16.1. PROBLEM SET I 197


 Mavis Welch
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1 6.. PROBLEM SET I 97 Answers: Problem set I. a In one dimension, the current operator is specified by ĵ = m ψ ˆpψ + ψˆpψ. Applied to the left hand side of the system outside the region of the potential, where ψx =Ae ikx + Be ikx, we have j left = k m A B ; similarly j right = k m C D. Current conservation demands that j left = j right, i.e. A + D = C + B all of the incoming beam is transferred to the outgoing beam, a statement of particle conservation. This conservation of current can be written as Ψ in Ψ in = Ψ out Ψ out. Therefore, using the expression for ψ out, we have Ψ in Ψ in = Ψ out Ψ out = Ψ in }{{} S S Ψ in, leading to the unitarity condition.! =I b From the unitarity condition, it follows that S t S = I = + r rt + r t rt + r t t + r Comparing the matrix elements, we obtained the required result. c For the δfunction scattering problem, the wavefunction is given by { e ψx = ikx + re ikx x<0 te ikx x>0 For the δfunction potential, the wavefunction must remain continuous as x = 0, and x ψ +ɛ x ψ ɛ = mav 0 ψ0. This translates to the conditions + r = t, and ikt +r = mav0 t. As a result, we obtain ik t =, ik + mav0 as required. Using the result from b, since the potential is symmetric, we have r = t t t = ik + γ k ik + γ k + γ = ik + γ γ ik + γ k + γ = γ ik + γ As a result, we obtain the required expression for r.. a We are told that Ĥ ψ = E ψ and Ĥ ψ = E ψ, where E E. Therefore ψ Ĥ ψ = ψ Ĥψ dx = ψe ψ dx = E ψ ψ. Since Ĥ is Hermitian, we can write ψ Ĥ ψ = Ĥψ ψ dx = E ψ ψ dx = E ψ ψ = E ψ ψ, since E and E are real. Therefore, E E ψ ψ = 0 and, if E E then ψ ψ = 0, i.e. ψ and ψ are orthogonal. b If Â ψ = ψ and Â ψ = ψ, then adding them, Â ψ + ψ = ψ + ψ and subtracting, Â ψ ψ = ψ ψ = ψ ψ. Hence we have an eigenvector of a = + corresponding to a normalized eigenvector ψ + ψ and an eigenvalue a = corresponding to eigenvector ψ ψ.
2 6.. PROBLEM SET I 98 c The timedependent Schrödinger equation is Ĥψ = Eψ = i tψ, hence ψt =ψt = 0e iet/. Since ψ and ψ are eigenstates of the Hamiltonian Ĥ then we can write, ψt = [ ψ e iet/ ψ e iet/ ]. P = ψt = 0 ψt = [ ψ ψ ][ ψ e iet/ ψ e iet/ ] 4 = [ + cose E t/] = cos E E t/ 3. a Differentiating the left hand side of the given expression with respect to β, one obtains e βa [a, a ] e βa =. }{{} = Integrating, we therefore have that e βa ae βa = β+ integration constant. By setting β = 0 we can deduce that the constant must be a yielding the required result. Using this result, we have that e βa a β = e } βa {{ ae βa } 0 =β + a 0 = β 0. =β+a Lastly, to obtain the normalization, we have that β β = N 0 e β a β = N β a n 0 β n! i.e. N = e β / as required. n=0 n=0 = N β β n 0 β = N e β! =. n! b For the harmonic oscillator, the creation and annihilation operators are related to the phase space operators by ˆx = mω a + a, and ˆp = mω i a a. Therefore, we have mω ˆx = mω β + β, ˆp = i β β. Then, using the identity x = x x = x x, we have x = mω β a + aa + a a +a β = mω + β + β, ˆp = mω β a aa a a +a β = mω +β β. As a result, we find that x = mω mω and p = leading to the required expression. c The equation follows simply from the definion of the operator a and the solution may be checked by substitution. d Using the timeevolution of the stationary states, nt = e ient/ n0, where E n = ωn +/, it follows that βt = e iωt/ e β / βn n! e inωt n = e iωt/ e iωt β. Therefore, during the timeevolution, the coherent state form is preserved but the centre of mass and momentum follow that of the classical oscillator, x 0 t =A cosϕ + ωt, p 0 t =mωa sinϕ + ωt. The width of the wavepacket remains constant.
