1
The Physics of Quantum Mechanics Solutions to starred problems 3.11∗
By expressing the annihilation operator A of the harmonic oscillator in the momentum representation, obtain p 0 . Check Check that your expression expression agrees with that obtained obtained from the Fourier Fourier transform of 2 2 1 h ¯ x0 = e−x /4ℓ , where ℓ . (3. (3.1) 2 1 4 / 2mω (2πℓ (2πℓ ) representation x = i¯ h∂/∂p so h∂/∂p so [x, p] = i¯h∂p/∂p = h∂p/∂p = i¯ h. Thus from Problem Soln: In the momentum representation x 3.8 x ℓ ℓp h ¯ ∂ A = + i p = i + 2ℓ h ¯ h ¯ 2ℓ ∂p 2 2 2 ℓp h ¯ ∂u 0 0 = Au 0 u 0 = u0 ( p) p) e− p ℓ /h¯ h ¯ 2ℓ ∂p Alternatively, Alternatively, transforming u transforming u 0 (x):
|
|
≡
⇒
p|0 =
1 dx p x x 0 = h
−
∞
| | √ −
1 = 2 (2πℓ (2πℓ h2 )1/4
dx e
dx exp
−∞
∝
2
−∞
∞
⇒
−i px/h ¯
2
e−x /4ℓ (2πℓ (2πℓ2 )1/4
x i pℓ i pℓ + 2ℓ h ¯
2
e
− p2 ℓ2 /h ¯2
√
2 2 2 2ℓ π = e− p ℓ /h¯ / 2 2 1 4 (2πℓ (2πℓ h )
3.13∗
A Fermi oscillator has Hamiltonian H Hamiltonian H = f † f , f , where f is f is an operator that satisfies † † 2 f = 0, f f + f f = 1. (3. (3.2) 2 Show that H = H , and thus find the eigenvalues of H . If th the ket ket 0 satisfies H 0 = 0 with † 0 0 = 1, 1 , what are the kets (a) a f 0 , and (b) b f 0 ? In quantum field theory the vacuum is pictured as an assembly of oscillators, one for each possible value of the momentum of each particle type. A boson is an excitation of a harmonic oscillator, oscillator, while a fermion fermion in an excitation excitation of a Fermi Fermi oscillator. Explain Explain the connection connection between between the spectrum of f f † f and f and the Pauli principle.
|
| ≡ |
|
| ≡ |
|
Soln:
H 2 = f † f f † f = f † (1 f † f ) f )f = f † f = H Since eigenvalues have to satisfy any equations satisfied by their operators, the eigenvalues of H must satisfy λ2 = λ, which which restricts them to the numbers numbers 0 and 1. The Fermi Fermi exclusion exclusion principle principle says there can be no more than one particle in a single-particle state, so each such state is a Fermi oscillator that is either excited once or not at all.
−
Moreover,
this ket ket vani vanishe shess. ||a|2 = 0|f †f |0 = 0 so this † † 2 f )|0 = 1 so |b is more interesting. interesting. ||b| = 0|f f |0 = 0|(1 − f f ) H |b = f † f f † |0 = f † (1 − f † f ) f )|0 = f † |0 = |b
so b is the eigenket with eigenvalue 1.
|
3.15∗
P P is the probabil probabilit ity y that that at the end of the experimen experimentt descri described bed in Proble Problem m 3.1 3.14, 4, the 1 oscillator oscillator is in its second second excited state. state. Show that when f = 2 , P = 0.144 144 as follows. follows. First show show that the annihilation operator of the original oscillator A = 12 (f −1 + f ) f )A′ + (f (f −1 f ) f )A′† , (3. (3.3)
−
where A′ and A′† are the annihilation annihilation and creation creation operators of the final oscillator. oscillator. Then writing writing the ground-state ket of the original oscillator as a sum 0 = n cn n′ over the energy eigenkets of the final oscillator, show that the condition A condition A 0 = 0 yields the recurrence relation f −1 f n cn+1 = c n−1 . (3. (3.4) − 1 f + f n + 1
| − −
|
|
2 Finally using the normalisation of 0 , show numerically that c2 0. 0 .3795 3795.. What What value alue do you get get for the probability of the oscillator remaining in the ground state? Show that at the end of the experiment the expectation value of the energy is 0 is 0..2656¯hω. hω. Explain 1 physically why this is less than the original ground-state energy 2 hω. ¯hω. This example contains the physics behind the inflationary origin of the universe: gravity explosively enlarges the vacuum, which is an infinite collection of harmonic oscillators (Problem 3.13). Excitations Excitations of these oscillators oscillators correspond to elementar elementary y particles. particles. Before Before inflation inflation the vacuum is unexcited unexcited so every oscillator oscillator is in its ground state. state. At the end of inflation, there is non-negligible non-negligible probability of many oscillators being excited and each excitation implies the existence of a newly created particle. have Soln: From Problem 3.6 we have mf 2 ωx + i p p mωx + i p p A′ A 2mhf ¯hf 2 ω 2mhω ¯hω x iℓ i ℓ fx iℓ = + p = + p 2ℓ h ¯ 2ℓ f ¯ f h ¯ Hence f 2iℓ 2iℓ 1 f A′ + A′† = x A′ A′† = f p so A = (A′ + A′† ) + (A′ A′† ) ¯ 2f 2 ℓ f ¯ f h
| |
≃
≡
≡ √
−
0 = A 0 =
|
=
1 2
1 2
−
(f −1 + f ) f )ck A′ k′ + (f (f −1
k
−1
(f
|
√
+ f ) f ) kck k
k
Multiply through by n′ :
|
′
f )ck A′† |k ′ − f ) −1 − √ |
| − 1 + (f (f
f ) f ) k + 1c 1 ck k + 1
√
′
− √
0 = (f −1 + f ) f ) n + 1c 1cn+1 + (f ( f −1 f ) f ) ncn−1 , which is a recurrence relation from which all non-zero cn can be determined in terms of c0 . Put c0 = 1 and solve for the c the c n . Then evaluate S evaluate S cn 2 and renormalise: cn cn / S . 2 The probability of remaining in the ground state is c0 = 0. 0 .8. E = n cn 2 (n + 12 )¯ hf hf 2 ω . It is less than the original energy because of the chance that energy is in the spring when the stiffness is reduced.
≡ ≡ | |
| |
→ √ | |
3.16∗
In terms of the usual ladder operators A, A , A † , a Hamiltonian can be written H = µA† A + λ(A + A† ). (3. (3.5) What restrictions on the values of the numbers µ and λ follow from the requirement for H H to be Hermitian? Show that for a suitably chosen operator B operator B,, H can H can be rewritten † H = µB B + constant + constant,, (3. (3.6) † where [B, [ B, B ] = 1. Hence determine the spectrum of H of H .. real. Definin Definingg B = A + a + a with a a number, we have Soln: Hermiticity requires µ and λ to be real. † [B, B ] = 1 and H = µ( µ(B † a∗ )(B )(B a) + λ(B a + B † a∗ ) = µB † B + (λ µa∗ )B + (λ µa) µa)B † + ( a 2 µ λ(a + a∗ )). )). We dispose of the terms linear in B by setting a = λ/µ, real number. number. Then Then H = µB † B λ2 /µ. λ/µ , a real /µ. From the theory of the harmonic oscillator we know that the spectrum of B † B is 0, 1, . . ., ., so the spectrum of H H is nµ is nµ λ2 /µ. /µ.
−
3.17∗
−
−
−
−
−
−
|| −
−
Numericall Numerically y calculate calculate the spectrum spectrum of the anharmonic oscillator oscillator shown in Figure 3.2. From it estimate the p eriod at a sequence of energies. Compare Compare your quantum quantum results with the equivalen equivalentt classical results. Soln:
3 3.18∗
Let B = cA + cA + sA sA† , where c cosh θ, s usual ladder operators. Show that [ that [B, B, B † ] = 1. Consider the Hamiltonian
≡ sinh θ with θ a real constant and A, A† are the
≡
H = ǫA† A + 12 λ(A† A† + AA AA)),
(3. (3.7)
where ǫ ǫ and λ and λ are real and such that ǫ > λ > 0 > 0.. Show that when ǫc
λs = E c, − λs =
λc
ǫs = E Ess − ǫs =
(3. (3.8)
with E with E a constant, [ constant, [B, B, H ] = E B . Hence determine the spectrum of H of H in in terms of ǫ ǫ and a nd λ. λ . Soln:
[B, B † ] = [cA + sA† , cA† + sA] sA] = (c2
− s2)[A, )[A, A† ] = 1
[B, H ] = [cA + sA† , ǫA† A + 12 λ(A† A† + AA AA)] )] = c[ c[A,ǫA† A + 12 λA† A† ] + s[A† , ǫA† A + 12 λAA] λAA] = c( c (ǫA + λA† )
λA)) = cEA cE A + sEA † = EB E B − s(ǫA† + λA as required. Let H Let H |E 0 = E 0 |E 0 . Then multiplying through by B by B E 0 B |E 0 = BH B H |E 0 = (H B + [B, [ B, H ]) ])|E 0 = (H B + EB) EB )|E 0 So H So H ((B |E 0 ) = (E ( E 0 − E )(B )(B |E 0 ), which says the B |E 0 is an eigenket for eigenvalue E 0 − E . We assume that the sequence of eigenvalues E 0 , E 0 −E, E 0 −2E , . . . terminates because B |E min min = 0. Mod-squaring this equation we have
† † 0 = E min sA)(cA cA + + sA† ) E min min B B E min min = E min min (cA + sA)( min
| | | | 2 2 2 † = E min cs(A† A† + AA AA))}|E min min|{(c + s )A A + s + cs( min = cs E min s/c)A† A + s/c + (A (A† A† + AA AA))}|E min min |{(c/s + s/c) min
But eliminating E from E from the given equations, we find λ(c/s + c/s + s/c s/c)) = 2ǫ. Puttin Putting g this into the last equation 2ǫ † 0 = E min A A + s/c + (A (A† A† + AA AA)) E min min min λ
|
|
Multiplying through by λ/ by λ/2 2 this becomes 0 = E min + sλ/2 sλ/2c E min min H + min
so E so E min min =
|{
}|
sλ/2c. Finally, x Finally, x = s/c = s/c satisfies the quadratic −sλ/2 ǫ ǫ ǫ2 − 1. x2 − 2 x + 1 = 0 ⇒ x = ± λ λ λ2 Also from the above E above E = = ǫ − λx so λx so the general eigenenergy is E n = E = E min = − 12 λx + nǫ − nλx = nλx = nǫ nǫ − (n + 12 )λx = λx = nǫ nǫ − (n + 12 ) ǫ ± min + nE = = − 21 ǫ ∓ (n + 12 ) ǫ2 − λ2
−
ǫ2
λ2
We have to choose the plus sign in order to achieve consistency with our previously established value of E E min min ; thus finally E n = 12 ǫ + (n (n + 12 ) ǫ2 λ2
−
−
4.2∗
Show that the vector product a b of two classical vectors transforms like a vector under rotations. rotations. Hint: Hint: A rotation rotation matrix R satisfies the relations R RT = I and det( det(R) = 1, which in tensor notation read p Rip Rtp = δ it and ǫ R R R = ǫ . it rst ijk ijk ir js kt
×
·
4 Soln:
Let the rotated rotated vector vectorss be a′ = Ra and b′ = Rb . Then (a′ b′ )i = ǫijk Rjl al Rkm bm
×
jklm
=
δ it it ǫtjk Rjl Rkm al bm
tjklm
=
Rip Rtp ǫtjk Rjl Rkm al bm
ptjklm
=
Rip ǫ plm al bm = (Ra
plm
4.3∗
× b)i.
