Problem Solutions for Modern Quantum Mechanics, 2nd Edition J.J. Sakurai and Jim Napolitano

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1 Problem Solutions for Modern Quantum Mechanics, nd Edition J.J. Sakurai and Jim Napolitano Jim Napolitano January 7, 4 Contents Fundamental Concepts Quantum Dynamics 5 3 Theory of Angular Momentum 34 4 Symmetry in Quantum Mechanics 55 5 Approximation Methods 59 6 Scattering Theory 86 7 Identical Particles 97 8 Relativistic Quantum Mechanics

2 Copyright, Pearson Education. Chapter One. AC{D, B} = ACDB + ACBD, A{C, B}D = ACBD + ABCD, C{D, A}B = CDAB + CADB, and {C, A}DB = CADB+ACDB. Therefore AC{D, B}+A{C, B}D C{D, A}B+ {C, A}DB = ACDB + ABCD CDAB + ACDB = ABCD CDAB = [AB, CD] In preparing this solution manual, I have realized that problems and 3 in are misplaced in this chapter. They belong in Chapter Three. The Pauli matrices are not even defined in Chapter One, nor is the math used in previous solution manual. Jim Napolitano. a TrX = a Tr + l Trσ la l = a since Trσ l =. Also Trσ k X = a Trσ k + l Trσ kσ l a l = l Trσ kσ l + σ l σ k a l = l δ kltra l = a k. So, a = TrX and a k = Trσ kx. b Just do the algebra to find a = X + X /, a = X + X /, a = i X + X /, and a 3 = X X /. 3. Since detσ a = a z a x + a y = a, the cognoscenti realize that this problem really has to do with rotation operators. From this result, and 3..44, we write [ ] iσ ˆnφ φ φ det exp ± = cos ± i sin and multiplying out determinants makes it clear that detσ a = detσ a. Similarly, use to explicitly write out the matrix σ a and equate the elements to those of σ a. With ˆn in the z-direction, it is clear that we have just performed a rotation of the spin vector through the angle φ. 4. a TrXY a a XY a = a b a X b b Y a by inserting the identity operator. Then commute and reverse, so TrXY = b a b Y a a X b = b b Y X b = TrY X. b XY α = X[Y α ] is dual to α XY, but Y α β is dual to α Y β and X β is dual to β X so that X[Y α ] is dual to α Y X. Therefore XY = Y X. c exp[ifa] = a exp[ifa] a a = a exp[ifa] a a d a ψ ax ψ a x = a x a x a = a x a a x = x x = δx x 5. For basis kets a i, matrix elements of X α β are X ij = a i α β a j = a i α a j β. For spin-/ in the ± z basis, + S z = / =, S z = / =, and, using.4.7a, ± S x = / = /. Therefore S z = / S x = /. = 6. A[ i + j ] = a i i + a j j [ i + j ] so in general it is not an eigenvector, unless a i = a j. That is, i + j is not an eigenvector of A unless the eigenvalues are degenerate.

3 Copyright, Pearson Education Since the product is over a complete set, the operator a A a will always encounter a state a i such that a = a i in which case the result is zero. Hence for any state α A a α = A a a i a i α = a i a a i a i α = = a a i i a i If the product instead is over all a a j then the only surviving term in the sum is a j a a i a i α a and dividing by the factors a j a just gives the projection of α on the direction a. For the operator A S z and { a } { +, }, we have A a = S z S z + a A and a = S z + / for a = + a a a a or = S z / for a = It is trivial to see that the first operator is the null operator. For the second and third, you can work these out explicitly using.3.35 and.3.36, for example S z + / = [S z + ] = [ ] = + + which is just the projection operator for the state I don t see any way to do this problem other than by brute force, and neither did the previous solutions manual. So, make use of + + = = and + = = + and carry through six independent calculations of [S i, S j ] along with [S i, S j ] = [S j, S i ] and the six for {S i, S j } along with {S i, S j } = +{S j, S i }. 9. From the figure ˆn = î cos α sin β + ĵ sin α sin β + ˆk cos β so we need to find the matrix representation of the operator S ˆn = S x cos α sin β + S y sin α sin β + S z cos β. This means we need the matrix representations of S x, S y, and S z. Get these from the prescription.3.9 and the operators represented as outer products in.4.8 and.3.36, along with the association.3.39a to define which element is which. Thus S x. = S y. = i i S z. = We therefore need to find the normalized eigenvector for the matrix cos β cos α sin β i sin α sin β cos β e = iα sin β cos α sin β + i sin α sin β cos β e iα sin β cos β

