Download Transmission Coefficients and Perturbation Theory in Quantum Mechanics and more Study notes Quantum Mechanics in PDF only on Docsity! PHY662, Spring 2004, Feb. 26, 2004 4th March 2004 1 Miscellaneous 1. HWK #8 due Tuesday, Mar. 16. 2. Continue to read Ch. 17 Shankar (or Griffiths Ch. 6), especially for 2nd-order perturbation theory. Start Ch. 18 for time-dependent perturbation theory for Tuesday. 3. Today: Transmission coefficients in 1D, 2nd-order perturbation theory, degener- ate perturbation theory. 2 Transmission coefficients Last time, we reviewed the problem of a plane wave Aeikx incident upon a step in the potential, ∆V < h̄ 2k2 2m . Remember that we found the amplitude for the transmitted wave: Ceik ′x, k′ = √ k2 − 2m h̄2 ∆V , with C = ( 2k k+k′ ) A. What is the probability of transmission through the step? 3 2nd order The second order corrections to the energy require the first order corrections to the wavefunction. (Note that the wavefunctions calculated using perturbation theory are generally less reliable than the energy estimates.) In any case, the first order corrections to the wavefunction are found by computing the inner product of the first order terms in the Schrodinger equation for eigenstate n with the unperturbed eigenstate m: 〈ψ0m|H0|ψ1n〉 + 〈ψ0m|H ′|ψ0n〉 = 〈ψ0m|E0n|ψ1n〉 + 〈ψ0m|E1n|ψ0n〉 E0m〈ψ0m|ψ1n〉 + 〈ψ0m|H ′|ψ0n〉 = E0n〈ψ0m|ψ1n〉 + E1n〈ψ0m|ψ0n〉 〈ψ0m|H ′|ψ0n〉 = ( E0n − E0m ) 〈ψ0m|ψ1n〉 〈ψ0m|ψ1n〉 = 〈ψ0m|H ′|ψ0n〉 E0n − E0m . 1 Since the set of all |ψ0n〉 form a complete basis, |ψ1n〉 can be written as a sum over c1m,n|ψ0n〉. Clearly, c1m,n = 〈ψ0m|H ′|ψ0n〉/(E0n − E0m), so |ψ1n〉 = ∑ m6=n 〈ψ0m|H ′|ψ0n〉 E0n − E0m |ψ0m〉 . Why did we leave |ψ0n〉 out of the sum? Well, the first order approximation to |ψn〉 is |ψ0n〉 + |ψ1n〉, so that any part of |ψ1n〉 that is proportional to |ψ0n〉 is redundant. In fact, this choice also keeps this first order wave function normalized, at least to first order: 〈ψn|ψn〉 = (〈ψ0n| + 〈ψ1n|)(|ψ0n〉 + |ψ1n〉) = 1 + 0 + 0 + (terms second order in H’) . Given this, we can now look at the second order correction to the energy, by taking the inner product of the second order part of the Schrodinger equation with 〈ψ0n|, imme- diately canceling the first terms on each side, due to the same trick as for first order (〈ψ0n|H0 = 〈ψ0n|E0): 〈ψ0n|H ′|ψ1n〉 = E1n〈ψ0n|ψ1n〉 + E2n E2n = ∑ m6=n 〈ψ0n|H ′|ψ0m〉〈ψ0m|H ′|ψ0n〉 (E0n − E0m) − 0 E2n = ∑ m6=n ∣∣〈ψ0n|H ′|ψ0m〉∣∣ E0n − E0m , an expression that again uses only matrix elements of H ′. 3.1 Application Here is a sample application (taken from problems 6.1 and 6.3 in Griffiths). Let H0 be the square well potential for a particle confined to 0 < x < a, so that the normalized unperturbed wave-functions areψ0n(x) = √ 2 a sin( nπx a ) withE 0 n = h̄2 2m n2π2 a2 2 a ∫ a 0 sin2(nπxa ) dx = n2π2h̄2 2ma2 . If the perturbing potential is a delta-function at x = a/2, H ′ = αδ(x − a2 ), then E1n = ∫ a 0 (ψ0n(x)) ∗αδ(x− a 2 )ψ0n(x) dx = 2α a ∫ a 0 sin2( nπx a )δ(x− a 2 ) = { 2α a , n odd 0, n even . The second order correction is E2n = ∑ m6=n 4α2 a2 [∫ a 0 dx sin(nπxa ) sin( mπx a )δ(x− a 2 ) ]2 π2h̄2 4m2a4 (n 2 −m2) 2