We have shown that [vi , J j ] = i k ǫijk vk for any operator whose components vi form a vecto vector. r. The expectati expectation on value value of this this operato operatorr relati relation on in any any state state ψ is then ψ [vi , J j ] ψ = i k ǫijk ψ vk ψ . Check that with U with U ((α) = e−i ·J this relation is consistent under a further rotation ψ ψ ′ = U ( U (α) ψ by evaluating both sides separately. the further further rotation the LHS ψ U † [vi , J j ]U ψ . Now Soln: Under the
| | | →| | †
†
†
→ |
[Rik vk , Rjl J l ] =
kl
Similar ψ
|
|
|
( U vi U )( U )(U U J j U ) U ) − (U † J j U )( U )(U U † vi U ) U ) − − U J j viU = (U
U [vi , J j ]U = U vi J j U =
| |
α
†
†
Rik Rjl [vk , J l ].
kl
| → U |ψ on the RHS yields i
Rkm ǫijk ψ vm ψ .
| |
km
We now multiply each side by Ris Rjt and sum over i and j . On the LHS this operati operation on yields yields [vs , J t ]. On the right it yields i
Ris Rjt Rkm ǫijk ψ vm ψ = i
| |
ijkm
ǫstm ψ vm ψ ,
| |
m
which is what our original equation would give for [v [vs , J t ]. 4.4∗
The matrix for rotating an ordinary vector by φ by φ around the z z -axis is cos φ sin φ 0 sin φ cos φ 0 . (4. (4.1) R(φ) 0 0 1 By considering the form taken by R φ calculate from R the matrix z that appears R for infinitesimal φ calculate in R (φ) = exp( i z φ). Introduce Introduce new coordinates coordinates u u1 ( x + iy )/ 2, u 2 = z and z and u u 3 (x + iy )/ 2. Write down the matrix M x,y,z ) ] and a nd show that it is unitary. M that appears in u = M x [where x x (x,y,z) Then show that ′ † (4. (4.2) M z M z is identical identical with S with S z in the set of spin-one Pauli analogues 0 1 0 0 i 0 1 0 0 1 1 S x = 1 0 1 , S y = i 0 i , S z = 0 0 0 . (4. (4.3) 2 2 0 1 0 0 i 0 0 0 1 Write down the matrix x whose exponential generates rotations around the x-axis, calculate calculate x′ by analogy with equation ( 4.2) 4.2) and check that your result agrees with S x in the set ( 4.3). 4.3). Explain Explain as fully as you can the meaning of these calculations. infinitesimal rotation angle δ angle δφ φ to first order in δ in δφ φ we have Soln: For an infinitesimal 1 δφ 0 1 i z δφ = δφ = R (δφ) δφ) = δφ 1 0 0 0 1
≡
− JJ
√
−
≡ −
· JJ ≡ · JJ ·
J JJ
− JJ
√
−
JJ ≡
√
≡
− −
−
√
J JJ
5 comparing coefficients of δφ δ φ we find
− JJ − √ √ − − √ − √ − − − √ − − JJ − − √ − − √ − √ √ − √ √ √ √ z
In components u = M x reads
·
u1 u2 u3
0 1 0
=i
1 0 0 0 0 0
1 i 0 0 1 i
1 = 2
0 2 0
x y z
so M is the matrix above. We show that M is unitary by calculating the product MM†. Now we have 1 i 0 0 i 0 1 0 1 ′ 1 0 0 2 i 0 0 i 0 i z = 2 1 i 0 0 0 0 0 2 0
JJ
= Similarly, we have
1 2
− −
1 i 0 0 1 i
0 2 0
x =
so ′ 1 x = 2
JJ
− −
1 i 0 0 1 i
0 2 0
1 0 i 0 0 0
1 i 0
0 0 0 0 0 1
0 1 0
i
0 0 0 0 0 i
0 i 0
=
1 i 0
2 0 0 0 0 0
1 2
0 0 2
0 0 2
1 i 0
1 i 0 0 0 0 0 2 0 1 = 0 0 2 0 i 2 0 =2 2 0 2 1 i 0 1 0 1 0 2 0 These results show that the only difference between the generators of rotations of ordinary 3d vectors and the spin-1 representations of the angular-momentum operators, is that for conventional vectors we use a different coordinate system than we do for spin-1 amplitudes. Apart from this, the three amplitudes for the spin of a spin-1 particle to point in various directions are equivalent to the components of a vector, and they transform among themselves when the particle is reoriented for the same reason that the rotation of a vector changes its Cartesian components. 1 2
4.6∗ Show that if α α and β are non-parallel vectors, α is not invariant under the combined rotation R(α)R(β). Hence show that RT (β)RT (α)R(β)R(α)
is not the identity operation. Explain the physical significance of this result. because a rotatio rotation n leave leavess its axis inv invarian ariant. t. But the only only vecto vectors rs that are R(α)α = α because invariant invariant under R (β) are multiples of the rotation axis β . So R (β)α is not parallel to α. If RT (β)RT (α)R(β)R(α) were the identity, we would have RT (β)RT (α)R(β)R(α)α = α R(β)R(α)α = R (α)R(β)α R(β)α = R (α)(R(β)α) which would imply that R (β)α is invariant under R (α). Consequently we would have R (β)α = α . But this is true only if α is parallel to β. So our original original hypothesis hypothesis that RT (β)RT (α)R(β)R(α) = I is wrong. wrong. This This demons demonstra trates tes that when you you rotate rotate about two non-paral non-parallel lel axes and then then do the reverse rotations in the same order, you always finish with a non-trivial rotation. Soln:
⇒
⇒
4.7∗
In this problem you derive the wavefunction (4. (4.4) x p = e ip·x/h¯ of a state of well-defined momentum from the properties of the translation operator U operator U ((a). The state momentum ¯hk. How would you characterise the state k′ k is one of well-defined momentum ¯ U ( U (a) k ? Show
|
|
| ≡
|
6
Figure 5.0 The real part of the wavefunction when a free particle of energy E is E is scattered by a classically forbidden
square square barrier barrier (top) (top) and a potential potential well well (bottom). The upper panel is for a barrier barrier of height height V 0 = E /0.7 and half-width a half-width a such that 2mEa 2mEa 2 /h ¯ 2 = 1. The lower panel is for a well of depth V 0 = E /0.2 and half-width a half-width a such that 2mEa 2 /h ¯ 2 = 9. In both panels (2mE/ (2mE/¯ h ¯ 2 )1/2 = 40.
Figure 5.1 A triangle for Prob-
lem 5.10
that the wavefunctions of these states are related by u by u k (x) = e−ia·k uk (x) and a nd uk (x) = u k (x a). Hence obtain equation ( 4.4). 4.4). U (a) k is the result of translating a state of well-defined momentum by k. Mo Movin ving g to the Soln: U ( position representation uk (x) = x U ( U (a) k = k U † (a) x ∗ = k x a ∗ = u k (x a) Also U (a) k = x e−ia·p/h¯ k = e−ia·k x k = e−ia·k uk (x) x U ( Putting these results together we have uk (x a) = e−ia·kuk (x). Settin Settingg a = x we find uk (x) = eik·x uk (0), as required. ′
−
′
|
′
| | | | | | | | −
| − |
−
5.13∗
This problem is about the coupling of ammonia molecules to electromagnetic waves in an ammonia maser. Let + be the state in which the N atom lies above the plane of the H atoms and be the state in which the N lies below the plane. Then when there is an oscillating electric field cos ωt directed ωt directed perpendicular to the plane of the hydrogen atoms, the Hamiltonian in the basis becomes E + + q s cos ωt A H = = . (5. (5.1) A E q s cos ωt
|− E
|
|±
E E − − − E E Transform this Hamiltonian from the |± to the basis provided by the states of well-defined |± basis √ parity | |e and | |o (where | |e = (|+ + |−)/ 2, etc). Writing (5. (5.2) |ψ = ae(t)e−iE t/h¯ |e + ao(t)e−iE t/h¯ |o, e
o
7 show that the equations of motion of the expansion coefficients are dae = iΩa iΩao (t) ei(ω−ω0)t + e−i(ω+ω0 )t dt (5. (5.3) dao −i(ω −ω0)t i(ω +ω0 )t = iΩa iΩae (t) e +e , dt where Ω q s/2¯ s/2¯h and ω0 = (E o E e )/h ¯ . Explai Explain n why in the case of a maser maser the exponenti exponentials als involving ω + ω + ω0 can be neglected so the equations of motion become dae dao = iΩa iΩao (t)ei(ω−ω0 )t , = iΩa iΩae (t)e−i(ω−ω0 )t . (5. (5.4) dt dt Solve the equations by multiplying the first equation by e−i(ω−ω0)t and differentiating the result. Explain how the solution describes the decay of a population of molecules that are initially all in the higher energy level. Compare your solution to the result of setting ω = ω = ω 0 in ( 5.4). 5.4). Soln: We have e H e = 12 ( + + ) H ( ( + + )
− − −
≡ E E
−
−
| |
| −| | |− = 12 ( +|H |+ + −|H |− |− + −|H |+ + +|H |− |−) = E − A = E = E e o|H |o = 12 (+|−−|) H ( (|+−|−) = 12 ( +|H |+ + −|H |−−−| | −−−|H |+ − +|H |− |−) = E + A + A = = E E o
o|H |e = e|H |o = 12 (+| + −|) H ( (|+−|−) |− + −|H |+ − +|H |− |−) = 12 ( +|H |+−−|H |− E s cos(ωt = q E cos(ωt)) Now we use the to calculate the evolution of | |ψ = aee−iE t/h¯ |e + aoe−iE t/h¯ |o: ∂ |ψ i¯h = i¯ ha˙ e e−iE t/h¯ |e + ae E e e−iE t/h¯ |e + i¯ha˙ o e−iE t/h¯ |o + ao E o e−iE t/h¯ |o ∂t = a e e−iE t/h¯ H |e + ao e−iE t/h¯ H |o We now multiply through by first e| and then o|. After After dividi dividing ng throug through h by some exponenti exponential al e
tdse
e
e
e
o
o
o
o
factors to simplify, we get
i¯ha˙ e + ae E e = a e e H e + ao ei(E e−E o )t/h¯ e H o
| |
| | i¯ha˙ o + ao E o = a e ei(E −E )t/h¯ o|H |e + ao o|H |o o
e
With the results derived above i¯ha˙ e + ae E e = ae E e + ao ei(E e −E o )t/h¯ q s cos(ωt cos(ωt))
E E
i¯ha˙ o + ao E o = ae ei(E o −E e )t/h¯ q s cos(ωt cos(ωt)) + ao E o
E E
After cancelling terms in each equation, we obtain the desired equations of motion on expressing the cosines in terms of exponentials and using the new notation. The exponential with frequency ω frequency ω + + ω0 oscillates so rapidly that it effectively averages to zero, so we can drop it. Multiplying the first eqn through by e−i(ω−ω0)t and differentiating gives d −i(ω−ω0)t e a˙ e = e −i(ω−ω0 )t [ i(ω i(ω ω0 )a˙ e + ¨ae ] = Ω2ae e−i(ω−ω0)t dt The exponentials cancel leaving a homogeneous second-order o.d.e. with constant coefficients. Since initially all molecules are in the higher-energy state o , we have to solve subject to the boundary
− −
|
−
8
Figure 5.2 The symbols show the
ratio of the probability of reflection to the probability of transmission when particles move from x = −∞ in the potential (5.69) with energy E = h ¯ 2 k2 /2m and V 0 = 0.7E . The The dotted line is the value obtained for a step change in the potential
condition a condition a e (0) = 0. With a With a 0 (0) = 1 we get from the original equations the second initial condition a˙ e (0) = iΩ. For trial solution a solution a e eαt the auxiliary eqn is
−
∝
α2
− i(ω i(ω − ω0 )α + Ω2 = 0 ⇒ α = 12 i(ω i(ω − ω0 ) ± −(ω − ω0 )2 − 4Ω2 = iω i ω± with ω ± = 12 (ω − ω0 ) ± (ω − ω0 )2 + 4Ω2 . When ω When ω ≃ ω0 , these frequencies both lie close to Ω. From the condition ae (0) = 0, the required solution is ae (t) ∝ (eiω t − eiω t ) and the constant of proportionalit proportionality y follows follows from the second second initial initial condition, condition, so finally finally −Ω ae (t) = (eiω t − eiω t ) (∗) 2 2 (ω − ω0 ) + 4Ω The probability oscillates between the odd and even states. First the oscillating field stimulates emission of radiation and decay from |o to |e. Later the field excites molecules in the ground state to move back up to the first-excited state |o. If we solve the original equations (1) exactly on resonance (ω (ω = ω = ω 0 ), the relevant solution is iΩt 1 −iΩt − e ), ae (t) = 2 (e which is what our general solution ( ∗) reduces to as ω as ω → ω0 . Particles of mass m m and momentum ¯ momentum ¯hk hk at x at x < −a move in the potential 5.15∗ Particles 0 for x x < −a 1 V ( V (x) = V 0 2 [1 + sin(πx/ sin(πx/22a)] for | (5. (5.5) |x| < a
+
+
−
−
1 for x x > a, a, where V 0 < ¯h2 k 2 /2m. Numericall Numerically y reproduce the reflection probabilitie probabilitiess plotted in Figure 5.20 as follows. Let ψ Let ψ i ψ (xj ) be the value of the wavefunction at x j = j ∆, where ∆ ∆ is a small increment in the x coordinate. From the tise show that ψj (2 ∆2 k 2 )ψj+1 ψj+2 , (5. (5.6)
≡
≡
≃ −
−
where k k 2m(E V ) V )/h ¯ . Determine Determine ψ ψ j at the two grid points with the largest values of x from a suitable boundary condition, and use the recurrence relation ( 5.6) 5.6) to determine ψj at all other grid points. points. By matching matching the values values of ψ at the points with the smallest values of x to a sum of sinusoidal waves, determine the probabilities required for the figure. Be sure to check the accuracy of your code when V 0 = 0, and in the general case explicitly check that your results are consistent with equal fluxes of particles towards and away from the origin. Equation ( 12.40) 12.40) gives an analytical approximation for ψ in the case that there is negligible reflection. reflection. Compute Compute this approximate approximate form of ψ and compare it with your numerical results for larger values of a. a .
