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Solution Manual for Modern Quantum Mechanics, 2nd Edition

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    2 Chapter One 1. 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 2 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 2. (a) Tr ( X ) = a 0 Tr (1) + ￿ ￿ Tr ( σ ￿ ) a ￿ = 2 a 0 since Tr ( σ ￿ ) = 0. Also Tr ( σ k X ) = a 0 Tr ( σ k ) + ￿ ￿ Tr ( σ k σ ￿ ) a ￿ = 1 2 ￿ ￿ Tr ( σ k σ ￿ + σ ￿ σ k ) a ￿ = ￿ ￿ δ k ￿ Tr (1) a ￿ = 2 a k . So, a 0 = 1 2 Tr ( X ) and a k = 1 2 Tr ( σ k X ). (b) Just do the algebra to find a 0 = ( X 11 + X 22 ) / 2, a 1 = ( X 12 + X 21 ) / 2, a 2 = i ( − X 21 + X 12 ) / 2, and a 3 = ( X 11 − X 22 ) / 2. 3. Since det( σ · a ) = − a 2 z − ( a 2 x + a 2 y ) = − | a | 2 , the cognoscenti realize that this problem really has to do with rotation operators. From this result, and (3.2.44), we write det ￿ exp ￿ ± i σ · ˆ n φ 2 ￿￿ = cos ￿ φ 2 ￿ ± i sin ￿ φ 2 ￿ and multiplying out determinants makes it clear that det( σ · a ￿ ) = det( σ · a ). Similarly, use (3.2.44) to explicitly write out the matrix σ · a ￿ and equate the elements to those of σ · a . With ˆ n in the z -direction, it is clear that we have just performed a rotation (of the spin vector) through the angle φ . 4. (a) Tr ( XY ) ≡ ￿ a ￿ a | XY | a ￿ = ￿ a ￿ b ￿ a | X | b ￿￿ b | Y | a ￿ by inserting the identity operator. Then commute and reverse, so Tr ( XY ) = ￿ b ￿ a ￿ b | Y | a ￿￿ a | X | b ￿ = ￿ b ￿ b | Y X | b ￿ = Tr ( Y 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[ if ( A )] = ￿ a exp[ if ( A )] | a ￿￿ a | = ￿ a exp[ if ( a )] | a ￿￿ a | (d) ￿ a ψ ∗ a ( x ￿ ) ψ a ( x ￿￿ ) = ￿ a ￿ x ￿ | a ￿ ∗ ￿ x ￿￿ | a ￿ = ￿ a ￿ x ￿￿ | a ￿￿ a | x ￿ ￿ = ￿ x ￿￿ | x ￿ ￿ = δ ( x ￿￿ − 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-1/2 in the | ± z ￿ basis, ￿ + | S z = ¯ h/ 2 ￿ = 1, ￿− | S z = ¯ h/ 2 ￿ = 0, and, using (1.4.17a), ￿ ±| S x = ¯ h/ 2 ￿ = 1 / √ 2. Therefore | S z = ¯ h/ 2 ￿￿ S x = ¯ h/ 2 | . = 1 √ 2 ￿ 1 1 0 0 ￿ 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.

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    3 7. 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 − a ￿ ) | α ￿ = ￿ a ￿ ( A − a ￿ ) ￿ i | a i ￿￿ a i | α ￿ = ￿ i ￿ a ￿ ( a i − a ￿ ) | a i ￿￿ a i | α ￿ = ￿ i 0 = 0 If the product instead is over all a ￿ ￿ = a j then the only surviving term in the sum is ￿ a ￿ ( a j − a ￿ ) | a i ￿￿ a i | α ￿ 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 − a ￿ ) = ￿ S z − ¯ h 2 ￿ ￿ S z + ¯ h 2 ￿ and ￿ a ￿ ￿ = a ￿￿ A − a ￿ a ￿￿ − a ￿ = S z + ¯ h/ 2 ¯ h for a ￿￿ = + ¯ h 2 or = S z − ¯ h/ 2 − ¯ h for a ￿￿ = − ¯ h 2 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 (1.3.35) and (1.3.36), for example S z + ¯ h/ 2 ¯ h = 1 ¯ h ￿ S z + ¯ h 2 1 ￿ = 1 2 [( | + ￿￿ + | ) − ( | −￿￿− | ) + ( | + ￿￿ + | ) + ( | −￿￿− | )] = | + ￿￿ + | which is just the projection operator for the state | + ￿ . 8. 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 ￿ + | + ￿ = 1 = ￿− | −￿ and ￿ + | −￿ = 0 = ￿− | + ￿ 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 ˆ n = ˆ i cos α sin β + ˆ j sin α sin β + ˆ k cos β so we need to find the matrix representation of the operator S · ˆ n = 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 (1.3.19) and the operators represented as outer products in (1.4.18) and (1.3.36), along with the association (1.3.39a) to define which element is which. Thus S x . = ¯ h 2 ￿ 0 1 1 0 ￿ S y . = ¯ h 2 ￿ 0 − i i 0 ￿ S z . = ¯ h 2 ￿ 1 0 0 − 1 ￿ We therefore need to find the (normalized) eigenvector for the matrix ￿ cos β cos α sin β − i sin α sin β cos α sin β + i sin α sin β − cos β ￿ = ￿ cos β e − i α sin β e i α sin β − cos β ￿

