Physics 215b: Problem Set 5

Transcription

1 Physics 25b: Problem Set 5 Prof. Matthew Fisher Solutions prepared by: James Sully April 3, 203 Please let me know if you encounter any typos in the solutions. Problem 20 Let us write the wavefunction ψx in the region x << 0 where V 0 as ψx < expikx + exp ikx and in the region x >> 0 again where V 0 as ψx > exp ikx + expikx. 2 The coefficients are related by the S-matrix S 2 S a Because we are looking at the steady-state, late-time behaviour, the probability density is constant in time and the continuity equation reduces to x J x Im ψ x ψ 0. 4 Since J is constant in space, we have that J < J >, which after some quick algebra translates to

2 We can rewrite the RHS using the defining S-matrix equation above to find A in Bin A in B in S S 6 Since this must be true for any choice of,, we must have S S I, 7 ie. S is unitary. A 2 2 complex matrix has 8 real degrees of freedom. We have four constraint equations from equating the matrix elements such that S S I, although only 3 of them are independent as the off-diagonal equations are conjugate to each other. The constraints are s 2 + S 2 2 s S 2 2 s s 2 + s 2 s The first two equations each give one real constraint because they only constrain the magnitude. The last equation constrains the real and imaginary parts and so gives two real constraints. We are left with 4 real degrees of freedom. b For a particle incident from the left, the transmission coefficient is given by T LR s 2 2 and for a particle incident from the right, the transmission coefficient is given by T RL s 2 2. The last constraint equation, when we take the magnitude, gives s 2 s 2 2 s 2 2 s which, substituting in the first two constraints, becomes s 2 2 s 2 2 s 2 2 s 2 2 s 2 2 s Thus T LR T RL. c If the system is time reversal invariant, then it leaves the S-matrix invariant. On the other hand, the action of time-reversal Θ on our wavefunctions is given by complex conjugation so that ψ < Θψx < A in exp ikx + A out expikx 2

3 and so that ψ > Θψx > B in expikx + B out exp ikx. 2 Ã in A out, Ã out A in, Bin B out, Bout B in. 3 Because the S-matrix is invariant, we have Ãout S 2 S 22 Ãin A in B in S 2 S 22 Conjugating this equation, and then using the S-matrix a second time, gives S S A out B out 4 5 so that S S I. Since the inverse of a matrix is unique, we conclude S S and hence that S is symmetric. Symmetry gives s 2 s 2. This is only one new constraint on the phase because unitarity already implied, as we showed above, that s 2 s 2. The most general symmetric unitary matrix then only has 3 real degrees of freedom. d If the Hamiltonian is parity invariant, then a parity transformation again leaves the S-matrix unchanged. The action of parity on the wavefunction exchanges the left and right asymptotic regions as well as leftmovers with rightmovers, so that ψ < P ψx > expikx + exp ikx 6 and ψ > P ψx < exp ikx + expikx. 7 Again we find Ãout S 2 S 22 Ãin S 2 S 22 8 We rewrite this as S 22 S 2 S 2 S 9 and hence conclude that s s 22. As the previous constraints already required these to have the same magnitude, this gives one additional constraint on the phase. The most general such unitary matrix has the form sin θe iφ i cos θe iφ 20 i cos θe iφ sin θe iφ 3

4 Problem 2 20 a An energy eigenstate must have a wavefunction of the form expikx + exp ikx, if x < 0 ψx expikx + exp ikx, if x > 0 2 as this wavefunction solves Schrodinger s equation on any open interval that does not contain the origin. It remains only to constrain the coefficients by the appropriate boundary conditions at the origin. These are. Continuity: This implies ψ0 ψ Equation of Motion: Integrating Schrodinger s equation over an arbitrarily small interval about the origin implies that [ dψx dx ] ɛ ɛ 2m 2 ɛ ɛ V xψx 2mV 0ψ The first constraint gives while the second gives iα + 24 where α mv0 k 2. While these constraints do not completely determine the coefficients, we will see that they do completely determine the S-matrix elements. We can determine the S-matrix elements most easily by an appropriate choice of in conditions. 0, then s / and s 2 /. Likewise, if 0, then s 2 / and s 22 /. So, first take 0. Then the constraints reduce to + 25 If and 2iα + 26 We rearrange to constraints to give and iα iα 27 + iα 28 This gives s iα/ + iα and s 2 / + iα. Next, take 0 and repeat to find s 22 iα/ + iα and s 2 / + iα. We conclude iα/ + iα / + iα S / + iα iα/ + iα k 2 + imv 0 imv 0 k 2 k 2 imv