−
Soln:
We discretise the tise h ¯ 2 d2 ψ + V ψ = E = Eψ ψ 2m dx2
−
2
by
− 2h¯m ψj+1 + ψ∆j2−1 − 2ψj + V j ψj = Eψ E ψj
9 which which readily readily yields the required required recurrence recurrence relation. relation. At the right-hand right-hand boundary boundary we require require a pure outgoing wave, so ψ so ψ j = exp(i jK ∆) jK ∆) gives ψ gives ψ at the two last grid points. From the recurrence recurrence relation relation we obtain ψ obtain ψ elsewhere. At the left boundary we solve for A for A + and A and A − the equations A+ exp(i0k exp(i0k∆) + A− exp( i0k i0k ∆) = ψ 0 A+ exp(i1k exp(i1k∆) + A− exp( i1k i1k ∆) = ψ 1
− −
The transmission probability is (K/k (K/k))/ A+ 2 . The code must must reprodu reproduce ce the result result of Problem Problem 5.4 in the appropriate limit.
| |
5.16∗
In this problem we obtain an analytic estimate of the energy difference between the evenand odd-parity states of a double square well. Show that for large θ, θ , coth θ tanh θ 4e−2θ . Next letting δ δ k be the difference between the k values that solve
−
− − − − − ≡ ≡ − − − − − − ≃− tan[rπ tan[rπ
where
≃
k (b
a)]
W 2
1=
(ka) ka )2
coth
W 2
(ka) ka )2
even parity
tanh
W 2
(ka) ka )2
odd parity parity,
(5. (5.7a)
2mV 0 a2 h ¯2 for given r given r in the odd- and even-parity cases, deduce that W
W 2 (ka) ka )2
1/2
1
+
W 2 (ka) ka )2
−1/2
1
(b
a) +
1 k
(ka) ka )2 W 2
1
2 W 2
4exp
(5. (5.7b)
−1
δk (5. (5.8)
(ka) ka )2 .
Hence show that when W when W excited states is
≫ 1 the fractional difference between the energies of the ground and first ≫ δE −8a e−2W √ 1−E/V . (5. (5.9) ≃ E W ( W (b − a) Soln: First eθ + e−θ e θ − e−θ 1 + e −2θ 1 − e−2θ coth θ − tanh θ = θ = (1 + 2e−2θ ) − (1 − 2e−2θ ) = 4e−2θ − − ≃ − θ θ − θ − θ − θ 2 2 e −e e +e 1−e 1+e So when W ≫ ≫ 1 the difference in the right side of the equations for k in the cases of even and odd 0
parity parity is small and we may estimate estimate the difference difference in the left side by its derivative derivative w.r.t. w.r.t. k times the difference δk in the solutions. That is 2
)](b − a)δk − s [rπ − k(b − a)](b
W 2 (ka) ka )2
√ W 2 /(ka) ka )2 δk/k − −2 W −(ka) tan[rπ − k (b − a)] − 1 + tan[rπ ≃ 4e W 1 − (ka) 2
2
2
2
In the case of interest the right side of the original equation is close to unity, so we can simplify the last equation by using W 2 tan[rπ tan[rπ k (b a)] 1 1 (ka) ka )2
−
−
− ≃
With the help of the identity s2 θ = 1+ tan tan2 θ we obtain the required relation. We now approximate the left side for W for W ka. ka. This yields W 2 (b a)δk 4e−2W 1−(ka/W ) ($) ka Since E = h ¯ 2 k 2 /2m, δE/E = = 2δk/k and
≫ ≫
−
√
≃ −
2mEa 2 h ¯2 = E/V E /V 0 . 2mV 0 a2 h ¯2 The required relation follows when we use these relations in ($). (ka/W ) ka/W )2 =
×
10 6.11∗
Show that when the density operator takes the form ρ = ψ ψ , the expression Q = Tr Qρ for Qρ for the expectation value of an observable can be reduced to ψ Q ψ . Explain Explain the physical physical significance significance of this result result.. For the given given form of the densit density y operato operator, r, show that the equatio equation n of motion of ρ yields ∂ ψ φ ψ = ψ φ where φ i¯ h H ψ . (6. (6.1) ∂t Show from this equation that φ = a ψ , where a a is real. Hence Hence determine the time evolution evolution of ψ given the at t = 0, ψ = E is an eigenket of H . Explain Explain why ρ why ρ does not depend on the phase of ψ and relate this fact to the presence of a a in your solution for ψ, t .
| | | |
| | | | | | | |
|
| − |
| ≡
|
| |
Soln:
Tr(Qρ r(Qρ)) =
| | | nQψ ψn
n
We choose a basis that ψ is a mem member. ber. Then Then there there is only only one non-v non-vanishi anishing ng term in the sum, when n = ψ , and the right side reduces to ψ Q ψ as required. required. This result shows shows that density density operators recover standard experimental predictions when the system is in a pure state. Differentiating the given ρ given ρ we have dρ ∂ ψ ∂ ψ 1 = ψ + ψ = (H ( H ψ ψ ψ ψ H ) dt ∂t ∂t i¯h Gathering the terms proportional to ψ on the left and those proportional to ψ on the right we obtain the required expression. Now φ ψ = ψ φ φ ψφ = ψ φφ , which establishes that φ ψ . We define a as the constant constant of proportionalit proportionality y. Using φ = a ψ ∗ in φ ψ = ψ φ we learn that a = a = a so a so a is real. Returning to the definition of φ we now have ∂ ψ i¯h = (H a) ψ . ∂t This differs from the tdse in having the term in a in a.. If ψ is an eigenfunction of H of H ,, we find that its time dependence is ψ, t = ψ, 0 e−i(E −a)t/h¯ rather than the expected result ψ, t = ψ, 0 e−iEt/h¯ . We cannot determine a from the density-matrix formalism because ρ is invariant under the transformation ψ e−iχ ψ , where χ where χ is any real number.
