ = while perturbation theory tells us that (1) 0 10 2 qF hα ν − =Φ Φ Show that these two results are consistent. Hint: expand the exact solution in a power series in the electric field. J. F. Harrison Michigan State University 9/19/2006 6