Spring /2/ pts 1 point per minute

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1 Physics 519 MIDTERM Name: Spring 014 6//14 80 pts 1 point per minute Exam procedures. Please write your name above. Please sit away from other students. If you have a question about the exam, please ask. This is a closed book exam. All relevant formulae are given on the exam. If you think a needed equation is not provided, please ask. Note that questions extend to the back of the sheets. Write your answers on the exam. I have tried to leave ample space, but if you need more, use additional paper and be sure to write 519 Midterm and your name on the top of each page. Unless stated otherwise, it is necessary for full credit that you explain the logic of your calculation, deriving any results that you use. Standard results given in the equations sheet do not, however, require explanation. 1

2 Some useful formulas regarding 3d scattering Asymptotic form of scattering wavefunction: ψ e ikz + f(θ, φ) eikr r Radial Schrödinger equation: ( m u + l(l + 1) V (r) + h r ) u = k u, u = Rr Asymptotic form of the radial wavefunction in central potential: Definition of phase shift: R l 1 (η l (k)e i(kr lπ/) e i(kr lπ/)) ikr η l = e iδ l(k). Alternate asymptotic form of the radial wavefunction in central potential: Asymptotic Form of spherical Bessel functions: R l (cos(δ l )j l (kr) sin(δ l )n l (kr)) j l (ρ) ρ 1 sin(ρ lπ/) as ρ ; j l (ρ) ρ l (l + 1)!! as ρ 0 n l (ρ) ρ 1 cos(ρ lπ/) as ρ ; n l (ρ) (l 1)!!ρ l 1 as ρ 0 Partial wave cross sections: σ T = l σ l, σ l = 4π(l + 1) f l Partial wave amplitudes: kf l = eiδ l 1 i = e iδ l sin(δ l ), f(θ) = l (l + 1)f l P l (cos θ) Lippmann-Schwinger (formal): k (+) = k + 1 E H 0 + iɛ V k (+) Lippmann-Schwinger (in position space): ψ( r) = e i k r + m h d 3 r G( r r )V ( r )ψ( r ) where Born Approximation: f( k, k ) = m 4π h G( r) = e+ik r 4π r. d 3 x e i( k k ) x V ( x).

3 1. Bosons and Fermions (40 pts) Four non-interacting equal-mass particles are placed in a one-dimensional infinite square well potential of width L. Two of the particles are identical spin-0 bosons while the other two are identical spin 1/ fermions. (a) (5 pts) Find the ground state energy and the corresponding wave function (wave functions if there is degeneracy). The energy levels of the infinite square well are given by with E n = E 0 n E 0 = h π ml. The corresponding eigenfunctions are ( nπx ) ψ n = L sin. L The lowest energy we can imagine is to have all 4 particles in the n = 1 state for a total energy of E GS = 4E 0. We should confirm that this wavefunction is possible. All 4 particles have to have the same spatial wavefunction in this state. This is symmetric under exchange of the spatial coordinates. For the bosons this is just what we need. For the fermions this needs to be accompanied with an antisymmetric wavefunction in spin. The corresponding wavefunction is unique and can be written as ψ(x 1, x, x 3, x 4 ) = ψ 1 (x 1 )ψ 1 (x )ψ 1 (x 3 )ψ 1 (x 4 ) S = 0. Here the first two position entries are the boson positions, the second two the fermion positions and S = 0 = 1 ( ) is the spin wavefunction. (b) (5 pts) Find the energy, degenaracy and spin of the first excited states. For the first excited state one particle got to be in the n = level for a total energy of E 1st = ( )E 0 = 7E 0. If the one particle is a boson, the two fermions are still both in the n = 1 state and so their spin has to be anti-symmetric. That is there is one first excited S = 0 state with the excited boson. If the excited particle is one of the fermions, there is no more symmetry requirement on the fermion spin and so we get one S = 0 state and thre S = 1 states. So in total there are five first excited states, 3 with spin 1 and with spin 0. (c) (5 pts) Use first order perturbation techniques to calculate the change in the ground state energy of the system if the two fermions interact through the potential V 0 S1 S, where V 0 is a constant and S 1 and S are the spin operators of the respective fermions. As usual, we can rewrite the perturbation as V 0S1 S = V 0 ( S S 1 S ). 3