|
| |
| |
| | | | |
| | − | |
|
| | | | ⇒ | | | | | ∝ | | | | | | | | − | | | | | | → |
|
|
| |
7.9∗
Repeat the analysis of Problem 7.8 for spin-one particles coming on filters aligned successively along +z + z , 45 ◦ from z from z towards x x [i.e. along (1,0,1)], and along x. x . Use classical electromagnetic theory to determine the outcome in the case that the spin-one particles were photons and the filters were Polaroid. Why do you get a different answer? calculation of Problem 7.8 by replacing replacing the matrix for J for J x by that for n J = Soln: We adapt the calculation (J x + J z )/ 2. So if now (a,b,c (a,b,c)) is + n in the usual basis, we have b a = 2−1/2 12 0 a a 2 2 1 1 0 b = b 2 2 b 1 c c 0 2−1/2 c = 2 2+ 2 1 1 The normalisation yields b yields b = = 2 , so a so a = = 2 /(2 2) and the required probability is the square of this, 0.25 25//(6 4 2) 0.73. So the probability of getting through all three filters is 13 (0. (0.73)2 0.177. In electromagnetism just one of two polarisations gets through the first filter, so we must say that a photon has a probability probability of half of passing the first filter. Then we resolve resolve its field field along the direction of the second filter and find that the amplitude of falls falls by 1/ 1/ 2 on passing the second filter, so half the energy and therefore photons photons that pass the first filter pass the second. second. Of these just a half pass the third filter. Hence in total 18 = 0.125 of the photons get right through. Although Although photons photons are spin-one particles, particles, there are two major difference difference between the two two cases. Most obviously, polaroid selects for linear polarisation rather than circular polarisation, and a photon with well-defined well-defined angular momentum momentum is circularly polarised. polarised. The other difference difference is that a photon can be in the state + z or z but not the state 0z , where the z the z -axis is parallel to the photon’s
√
|
−
− √ ≃
·
⇒
− √
E E
| | −
|
− √ √
×
√
≃
E
11 motion motion.. This This fact fact arises arises because because emag waves waves are transvers transversee so they they do not drive drive motion motion in the direction direction of propagation propagation k ; an angular momentum vector perpendicular to k would require motion along k. Our theory theory does not allow allow for this this case case because because it is non-relat non-relativi ivisti stic, c, whereas whereas a photon photon,, having zero rest mass, is an inherently relativistic object; we cannot transform to a frame in which a photon is at rest so all three directions would be equivalent. 7.13∗
Write a computer computer program that determines determines the amplitudes amplitudes a am in s
|n; s, s =
am s, m
| m=−s where n = (sin θ, 0, cos θ) with θ any angle and | | n; s, s is the ket that solves the equation (n · S)|n; s, s = s |n; s, s. Explain physically the nature of this state. S x and S x2 for this state and hence show Use your am to evaluate the expectation values
that the rms fluctuation in measurements of S S x will be s/2cos s/2cos θ. routinee tridiag() that computes the e-values and e-kets of a real symmetric Soln: We use a routin tri-diagonal matrix – the routine tqli() in Numerical Recipies by by Press et al. is suitable. #defin #define e J 100 #def #defin ine e NT 3 double tridiag(doub tridiag(double*,d le*,doubl ouble*,in e*,int,dou t,double** ble**)// )// evaluates evaluates & ekets of real, // symmetric tridiagonal matrix double double alphap alphap(in (int t j,int j,int m) if(m>=j)return 0; return sqrt((double)(j*(j+1)-m*(m+1))) sqrt((double)(j*(j+1)-m*(m+1))); ;
{
}
{
double double alpham alpham(in (int t j,int j,int m) if(m<=-j) =-j) return return 0; return sqrt((double)(j*(j+1)-m*(m-1))) sqrt((double)(j*(j+1)-m*(m-1))); ;
}
void expect(doubl expect(double e *a,int j,double j,double st) //eval //evaluat uate e and double s1=0,s2=0; s1=0,s2=0; for(int for(int n=-j;n<=j;n++) int nm2=n-2,nm1=n-1,np1=n+1,np2=n+ nm2=n-2,nm1=n-1,np1=n+1,np2=n+2; 2; if(nm2>=-j) s2+=alpham(j,n)*alpham(j,nm1)*a[nm s2+=alpham(j,n)*alpham(j,nm1)*a[nm2]*a[n]; 2]*a[n]; if(np2<=j) s2+=alphap(j,n)*alphap(j,np1)* s2+=alphap(j,n)*alphap(j,np1)*a[np2]*a[n]; a[np2]*a[n]; s2+=(alphap(j,nm1)*alpham(j,n)+alpham(j,np1)*alph s2+=(alphap(j,nm1)*alpham(j,n)+a lpham(j,np1)*alphap(j,n))*pow(a[n ap(j,n))*pow(a[n],2); ],2); if(nm1>=-j) s1+=alpham(j,n)*a[nm1]*a[n]; if(np1<=j) s1+=alphap(j,n)*a[np1]*a[n];
{
{
}
s1*=.5; s1*=.5; s2*=.25; s2*=.25; printf printf("%f ("%f %f %f %f n",s1,j*st,s2,.5*j*(1-st*st)+pow(j*st,2));
\
}
{
int main(void) main(void) 80,, 120 120,, 30 ; double pi=acos(-1),theta[NT]= 80 double double *D = new double double[2* [2*J+1] J+1]; ; double double *E = new double double[2* [2*J+1] J+1]; ; double double **Z = new double* double*[2* [2*J+1 J+1];/ ];//al /alloc locate ate storage storage for square square matrix matrix for(int for(int i=0; i<2*J+ i<2*J+1; 1; i++) i++) Z[i] Z[i] = new double double[2* [2*J+1 J+1]; ]; for(int for(int it=0; it=0; it<3; it++) it++) theta[it]=theta[it]*pi/180; double ct=cos(theta[it]), st=sin(theta[it]); for(in for(int t m=-J; m=-J; m<=J; m++) m++) D[J+m]=m*ct; D[J+m]=m*ct;//dia //diagona gonal l elements elements of matrix if(m>-J) E[J+m]=st*.5*a lpham(J,m);//sub-diagonal lpham(J,m);//sub-diagonal elements
{
}
{
{
}
tridiag(D,E, tridiag(D,E,2*J+1 2*J+1,Z);/ ,Z);//find /finds s evalues evalues & ekets of tridiagonal tridiagonal matrix int mm; for(in for(int t i=0; i=0; i<2*J+1; i++) if(fabs(D[i]-J)<.05) .05) mm=i; mm=i; // identif identify y eket eket m=J
{
12
}
expect(Z[mm]+J,J,st);
}
7.14∗
} We have that
∂ ∂ i Ly = eiφ + i cot cot θ . (7. (7.1) ≡ Lx + iL ∂θ ∂φ From the Hermitian nature of Lz = −i∂/∂φ we ∂/∂φ we infer that derivative operators are anti-Hermitian. † † †
L+
So using the rule (AB ( AB)) = B A on equation ( 7.1), 7.1), we infer that ∂ ∂ L− L†+ = + i cot θ e−iφ . ∂θ ∂φ This argument and the result it leads to is wrong. Obtain the correct result by integrating by parts dθ sin θ dφ (f ∗ L+g ), where f and g are arbitrary functions of θ and φ. What What is the fallacy fallacy in the given argument?
−
≡
Soln:
dθ sin θ
− − − − − − −
∗
dφ (f L+ g ) =
dθ sin θ
=
dφ eiφ
=
dφ eiφ
∂g ∂g + i cot cot θ ∂θ ∂φ ∂g ∂g dθ sin θf ∗ + i dθ cos θ dφ f ∗ eiφ ∂θ ∂φ ∗ ∂ (sin (sin θf ) [sin θ f ∗ g ] dθ g ∂θ dφ f ∗ eiφ
∂ (f ∗ eiφ ) ∂φ The square brackets vanish so long f , g are periodic in φ. φ . Differentiating out the products we get ∂f ∗ iφ ∗ dθ sin θ dφ (f L+ g ) = dφ e dθ sin θg + dθ cos θgf ∗ ∂θ ∗ iφ ∂f + i dφ eiφ gf ∗ i dθ cos θ dφ e g ∂φ ∗ The two integrals containing f containing f g cancel as required leaving us with ∂f ∗ ∂f ∗ dθ sin θ dφ (f ∗ L+ g ) = dθ sin θ dφ g eiφ + i cot cot θ = dθ sin θ dφ g (L− f ) f )∗ ∂θ ∂φ where ∂ ∂ L− = e−iφ icot θ . ∂θ ∂φ The fallacy is the proposition that that ∂/∂θ ∂/∂θ is anti-Hermitia anti-Hermitian: n: the inclusion of the factor sin θ in the integral integral prevents prevents this being b eing so. +i
7.15∗
By writing writing ¯ ¯h2 L2 = (x
× p) · (x × p) =
· − ˆr · p = −
dφ g
ijklm ǫ ijk xj pk ǫilm xl pm show
h ¯ 2 L2 1 + 2 (r p)2 i¯hr p . 2 r r 2i¯h/r, h/r, obtain r p = rp = rp r + i¯h. Hence obtain
p2 = By showing that p ˆr
dθ cos θ [f ∗ eiφ g ]
·
·
−
·
h ¯ 2 L2 . r2 Give a physical interpretation of one over 2m 2 m times this equation. p2 = p 2r +
that (7. (7.2)
(7. (7.3)
13 Soln:
From the formula for the product of two epsilon symbols symbols we have h ¯ 2 L2 =
(δ jl jl δ km km
jklm
=
− δ jm jm δ kl kl )xj pk xl pm
xj pk xj pk
jk
The first term is
xj pk xj pk =
jk
− xj pk xk pj
xj (xj pk + [ p [ pk , xj ]) pk =
jk
.
xj (xj pk
jk
= r 2 p2
The second term is
− i¯hr · p.
xj pk xk pj =
jk
xj (xk pk
jk
=
− i¯h) pj
xj ( pj xk pk + i¯hδ hδ jk jk pk )
jk
− 3i¯h
= (r p)(r p)
· − 2i¯h(r · p).
·
hδ jk − i¯hδ jk ) pk
xj pj
j
When these relations relations are substituted substituted above, the required required result follows. follows. Using the position representaion 3i¯h 3i¯h ∂r −1 h∇ (r/r) /r) = i¯hr ∇(1/r (1/r)) = i¯hr hr = p ˆ r ˆ r p = i¯ r r ∂r Using this relation and the definition of pr r r 2i¯ 2 i¯h rpr = (ˆ (ˆr p + p ˆr) = 2ˆr p = r p i¯h 2 2 r
· − ·
−
·
−
·
−
·
·
−
−
· −
Substituting this into our expression for p for p 2 we have
1 hr 2 − 3i¯rh + i¯hr r
· −
h ¯ 2 L2 1 p = 2 + 2 ((rp (( rpr + i¯ h)(rp )(rpr + i¯ h) i¯h(rpr + i¯ h)) r r When we multiply out the bracket, we encounter rpr rpr = r2 p2r + r[ r [ p pr , r] p pr = r2 p2r i¯hrp hrpr . No Now w when we clean up we find that all terms in the bracket that are proportional to ¯h cancel and we have desired result. This equation divided divided by 2m 2m expresses the kinetic energy as a sum of tangetial and radial KE. 2
−
−
definition 7.20∗ Show that [J i , Lj ] = i k ǫijk Lk and [J i , L2 ] = 0 by eliminating Li using its definition −1 ¯ x p, and then using the commutators of J of J i with x and p. L = h Soln:
×
h ¯ [J i , Lj ] = ǫ jkl [J i , xk pl ] = ǫ jkl ([J ([J i , xk ] p pl + xk [J i , pl ]) = ǫ jkl (iǫ (iǫikm xm pl + iǫ i ǫiln xk pn ) = i(ǫ i(ǫklj ǫkmi xm pl + ǫljk ǫlni xk pn ) = i(δ i(δ lm δ li i( δ jn δ ji lm δ ji ji li δ jm jm )xm pl + i(δ jn δ ki ki ji δ kn kn )xk pn = i(x pδ ij xj pi + x + xi pj x pδ ij i(xi pj xj pi ) ij ij ) = i(x
But
− · −
− ·
−
i¯hǫ hǫijk Lk = iǫijk ǫklm xl pm = iǫkij ǫklm xl pm = i(δ i(δ il il δ jm jm 7.21∗
−
i(xi pj − xj pi ) − δ im im δ jl jl )xl pm = i(x
In this problem you show that many matrix elements of the position operator x x vanish when states of well-defined l, l, m are used as basis states. These results will lead to selection rules for electric
14 dipole radiation. First show that [ that [L L2 , xi ] = i using this result derive [L2 , [L2 , xi ]] = i
−
jk ǫ jik (Lj xk +
xk Lj ). Then show that L x = 0 and
·
ǫjik Lj [L2 , xk ] + [L [ L2 , xk ]Lj = 2(L 2(L2xi + xi L2 ).
jk
(7. (7.4)
By squeezing this equation between angular-momentum eigenstates l, m and l ′ , m′ show that 0 = (β β ′ )2 2(β 2(β + + β ′ ) l, m xi l ′ , m′ , where β l(l + 1) and β ′ = l′ (l ′ + 1). 1) . By equatin equating g the factor in front front of l, m xi l ′ , m′ to zero, and treating the resulting equation as a quadratic equation for β for β given given β β ′ , show that l, m xi l ′ , m′ must vanish unless l + l + l ′ = 0 or l or l = l ′ 1. Explain why the matrix element must also vanish when ′ l = l = l = 0. 0.
≡ ≡
−
| | |
| |
| |
±
Soln:
[Lj2 , xi ] =
j
h ¯ L x =
·
(Lj [Lj , xi ] + [L [ Lj , xi ]Lj ) = i
j
| |
ǫjik (Lj xk + xk Lj )
jk
ǫijk xj pk xi =
ijk
ǫijk (xj xi pk + xj [ p pk , xi ]) =
ijk
ǫijk (xj xi pk
ijk
− i¯hx hxj δ ki ki )
Both terms on the right side of this expression involve ik ǫijk S ik ik where S ik ik = S ki ki so they vanish by Problem 7.3. Hence x L = 0 as in classical physics. Now 2 [L , [L2 , xi ]] = i ǫjik [L2 , (Lj xk + xk Lj )] = i ǫjik (Lj [L2 , xk ] + [L [ L2 , xk ]Lj )
·
− { } { } − − { } { { − } { } − { − − − · − − − − jk
jk
=
ǫjik ǫlkm (Lj Ll xm + xm Ll + Ll xm + xm Ll Lj )
jklm
=
(δ jm jm δ il il
δ jl )(Lj Ll xm + xm Ll + Ll xm + xm Ll Lj ) jl δ im im )(L
jlm
=
(Lj Li xj + xj Li + Li xj + xj Li Lj
}
Lj Lj xi + xi Lj
j
=
(Lj Li xj + xj Li Lj )
L2 xi
(Lj xi Lj + Lj xi Lj )
j
j
} − {Lj xi + xiLj }Lj )
− xi L2
where to obtain the last line we have identified occurrences of L x and x L. Now Lj Li xj = (Lj xj Li + Lj [Li , xj ]) = i ǫijk Lj xk j
Similarly,
j x j Li Lj
j
=i
jk ǫjik xk Lj .