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    4 with eigenvalue +1. If the upper and lower elements of the eigenvector are a and b , respec- tively, then we have the equations | a | 2 + | b | 2 = 1 and a cos β + be − i α sin β = a ae i α 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 2 (1 − cos β ) 2 = | b | 2 sin 2 β = (1 − a 2 ) sin 2 β 4 a 2 sin 4 ( β / 2) = (1 − a 2 )4 sin 2 ( β / 2) cos 2 ( β / 2) a 2 [sin 2 ( β / 2) + cos 2 ( β / 2)] = cos 2 ( β / 2) a = cos( β / 2) and so b = ae i α 1 − cos β sin β = cos( β / 2) e i α 2 sin 2 ( β / 2) 2 sin( β / 2) cos( β / 2) = e i α sin( β / 2) which agrees with the answer given in the problem. 10. 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 2 = − 2 a 2 + E 2 = 0 or E = ± a √ 2. Let x 1 and x 2 be the two elements of the eigenvector. For E = + a √ 2 ≡ E (1) , (1 − √ 2) x (1) 1 + x (1) 2 = 0, and for E = − a √ 2 ≡ E (2) , (1 + √ 2) x (2) 1 + x (2) 2 = 0. So the eigenstates are represented by | E (1) ￿ . = N (1) ￿ 1 √ 2 − 1 ￿ and | E (2) ￿ . = N (2) ￿ − 1 √ 2 + 1 ￿ where N (1) 2 = 1 / (4 − 2 √ 2) and N (2) 2 = 1 / (4 + 2 √ 2). 11. It is of course possible to solve this using simple matrix techniques. For example, the characteristic equation and eigenvalues are 0 = ( H 11 − λ )( H 22 − λ ) − H 2 12 λ = H 11 + H 22 2 ± ￿ ￿ H 11 − H 22 2 ￿ 2 + H 2 12 ￿ 1 / 2 ≡ 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 1 + B σ z + C σ x

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    5 where A ≡ ( H 11 + H 22 ) / 2, B ≡ ( H 11 − H 22 ) / 2, and C ≡ H 12 . The eigenvalues of the first term are both A , and the eigenvalues for the sum of the second and third terms are those of ± (2 / ¯ h ) times a spin vector multiplied by √ B 2 + C 2 . In other words, the eigenvalues of the full Hamiltonian are just A ± √ B 2 + C 2 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 2 | 1 ￿ + sin β 2 | 2 ￿ and | λ − ￿ = − sin β 2 | 1 ￿ + cos β 2 | 2 ￿ where α = 0, tan β = C/B = 2 H 12 / ( H 11 − H 22 ), and we do β → π − β to “flip the spin.” 12. Using the result of Problem 9, the probability of measuring +¯ h/ 2 is ￿ ￿ ￿ ￿ ￿ 1 √ 2 ￿ + | + 1 √ 2 ￿− | ￿ ￿ cos γ 2 | + ￿ + sin γ 2 | −￿ ￿ ￿ ￿ ￿ ￿ 2 = 1 2 ￿￿ 1 + cos γ 2 + ￿ 1 − cos γ 2 ￿ 2 = 1 + sin γ 2 The results for γ = 0 (i.e. | + ￿ ), γ = π / 2 (i.e. | S x + ￿ ), and γ = π (i.e. | −￿ ) are 1/2, 1, and 1/2, as expected. Now ￿ ( S x − ￿ S x ￿ ) 2 ￿ = ￿ S 2 x ￿ − ￿ S x ￿ 2 , but S 2 x = ¯ h 2 / 4 from Problem 8 and ￿ S x ￿ = ￿ cos γ 2 ￿ + | + sin γ 2 ￿− | ￿ ¯ h 2 [ | + ￿￿− | + | −￿￿ + | ] ￿ cos γ 2 | + ￿ + sin γ 2 | −￿ ￿ = ¯ h 2 ￿ cos γ 2 ￿− | + sin γ 2 ￿ + | ￿ ￿ cos γ 2 | + ￿ + sin γ 2 | −￿ ￿ = ¯ h cos γ 2 sin γ 2 = ¯ h 2 sin γ so ￿ ( S x − ￿ S x ￿ ) 2 ￿ = ¯ h 2 (1 − sin 2 γ ) / 4 = ¯ h 2 cos 2 γ / 4 = ¯ h 2 / 4 , 0 , ¯ h 2 4 for γ = 0 , π / 2 , π . 13. 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 2 | + ￿ + sin β 2 | −￿ ￿ ￿ cos β 2 ￿ + | + sin β 2 ￿− | ￿ and the third apparatus projects out | −￿ . Therefore, the probability of measuring − ¯ h/ 2 after the third apparatus is P ( β ) = | ￿ + | S n + ￿￿ S n + | −￿ | 2 = cos 2 β 2 sin 2 β 2 = 1 4 sin 2 β The maximum transmission is for β = 90 ◦ , when 25% of the atoms make it through. 14. The characteristic equation is − λ 3 − 2( − λ )(1 / √ 2) 2 = λ (1 − λ 2 ) = 0 so the eigenvalues are λ = 0 , ± 1 and there is no degeneracy. The eigenvectors corresponding to these are 1 √ 2   − 1 0 1   1 2   1 √ 2 1   1 2   1 − √ 2 1   The matrix algebra is not hard, but I did this with matlab using