3 6.. PROBLEM SET I a Starting with the electron Hamiltonian Ĥ = m ˆp qa, substitution of the expression for the vector potential leads to the required result. b Setting ω = m = =, we immediately obtained the required dimensionless form of the Hamiltonian. c Straightforward substitution of the differential operators leads to the required identities. As a result, we can confirm that [a, a ]= 4 [ z, z] [z, z ] =, and similarly [b, b ] =. In the complex coordinate representation, Ĥ = i z + z i 4 z z + z z + 4 z + z = z z + 8 zz + z z z z = z + z z + z = a a + d The angular momentum operator is given by = ˆL z =x ˆp z = ix y y x = i z + z z z z z z + z = z + z z + z + z + z z + z = a a b b e In the coordinate representation, the condition a 0, 0 = 0 translates to the equation z + z r 0, 0 =0 4 We thus obtain the Gaussian expression for the wavefunction. Then, using the relation 0,m = b m m! 0, 0, we have r 0,m = m/ z + z m e zz/4 = m! 4 π πm m! zm e zz/4, as required. f If we populate the states of the lowest Landau with electrons, starting from states of the lowest angular momentum m, the wavefunction is a Slater determinant involving entries φ m r j =zj m /4 e zj. Taking the determinant, and making use of the Vandemonde determinant identity, one obtains the required manyelectron wavefunction. 5. The spin operator in the θ, φ direction, Ŝθφ, can be found by forming the scalar product of the spin operator Ŝ with a unit vector in the θ, φ direction, sin θ cos φ, sin θ sin φ, cos θ. Therefore Ŝ θφ = cos θ e iφ sin θ e iφ. sin θ cos θ We need the eigenvalues of the matrix, i.e. cos θ sin θe iφ u sin θe iφ cos θ v u = λ v. Eliminating u and v, we find λ = and hence the eigenvalues of Ŝθφ are ±/, as expected. Substituting the values λ = ± back into the equations relating u and v, we can infer the ratios, u v = e iφ cotθ/ and e iφ tanθ/. So, in matrix notation, the eigenstates are cosθ/ e iφ sinθ/ and sinθ/ e iφ cosθ/,
4 6.. PROBLEM SET I 00 for eigenvalues +/ and / respectively. The spin states in the xdirection are obtained by setting θ = π/, φ = 0, and the spin states in the ydirection are obtained by setting φ = π/4 in these general formulae. 6. The four states are given by φ = χ + χ +, φ = χ χ, φ 3 = [χ + χ +χ χ + ], and φ 4 = [χ + χ χ χ + ], where φ, φ and φ 3 are symmetric under particle interchange and φ 4 is antisymmetric. Since Ŝ+Ŝ = Ŝ x + Ŝ y + iŝyŝx ŜxŜy =Ŝ Ŝ z + Ŝz, Ŝ = Ŝ+Ŝ + Ŝ z Ŝz. 6. Noting that Ŝz = Ŝ z + Ŝ z, and likewise for Ŝ±, and using the basic results for the operation of the spin operators on the single particle states, Ŝ z χ ± = ± χ ±, Ŝ χ ± = χ, Ŝ ± χ ± = 0, we can now calculate the effect of applying Ŝ to each state. For example Ŝ φ = [χ + χ + + χ + χ + χ + χ + ] = φ, where the three terms in square brackets correspond to the three operators on the right hand side of Eq. 6.. Likewise φ. Clearly Ŝzφ 3 =0=Ŝzφ 4, so only the Ŝ+Ŝ term need be considered. We find Ŝ + Ŝ χ + χ = Ŝ+χ χ = χ + χ + χ χ + φ 3. Similarly Ŝ+Ŝ χ χ + = φ 3 so that Ŝ φ 3 = φ 3, Ŝ φ 4 =0. Thus, φ, φ, φ 3 all have eigenvalue and hence S =, while φ 4 has S = 0. The state ψ is clearly an eigenstate of Ŝz with eigenvalue 0, and must therefore be a linear combination of φ 3 and φ 4. The S = component is thus the φ 3 term in the wavefunction with amplitude c 3 = φ 3 ψ = 3 χ +χ + = /3+ /6 3 χ χ + The probability of S = is therefore c 3 = 3+ 6 =0.97. χ +χ + χ χ + 7. Taking as a basis the states of the Ŝz operator, we can deduce the form of the Ŝ x operator most easily from the action of the spin raising and lower operators. Noting that, for spin S =, Ŝ± S =,m = [ mm + ] /,m±, we can construct the matrix elements of Ŝ±. The latter are given by Ŝ + = , Ŝ = Ŝ +. Then, using the relation, Ŝx = Ŝ+ Ŝ, we obtained the matrix elements of the operator and corresponding eigenstates, Ŝ x = , m x = m x =0 m x = 0