Lj xi Lj =
j
jk
Moreover
ǫjik xk Lj + xi L2
([L ([Lj , xi ]Lj + xi Lj Lj ) = i
j
=
·
jk
ǫijk Lj xk + L2 xi
(Lj [xi , Lj ] + Lj Lj xi ) = i
j
jk
Assembling these results we find [L2 , [L2 , xi ]] =
i
ǫijk [Lj , xk ]
jk
L2 xi
i
ǫjik [xk , Lj ]
jk
xi L2
− L2xi − xi L2
= 2(L 2(L2 xi + xi L2 ) as required. The relevant matrix element is lm [L2 , [L2 , xi ]] l′ m′ = lm (L2 L2 xi 2L2 xi L2 + xi L2 L2 ) l ′ m′ = 2 lm (L2 xi + xi L2 ) l′ m′
|
|
|
−
|
|
|
15 which implies β 2 lm xi l′ m′ 2β lm xi l ′ m′ β ′ + lm xi l ′ m′ β ′2 = 2β lm xi l ′ m′ + 2 lm xi l ′ m′ β ′ Taking out the common factor we obtain the required result. The quadratic for β for β ((β ′ ) is β 2 2(β 2(β ′ + 1)β 1)β + + β ′ (β ′ 2) = 0 so β = β = β ′ + 1 (β ′ + 1)2 β ′ (β ′ 2) = β ′ + 1 4β ′ + 1
| |
− | |
| |
−
±
= l ′ (l′ + 1) + 1
±
| |
| |
−
−
−
±
4l′2 + 4l 4l ′ + 1 = l ′ (l′ + 1) + 1
= l ′2 + 3l 3l′ + 2 or l′2 l′ We now have two quadratic equations to solve l 2 + l (l ′2 + 3l 3l ′ + 2) = 0
−
± (2l (2l′ + 1)
− ± (2l (2l ′ + 3)] l2 + l − (l ′2 − l′ ) = 0 l = 21 [−1 ± (2l (2l ′ − 1)] Since l Since l,, l ′ ≥ 0, the only acceptable solutions are l are l + l ′ = 0 and l and l = l = l ′ ± 1 as required. However, when ′ −
⇒ ⇒
l = 12 [ 1
l = l = 0 the two states have the same (even) parity so the matrix element vanishes by the proof given in eq (4.42) of the book.
7.22∗
Show that l excitations can be divided amongst the x, x, y or y or zz oscillators of a three-dimensional harmonic oscillator in ( in ( 12 l +1)(l +1)(l + 1) ways. 1) ways. Verify in the case l = l = 4 that this agrees with the number of states of well-defined angular momentum and the given energy. assign n x of the l the l excitations excitations to the x the x oscillator, oscillator, we can assign 0, 0 , 1, . . . , l nx excitations Soln: If we assign n to the y the y oscillator [(l [(l nx + 1) possibilities], and the remaining excitations go to z . So the number of ways is
−
−
l
l
≡ ≡
S
(l
nx =0
− nx + 1) =
l
(l + 1)
nx =0
−
nx = (l + 1) 2
nx =1
− 12 l(l + 1) = (l(l + 1)( 12 l + 1)
In the case of 4 excitations, the possible values of l are 4, 2 and 0, so the number of states is (2 4 + 1) + (2 2 + 1) + 1 = 15, which is indeed equal to (4 + 1) (2 + 1).
∗
∗
7.23
Let
∗
∗
1 (l ( l + 1)¯h Al i p pr + mωr . (7. (7.5) r 2mhω ¯hω be the ladder operator of the three-dimensional harmonic oscillator and E, l be the stationary state of the oscillator that has energy E and E and angular-momentum quantum number l. Show Show tha thatt if we write A A l E, l = α − E hω,l ¯h ω,l + 1 , then α then α − = l, where is the angular-momentum quantum number of a circular orbit of energy E energy E .. Show similarly that if A A †l−1 E, l = α + E + + hω,l h ¯ ω,l 1 , then α+ = l + 2. 2. mod-square of each each side of A A l E, l = α − E hω,l + ¯h ω,l + 1 we find Soln: Taking the mod-square H l E α− 2 = E, l A†l Al E, l = E, L (l + 32 ) E, l = (l + 32 ). hω ¯hω hω ¯hω
≡ √
| √ L −
| −
−
√ L −
L L
| | −
|
|
|
−
| − | | | | | − In the case l = L, |α− |2 = 0, so L = (E/¯ E/ hω) ¯hω) − 32 and therefore |α− |2 = L − l as required. required. We can
choose the phase of α α − at our convenience. Similarly
α2+ = E, l Al−1 A†l−1 E, l = E, l (A†l−1 Al−1 + [A [ Al−1 , A†l−1 ]) E, l
| = E, l|
| | | H l−1 H l − H l−1 E (l + 12 ) + + 1 |E, l = − − l + 12 = L − l + 2 hω ¯hω hω ¯hω hω ¯hω
16 7.24∗
Show Show that that the probab probabili ility ty distribut distribution ion in radius radius of a partic particle le that that orbits orbits in the threethreedimensional dimensional harmonic oscillator potential potential on a circular circular orbit with angular-mome angular-momentum ntum quantum quantum number l number l peaks at r at r/ℓ /ℓ = = 2(l 2(l + 1), 1), where
ℓ
≡
h ¯ . 2mω
(7. (7.6)
Derive the corresponding classical result. wavefunctions tions of circular orbits are annihilated annihilated by Al , so Al E, l = 00.. In the the Soln: The radial wavefunc position representation this is ∂ 1 l + l + 1 r + + 2 u(r) = 0 ∂r r r 2ℓ Using the integrating factor, l r exp dr + 2 = r −l exp r2 /4ℓ2 , (7. (7.7) r 2ℓ
|
− − 2
2
to solve the equation, we have u rl e−r /4ℓ . The radial distribution is P ( P (r) r2 u 2 = r 2(l+1) e−r Differentiating to find the maximum, we have 2(l 2(l + 1)r 1) r2l+1 r2(l+1) r/ℓ2 = 0 r = 2(l 2(l + 1) 1/2 a For the classical result we have mv2 lh ¯ mrv = mrv = l l¯h ¯ and = mω 2 r r = v/ω = v/ω = r mrω /2 /2 1 1 so r so r = (lh/mω) ¯h/mω ) = (2l (2 l ) ℓ in agreement with the QM result when l 1.
∝
−
∝ ||
2
/2ℓ2
.
√
⇒
⇒
≫
7.25∗
A particle moves in the three-dimensional harmonic oscillator potential with the second largest angular-momentum quantum number possible at its energy. Show that the radial wavefunction is 2l + 1 −x2 /4 h ¯ u1 xl x e where x r/ℓ with ℓ . (7. (7.8) x 2mω How many radial nodes does this wavefunction have? wavefunction on of the circular orbit with angular moSoln: From Problem 7.24 we have that the wavefuncti l −r 2 /4ℓ2 mentum l mentum l is r E, l re . So the required radial wavefunction is
−
∝
| ∝ + ¯hω,l h ω,l − 1 ∝ r|A†l−1 |E, l r|E + l + 1 r + 2 ∝ − ∂r∂ − l + r 2ℓ
l −r 2 /4ℓ2
= r e
r ℓ2
−
≡
l −r 2 /4ℓ2
re
2l + 1 r
∝
This wavefunction clearly has one node at x = 7.28∗
=
xl e−x
2
rl+1 lr + 2 2ℓ 2l + 1 x x
− − − l−1
/4
√ 2l + 1.
≡
(l + 1)r 1) r
l−1
r l+1 + 2 2ℓ
e−r
2
/4ℓ2
The interaction between neighbouring spin-half atoms in a crystal is described by the Hamil-
tonian H = K
S(1) S(2)
·
a
−
(S(1) a)(S(2) a) 3 a3
·
·
,
(7. (7.9)
where K is K is a constant, a is the separation of the atoms and S(1) is the first atom’s spin operator. (1) (2) (1) (2) (1) (2) Explain what physical idea underlies this form of H of H .. Show that S that S x S x + S y S y = 12 (S + S − + (1) (2) S − S + ). Show that the mutual eigenkets of the total spin operators S S 2 and S S z are also eigenstates of H H and find the corresponding eigenvalues. At time t t = 0 particle 1 has its spin parallel to a, while the other particle’s spin is antiparallel a . Find the time required for both spins to reverse their orientations. to a
17 Hamiltonian recalls recalls the mutual mutual potential energy energy V V of two classical magnetic dipoles Soln: This Hamiltonian µ(i) that are separated by the vector a, which we can calculate by evaluating the magnetic field B that the first dipole creates at the location of the second and then recognising that V = µ B.
− ·
(1) (2)
S + S − = (S x(1) + iS iS y(1) )(S )(S x(2) iS y(2) ) = S x(1) S x(2) + S y(1) S y(2) + i(S i(S y(1) S x(2) S x(1) S y(2)) Similarly, (1) (2) S − S + = S x(1) S x(2) + S y(1) S y(2) i(S i(S y(1) S x(2) S x(1)S y(2) ) Adding these expressions we obtain the desired relation. We choose to orient the z the z -axis along a . Then H Then H becomes K 1 (1) (2) (1) (2) H = = (S S + S − S + ) + S z(1)S z(2) 3S z(1) S z(2) . (7. (7.10) a 2 + − The eigenkets of S S 2 and S and S z are the three spin-one kets 1, 1 , 1, 0 and 1, 1 and the single spin-zero ket 0, 0 . We multiply each of these kets in turn by H by H :: K 1 (1) (2) (1) (2) H 1, 1 = H + + = (S + S − + S − S + ) 2S z(1)S z(2) + + 2 a K = 1, 1 2a (i) which uses the fact that S that S + + = 0. Similarly H Similarly H 1, 1 = H = (K/2 K/2a) 1, 1 . 1 K 1 (1) (2) (1) (2) H 1, 0 = H ( + + + )= (S + S − + S − S + ) 2S z(1)S z(2) ( + + +) 2 2 2a K 1 K = + 1 ( + + + ) = 1, 0 a 2a 2 where we have used S used S + = + , etc. Finally 1 K 1 (1) (2) (1) (2) H 0, 0 = H ( + + )= 2S z(1)S z(2) ( + +) 2 (S + S − + S − S + ) 2 2a K 1 1 = + ) =0 2 + 2 ( + 2a The given initial condition 1 ψ = + = ( 1, 0 + 0, 0 ), 2 which is a superposition of two stationary states of energies that differ by K/a. K/a . By analogy with the symmetrical-well problem, we argue that after time π h/∆ ¯h/∆E = π ha/K the ¯ha/K the particle spins will have reversed.