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    6 M=[[0 1 0];[1 0 1];[0 1 0]]/sqrt(2) [V,D]=eig(M) 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. 15. The answer is yes . The identity operator is 1 = ￿ a ￿ ,b ￿ | a ￿ , b ￿ ￿￿ a ￿ , b ￿ | so AB = AB 1 = 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. 16. 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 = 0 or b = 0 or both. 17. 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 1 ] = 0, we must have HA 1 | n ￿ = A 1 H | n ￿ = E n A 1 | n ￿ . That is, A 1 | n ￿ is an eigenstate of energy with eigenvalue E n . Since H and A 1 commute, though, they may have simultaneous eigenstates. Therefore, A 1 | n ￿ = a 1 | n ￿ since there is only one energy eigenstate. Similarly, A 2 | n ￿ is also an eigenstate of energy with eigenvalue E n , and A 2 | n ￿ = a 2 | n ￿ . But A 1 A 2 | n ￿ = a 2 A 1 | n ￿ = a 2 a 1 | n ￿ and A 2 A 1 | n ￿ = a 1 a 2 | n ￿ , where a 1 and a 2 are real numbers. This cannot be true, in general, if A 1 A 2 ￿ = A 2 A 1 so our assumption of “no degeneracy” must be wrong. There is an out, though, if a 1 = 0 or a 2 = 0, since one operator acts on zero. The example given is from a “central forces” Hamiltonian. (See Chapter Three.) The Hamil- tonian commutes with the orbital angular momentum operators L x and L y , but [ L x , L y ] ￿ = 0. 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. 18. The positivity postulate says that ￿ γ | γ ￿ ≥ 0, 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 ￿ | 2 , from which one derives the generalized uncertainty relation (1.4.53), namely ￿ ( ∆ A ) 2 ( ∆ B ) 2 ￿ ≥ 1 4 | ￿ [ A, B ] ￿ | 2 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 ￿ = − 2 λ ￿ α | ( ∆ B ) 2 | α ￿

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    7 and the equality is clearly satisfied in (1.4.53). 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 ￿ = ¯ h i d dx ￿ ￿ x ￿ | α ￿ − ￿ p ￿￿ x ￿ | α ￿ with ￿ x ￿ | α ￿ = (2 π d 2 ) − 1 / 4 exp ￿ i ￿ p ￿ x ￿ ¯ h − ( x ￿ − ￿ x ￿ ) 2 4 d 2 ￿ to get ¯ h i d dx ￿ ￿ x ￿ | α ￿ = ￿ ￿ p ￿ − ¯ h i 1 2 d 2 ( x ￿ − ￿ x ￿ ) ￿ ￿ x ￿ | α ￿ and so ￿ x ￿ | ∆ p | α ￿ = i ¯ h 2 d 2 ( 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 (2.3.24) and (2.3.25). Clearly ￿ x ￿ = 0 = ￿ p ￿ for any eigenstate | n ￿ , and x | 0 ￿ is proportional to p | 0 ￿ , with a proportionality constant that is purely imaginary. 19. Note the obvious typographical error, i.e. S x 2 should be S 2 x . Have S 2 x = ¯ h 2 / 4 = S 2 y = S 2 z , also [ S x , S y ] = i ¯ hS z , all from Problem 8. Now ￿ S x ￿ = ￿ S y ￿ = 0 for the | + ￿ state. Then ￿ ( ∆ S x ) 2 ￿ = ¯ h 2 / 4 = ￿ ( ∆ S y ) 2 ￿ , and ￿ ( ∆ S x ) 2 ￿ ( ∆ S y ) 2 = ¯ h 4 / 16. Also | ￿ [ S x , S y ] ￿ | 2 / 4 = ¯ h 2 | ￿ S z ￿ | 2 / 4 = ¯ h 4 / 16 and the generalized uncertainty principle is satisfied by the equality. On the other hand, for the | S x + ￿ state, ￿ ( ∆ S x ) 2 ￿ = 0 and ￿ S z ￿ = 0, and again the generalized uncertainty principle is satisfied with an equality. 20. Refer to Problems 8 and 9. Parameterize the state as | ￿ = cos β 2 | + ￿ + e i α sin β 2 | −￿ , so ￿ S x ￿ = ¯ h 2 ￿ cos β 2 ￿ + | + e − i α sin β 2 ￿− | ￿ [ | + ￿￿− | + | −￿￿ + | ] ￿ cos β 2 | + ￿ + e i α sin β 2 | −￿ ￿ = ¯ h 2 sin β 2 cos β 2 ( e i α + e − i α ) = ¯ h 2 sin β cos α ￿ ( ∆ S x ) 2 ￿ = ￿ S 2 x ￿ − ￿ S x ￿ 2 = ¯ h 2 4 (1 − sin 2 β cos 2 α ) (see prob 12) ￿ S y ￿ = i ¯ h 2 ￿ cos β 2 ￿ + | + e − i α sin β 2 ￿− | ￿ [ − | + ￿￿− | + | −￿￿ + | ] ￿ cos β 2 | + ￿ + e i α sin β 2 | −￿ ￿ = i ¯ h 2 sin β 2 cos β 2 ( e i α − e − i α ) = − ¯ h 2 sin β sin α ￿ ( ∆ S y ) 2 ￿ = ￿ S 2 y ￿ − ￿ S y ￿ 2 = ¯ h 2 4 (1 − sin 2 β sin 2 α )