4 Copyright, Pearson Education. 4 with eigenvalue +. If the upper and lower elements of the eigenvector are a and b, respectively, then we have the equations a + b = and a cos β + be iα sin β = a ae iα sin β b cos β = b Choose the phase so that a is real and positive. Work with the first equation. The two equations should be equivalent, since we picked a valid eigenvalue. You should check. Then a cos β = b sin β = a sin β 4a sin 4 β/ = a 4 sin β/ cos β/ a [sin β/ + cos β/] = cos β/ a = cosβ/ iα cos β and so b = ae sin β = e iα sinβ/ which agrees with the answer given in the problem. = cosβ/e iα sin β/ sinβ/ cosβ/. Use simple matrix techniques for this problem. The matrix representation for H is H =. [ ] a a a a Eigenvalues E satisfy a E a E a = a + E = or E = ±a. Let x and x be the two elements of the eigenvector. For E = +a E, x + x =, and for E = a E, + x + x =. So the eigenstates are represented by E =. [ ] N and E. [ ] = N + where N = /4 and N = /4 +.. It is of course possible to solve this using simple matrix techniques. For example, the characteristic equation and eigenvalues are = H λh λ H λ = H + H ± [ H H + H ] / λ ± You can go ahead and solve for the eigenvectors, but it is tedious and messy. However, there is a strong hint given that you can make use of spin algebra to solve this problem, another two-state system. The Hamiltonian can be rewritten as H. = A + Bσ z + Cσ x

5 Copyright, Pearson Education. 5 where A H + H /, B H H /, and C H. The eigenvalues of the first term are both A, and the eigenvalues for the sum of the second and third terms are those of ±/ times a spin vector multiplied by B + C. In other words, the eigenvalues of the full Hamiltonian are just A ± B + C in full agreement with what we got with usual matrix techniques, above. From the hint or Problem 9 the eigenvectors must be λ + = cos β + sin β and λ = sin β + cos β where α =, tan β = C/B = H /H H, and we do β π β to flip the spin.. Using the result of Problem 9, the probability of measuring +/ is [ + + ] [ cos γ + + sin γ ] [ = ] + cos γ cos γ + = + sin γ The results for γ = i.e. +, γ = π/ i.e. S x +, and γ = π i.e. are /,, and /, as expected. Now S x S x = Sx S x, but Sx = /4 from Problem 8 and S x = [cos γ + + sin γ ] [cos [ ] γ + + sin γ ] = [cos γ + sin γ ] [cos + γ + + sin γ ] = cos γ sin γ = sin γ so S x S x = sin γ/4 = cos γ/4 = /4,, 4 for γ =, π/, π. 3. All atoms are in the state + after emerging from the first apparatus. The second apparatus projects out the state S n +. That is, it acts as the projection operator S n + S n + = [cos β + + sin β ] [cos β + + sin β ] and the third apparatus projects out. Therefore, the probability of measuring / after the third apparatus is P β = + S n + S n + = cos β sin β = 4 sin β The maximum transmission is for β = 9, when 5% of the atoms make it through. 4. The characteristic equation is λ 3 λ/ = λ λ = so the eigenvalues are λ =, ± and there is no degeneracy. The eigenvectors corresponding to these are The matrix algebra is not hard, but I did this with matlab using