−
−
−
− | | | −
|
|
|
| | − | | √ | |− |−| √
√
|
| |−
√
| |
−
| −
|−|
|− | √ | |−−|−|
√ −
−
|−|− − −
| | −|−
|−|
|
−
| |−−|−|
| |−−|−|
| | |− √ | |
8.10∗
A spherical potential well is defined by 0 for r r < a V ( V (r) = (8. (8.1) V 0 otherwise, where V V 0 > 0. 0 . Consider Consider a stationary stationary state with angular-momen angular-momentum tum quantum number number l. l . By writing the wavefunction ψ (x) = R(r)Ylm (θ, φ) and using p2 = p2r + h ¯ 2 L2 /r2 , show that the state’s radial wavefunction R( R (r) must satisfy
2
h ¯2 d 1 l( l (l + 1)¯h2 + R + R + V ( V (r)R = E = E R. 2m dr r 2mr2 Show that in terms of S ( S (r) rR( rR(r), this can be reduced to d2 S S 2m 2 m l ( l + 1) + (E V ) V )S = 0. 0. dr2 r2 h ¯2
−
≡
−
−
(8. (8.2)
(8. (8.3)
18 Assume that V that V 0 > E > 0. 0 . For the case l = l = 0 write down solutions to this equation valid at (a) r < a and (b) r > a. Ensure Ensure that that R does not diverge at the origin. What conditions conditions must S S satisfy at r = a? a ? Show that these these conditions can be simultaneous simultaneously ly satisfied satisfied if and only if a solution solution can be 2 2 2 2 found to k cot ka = ka = K , where h ¯ k = 2mE and h ¯ K = 2m(V 0 E ). Show graphica graphically lly that the equation can only be solved when 2mV 0 a/¯ a/h ¯ > π/2 π/2. Compare Compare this result with with that obtained for the corresponding one-dimensional potential well. The deuteron deuteron is a bound state of a proton and a neutron neutron with zero angular momentum. momentum. Assume Assume that the strong force that binds them produces a sharp potential step of height V 0 at interparticle distance a = 2 10−15 m. Determine Determine in MeV the minimum value value of V 0 for the deuteron to exist. Hint: remember to consider the dynamics of the reduced particle. representation tation pr = i¯h(∂/∂r + ∂/∂r + r r −1 ), so in this representation and for an Soln: In the position represen 2 eigenfunction of L L we get the required form of E E E = H E = ( p2 /2m + V ) V ) E . Writing R Writing R = S/r = S/r we have 2 d 1 d 1 S 1 dS d 1 d 1 1 dS 1 d2 S + R = + = + R = + = dr r dr r r r dr dr r dr r r dr r dr2 Inserting this into our tise and multiplying through by r by r,, we obtain the required expression. When l When l = = 0 the equation reduces to either exponential decay or shm, so with the given condition on E on E we we have cos kr or or sin kr at r < a S Ae−Kr at r > a 2 2 2 2 where k = 2mE/¯ mE/ h ¯ and K = 2m(V 0 E )/h ¯ . At r < a we must chose S sin kr because we require R require R = S/r = S/r to be finite at the origin. We require S and S and its first derivative to be continuous at r = a = a,, so sin(ka sin(ka)) = Ae Ae−Ka K cot(ka cot(ka)) = = W 2 /(ka) ka )2 1 −Ka k k cos(ka cos(ka)) = K Ae
−
−
√
×
−
|
⇒
∝ ∝
≡ ≡
|
|
−
∝ ∝
⇒
−
−
−
−
with W with W 2mV 0 a2 /h ¯ 2 . In a plot of each side against k against ka a, the right side starts at when k when ka a = 0 and rises towards the x axis, where it terminates when ka = W . W . The The left side side starts starts at and becomes negative when k a = π/ = π/2. 2. There is a solution iff the right side has not already terminated, i.e. iff W W > π/ π/2. We obtain the minimum value of V V 0 for W for W = (a/¯ a/h ¯ ) 2mV 0 = π/2, π/ 2, so
−∞
∞
√
V 0 = where m where m ∗
8.13
π2 h ¯2 (π h/a) ¯h/a)2 = = 25 25..6MeV 8ma2 4m p
≃ 12 mp is the reduced mass of the proton.
From equation ( 8.50) 8.50) show that l that l
′
+ 12
=
(l + 12 )2
and that the increment ∆ increment ∆ in in l l ′ when − β and
l is increased by one satisfies ∆2 + ∆(2l ∆(2l′ + 1) = 2(l 2(l + 1). 1) . By considering the amount by which the ′ solution of this equation changes when l changes from l as a result of β increasing β increasing from zero to a small number, show that 2β ∆=1+ 2 + O(β O(β 2 ). (8. (8.4) 4l 1 Explain the physical significance of this result. Soln: The given given eqn is a quadratic quadratic in l in l ′ :
−
1 + 4l 4 l (l + 1) 4β l′ + 12 = (l + 12 )2 2 where we’ve chosen the root that makes makes l l ′ > 0. Squaring up this equation, we have (l′ + 12 )2 = (l + 12 )2 β (l ′ + ∆ + 12 )2 = (l + 32 )2 β Taking the first eqn from the second yields ∆2 + 2(l 2(l ′ + 12 )∆ = (l (l + 32 )2 (l + 12 )2 = 2(l 2(l + 1) l ′2 + l′ l (l +1)+β +1)+ β = = 0
−
⇒ l′ = −1 ± −
−
⇒
⇒
−
−
− β,
(8. (8.5)
19 This is a quadratic equation for ∆, which is solved by ∆ = 1 when l′ = l. We are interest interested ed in the small change δ ∆ in this solution when l′ changes by a small amount δl ′ . Different Differentiating iating the equation, we have 2∆δl 2∆ δl ′ 2∆δ 2∆ δ ∆ + 2∆δl 2∆δl ′ + (2l (2l ′ + 1)δ 1)δ ∆ = 0 δ ∆ = 2∆ + 2l 2l ′ + 1 ′ Into this we put ∆ = 1, l 1, l = l, l , and by binomial expansion of (8.5) β δl ′ = 2l + 1 and have finally 2β δ ∆ = (2l (2l + 1)(2l 1)(2l + 3) Eq (8.55) gives the energy of a circular orbit as Z 02 e2 = E = , 8πǫ0 a0 (l′ (l ) + k + 1) 2 with k with k the number of nodes in the radial wavefunction. This differs from Rydberg’s formula in that (l ′ (l) + k +1) is not an integer n integer n.. Crucially l Crucially l ′ (l ) + k does not stay the same if k in k in decreased by unity and l increased by unity – in fact these changes (which correspond to shifting to a more circular orbit) cause l′ (l) + k to increase slightly and therefore E E to decrease slightly slightly:: on a more circular circular orbit, the electron is more effectively screened from the nucleus. So in the presence of screening the degeneracy in H under which at the same E same E there there are states of different angular momentum is lifted by screening.
⇒
−
−
−
−
8.15∗
(a) A particle of mass m moves in a spherical potential V ( V (r). Show Show that that accor accordi ding ng to classical mechanics d dV der (p Lc ) = mr 2 , (8. (8.6) dt dr dt where Lc = r p is the classical angular-momentum vector and er is the unit vector in the radial direct direction ion.. Hence Hence show that when V ( V (r) = K/r, K/r, with K K a constant, the Runge–Lenz vector vector constant of motion. Deduce Deduce that Mc lies in the orbital plane, and that Mc p Lc mK er is a constant for an elliptical orbit it points from the centre of attraction to the pericentre of the orbit, while it vanishes for a circular orbit. (b) Show that in quantum mechanics ( ( p L)† p L = 2ip. Hence explain why in quantum mechanics we take the Runge–Lenz vector operator to be 1 ¯ N mK er where N p L L p. (8. (8.7) M 2h Explain why we can write down the commutation relation [L [ Li , M j ] = i k ǫijk M k . (c) Explain why [ p [ p2 , N ] N ] = 0 and why [1/r, [1/r, p L] = [1/r, [1/r, p] L. Hence show that 1 1 [1/r, [1/r, N] = i 3 (r2 p x x p) (8. (8.8) pr2 p x x 3 . r r (d) Show that 1 1 xj xj [ p p2 , er ] = i¯h pj 3 x + x 3 pj . (8. (8.9) p + p + r r r r j
×
× ≡ × −
−
×
≡
−
−
− ×
−
≡ × − × × × − · − − ·
(e) Hence show that [ that [H, H, M] = 0. What is the physical significance of this result? (f) Show that (i) [M i , L2 ] = i jk ǫijk (M k Lj + Lj M k ), (ii) [Li , M 2 ] = 0, where M 2 M y2 + M z2 . What are the physical implications of these results? (g) Show that [N i , N j ] = 4i ǫiju p2 Lu
−
u
≡ M x2 + (8. (8.10)
20 and that [N i , (er )j ]
− [N j , (er )i] = − 4i¯rh
and hence that
−2i¯h2mH
[M i , M j ] =
What physical implication implication does this equation equation have? have? Soln: (a) Since L c is a constant of motion
ǫijt Lt
(8. (8.11)
t
ǫijk Lk .
(8. (8.12)
k
d ∂V dV (p Lc ) = p˙ Lc = Lc = e r Lc , dt ∂ x dr where we have used Hamilton’s equation p˙ = ∂H/∂ x and ∂r/∂ x = e r . Also
×
×
− ×
−
(8. (8.13)
×
−
der = ω er , dt where ω = Lc /mr2 is the particle’s particle’s instantaneous instantaneous angular velocity velocity.. So er Lc = mr2 ω er = 2˙ mr er . Using Using this equati equation on to elimin eliminate ate er Lc from (8.13), we find that when dV dV /dr = K r2 , the right side becomes mK ˙ mK e˙ r , which is a total time-derivative, and the invariance of Mc follows. Dotting M c with Lc we find that M c is perpendicular to L c so it lies in the orbital plane. Also
×
−
×
×
−
×
× (r × p) = p2 r − p · r p. Evaluati Evaluating ng the right side at pericentre, pericentre, where p · r = 0, we have Mc = ( p2 r − mK )er . Mc + mK er = p
In the case of a circular orbit, by centripetal balance p balance p 2 /mr = /mr = K/r K/r 2 and M c = 0. At pericentre, the particle is moving faster than the circular speed, so p 2 > mK/r and mK/r and the coefficient of e e r is positive, so M c points to pericentre. (b) Since both p and L are Hermitian, (p
× L)†i = =
ǫijk ( pj Lk )† =
jk
jk
[ Lk , pj ]) = ǫijk ( pj Lk + [L
jk
= (p
ǫijk Lk pj
ǫijk pj Lk + i
jk
× L)i − 2i p pi.
ǫkjm pm
m
We want the Runge–Lenz vector to be a Hermitian operator, so we apply the principle that BA) BA ) is Hermitian even when [A, [A, B ] = 0 and write M i = 12 h ¯
ǫijk ( pj Lk + Lk pj )
jk
1 2 (AB +
− mK er = 12 h¯ (p × L − L × p) − mK er
M is a (pseudo) vector operator, so its components have the standard commutation relations with the components of L L . 2 (c) p (c) p is a scalar so it commutes with L , and of course it commutes with p , so it must commute with both p L and L p. As a scalar 1/r 1 /r commutes commutes with L, so
×
×
[1/r, [1/r, p
i¯h i¯h [1/r, p] × L = − 3 [r2 , p] × L = − 3 x × L. × L] = [1/r, 2r r
Similarly, [1/r, [1/r, L
× p] = −i¯hL × x 1r3 .