5 and note that it takes precisely the form we described in Problem. b The Landauer formular gives the conductance as G e2 s 2 2 e2 + α 2 e 2 k 2 3 k m 2 V c Rather than solve for the coefficients in all three regions, it is more useful to solve for a transfer matrix T such that T so that we can determine both ingoing and outgoing data in one region from the corresponding data in another. Using the S-matrix, one finds s 2 s s 2 s 2 s s 22 s 22 In part a we calculated the s-matrix for a delta function potential at x 0. A potential at a general location x a is found simply by shifting the coordinate x y a. Under the shift we find the coefficients shift by A a,in e ika, A a,out e ika, B a,in e ika, and B a,out e ika. We absorb the shifts into the S-matrix to find S a k 2 + imv 0 imv 0 e i2ka k 2 k 2 imv 0 e 2ika Now, for two potential barriers, since the T -matrix relates one region to another, they compose by multiplication so that T tot T a T a 34 where we have shown that iα iαe 2ika T a iαe 2ika + iα 35 This gives e 4iak α + i 2 2iα sin2ak + cos2ak T tot 2iα sin2ak + cos2ak e 4iak + + iα 2 36 We now simply invert the previous process to find the S-matrix from the T-matrix: S t 2 t 22 t

6 so that S tot 2iα sin2ak + cos2ak e 4iak + + iα 2 2iα sin2ak + cos2ak 38 where α mv 0 /k 2. d The Landauer formula gives the conductance as G e2 s 2 2 e2 α 4 + 2α α 2 cos4ak + 4α sin4ak 39 where α mv 0 /k 2. The resistance do not add because of resonant behaviour when the inter-potential distance is commensurate with the wavelength of the electron. Rather than plot a few values here, check out the dynamic plot at Exciting! Problem 3 20 We have a particle of mass m, whose position is given by the operator ˆX, and who is confined to a harmonic potential of natural frequency ω 0, which interacts with a free particle of mass M whose position is given by the operator ˆR, via a potential V ˆX ˆR V 0 exp 2a 2 ˆX ˆR 2 We wish to calculate the cross section to find an outgoing free particle of energy E 2 k 2 /2M from an ingoing free particle with energy E 2 k 2 /2M 40 d 2 σ dωde Γ2π3 ρe k /M 4 where the density of states is ρe Mk/ 2 and the amplitude Γ is given by Γ f 2π f V i 2 δe f E i 42 where we sum over all possible final states of the harmonic oscillator. We denote the total energy of the final state of the particle and the oscillator to be E f E + N + 3/2ω and the total energy of the initial state E i E + 3/2ω. If you re confused where this comes from, consult the first few pages of Sakurai Chapter 6. Relative to the notation in the book, I have simply chosen the normalization L 2π. 6

7 a Let us begin to compute. Our first step is to find the matrix elements f V i 2 : Let a state with momentum k and oscillator level n have the wavefunction Then we have x, r n, k k, n V 0, k 2π 3 Doing the fourier transform and setting q k k gives k, n V 0, k V 0 2π 3 2π3/2 a 3 e a 2 2π 3/2 eikx ψ n r 43 d 3 x d 3 r e ixk k e 2a 2 x r2 ψ nrψ 0 r 44 2 q2 We then need to compute the harmonic oscillator matrix element d 3 r e irq ψ nrψ 0 r 45 n exp iq ˆR 0 46 Using the hint given to you, we rewrite the exponential as exp iq 2mω a + a exp q 2 8mω exp i 2mω qa exp i 2mω qa 47 where a a x, a y, a z. Only the first term of the annihilation operator exponential survives. The creation operator exponential gives n exp iq 2mω a 0 i where n tot n x + n y + n z. So we have that 2mω ntot qx nx nx! q ny y q nz z ny! nz! 48 k, n V 0, k 2 V 2 0 a 6 2 2π 3 e a q 2 e 4mω q2 2mω ntot q 2nx x n x! q 2ny y n y! q 2nz z n z! 49 and hence that Γ V 0 2 a 6 2 2π 2 e a q 2 e 4mω q2 n ntot qx 2nx 2mω n x! q 2ny y n y! q 2nz z n z! δ E f E i. 50 Now E i E + 3/2ω and E f E + ωn tot + 3/2 so that the delta function enforces that n tot E E ω N 5 7

8 the delta function is only satisfied when this is an integer. So we can truncate the sum to n x,n y,n z n totn q 2nx x n x! q 2ny y n y! q 2nz z n z! 52 N! q2 N 53 So at last we write the amplitude as where N is an integer. b Γ V 0 2 a 6 N 2 2π 2 N! e a q 2 e 4mω q2 q2 δ E + Nω E 54 2mω At last we put this together to find the cross section d 2 σ dωde 2πM 2 V 2 N! 4 0 a 6 N E 2 E e a q 2 e 4mω q2 q2 δ E + Nω E 55 2mω The oscillator is in its highest possible state when there is minimal energy carried by the scattered particle. Because the energy of the oscillator is quantized, this does not necessarily happen when E 0, but when the integer part. E N ω 56 8