Quantum Physics 2011/12

For m even, no such argument applies and we have to work hard to evaluate therequired integral!〈m|∆ ˆV |1〉 = V 0a= 2V 0π= 4V 0π∫ a( ) mπxsin sin2a−a∫ +π/2−π/2∫ π/20( ) πxacossin mθ sin 2θ cos θ dθsin mθ sin 2θ cos θ dθ( ) πx2aWe now use the trigonometric identities (from cos A cos B etc.) :to write〈m|∆ ˆV |1〉 = V 0π= V 0πdxwhere θ ≡ πx2asin mθ sin 2θ ≡ 1 [cos(m − 2)θ − cos(m + 2)θ]2cos(m − 2)θ cos θ ≡ 1 [cos(m − 1)θ + cos(m − 3)θ]2cos(m + 2)θ cos θ ≡ 1 [cos(m + 3)θ + cos(m + 1)θ]2∫ π/20[cos(m − 1)θ + cos(m − 3)θ − cos(m + 3)θ − cos(m + 1)θ] dθ[ sin(m − 1)π/2+(m − 1)Recall that m is even and ≥ 2 so thatsin(m − 3)π/2(m − 3)−sin(m + 3)π/2(m + 3)sin(m − 1)π/2 = sin(m + 3)π/2 = (−1) m 2 +1sin(m − 3)π/2 = sin(m + 1)π/2 = (−1) m 2−sin(m + 1)π/2(m + 1)]Thus〈m|∆ ˆV |1〉 = (−1) m/2 V 0π[− 1](m − 1) + 1(m − 3) + 1(m + 3) − 1(m + 1)Putting everything over a common denominator yields〈m|∆ ˆV |1〉 = (−1) m/2 V 0π16m(m 2 − 9)(m 2 − 1)Thus〈2|∆ ˆV |1〉 = 32V 015π ; 〈4|∆ ˆV |1〉 = 64V 0105πNow for the unperturbed Hamiltonianetc(E 1 − E 2 ) = π2¯h 22(1 − 4) = −3π2¯h8ma2 8ma 2and so we find that the second order energy shift is an infinite sum of the even mterms, changing the index so that n = 2m ...