21 Now (x
× L)i = h¯1 =
1 h ¯
ǫijk ǫklm xj xl pm =
jklm
1 (xi x p h ¯
ǫijk ǫlmk xj xl pm =
jklm
1 h ¯
(δ il il δ jm jm
− δ im im δ jl jl )xj xl pm
1 h ¯
(δ kl kl δ im im
− δ km km δ il il )xl pm xk
jlm
· − r2 pi)
and (L
× x)i = h¯1 =
−
ǫijk ǫjlm xl pm xk =
jklm
1 h ¯
(xk pi xk
k
1 = pi r2 h ¯ Hence [1/r, [1/r, N] = [p
1 h ¯
ǫjki ǫjlm xl pm xk =
jklm
− xi pk xk ) = h¯1
p x xi
·
( pi xk xk + i¯hδ hδ ik ik xk
k
klm
hδ ki − pk xixk − i¯hδ ki xk )
1 1 /r] − [L × p, 1/r] /r] = i − 3 (xi x · p − r2 pi ) + pi r2 − p · x xi 3 } × L, 1/r] r r
(d) [ p p2 , (er )n ] = [ p p2 , xn /r] /r] =
(8. (8.14)
( pj [ p pj , xn /r] /r] + [ p [ pj , xn /r] /r] p pj )
j
=
− −
( pj [ p pj , xn ]/r + /r + p pj xn [ p pj , 1/r] /r] + [ p [ pj , xn ]/rpj + xn [ p pj , 1/r] /r] p pj )
j
δ jn xj jn pj + pj xn 3 r r
= i¯h
j
= i¯h (e)
−
δ jn xj jn pj + xn 3 pj r r
1 1 pn + pn + r r
pj
j
xj xj xn + xn 3 pj 3 r r
p2 K 1 , h ¯ p L L p mK er 2m r 2 The results we have in hand imply that when we expand this commutator, there are only two non-zero non-zero terms, terms, so 1 1 [H, M] = 12 K p2 , er ¯hK , p L L p 2 hK r 1 1 xj xj 1 1 = 12 i¯hK hK p + p pj 3 x + x 3 pj + 3 x x p r2 p pr2 p x x 3 r r r r r r j [H, M] =
−
−
− × −
{ × − × }−
−
×
· −
−
− ·
=0 This result shows: shows: (i) that the eigenv eigenvalues of the M i are good quantum numbers – if the particle starts in an eigenstate of M i , it will remain in that state; (ii) the unitary transformations U transformations U i (θ) exp( iθM i ) are dynamic dynamical al symmetrie symmetriess of a hydro hydrogen gen atom. In particula particular, r, these these operato operators rs turn stationary states into other stationary states of the same energy. (f) (i)
≡
−
[M i , L2 ] =
[M i , Lj2 ] =
j
([M ([M i , Lj ]Lj + Lj [M i , Lj ]) = i
j
jk
ǫijk (M k Lj + Lj M k ) = 0.
so we do not expect to know the total angular momentum when the atom is in an eigenstate of any of the M i .
22 (ii) [L [Li , M 2 ] = j [Li , M j2 ] = i mutual eigenstates of L L 2 , Lz and M and M 2 . (g) [(p
× L)i, pm ] =
ǫijk [ p pj Lk , pm ] =
jk
=i
ǫijk pj [Lk , pm ] = i
ǫijk ǫkmn pj pn = i
jkn
ǫkij ǫkmn pj pn
jkn
2 − δ inin δ jm jm ) pj pn = i( p δ im im − pi pm )
(δ im im δ jn jn
× p)i , pm] = −i( p2δ im im − pi pm ), so we have shown that [N i , pm] = 2i( p2 δ im im − pi pm ).
Moreover, since N is a vector, [N [ N i , Lm ] = i [N i , N s ] =
+ M j M k ) = 0, so there is a complete set of
jk
nj
Similarly [(L
jk ǫijk (M k M j
ǫstu [N i , pt Lu
tu
ǫstu 2( p 2( p2 δ it it
=i
tu
ǫsiu p2 Lu
= 2i
u
+i
− Lt pu] =
− 2i
−i
u
tn
= 4i
2
ǫsiu p Lu
u
= 4i
− 2i
− 2i
2
ǫsiu p Lu + i
ǫstu ( pi pt Lu
tu
(ǫiun pt N n
n
− ǫitnN n pu)
− Lt pi pu)
tun
− 2i
ǫstu ( pi pt Lu
− Lt pi pu )
− δ uiuiδ sn sn )N n pu
tu
(δ un un δ si si
nu
ǫstu ( pi pt Lu
tu
−
− [N i, Lt] p pu − Lt[N i, pu]
ǫstu ǫitn N n pu
− δ sisi δ ntnt) pt N n − i
u
+ pi (p
tu
ǫsti Lt p2
ǫsiu ( p2 Lu + Lu p2 ) (δ sn sn δ ti ti
ǫstu [N i , pt ]Lu + p + pt [N i , Lu ]
t
ǫstu ǫiun pt N n
+i
so
− pi pt )Lu − 2Lt( p2δ iuiu − pi pu) +
tun
= 2i
n ǫimn N n ,
2 pi (p
− Lt pi pu ) + i( pi N s + N s pi) − i(p · N + N · p)δ isis
× L)s + 2( L × p)s pi
× L)s − pi(L × p)s + ( p × L)s pi − (L × p)s pi − i(p · N + N · p)δ isis,
(8. (8.15) where we have used the fact that u ] = 0. We show that the terms with cross products sum to zero by first ensuring that all terms with pi on the left contain p L and all terms with pi on the right contain L p. We have to amend two terms to achieve this standardisation: [ p2 , L [ p
×
×
− pi(L × p)s + (p × L)s pi = =
− −
ǫsjk ( pi Lj pk + p + pj Lk pi )
jk
ǫsjk
pi pk Lj + i
ǫjkn pn + Lk pj + i
n
jk
= p i (p
× L)s − (L × p)s pi
ǫjkn pn pi
n
(8. (8.16)
The standardised sum of cross products in equation (8.15) is now
−
i
2 pi (p
+ pi (p × L)s + p + pi (p × L)s − (L × p)s pi − (L × p)s pi × L)s + 2( L × p)s pi + p
and is manife manifestl stly y zero. zero. The last term in (8.15) has to vanish because it alone is symmetric in is, is, and it’s not hard to show that it does: ǫijk pi ( pj Lk Lj pk ) + ( p ( pj Lk Lj pk ) pi p N + N p =
·
·
−
ijk
−
The first and last terms trivially vanish because they are symmetric in ij and i and ik k,respectively. ,respectively. The remaining terms can be written
−
ijk
ǫijk pi Lj pk +
jki
ǫijk pj Lk pi
23 and they cancel. Since er = x/r and in (8.8) we already have [1/r, [1/r, N] we prepare for calculating [N [ N i , er ] by calculating [(p
× L)i, xj ] =
ǫist [ p ps Lt , xj ] =
st
=i
(δ ij ij δ sn sn
[(L
− − · − − − · − − · − −
ǫist ( ps [Lt , xj ] + [ p [ ps , xj ]Lt ) = i
st
sn
Similarly
× p)i, xj ] =
ǫtjn xn
st
− δ ininδ sjsj ) psxn
so
ǫist ps
h ¯
ǫijt Lt
n
= i p x δ ij ij
pj xi
h ¯
ǫijt Lt
t
t
i x p δ ij ij
xi pj
h ¯
ǫijt Lt
t
[N i , xj ] = i (p x + x p)δ ij ij
·
hδ ¯hδ sj sj Lt
( pj xi + xi pj )
2¯h
ǫijt Lt
t
Now we can compute [N i , (er )j ] = [N [ N i , xj /r] /r] = [N i , xj ]/r + /r + xj [N j , 1/r] /r]
δ ij ij = i (p x + x p) r
1 ( pj xi + xi pj ) r
2¯ 2 h ¯ r
· · − − + xj [(p × L)i , 1/r] /r] − xj [(L × p)i , 1/r] /r] δ ij 1 2¯ 2 h ¯ ij = i (p · x + x · p) ( pj xi + xi pj ) − − r r r
ǫijt Lt
t
ǫijt Lt
t
x j 1 2 2 (x ( x r p ) x ( p r x ) x p p x i i j i i r3 r3 when we calculate [N [N i , (er )j ] [N j , (er )i ] all terms above that are symmetric in ij i j and will vanish and we find 1 4¯h [N i , (er )j ] [N j , (er )i ] = i ( pj xi + xi pj pi xj xj pi ) ǫijt Lt r r t +
· −
−
− ·
−
−
−
1 (xj pi r 4¯h =i r
−
−
−
−
−
1 1 xi pj ) (xj pi xi pj ) + (x (xj p x xi xi p x xj ) 3 r r 1 1 ǫijt Lt ( pj xi pi xj ) (xj pi xi pj ) + (x ( xj p x xi r r t
−
−
−
−
·
−
−
·
−
−
·
xi xj pk xk
hxi xj = − i¯hx
−
1 xi p xxj ) 3 r (8. (8.17)
·
Now
xi pk xk xj =
k
xi (xj pk
k
=
hδ jk − i¯hδ jk )xk =
xj pk xk xi
k
xj ( pk xi + i¯hδ hδ ki ki )xk
k
hxi xj − i¯hx
k
so the terms with dot products in (8.17) cancel. Finally [1 /r,pj ] = i¯hx hxj /r3 so 1 1 (xj pi xi pj ) = x j ( pi /r i¯hx hxi /r3 ) xi ( pj /r i¯hx hxj /r3 ) = (xj pi xi pj ) r r so the terms with factors 1/r 1 /r in in (8.17) cancel and we are left with 4i¯h [N i , (er )j ] [N j , (er )i ] = ǫijt Lt r t
−
−
−
−
−
−
−
−
From the definition of M we have [M i , M j ] = [ 12 hN ¯hN i mK (er )i , 12 hN ¯hN i mK (er )i ] = 14 h ¯ 2 [N i , N j ]
−
−
(8. (8.18)
([N i , (er )j ] + [(er )i , N j ]) − 12 mK h¯ ([N
= 14 h ¯ 2 [N i , N j ] 12 mK h ¯ [N i , (er )j ] [N j , (er )i ] . since the components of e e commute with each other. We obtain the required result on substituting from equations (8.10) and (8.11). A physical consequence of (8.12) is that we will not normally be able to know the values of more than one component of M – but we can in the exceptional case of completely radial orbits (L2 ψ = 0).
|
−
−
24 10.8∗
The Hamiltonian of a two-state system can be written A1 + B1 ǫ B2 ǫ H = , B2 ǫ A2
(10. (10.1)
where all quantities are real and ǫ is a small small paramete parameter. r. To first first order order in ǫ, what are the allowed energies in the cases (a) A 1 = A 2, and (b) A (b) A 1 = A = A 2 ? Obtain the exact eigenvalues and recover the results of perturbation theory by expanding in powers of ǫ. ǫ . When A1 = A 2 , the eigenvectors of H H 0 are (1, (1, 0) and (0, (0, 1) so to first-order in ǫ in ǫ the the perturbed Soln: When A energies are the diagonal elements of H of H ,, namely A namely A 1 + B1 ǫ and A 2 . When A When A 1 = A2 the unperturbed Hamiltonian is degenerate and degenerate perturbation theory applies: applies: we diagonalise diagonalise the perturbation perturbation B1 ǫ B2 ǫ B1 B2 H 1 = = ǫ B2 ǫ 0 B2 0
The eigenvalues λ eigenvalues λ of the last matrix satisfy 2
λ
− B1 λ −
B22
=0
⇒
and the perturbed energies are
± ± λ =
A1 + λǫ = λǫ = A A1 + 12 B1 ǫ
1 2
B12 +
B1
1 2
4B22
B12 + 4B22 ǫ
Solving for the exact eigenvalues of the given matrix we find λ = 12 (A1 + A2 + B1 ǫ)
± 12 = 12 (A1 + A2 + B1 ǫ) ± 12
If A A 1 = A2 this simplifies to
(A1 + A2 + B1 ǫ)2
− 4A2(A1 + B1ǫ) + 4B 4 B2 ǫ2 (A1 − A2 )2 + 2(A 2(A1 − A2 )B1 ǫ + (B (B12 + 4B22 )ǫ2
± ± −
λ = A = A 1 +
1 2 B1 ǫ
1 2
+
B12 + 4B22 ǫ
in agreement agreement with p erturbation erturbation theory. theory. If A A 1 = A2 we expand the radical to first order in ǫ B1 λ = 12 (A1 + A2 + B1 ǫ) 12 (A1 A2 ) 1 + ǫ + O(ǫ O(ǫ2 ) A1 A2
−
A1 + B1 ǫ if + A2 if again in agreement with perturbation theory =
− −
10.9∗
For the P states of hydrogen, obtain the shift in energy caused by a weak magnetic field (a) by evaluating evaluating the Land´ Land´e g factor, factor, and (b) by use equation equation ( 10.28) 10.28) and the Clebsch–Gordan coefficients calculated in 7.6.2. From l l = 1 and s = 12 we can construct j construct j = 32 and 12 so we have to evaluate two values Soln: (a) From of g g L. When j When j = 32 , j ( j + j + 1) = 15/ 15/4, and when j when j = 12 , j ( j + j + 1) = 3/ 3/4, so
§
gL = So
3 2
−
1 2
l (l + 1) s(s + 1) = j( j ( j + j + 1)
−
4 3 2 3
for j for j = for j for j =
for j = 32 , m = 2 E B /(µB B ) = mg L = for j for j = 32 , m = 3 1 for j for j = 12 , m = 3 with the values for negative m negative m being minus the values for positive m positive m..