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    8 Therefore, the left side of the uncertainty relation is ￿ ( ∆ S x ) 2 ￿￿ ( ∆ S y ) 2 ￿ = ¯ h 4 16 (1 − sin 2 β cos 2 α )(1 − sin 2 β sin 2 α ) = ¯ h 4 16 ￿ 1 − sin 2 β + 1 4 sin 4 β sin 2 2 α ￿ = ¯ h 4 16 ￿ cos 2 β + 1 4 sin 4 β sin 2 2 α ￿ ≡ P ( α , β ) which is clearly maximized when sin 2 α = ± 1 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 ∝ − 2 cos β sin β + sin 3 β cos β (1) = cos β sin β ( − 2 + sin 2 β ) = 0 or sin β = ± 1 / √ 2, that is, 45 ◦ with respect to the z -axis. It all hangs together. The maximum uncertainty product is ￿ ( ∆ S x ) 2 ￿￿ ( ∆ S y ) 2 ￿ = ¯ h 4 16 ￿ 1 2 + 1 4 1 4 ￿ = 9 256 ¯ h 4 The right side of the uncertainty relation is | ￿ [ S x , S y ] ￿ | 2 / 4 = ¯ h 2 | ￿ S z ￿ | 2 / 4, so we also need ￿ S z ￿ = ¯ h 2 ￿ cos 2 β 2 − sin 2 β 2 ￿ = ¯ h 2 cos β so the value of the right hand side at maximum is ¯ h 2 4 | ￿ S z ￿ | 2 = ¯ h 2 4 ¯ h 2 4 1 2 = 8 256 ¯ h 4 and the uncertainty principle is indeed satisfied. 21. The wave function is ￿ x | n ￿ = ￿ 2 /a sin( n π x/a ) for n = 1 , 2 , 3 , . . . , so we calculate ￿ x | x | n ￿ = ￿ a 0 ￿ n | x ￿ x ￿ x | n ￿ dx = a 2 ￿ x | x 2 | n ￿ = ￿ a 0 ￿ n | x ￿ x 2 ￿ x | n ￿ dx = a 2 6 ￿ − 3 n 2 π 2 + 2 ￿ ( ∆ x ) 2 = a 2 6 ￿ − 3 n 2 π 2 + 2 − 6 4 ￿ = a 2 6 ￿ − 3 n 2 π 2 + 1 2 ￿ ￿ x | p | n ￿ = ￿ a 0 ￿ n | x ￿ ¯ h i d dx ￿ x | n ￿ dx = 0 ￿ x | p 2 | n ￿ = − ¯ h 2 ￿ a 0 ￿ n | x ￿ d 2 dx 2 ￿ x | n ￿ dx = n 2 π 2 ¯ h 2 a 2 = ( ∆ p ) 2

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    9 (I did these with maple .) Since [ x, p ] = i ¯ h , we compare ( ∆ x ) 2 ( ∆ p ) 2 to ¯ h 2 / 4 with ( ∆ x ) 2 ( ∆ p ) 2 = ¯ h 2 6 ￿ − 3 + n 2 π 2 2 ￿ = ¯ h 2 4 ￿ n 2 π 2 3 − 2 ￿ which shows that the uncertainty principle is satisfied, since n π 2 / 3 > n π > 3 for all n . 22. We’re looking for a “rough order of magnitude” estimate, so go crazy with the approx- imations. 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 2 and the torque is mgL sin θ where θ is the angle from the vertical. So mL 2 ¨ θ = mgL sin θ or ¨ θ = ￿ g/L sin θ . Since θ ￿ 0 as the pick starts to fall, take sin θ = θ so θ ( t ) = A exp ￿￿ g L t ￿ + B exp ￿ − ￿ g L t ￿ x 0 ≡ θ (0) L = ( A + B ) L p 0 ≡ m ˙ θ (0) L = m ￿ g L ( A − B ) L = ￿ m 2 gL ( A − B ) Let the uncertainty principle relate x 0 and p 0 , i.e. x 0 p 0 = ￿ m 2 gL 3 ( A 2 − B 2 ) = ¯ h . 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 1 ◦ = π / 180, so solve for a time T where θ ( T ) = π / 180. Then T = ￿ L g ln π / 180 A = ￿ L g ￿ 1 4 ln m 2 gL 3 ¯ h 2 − ln 180 π ￿ Take L = 10 cm, so ￿ L/g ≈ 0 . 1 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 ln(180 / 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 = 0 . 1 kg, find m 2 gL 3 / ¯ h 2 = 10 64 , and so T = 0 . 1 sec × (147 / 4 − 4) ∼ 3 sec. I’d say that’s a surprising and interesting result. 23. The eigenvalues of A are obviously ± a , with − a twice. The characteristic equation for B is ( b − λ )( − λ ) 2 − ( b − λ )( ib )( − ib ) = ( b − λ )( λ 2 − b 2 ) = 0, so its eigenvalues are ± b with b twice. (Yes, B has degenerate eigenvalues.) It is easy enough to show that AB =   ab 0 0 0 0 iab 0 − iab 0   = BA so A and B commute, and therefore must have simultaneous eigenvectors. To find these, write the eigenvector components as u i , i = 1 , 2 , 3. Clearly, the basis states | 1 ￿ , | 2 ￿ , and | 3 ￿ are eigenvectors of A with eigenvalues a , − a , and − a respectively. So, do the math to find