5 6.. PROBLEM SET I 0 When placed in a magnetic field, B, the molecules will acquire an energy µbm z, where µ is the magnetic moment of the molecule, which in this case equals twice the magnetic moment of the proton, and m z is the eigenvalue of Ŝ z. At t = 0, the molecules enter the magnetic field in the m x = state, after which their wavefunction evolves with time in the usual way, i.e. ψt =e iµbŝzt/ ψ0 = e iµbt e iµbt = e iµbt e iµbt Thus, if µbt = n + π, with n an integer, the molecules will be in a pure m x = state, and none will pass the second filter. The time is given by t = L/v = L m/e, where L = 0 mm and m and E are the mass and energy of the molecules respectively. We thus have / n + π E µ = = JT BL m and hence the proton magnetic moment is JT. Note that the result can also be obtained by treating the problem as one of classical precession. The couple = µb = LΩ, where L = is the angular momentum and Ω the angular frequency of precession. If Ωt = n + π, the molecules have precessed into the m x = state, and the result readily follows.. 8. Substituting for the definition of the spin raising and lowering operators using the HolsteinPrimakoff transformation, the commutator is obtained as S [Ŝ+, Ŝ ]= = a a S + a a a a S a a S a a + / {}}{ aa a a S a a + a a aa = a a S S. / a a a a S With Ŝz = S a a, we obtain the required commutation relation [Ŝ+, Ŝ ]= Ŝz. 9. For the case l =, l =, a table of possible ways of forming each value of M = m + m is shown right. The largest value of M is 3, so the largest value of L must be 3. There must also be a state with M = corresponding to L = 3, but we have two states with M =. Therefore there must be a state with L = as well. We need two states with M =, one for each of the L =3, multiplets, but we actually have three states with M =, so there must be an L = state as well. All of the M states are now accounted for. For the case l = 3, l =, we can again for a table see right. Following the same logic as before, we see that states with L =4, 3, just account for all the states. To construct the states explicitly, we start by writing the L =3M = 3 state, since there is only one way of forming M = 3, viz. 3, 3 =,,. We then operate with the lowering operator ˆL, which is simply the sum of the lowering operators for the two separate particles. Recalling that: ˆL l, m = ll + mm l, m, we obtain 6 3, =, 0, + 4,,, where the first term on the right hand side comes from lowering the l = state and the second from lowering the l = state. Hence 3, = /3, 0, + /3,,. The state, must be the orthogonal linear combination, i.e., = /3, 0, /3,,. Further states could be computed in the same way if required. M m,m 3,, 0,, 0 0,, 0, 0, 0,, 0,, 0 0,, 3, M m,m 4 3, 3 3, 0, 3,, 0,,, 0 0, 0, 0, 0, 0,, 0,,, 0 3, 3, 3, 0 4 3,
6 6.. PROBLEM SET I 0 0. The commutation relations are given by [J i,j j ]=iɛ ijk J k. a First define the eigenstates ψ m : J z ψ m = mψ m. To see if J ± ψ m is an eigenstate of J z, we need to look at J z J ± ψ m, which is equal to J ± J z ψ m [J ±,J z ]ψ m. The required commutator is [J ±,J z ] = [J x,j z ] ± i[j y,j z ], from the definition of J ±. From the basic commutators given at the start, this is [J ±,J z ]= ij y ± J x = ± J ± if treating ± like a number is confusing, do this separately for J + and J. Going back to J z J ± ψ m, we can now write this as J ± J z ψ m + ±J ± ψ m. The first term is just J ± mψ m, so this is m ± J ± ψ m. Thus, J ± ψ m is an eigenstate of J z, with eigenvalue m ±. This establishes the raising and lowering property of J ±. b Two electrons would have a total spin of S = or 0. Adding a third spin / particle creates total spin S =3/ or / from the S = twoparticle state. The S = 0 twoparticle state becomes S =/ only on adding the third particle, so total S =3/ or / are the only possibilities. c The states with welldefined values of m, m, and m 3 for the z spin components of all particles are the uncoupled basis. Where all particles are spin up, this state may be written as. This state is also the m S =3/ state of total S =3/ there is no other way to get m + m + m 3 =3/in the uncoupled basis. We can therefore write S =3/,m S = 3/ =. To get from here to S =3/,m S =/, we need to apply J = S + S + S 3. In other words, the total lowering operator is the sum of the lowering operator for each separate spin reasonably enough this follows from the definition of J and J x = S x + S x + S x 3, etc. Now, we need to use the given normalization result. This says that J S =3/,m S =3/ = 5/4 3/4 S =3/,m S =/ = 3 S =3/,m S =/. Notice that the total quantum number, J, is the same as the overall spin quantum number, S in this case. Therefore S = 3/,m S = / = / 3J S = 3/,m S = 3/. Using the given normalization result again for a single state, S /, / = 3/4+/4 /, /. This establishes the required result.
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