6 Copyright, Pearson Education. 6 M=[[ ];[ ];[ ]]/sqrt [V,D]=eigM These are the eigenvectors corresponding to the a spin-one system, for a measurement in the x-direction in terms of a basis defined in the z-direction. I m not sure if there is enough information in Chapter One, though, in order to deduce this. 5. The answer is yes. The identity operator is = a,b a, b a, b so AB = AB = AB a,b a, b a, b = A a,b b a, b a, b = a,b b a a, b a, b = BA Completeness is powerful. It is important to note that the sum must be over both a and b in order to span the complete set of sets. 6. Since AB = BA and AB a, b = ab a, b = BA a, b, we must have ab = ba where both a and b are real numbers. This can only be satisfied if a = or b = or both. 7. Assume there is no degeneracy and look for an inconsistency with our assumptions. If n is a nondegenerate energy eigenstate with eigenvalue E n, then it is the only state with this energy. Since [H.A ] =, we must have HA n = A H n = E n A n. That is, A n is an eigenstate of energy with eigenvalue E n. Since H and A commute, though, they may have simultaneous eigenstates. Therefore, A n = a n since there is only one energy eigenstate. Similarly, A n is also an eigenstate of energy with eigenvalue E n, and A n = a n. But A A n = a A n = a a n and A A n = a a n, where a and a are real numbers. This cannot be true, in general, if A A A A so our assumption of no degeneracy must be wrong. There is an out, though, if a = or a =, since one operator acts on zero. The example given is from a central forces Hamiltonian. See Chapter Three. The Hamiltonian commutes with the orbital angular momentum operators L x and L y, but [L x, L y ]. Therefore, in general, there is a degeneracy in these problems. The degeneracy is avoided, though for S-states, where the quantum numbers of L x and L y are both necessarily zero. 8. The positivity postulate says that γ γ, and we apply this to γ α + λ β. The text shows how to apply this to prove the Schwarz Innequality α α β β α β, from which one derives the generalized uncertainty relation.4.53, namely A B [A, B] 4 Note that [ A, B] = [A A, B B ] = [A, B]. Taking A α = λ B α with λ = λ, as suggested, so α A = λ α B, for a particular state α. Then α [A, B] α = α A B B A α = λ α B α

7 Copyright, Pearson Education. 7 and the equality is clearly satisfied in We are now asked to verify this relationship for a state α that is a gaussian wave packet when expressed as a wave function x α. Use x x α = x x α x x α = x x x α and x p α = x p α p x α = d i dx x α p x α [ ] i p x with x α = πd /4 exp x x 4d [ d to get i dx x α = p ] i d x x x α and so x p α = i d x x x α = λ x x α where λ is a purely imaginary number. The conjecture is satisfied. It is very simple to show that this condition is satisfied for the ground state of the harmonic oscillator. Refer to.3.4 and.3.5. Clearly x = = p for any eigenstate n, and x is proportional to p, with a proportionality constant that is purely imaginary. 9. Note the obvious typographical error, i.e. S x should be Sx. Have Sx = /4 = Sy = Sz, also [S x, S y ] = is z, all from Problem 8. Now S x = S y = for the + state. Then S x = /4 = S y, and S x S y = 4 /6. Also [S x, S y ] /4 = S z /4 = 4 /6 and the generalized uncertainty principle is satisfied by the equality. On the other hand, for the S x + state, S x = and S z =, and again the generalized uncertainty principle is satisfied with an equality.. Refer to Problems 8 and 9. Parameterize the state as = cos β + + eiα sin β, so S x = [cos β + + e iα sin β ] [ ] [cos β + + eiα sin β ] = sin β cos β eiα + e iα = sin β cos α S x = Sx S x = 4 sin β cos α see prob S y = i [cos β + + e iα sin β ] [ ] [cos β + + eiα sin β ] = i sin β cos β eiα e iα = sin β sin α S y = S y S y = 4 sin β sin α