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    10 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 | 1 ￿ ≡ | a, b ￿ , so just work with the reduced 2 × 2 basis of states | 2 ￿ 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 2 and u 3 . Then, for eigenvalue + b , − ibu 3 = + bu 2 and ibu 2 = + bu 3 both of which imply u 3 = iu 2 . For eigenvalue − b , − ibu 3 = − bu 2 and ibu 2 = − bu 3 both of which imply u 3 = − iu 2 . Choosing u 2 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 | 1 ￿ − a b 1 √ 2 ( | 2 ￿ + i | 3 ￿ ) − a − b 1 √ 2 ( | 2 ￿ − i | 3 ￿ ) 24. 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 2 and 3. Operating on the spinor representation of | + ￿ with (1 √ 2)(1 + i σ x ) gives 1 √ 2 ￿￿ 1 0 0 1 ￿ + i ￿ 0 1 1 0 ￿￿ ￿ 1 0 ￿ = 1 √ 2 ￿ 1 i i 1 ￿ ￿ 1 0 ￿ = 1 √ 2 ￿ 1 i ￿ So, for an operator U such that U . = (1 √ 2)(1 + i σ x ), we observe that U | + ￿ = | S y ; + ￿ , defined in (1.4.17b). Similarly operating on the spinor representation of | −￿ gives 1 √ 2 ￿￿ 1 0 0 1 ￿ + i ￿ 0 1 1 0 ￿￿ ￿ 0 1 ￿ = 1 √ 2 ￿ 1 i i 1 ￿ ￿ 0 1 ￿ = 1 √ 2 ￿ i 1 ￿ = i √ 2 ￿ 1 − i ￿ that is, U | −￿ = i | S y ; −￿ . This is what we would mean by a “rotation” about the x -axis by 90 ◦ . The sense of the rotation is about the + x direction vector, so this would actually be a rotation of − π / 2. (See the diagram following Problem Nine.) The phase factor i = e i π / 2 does not a ff ect this conclusions, and in fact leads to observable quantum mechanical e ff ects. (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 | −￿

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    11 Note that σ † x = σ x and σ 2 x = 1, so U † U . = (1 √ 2)(1 − i σ x )(1 √ 2)(1 + i σ x ) = (1 / 2)(1 + σ 2 x ) = 1 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 . = 1 √ 2 (1 − i σ x ) ¯ h 2 σ z 1 √ 2 (1 + i σ x ) = ¯ h 2 1 2 (1 − i σ x ) 2 σ z = − i ¯ h 2 σ x σ z = − i ¯ h 2 ￿ 0 1 1 0 ￿ ￿ 1 0 0 − 1 ￿ = ¯ h 2 ￿ 0 i − i 0 ￿ so S z . = ¯ h 2 ￿ 0 1 1 0 ￿ = ¯ h 2 σ x in the | S y ; ± ￿ basis. This can be easily checked directly with (1.4.17b), that is S z | S y ; ± ￿ = ¯ h 2 1 √ 2 [ | + ￿ ∓ i | −￿ = ¯ h 2 | S y ; ∓￿ There seems to be a mistake in the old solution manual, finding S z = (¯ h/ 2) σ y instead of σ x . 25. 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 24 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 ￿ = 1 2 π ¯ h ￿ x ￿ e i ( p ￿￿ − p ￿ ) x ￿ / ¯ h dx ￿ and then define q ≡ p ￿￿ − p ￿ and y ≡ x ￿ / ¯ h , then ￿ p ￿ | x | p ￿￿ ￿ = ¯ h 2 π i d dq ￿ e iqy dy = ¯ h i d dq δ ( q ) so you can also see that although x is real in the | x ￿ ￿ basis, it is not so in the | p ￿ ￿ basis. 26. From (1.4.17a), | S x ; ± ￿ = ( | + ￿ ± | −￿ ) / √ 2, so clearly U . = 1 √ 2 ￿ 1 1 1 − 1 ￿ = ￿ 1 / √ 2 1 / √ 2 0 0 ￿ + ￿ 0 0 1 / √ 2 − 1 / √ 2 ￿ = ￿ 1 / √ 2 1 / √ 2 ￿ [1 0] + ￿ 1 / √ 2 − 1 / √ 2 ￿ [0 1] = ⇒ = | S x : + ￿￿ + | + | S x : −￿￿− | . = ￿ r | b ( r ) ￿￿ a ( r ) |

