Solution to Problem Set No. 6: Time Independent Perturbation Theory

Transcription

1 Solution to Problem Set No. 6: Time Independent Perturbation Theory Simon Lin December, 17 1 The Anharmonic Oscillator 1.1 As a first step we invert the definitions of creation and annihilation operators to express ˆX, ˆP in terms of â, â ˆX mω â + â 1 mω ˆP i â â Hence ˆV λ ˆX 3 λ m ω â + â Since the first order wave function mix is proportional to the expectation value n ˆV m, we see from the from of ˆV that for m to mix with n, one need 1. n m n + 5 We now expand out and write all terms in normal ordering: ˆV λ ] m ω â + â 3 â + 6â â + â â 3 + â + 6â + 1â â + 6â The matrix element n ˆV can be read off as ˆV 3 λ m ω 7 ˆV 6 λ m ω 8 ˆV λ m ω 9 1

2 1.3 E n 1 n ˆV n λ m 3 + 1n + 6nn 1] 1 ω 1. n 1 n ˆV m m 11 E n E m m n nn 1n n 3 n + n 1 nn 1 n n + 5 n + 1n + n + 3n + n + 1n + n + n + λ m ω 3 1 for n < just ignore the negative eigenkets 1.5 E 1 ˆV 3λ m ω 3 13 E m m ˆV E Em Charged Particle in a Magnetic Field.1 1λ 3 8m ω 5 1 Assuming that the eigen wave function can be factorized into Ψ 1/ Le iky φx and plug it into the unperturbed Schödinger equation H Ψ EΨ we can show that k ebc ] x e iky ˆP x + φ n x MEe iky φ n x 15 We can factor e iky out and rewite the full eigen equation as ˆP x + k eb ] c ˆX φ n MEφ n 16 We see that if we make a shift x x x k c eb 17 x kl 18 then since it is just a shift on x, P X stays invariant. So we can cast the whole eigen equation in terms of x : ˆP x + l x ] φ n MEφ n 19

3 Hence the eigen function φ n must be a function of x x kl. Also note that this equation is just the equation of the 1D simple harmonic oscillator. We can directly write down its solution in terms of Hermite polynomials: The whole solution is thus x φ n x n n! πl ] 1/ H n l e x l Ψ n,k x, y 1 L e iky φ n x kl 1 The unperturbed energies are just the eigen energy of a simple harmonic oscillator E n,k eb Mc n + 1 ω c n We see that the Ψx, y is also a eigen function of ˆP y. If we demand periodic boundary condition in the y direction, then we can show that k πm/l, where m is an integer. The constraint from magnetic flux is more subtle. In order to do this we first consider the case where the length of strip W is finite and then take W. We know that the symmetric point x kl of the wave function has to lie inside the region W/ < x < W/, so kl < W πc ebl m < W m < ebw L N φ πc Since m can only take interger values, the allowed values of m is thus N φ 1 < m < N φ Since Ĥ, ˆP y ], the y-momentum is conserved and the inner products of different k eigenstates vanish by othogonality. If ˆV x is an operator only depending on x, we can factor out the x and y inner products and show that n, k ˆV n, k of different k vanish. More explicitly:.3 n, k ˆV x n, k k φ n,k ˆV x k φn,k 8 k k φ n,k ˆV x φ n,k 9 δ k,k n, k ˆV x n, k 3 We skip the y integrals here since the wave functions in y direction are othonomal. V k V πl V l 1 a + 1 e k l dx e x kl l e x a 31 a +l 3 3

4 V 11 k V πl dx x x kl l e l e x a 33 8V a 3 l + a k + 1 e k l a + l 5 a +l 3 V 1 k V 1 k V πl 8V a 3 k a + l 3 dx x x k l l e l e x a 35 e k l a +l 36. According to first order perturbation theory, the energy shift is proportional to the expetation value of the perturbative potential in the unperturbed eigenstates of which we just computed above. E k V k V l E 1 k V 11 k 1 e k l a + 1 a +l 37 8V a 3 l + a k + 1 e k l a + l 5 a +l 38 We plot the k-dependence of the energy diffference below: assuming V a l 1 We see that in the case of ground states, each k corresponds to a different shift E k, so the degenercies are completely removed. However in the first excited states, there is a region in which there are two different k s corresponding to the same energy shift. We still have a two-fold degenercies in first excited states in this region..5 To the first order, the perturbed wave function is given by Ψ 1 n Ψ n + m n n V m E n E m Ψ m 39