4 This perturbation is diagonal in the total spin basis. So the change in energy depends only on the total spin. For S = 0 the change in energy is Simlarly, for S = 1 we get S=0 = V 0 h S=1 = V 0 h ( ) = 3V 0 h 4 ( ) = V 0 h 4. As the groundstate has S = 0 it gets shifted down to E GS = 4E 0 + S=0.. (d) (5 pts) Use first order perturbation techniques to calculate the change in the energy of the first excited states if the two fermions interact through the same V 0 S1 S potential. Discuss the degree to which these energy shifts remove the degeneracy found in part (b). The S = 0 first excited states get shifted down to E 1st,A = 7E 0 + S=0. This state is still -fold degenerate (it can have an excited boson or an excited fermion). The 3 S = 1 first excited states get shifted up to This state is 3-fold degenerate. E 1st,B = 7E 0 + S=1. (e) (5 pts) What additional degeneracies are lifted if, in addition to the spin-spin interactions of part (c) and (d), a constant magnetic field in the z direction is turned on? An external magnetic field can break up the 3 S = 1 first excited states into the M = 1, M = 0 and M = 1 levels. The S = 0 first excited states remain degenerate. (f) (10 pts) How would the results of part (c) and (d) change if the spin-spin interaction is given by a contact interaction instead of the uniform spin-spin interaction we considered before. That is, take the interaction term in the Hamiltonian to be given by U 0 δ(x 1 x ) S 1 S, where x 1 and x are the fermion positions (replacing the interaction of part c and d). As the interaction as well as our wavefunctions factorize into a spin and a space part, we can calculate the matrix elements of the perturbation separately for spin and space. The spin part is as above (with U 0 replacing V 0 ). For fermions with symmetric spin wave function, (that is S = 1), one being in state n 1 and the other in state n, the spatial part is given by n 1 n δ(x 1 x ) n 1 n = 1 (ψ n1 (x 1 )ψ n (x ) ψ n1 (x )ψ n (x 1 )) δ(x 1 x ) = x 1,x = (ψ n1 (x 1 )ψ n (x 1 ) ψ n1 (x 1 )ψ n (x 1 )) = 0. x 1 That is the spin/spin interactions simply would have no effect for fermions. The change in energies is proportional to the likelihood that the two fermions are at the same location, which is zero due to the anti-symmetry of the wavefunction. So the only non-zero corrections are for the S = 0 states. The shift of the S = 0 groundstate energy is given by GS = U 0 S=0 dx 1 dx [ψ 1 (x 1 )ψ (x )] δ(x 1 x ) = 4U L 0 S=0 L dx sin 4 (πx/l) = 3U 0 S=0. L 4 0

5 The same shift also applies to the S = 0 first excited state with an excited boson. The shift of the S = 0 first excited state with an excited fermion would be given by = U 0 S=0 dx [ψ 1 (x)ψ (x)] = 8U 0 S=0 L L 0 dx sin (πx/l) sin (πx/l) = U 0 S=0. L Here we used that the spatial wavefunction evaluate on the same position for both fermions is ψ(x, x) = 1 (ψ 1 (x)ψ (x) + ψ (x)ψ 1 (x)) = ψ 1 (x)ψ (x).. 3d Scattering of Defects. (40 pts) (a) Double Delta Potential (0 pts) A particle of mass m initially traveling in the z-direction scatters from the potential V ( r) = V 0 δ (3) ( r aẑ) + V 0 δ (3) ( r + aẑ). Calculate the scattering amplitude f(k, θ, φ) in the Born approximation. Discuss the ka 1 and ka 1 limits (that is, simplify your answer in these limits and discuss whether the result agrees with general expectations). In the Born approximation we have f = m π h d 3 r V ( r)e i q r where q = k k. For the problem at hand the symmetries suggest to either use Cartesian or cylindrical coordinates. Here we will chose the former. f = mv 0 π h dxdydz e i(qxx+qyy+qzz) δ(x)δ(y) [δ(z a) + δ(z + a)] = mv 0 π h cos(q za). We can express our answer in terms of θ and φ by noting that q z = ẑ q = kẑ (ẑ ˆn) = k(1 cos θ) = k sin ( θ ). For ka 1 we see that q z a 1, so cos(q z a) = 1 and we get f = mv 0 π h. This is consistent with the expectation that low energy scattering should be s-wave dominated. For very large wavelength the incoming particle will not be able to resolve the distance between the two delta functions. For ka 1 we see that f is a rapidly oscillating function. So the scattering amplitude of any real wavepacket made by superposing scattering states of different k will vanish due to the large cancelations. This is true everywhere except for in the forward direction. To avoid cancelations we need that the argument of the cosine doesn t get large, so θ < ka. This is also consistent with the general expectation that the small momentum kicks provided by V in the high energy limit should not be able to significantly deflect the particle from the forward direction. 5

6 (b) Scattering off a Hard Sphere (0 pts) For the hard sphere potential (V (r) = 0 for r R, but infinite for r < R) derive the phase shifts δ l for all partial waves. Under what condition is scattering dominated by a single partial wave? Explain. The radial wavefunction reads u (r) + l(l + 1) r u = k u for r > R. The hard sphere imposes the boundary condition u(r) = 0. The most general solution to this equation is a linear superposition of spherical Bessel and Neumann functions u aj l (kr) + bn l (kr). The boundary condition fixes the relative size of the coefficients as Since (according to the formula sheet) we can directly read off that a/b = n l(kr) j l (kr). R l (cos(δ l )j l (kr) sin(δ l )n l (kr)) cot(δ l ) = a/b = n l(kr) j l (kr). 6