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    12 27. The idea here is simple. Just insert a complete set of states. Firstly, ￿ b ￿￿ | f ( A ) | b ￿ ￿ = ￿ a ￿ ￿ b ￿￿ | f ( A ) | a ￿ ￿￿ a ￿ | b ￿ ￿ = ￿ a ￿ f ( a ￿ ) ￿ 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 ￿ = 1 (2 π ¯ h ) 3 ￿ F ( r ￿ ) e i ( p ￿ − p ￿￿ ) · x ￿ / ¯ h d 3 x ￿ The angular parts of the integral can be done explicitly. Let q ≡ p ￿ − p ￿￿ define the “ z ”- direction. Then ￿ p ￿￿ | F ( r ) | p ￿ ￿ = 2 π (2 π ¯ h ) 3 ￿ dr ￿ F ( r ￿ ) ￿ π 0 sin θ d θ e iqr ￿ cos θ / ¯ h = 1 4 π 2 ¯ h 3 ￿ dr ￿ F ( r ￿ ) ￿ 1 − 1 dμ e iqr ￿ μ/ ¯ h = 1 4 π 2 ¯ h 3 ￿ dr ￿ F ( r ￿ ) ¯ h iqr ￿ 2 i sin( qr ￿ / ¯ h ) = 1 2 π 2 ¯ h 2 ￿ dr ￿ F ( r ￿ ) sin( qr ￿ / ¯ h ) qr ￿ 28. For functions f ( q, p ) and g ( q, p ), where q and p are conjugate position and momentum, respectively, the Poisson bracket from classical physics is [ f, g ] classical = ∂ f ∂ q ∂ g ∂ p − ∂ f ∂ p ∂ g ∂ q so [ x, F ( p x )] classical = ∂ F ∂ p x Using (1.6.47), then, we have ￿ x, exp ￿ ip x a ¯ h ￿￿ = i ¯ h ￿ x, exp ￿ ip x a ¯ h ￿￿ classical = i ¯ h ∂ ∂ p x exp ￿ ip x a ¯ h ￿ = − a exp ￿ ip x a ¯ h ￿ To show that exp( ip x a/ ¯ h ) | x ￿ ￿ is an eigenstate of position, act on it with x . So x exp ￿ ip x a ¯ h ￿ | x ￿ ￿ = ￿ exp ￿ ip x a ¯ h ￿ x − a exp ￿ ip x a ¯ h ￿￿ | x ￿ ￿ = ( x ￿ − a ) exp ￿ ip x a ¯ h ￿ | x ￿ ￿ In other words, exp( ip x a/ ¯ h ) | x ￿ ￿ is an eigenstate of x with eigenvalue x ￿ − a . That is exp( ip x a/ ¯ h ) | x ￿ ￿ is the translation operator with ∆ x ￿ = − a , but we knew that. See (1.6.36). 29. I wouldn’t say this is “easily derived”, but it is straightforward. Expressing G ( p ) as a power series means G ( p ) = ￿ nm ￿ a nm ￿ p n i p m j p ￿ k . Now [ x i , p n i ] = x i p i p n − 1 i − p n i x i = i ¯ hp n − 1 i + p i x i p n − 1 i − p n i x i = 2 i ¯ hp n − 1 i + p 2 i x i p n − 2 i − p n i x i . . . = ni ¯ hp n − 1 i so [ x i , G ( p )] = i ¯ h ∂ G ∂ p i

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    13 The procedure is essentially identical to prove that [ p i , F ( x )] = − i ¯ h ∂ F/ ∂ x i . As for [ x 2 , p 2 ] = x 2 p 2 − p 2 x 2 = x 2 p 2 − xp 2 x + xp 2 x − p 2 x 2 = x [ x, p 2 ] + [ x, p 2 ] x make use of [ x, p 2 ] = i ¯ h ∂ ( p 2 ) / ∂ p = 2 i ¯ hp so that [ x 2 , p 2 ] = 2 i ¯ h ( xp + px ). The classical Poisson bracket is [ x 2 , p 2 ] classical = (2 x )(2 p ) − 0 = 4 xp and so [ x 2 , p 2 ] = i ¯ h [ x 2 , p 2 ] classical when we let the (classical quantities) x and p commute. 30. This is very similar to problem 28. Using problem 29, [ x i , J ( l )] = ￿ x i , exp ￿ − i p · l ¯ h ￿￿ = i ¯ h ∂ ∂ p i exp ￿ − i p · l ¯ h ￿ = l i exp ￿ − i p · l ¯ h ￿ = l i J ( l ) We can use this result to calculate the expectation value of x i . First note that J † ( l ) [ x i , J ( l )] = J † ( l ) x i J ( l ) − J † ( l ) J ( l ) x i = J † ( l ) x i J ( l ) − x i = J † ( l ) l i J ( l ) = l i Therefore, under translation, ￿ x i ￿ = ￿ α | x i | α ￿ → ￿ α |J † ( l ) x i J ( l ) | α ￿ = ￿ α |J † ( l ) x i J ( l ) | α ￿ = ￿ α | ( x i + l i ) | α ￿ = ￿ x i ￿ + l i which is exactly what you expect from a translation operator. 31. This is a continued rehash of the last few problems. Since [ x , J ( dx ￿ )] = dx ￿ by (1.6.25), and since J † [ x , J ] = J † x J − x , we have J † ( dx ￿ ) x J ( dx ￿ ) = x + J † ( dx ￿ ) dx ￿ = x + dx ￿ since we only keep the lowest order in dx ￿ . Therefore ￿ x ￿ → ￿ x ￿ + dx ￿ . Similarly, from (1.6.45), [ p , J ( dx ￿ )] = 0, so J † [ p , J ] = J † p J − p = 0. That is J † p J = p and ￿ p ￿ → ￿ p ￿ . 32. 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 2 ) dx = ￿ π /a , and ￿ x 2 exp( − ax 2 ) dx = − ( d/da ) ￿ exp( − ax 2 ) dx = √ π / 2 a 3 / 2 . So, for the x -space case, ￿ p ￿ = ￿ ￿ α | x ￿ ￿￿ x ￿ | p | α ￿ dx ￿ = ￿ ￿ α | x ￿ ￿ ¯ h i d dx ￿ ￿ x ￿ | α ￿ dx ￿ = 1 d √ π ￿ ¯ hk exp ￿ − x ￿ 2 d 2 ￿ dx ￿ = ¯ hk ￿ p 2 ￿ = − ¯ h 2 ￿ ￿ α | x ￿ ￿ d 2 dx ￿ 2 ￿ x ￿ | α ￿ dx ￿ = − ¯ h 2 d √ π ￿ exp ￿ − ikx ￿ − x ￿ 2 2 d 2 ￿ d dx ￿ ￿ ￿ ik − x ￿ d 2 ￿ exp ￿ ikx ￿ − x ￿ 2 2 d 2 ￿￿ dx ￿ = − ¯ h 2 d √ π ￿ ￿ − 1 d 2 + ￿ ik − x ￿ d 2 ￿ 2 ￿ exp ￿ − x ￿ 2 d 2 ￿ dx ￿ = ¯ h 2 ￿ 1 d 2 + k 2 ￿ − ¯ h 2 d 5 √ π ￿ x ￿ 2 exp ￿ − x ￿ 2 d 2 ￿ dx ￿ = ¯ h 2 ￿ 1 d 2 + k 2 ￿ − ¯ h 2 2 d 2 = ¯ h 2 2 d 2 + ¯ h 2 k 2