8 Copyright, Pearson Education. 8 Therefore, the left side of the uncertainty relation is S x S y = 4 6 sin β cos α sin β sin α = 4 sin β sin4 β sin α = 4 6 cos β + 4 sin4 β sin α P α, β which is clearly maximized when sin α = ± for any value of β. In other words, the uncertainty product is a maximum when the state is pointing in a direction that is 45 with respect to the x or y axes in any quadrant, for any tilt angle β relative to the z-axis. This makes sense. The maximum tilt angle is derived from P β cos β sin β + sin3 β cos β = cos β sin β + sin β = or sin β = ±/, that is, 45 with respect to the z-axis. It all hangs together. The maximum uncertainty product is S x S y = = The right side of the uncertainty relation is [S x, S y ] /4 = S z /4, so we also need S z = [ cos β β ] sin = cos β so the value of the right hand side at maximum is 4 S z = 4 4 = and the uncertainty principle is indeed satisfied.. The wave function is x n = /a sinnπx/a for n =,, 3,..., so we calculate x x n = x x n = a a x = a 6 x p n = a n x x x n dx = a n x x x n dx = a 6 3 n x i x p n = a n π = a 6 d dx x n dx = 3 n π + 3 n π + n x d dx x n dx = n π a = p

9 Copyright, Pearson Education. 9 I did these with maple. Since [x, p] = i, we compare x p to /4 with x p = 3 + n π = n π which shows that the uncertainty principle is satisfied, since nπ /3 > nπ > 3 for all n.. We re looking for a rough order of magnitude estimate, so go crazy with the approximations. Model the ice pick as a mass m and length L, standing vertically on the point, i.e. and inverted pendulum. The angular acceleration is θ, the moment of inertia is ml and the torque is mgl sin θ where θ is the angle from the vertical. So ml θ = mgl sin θ or θ = g/l sin θ. Since θ as the pick starts to fall, take sin θ = θ so g θt = A exp L t x θl = A + BL p m θl g = m g + B exp L t L A BL = m gla B Let the uncertainty principle relate x and p, i.e. x p = m gl 3 A B =. Now ignore B; the exponential decay will become irrelevant quickly. You can notice that the pick is falling when it is tilting by something like = π/8, so solve for a time T where θt = π/8. Then L π/8 T = ln g A = L g 4 ln m gl 3 ln 8 π Take L = cm, so L/g. sec, but the action is in the logarithms. It is worth your time to confirm that the argument of the logarithm in the first term is indeed dimensionless. Now ln8/π 4 but the first term appears to be much larger. This is good, since it means that quantum mechanics is driving the result. For m =. kg, find m gl 3 / = 64, and so T =. sec 47/4 4 3 sec. I d say that s a surprising and interesting result. 3. The eigenvalues of A are obviously ±a, with a twice. The characteristic equation for B is b λ λ b λib ib = b λλ b =, so its eigenvalues are ±b with b twice. Yes, B has degenerate eigenvalues. It is easy enough to show that AB = ab iab iab = BA so A and B commute, and therefore must have simultaneous eigenvectors. To find these, write the eigenvector components as u i, i =,, 3. Clearly, the basis states,, and 3 are eigenvectors of A with eigenvalues a, a, and a respectively. So, do the math to find