5 Using our previous results we explicitly write down the perturbed wave functions for n and n 1: Ψ 1,k 1 e iky φ x kl V ] 1k φ 1 x kl L ω c Ψ 1 1,k 1 e iky φ 1 x kl + V ] 1k φ x kl 1 L ω c Where we ignored the contribution from all the higher energy levels. Now we want to compute the total current J where j n,k x, y is the current density mi dx L/ L/ dy j n,k x, y j n,k x, y Ψ n,k x, y Ψn,k x, y Ψ n,k x, y Ψ n,k x, y 3 We immediately see that j x is zero since φx is real everywhere even after the perturbation as we already know that wave functions of different k do not mix. The other component of current can be worked out to be j y x, y k ml φ1 n x kl After integrating we obtain the total current as a function of n, k: J n,k k m ŷ 5 The current is linear to k. This is because we are working with eigenstates of ˆPy and that the perturbation V does not depend on y so it preserves the y-symmetry..6 If we look at the figure of the first order shift of the energy, we see that there is a region where E > E 1. In this region if we tune V large enough we can actually see that the two perturbed energy levels starting cross each other. We can calculate the required V to be ω c V k E k E 1 k 3 l + a 5/ a 3 l + a k e k 8ω c l 6 a +l 7 We also see that E E 1 has maximum at k, so the mixing first occurs at k as one gradually increases V. To resolve the degeneracy we turn to almost degenerate perturbation theory and treat the two nearly degenerate states exactly while keeping states of other k perturbatively. This means that we need to solve for the matrix equation E k + V k V 1 k V 1 k E 1 k + V Ψ E Ψ 8 11k 5

6 Defining E n 1 E n + V nn we write the eigenvalues as E ± k E1 k + E1 1 k ± E 1 k E1 1 k + V1 k 9 Here since V 1 k, the energies will at least be split by V 1 k. We see that the energy level actually repels each other as we tune up V, and we can infer that there won t be any level crossing in the perturbative expansion. 3 Model of a Hydrogen-like Atom 3.1 Since Ŵ r is a function only of r, it commutes with the total angular momentum operator ˆL and the z-direction angular momentum operator ˆL z. The matrix elements of W can be decomposed into 3. n, l, m Ŵ n, l, m n Ŵ r n l, m l m 5 δ l,l δ m,m n, l, m Ŵ n, l, m 51 The perturbative potential W r is W r { V + e /r, r d, r > d 5 We now calculate the matrix elements of Ŵ : 1,, Ŵ 1,, d a 3 r dr V + e e r/a 53 r 1 a e a a + de d/a ] + V a a + a d + d e d/a ] 5 1 d,, Ŵ,, 8a 3 r dr V + e r e r/a 55 r a 1 e 8a a 3 a 3 + a d a d + d 3 e d/a ] +V 8a 8a + 8a 3 d + a d + d e d/a ] 56 1 d, 1, Ŵ, 1, a 3 r dr V + e r r a e r/a 57 1 e a 6a 3 6a 3 + 6a d + 3a d + d 3 e d/a ] +V a a + a 3 d + 1a d + a d 3 + d e d/a ] 58 6

7 1 d,, Ŵ 1,, a 3 r dr V + e ra e 3r/a 59 r 7a 3 e a a + 6a d + 9d e 3d/a ] + 9V d 3 e 3d/a 6, 1, ±1 Ŵ, 1, ±1, 1, Ŵ, 1, The shift in the ground state energy is just the matrix element 1,, Ŵ 1,, as we just calculated 3. E 1 1,, 1 a e a a + de d/a ] + V a a + a d + d e d/a ] 6 As seen from 56, 58 and 61, it is clear that this perturbation W r lifts the degeneracies between the,, and, 1, eigenstates while fail to resolve the 3-fold degeneracy of quantum number m among, 1, m states. The energy difference is given by E e d/a 1a e d 3a d + V d 3 a d ] 63 This follows from that the radial wave functions are completely determined by quantum number l in hydrogen atom solutions. As the perturbation does not have angular dependence this feature will continue to hold. 7