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    14 Using instead the momentum space wave function (1.7.42), we have ￿ p ￿ = ￿ ￿ α | p | p ￿ ￿￿ p ￿ | α ￿ dp ￿ = ￿ p ￿ | ￿ p ￿ | α ￿ | 2 dp ￿ = d ¯ h √ π ￿ p ￿ exp ￿ − ( p ￿ − ¯ hk ) 2 d 2 ¯ h 2 ￿ dp ￿ = d ¯ h √ π ￿ ( q + ¯ hk ) exp ￿ − q 2 d 2 ¯ h 2 ￿ dq = ¯ hk ￿ p 2 ￿ = d ¯ h √ π ￿ ( q + ¯ hk ) 2 exp ￿ − q 2 d 2 ¯ h 2 ￿ dq = d ¯ h √ π √ π 2 ¯ h 3 d 3 + (¯ hk ) 2 = ¯ h 2 2 d 2 + ¯ h 2 k 2 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 ￿ = 1 2 π ¯ h ￿ x ￿ exp ￿ − i ( p ￿ − p ￿￿ ) · x ￿ ¯ h ￿ dx ￿ = i ∂ ∂ p ￿ 1 2 π ￿ exp ￿ − i ( p ￿ − p ￿￿ ) · x ￿ ¯ h ￿ dx ￿ = i ¯ h ∂ ∂ p ￿ δ ( p ￿ − 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 ¯ h ∂ ∂ p ￿ ￿ δ ( p ￿ − p ￿￿ ) ￿ p ￿￿ | α ￿ dp ￿￿ = i ¯ h ∂ ∂ p ￿ ￿ p ￿ | α ￿ Given this, the next expression is simple to prove, namely ￿ β | x | α ￿ = ￿ dp ￿ ￿ β | p ￿ ￿￿ p ￿ | x | α ￿ = ￿ dp ￿ φ ∗ β ( p ￿ ) i ¯ h ∂ ∂ p ￿ φ α ( p ￿ ) using the standard definition φ γ ( p ￿ ) ≡ ￿ p ￿ | γ ￿ . Certainly the operator T ( Ξ ) ≡ exp( ix Ξ / ¯ h ) looks like a momentum translation operator. So, we should try to work out p T ( Ξ ) | p ￿ ￿ = p exp( ix Ξ / ¯ h ) | p ￿ ￿ and see if we get | p ￿ + Ξ ￿ . Take a lesson from problem 28, and make use of the result from problem 29, and we have p T ( Ξ ) | p ￿ ￿ = {T ( Ξ ) p + [ p, T ( Ξ )] }| p ￿ ￿ = ￿ p ￿ T ( Ξ ) − i ¯ h ∂ ∂ x T ( Ξ ) ￿ | p ￿ ￿ = ( p ￿ + Ξ ) T ( 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.)