10 Copyright, Pearson Education. the eigenvectors for B in this basis. Presumably, some freedom will appear that allows us to linear combinations that are also eigenvectors of A. One of these is obviously a, b, so just work with the reduced basis of states and 3. Indeed, both of these states have eigenvalues a for A, so one linear combinations should have eigenvalue +b for B, and orthogonal combination with eigenvalue b. Let the eigenvector components be u and u 3. Then, for eigenvalue +b, ibu 3 = +bu and ibu = +bu 3 both of which imply u 3 = iu. For eigenvalue b, ibu 3 = bu and ibu = bu 3 both of which imply u 3 = iu. Choosing u to be real, then No, the eigenvalue alone does not completely characterize the eigenket. we have the set of simultaneous eigenstates Eigenvalue of A B Eigenstate a b a b + i 3 a b i 3 4. This problem also appears to belong in Chapter Three. The Pauli matrices are not defined in Chapter One, but perhaps one could simply define these matrices, here and in Problems and 3. Operating on the spinor representation of + with + iσ x gives [ + i ] = i i = i So, for an operator U such that U =. +iσ x, we observe that U + = S y ; +, defined in.4.7b. Similarly operating on the spinor representation of gives [ ] + i = i = i = i i i that is, U = i S y ;. This is what we would mean by a rotation about the x-axis by 9. The sense of the rotation is about the +x direction vector, so this would actually be a rotation of π/. See the diagram following Problem Nine. The phase factor i = e iπ/ does not affect this conclusions, and in fact leads to observable quantum mechanical effects. This is all discussed in Chapter Three. The matrix elements of S z in the S y basis are then S y ; + S z S y ; + = + U S z U + S y ; + S z S y ; = i + U S z U S y ; S z S y ; + = i U S z U + S y ; S z S y ; = U S z U

11 Copyright, Pearson Education. Note that σ x = σ x and σ x =, so U U. = iσ x + iσ x = / + σ x = and U is therefore unitary. This is no accident, as will be discussed when rotation operators are presented in Chapter Three. Furthermore σ z σ x = σ x σ z, so U S z U =. iσ x σ z + iσ x = iσ x σ z = i σ xσ z = i = i i. so S z = = σ x in the S y ; ± basis. This can be easily checked directly with.4.7b, that is S z S y ; ± = [ + i = S y; There seems to be a mistake in the old solution manual, finding S z = /σ y instead of σ x. 5. Transforming to another representation, say the basis c, we carry out the calculation c A c = b b c b b A b b c There is no principle which says that the c b need to be real, so c A c is not necessarily real if b A b is real. The problem alludes to Problem 4 as an example, but not that specific question assuming my solution is correct. Still, it is obvious, for example, that the operator S y is real in the S y ; ± basis, but is not in the ± basis. For another example, also suggested in the text, if you calculate p x p = p x x x p dx = x p x x p dx = π and then define q p p and y x /, then p x p = d e iqy dy = d πi dq i dq δq x e ip p x / dx so you can also see that although x is real in the x basis, it is not so in the p basis. 6. From.4.7a, S x ; ± = + ± /, so clearly U =. [ ] [ ] [ / / = + [ ] / = / [ ] + ] / / ] [ ] [ / / = = S x : S x :. = r b r a r

12 Copyright, Pearson Education. 7. The idea here is simple. Just insert a complete set of states. Firstly, b fa b = a b fa a a b = a fa b a a b The numbers a b and b a constitute the transformation matrix between the two sets of basis states. Similarly for the continuum case, p F r p = p F r x x p d 3 x = F r p x x p d 3 x = F r e ip p x / d 3 x π 3 The angular parts of the integral can be done explicitly. Let q p p define the z - direction. Then p F r p π π = r dr F r sin θdθe iqr cos θ/ = π 3 4π 3 r dr F r dµ e iqr µ/ = 4π 3 r dr F r iqr i sinqr / = π r dr F r sinqr / qr 8. For functions fq, p and gq, p, where q and p are conjugate position and momentum, respectively, the Poisson bracket from classical physics is [f, g] classical = f g q p f g p q Using.6.47, then, we have [ ] [ ] ipx a ipx a x, exp = i x, exp classical so [x, F p x ] classical = F p x = i ipx a ipx a exp = a exp p x To show that expip x a/ x is an eigenstate of position, act on it with x. So [ ] ipx a x exp x ipx a ipx a = exp x a exp x = x ipx a a exp x In other words, expip x a/ x is an eigenstate of x with eigenvalue x a. That is expip x a/ x is the translation operator with x = a, but we knew that. See I wouldn t say this is easily derived, but it is straightforward. Expressing Gp as a power series means Gp = nml a nmlp n i p m j p l k. Now [x i, p n i ] = x i p i p n i p n i x i = ip n i + p i x i p n i p n i x i = ip n i + p i x i p n i p n i x i... = nip n i so [x i, Gp] = i G p i