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    15 Chapter Two 1. The equation of motion for an operator in the Heisenberg picture is given by (2.2.19), so ˙ S x = 1 i ¯ h [ S x , H ] = − 1 i ¯ h eB mc [ S x , S z ] = eB mc S y ˙ S y = − eB mc S x ˙ S z = 0 and ¨ S x,y = − ω 2 S x,y for ω ≡ eB/mc . Thus S x and S y are sinusoidal with frequency ω and S z is a constant. 2. The Hamiltonian is not Hermitian, so the time evolution operator will not be unitary, and probability will not be conserved as a state evolves in time. As suggested, set H 11 = H 22 = 0. Then H = a | 1 ￿￿ 2 | in which case H 2 = a 2 | 1 ￿￿ 2 | 1 ￿￿ 1 | = 0. Since H is time-independent, U ( t ) = exp ￿ − i ¯ h Ht ￿ = 1 − i ¯ h Ht = 1 − i ¯ h at | 1 ￿￿ 2 | even for finite times t . Thus a state | α , t ￿ ≡ U ( t ) | 2 ￿ = | 2 ￿ − ( iat/ ¯ h ) | 1 ￿ has a time-dependent norm. Indeed ￿ α | α ￿ = 1 + a 2 t 2 / ¯ h 2 which is nonsense. In words, it says that if you start out in the state | 2 ￿ , then the probability of finding the system in this state is unity at t = 0 and then grows with time. You can be more formal, and talk about an initial state c 1 | 1 ￿ + c 2 | 2 ￿ , but the bottom line is the same; probability is no longer conserved in time. 3. We have ˆ n = sin β ˆ x + cos β ˆ z and S . = (¯ h/ 2) σ , so S · n . = (¯ h/ 2)(sin βσ x + cos βσ z ) and we want to solve the matrix equation S · n ψ = (¯ h/ 2) ψ in order to find the initial state column vector ψ . This is, once again, a problem whose solution best makes use of the Pauli matrices, which are not introduced until Section 3.2. On the other hand, we can also make use of Problem 1.9 to write down the initial state. Either way, we find | α , t = 0 ￿ = cos ￿ β 2 ￿ | + ￿ + sin ￿ β 2 ￿ | −￿ so , | α , t ￿ = exp ￿ − i ¯ h eB mc tS z ￿ | α , t = 0 ￿ = e − i ω t/ 2 cos ￿ β 2 ￿ | + ￿ + e i ω t/ 2 sin ￿ β 2 ￿ | −￿ for ω ≡ eB/mc . From (1.4.17a), the state | S x ; + ￿ = (1 / √ 2) | + ￿ + (1 / √ 2) | −￿ , so | ￿ S x ; + | α , t ￿ | 2 = ￿ ￿ ￿ ￿ 1 √ 2 e − i ω t/ 2 cos ￿ β 2 ￿ + 1 √ 2 e i ω t/ 2 sin ￿ β 2 ￿￿ ￿ ￿ ￿ 2 = 1 2 cos 2 ￿ β 2 ￿ + 1 2 ￿ e i ω t + e − i ω t ￿ cos ￿ β 2 ￿ sin ￿ β 2 ￿ + 1 2 sin 2 ￿ β 2 ￿ = 1 2 + 1 2 cos( ω t ) sin β = 1 2 (1 + sin β cos ω t ) which makes sense. For β = 0, the initial state is a z -eigenket, and there is no precession, so you just get 1 / 2 for the probability of measuring S x in the positive direction. The same

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    16 works out for β = π . For β = π / 2, the initial state is | S x ; + ￿ so the probabilty is +1 at t = 0 and 0 at t = π / ω = T / 2. Now from (1.4.18a), S x = (¯ h/ 2)[ | + ￿￿− | + | −￿￿ + | ], so ￿ α , t | S x | α , t ￿ = ￿ e i ω t/ 2 cos ￿ β 2 ￿ ￿ + | + e − i ω t/ 2 sin ￿ β 2 ￿ ￿− | ￿ ¯ h 2 ￿ e i ω t/ 2 sin ￿ β 2 ￿ | + ￿ + e − i ω t/ 2 cos ￿ β 2 ￿ | −￿ ￿ = ¯ h 2 sin ￿ β 2 ￿ cos ￿ β 2 ￿ ￿ e i ω t + e − ω t ￿ = ¯ h 2 sin β cos ω t Again, this makes perfect sense. The expectation value is zero for β = 0 and β = π , but for β = π / 2, you get the classical precession of a vector that lies in the xy -plane. 4. First, restating equations from the textbook, | ν e ￿ = cos θ | ν 1 ￿ − sin θ | ν 2 ￿ | ν μ ￿ = sin θ | ν 1 ￿ + cos θ | ν 2 ￿ and E = pc ￿ 1 + m 2 c 2 2 p 2 ￿ Now, let the initial state | ν e ￿ evolve in time to become a state | α , t ￿ in the usual fashion | α , t ￿ = e − iHt/ ¯ h | ν e ￿ = cos θ e − iE 1 t/ ¯ h | ν 1 ￿ − sin θ e − iE 2 t/ ¯ h | ν 2 ￿ = e − ipct/ ¯ h ￿ e − im 2 1 c 3 t/ 2 p ¯ h cos θ | ν 1 ￿ − e − im 2 2 c 3 t/ 2 p ¯ h sin θ | ν 2 ￿ ￿ The probability that this state is observed to be a | ν e ￿ is P ( ν e → ν e ) = | ￿ ν e | α , t ￿ | 2 = ￿ ￿ ￿ e − im 2 1 c 3 t/ 2 p ¯ h cos 2 θ + e − im 2 2 c 3 t/ 2 p ¯ h sin 2 θ ￿ ￿ ￿ 2 = ￿ ￿ ￿ cos 2 θ + e i ∆ m 2 c 3 t/ 2 p ¯ h sin 2 θ ￿ ￿ ￿ 2 = cos 4 θ + sin 4 θ + 2 cos 2 θ sin 2 θ cos ￿ ∆ m 2 c 3 t 2 p ¯ h ￿ = 1 − sin 2 2 θ sin 2 ￿ ∆ m 2 c 3 t 4 p ¯ h ￿ Writing the nominal neutrino energy as E = pc and the flight distance L = ct we have P ( ν e → ν e ) = 1 − sin 2 2 θ sin 2 ￿ ∆ m 2 c 4 L 4 E ¯ hc ￿ It is quite customary to ignore the factor of c 4 and agree to measure mass in units of energy, typically eV.

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