13 Copyright, Pearson Education. 3 The procedure is essentially identical to prove that [p i, F x] = i F/ x i. As for [x, p ] = x p p x = x p xp x + xp x p x = x[x, p ] + [x, p ]x make use of [x, p ] = i p / p = ip so that [x, p ] = ixp+px. The classical Poisson bracket is [x, p ] classical = xp = 4xp and so [x, p ] = i[x, p ] classical when we let the classical quantities x and p commute. 3. This is very similar to problem 8. Using problem 9, [ ] ip l [x i, J l] = x i, exp = i ip l ip l exp = l i exp = l i J l p i We can use this result to calculate the expectation value of x i. First note that J l [x i, J l] = J lx i J l J lj lx i = J lx i J l x i = J ll i J l = l i Therefore, under translation, x i = α x i α α J lx i J l α = α J lx i J l α = α x i + l i α = x i + l i which is exactly what you expect from a translation operator. 3. This is a continued rehash of the last few problems. Since [x, J dx ] = dx by.6.5, and since J [x, J ] = J xj x, we have J dx xj dx = x+j dx dx = x+dx since we only keep the lowest order in dx. Therefore x x + dx. Similarly, from.6.45, [p, J dx ] =, so J [p, J ] = J pj p =. That is J pj = p and p p. 3. These are all straightforward. In the following, all integrals are taken with limits from to. One thing to keep in mind is that odd integrands give zero for the integral, so the right change of variables can be very useful. Also recall that exp ax dx = π/a, and x exp ax dx = d/da exp ax dx = π/a 3/. So, for the x-space case, p = α x x p α dx = α x d i dx x α dx = d k exp x dx = k π d p = α x d x α dx dx [ ] = d exp ikx d x ik x exp ikx x dx π d dx d d [ = d ] ik π d + x exp x dx d d [ ] [ ] = d + k d 5 x exp x dx = d π d + k d = d + k

14 Copyright, Pearson Education. 4 Using instead the momentum space wave function.7.4, we have p = α p p p α dp = p p α dp ] d = p exp [ p k d π dp = d ] q + k exp [ q d π dq = k ] p d = q + k exp [ q d π dq = d π 3 π d + 3 k = d + k 33. I can t help but think this problem can be done by creating a momentum translation operator, but instead I will follow the original solution manual. This approach uses the position space representation and Fourier transform to arrive the answer. Start with p x p = p x x x p dx = x p x x p dx = ] x exp [ i p p x dx = i ] exp [ i p p x dx π p π = i p δp p Now find p x α by inserting a complete set of states p, that is p x α = p x p p α dp = i δp p p α dp = i p p p α Given this, the next expression is simple to prove, namely β x α = dp β p p x α = dp φ βp i p φ αp using the standard definition φ γ p p γ. Certainly the operator T Ξ expixξ/ looks like a momentum translation operator. So, we should try to work out pt Ξ p = p expixξ/ p and see if we get p + Ξ. Take a lesson from problem 8, and make use of the result from problem 9, and we have { pt Ξ p = {T Ξp + [p, T Ξ]} p = p T Ξ i } x T Ξ p = p + ΞT Ξ p and, indeed, T Ξ p is an eigenstate of p with eigenvalue p + Ξ. In fact, this could have been done first, and then write down the translation operator for infinitesimal momenta, and derive the expression for p x α the same way as done in the text for infinitesimal spacial translations. I like this way of wording the problem, and maybe it will be changed in the next edition.