2
3 2 1 2 1 2
3 2 1 2
26 ∂ H h ¯ 2 7 −3 1 0= = a + 5 mω2 a 5 ∂a m 2 h ¯ 7 a4 = 7 a = 71/4 2 ℓ H = hω h ¯ω mω 5
−
√
⇒
√
2
2
Using the result proved in Problem 10.13, show that the trial wavefunction ψ b = e−b r /2 yields 8/(3π (3π) as an estimate of hydrogen’s ground-state energy, where is the Rydberg constant. 2 2 2 2 ψ/ dr = b2 re−b r /2 , so Soln: With ψ = e−b r /2 , dψ/d 10.14∗
−
R
−
− − − √ − √ − √ √ − −
H Now
=
h ¯ 2 b4 2m
=
h ¯2 2mb
e2 4πǫ 0
2 2
dr r4 e−b
r
dx x e
xe−x = 2
2
dx x2 e−x dx x4 e−x
h ¯2 3 π 2mb 8
H =
=
0
2
dx xe−x
2 2
dr r2 e−b
1 b3
2
r
dx x2 e−x
2
1 2
∞
+
0
x3 e−x = 2
2
r2
∞
2
e−x = 2
dx xe−x
2
dr re−b
e2 4πǫ0 b2
4 −x2
2
so
R
2
2
dx e−x =
1 2
π 4
∞
+
2
dx x2 e−x =
3 2
0
e2 1 4πǫ 0 b2 2
π 3¯ h2 b2 = 4b3 4m
3 π 8
e2 b 2π 3/2 ǫ0
At the stationary point of H b = b = me me 2 /(3π (3π 3/2 ǫ0 h ¯ 2). Plugging this into H we find
3¯h2 m2 e4 H = 4m 9π 3 ǫ20 h ¯4
−
e2
me2 = 2π3/2 ǫ0 3π3/2 ǫ0 h ¯2
−
8 m 3π 2
e2 4πǫ0
2
=
8 3π
R
10.18∗
A particle of mass m is initially trapped by the well with potential V ( V (x) = V δ δ (x), where V V δ > 0. 0 . From t rom t = 0 it is disturbed by the time-dependent potential v(x, t) = F xe−iωt . Its subsequent wavefunction can be written
−
¯ −iE 0 t/h
−
|0 + dk {bk(t)|k, e + ck (t)|k, o} e−iE t/h¯ , (10. (10.2) where E E 0 is the energy of the bound state |0 and E E k ≡ h ¯ 2 k 2 /2m and |k, e and |k, o are, respectively |ψ = a( a (t)e
k
the even- and odd-parity states of energy E energy E k (see Problem 5.17). Obtain the equations of motion
|
i¯h a˙ 0 e
−iE 0t/h ¯
+
|
= v a 0 e
dk b˙ k k, e + c˙k k, o
−iE 0t/h ¯
+
|
|
e
−iE k t/h ¯
−iE k t/h ¯
dk (bk k, e + ck k, o ) e
(10. (10.3)
. | | Given that the free states are normalised such that k ′ , o|k, o = δ (k − k ′ ), show that to first order in v in v , b k = 0 for all t, t , and that iF sin(Ωk t/2) t/2) E k − E 0 ck (t) = k, o|x|0 eiΩ t/2 , where Ωk ≡ (10. (10.4) − ω. k
h ¯ Ωk /2 h ¯ Hence show that at late times the probability that the particle has become free is 2πmF 2 t k, o x 0 2 ( ) = P fr . fr t k h ¯3 Ω =0
| | | |
k
(10. (10.5)
27 Given that from Problem 5.17 we have
x|0 = √ K e−K |x|
where K =
show that
mV δ h ¯2
and
sin(kx)), x|k, o = √ 1π sin(kx
(10. (10.6)
K 4kK . π (k 2 + K 2 )2 Hence show that the probability of becoming free is
k, o|x|0 =
(10. (10.7)
8¯ hF hF 2 t E f f / E 0 P fr , (10. (10.8) fr (t) = 2 mE 0 (1 + E f f / E 0 )4 where E f f > 0 is the final energy energy.. Check Check that this this expres expression sion for P fr fr is dimensionless and give a physical explanation of the general form of the energy-dependence of P P fr fr (t) given expansion of ψ in stationary states of the unperturbed HamilSoln: When we substitute the given tonian H 0 into the tise, the terms generated by differentiating the exponentials in time cancel on H 0 ψ . The given expression contains the surviving terms, namely the derivatives of the amplitudes a, bk and ck on the left and on the right v ψ . In the first order approximati approximation on we put a = 1 and ′ ′ bk = ck = 0 on the right. Then we bra through with k , e and k , o and exploit the orthonormality of the stationary states to obtain equations for b˙ k (t) and c˙ k (t). The equation for b˙ k is proportional to the matrix element k, e v 0 , which vanishes by parity because v because v is an odd-parity operator. Then we replace v replace v by xF e xF e−iωt and have t t iF iF eiΩ t 1 ck (t) = dt′ c˙k = k, o x 0 dt′ ei[(E −E 0 )/h¯ −ω]t = k, o x 0 h ¯ h ¯ iΩk 0 0 iF sin(Ωk t/2) t/2) = k, o x 0 eiΩ t/2 . h ¯ Ωk /2 The probability that the particle is free is F 2 sin2 (Ωk t/2) t/2) P fr dk ck 2 = 2 dk k, o x 0 2 . fr (t) = (Ωk /2)2 h ¯
| | | |
|
|
|
−
|
F 2 P fr fr (t) = 2 h ¯ 1 ¯hk 2 /m + 2 hk
||
′
k
k
| |
−
k
| → |
→ ∞ we have sin2 xt/x2
Moreover, Ω k =
||
| |
As t As t
|
|
| | | |
πtδ (x), so at large t large t
F 2 k, o x 0 2 πt dk k, o x 0 2 πtδ (Ω (Ωk /2) = 2 2)/dk h ¯ d(Ωk /2)/
| | | |
| | |
constant, so dΩk /dk = hk/m and ¯ k/m and therefore h
2πmF 2 t k, o x 0 2 ( ) = P fr t . fr k h ¯3 Ω =0 Evaluating k, o x 0 in the position representation, we have
| | | |
Ωk =0
k
| | ∞ √ k, o|x|0 = 2 dx sin√ kx x K e−Kx = 2 π
0
K 1 π 2i
∞
dx x e
0
(ik−K )x
−e
−(ik+K )x
K 1 1 K 4kK = . π (ik (ik K )2 (ik (ik + K )2 π (k 2 + K 2 )2 The probability of becoming free is therefore 2πmF 2 t K 16 16kK kK 2 32 32mF mF 2t k/K P fr = (10. (10.9) fr (t) = π (k 2 + K 2 )4 h ¯3 h ¯ 3 K 4 (k 2 /K 2 + 1)4 The requir required ed result result follo follows ws when when we substi substitut tutee into into the above above k 2 /K 2 = E f f / E 0 and h ¯ 4 K 2 = (2mE (2mE 0 )2 . =
−i
−
−
| |
28 Regarding dimensions, [F [F ]] = E/L E /L and and [¯h] = ET E T ,, so (E/L) E/L )2 ET T ET 2 M L2 T −2 T 2 = = . M E 2 M L2 M L2 P fr E because at such energies the free state, which always has a node at fr (t) is small for small E because the location of the well, has a long wavelength, so it is practically zero throughout the region of scale 2/K within within which the bound particle is trapped. Consequently for small E small E the the coupling between the bound and free state is small. At high E high E the the wavelength of the free state is much smaller than 2/K 2/K and the positive and negative contributions from neighbouring half cycles of the free state nearly cancel cancel,, so again the coupling coupling between between the bound and free free states states is small. small. The couplin couplingg is most effective when the wavelength of the free state is just a bit smaller than the size of the bound state. [P fr fr ] =
10.19∗
A particle travelling with momentum p = hk h ¯ k > 0 from encounters the steep-sided potential well V ( V (x) = V 0 < 0 for x < a. Use the Fermi golden rule to show that the probability that a particle will be reflected by the well is
−
−∞
| | |
2
≃ 4V E 0 2 sin 2(2ka (2ka)), where E E = p 2 /2m. Show that in the limit E limit E ≫ ≫ V 0 this result is consistent with the exact reflection P reflect reflect
probability derived in Problem 5.10. Hint: adopt periodic boundary conditions so the wavefunctions wavefunctions of the in and out states can be normalised. length L of the x axis where L a and k = 2nπ/L, nπ/L, where n 1 is an Soln: We consider a length integer. Then correctly normalised wavefunctions of the in and out states are 1 ikx 1 −ikx ψin (x) = e ; ψout (x) = e L L The required matrix element is
≫
√
1 L
L/2
−L/2
dx e
ikx
≫
√
ikx
V ( V (x)e
a
=
−V 0
dx e2ikx =
−a
sin(2ka)) −V 0 sin(2ka Lk
so the rate of transitions from the in to the out state is 2 2π 2π (2ka)) 2 2 sin (2ka ˙ P = g (E ) out V in = g (E )V 0 2 h ¯ h ¯ L k2 Now we need the density of states g( g (E ). ). E = p 2 /2m = h ¯ 2 k 2 /2m is just kinetic energy. energy. Eliminating Eliminating k in favour of n, n , we have L n = 2mE 2π h ¯ As n As n increases by one, we get one extra state to scatter into, so
| | | |
√
dn L 2m g = = . dE 4πh ¯ E Substituting this value into our scattering rate we find 2 2m sin2 (2ka (2ka)) ˙ = V 0 P E Lk 2 2¯h2 This vanishes as L as L because the fraction of the available space that is occupied by the scattering potential is 1/L. /L. If it is not scattered scattered,, the particl particlee covers covers distanc distancee L in a time τ = L/v = L/ 2E/m. E/m . So the probability that it is scattered on a single encounter is
∼
→∞
V 2 m sin2 (2ka (2ka)) V 02 P˙ τ = 0 2 = sin 2 (2ka (2ka)) 2 2 k 4 E 2E h ¯
29 Equation Equation (5.78) gives the reflection reflection probability probability as (K/k k/K ) k/K )2 sin2 (2K (2K a) P = 2 2 (K/k + K/k + k/K ) k/K ) sin (2K (2K a) + 4 cos cos2 (2K (2K a)
−
When V 0 E , E , K 2 k2 = 2mV 0 /h ¯ 2 k 2 , so we approximate K a with ka and, using K/k the denominator, the reflection probability becomes
≪
−
≪
2
≃ 1 in
2
K 2 k2 2mV 0 V 02 2 P sin2 (2ka (2ka)) sin (2ka (2 ka) ) = sin 2 (2ka (2ka)), 2 2 2 2kK 4 E 2¯h k which agrees with the value we obtained from Fermi’s rule.
≃ ≃
−
≃
Show that the number of states g g (E ) dE d d2 Ω with energy in (E, ( E, E + + dE d E ) and momentum in the solid angle d d 2 Ω around p ¯ k of a particle of mass m m that moves freely subject to periodic p = h boundary conditions on the walls of a cubical box of side length L is 10.20∗
3
√
L m3/2 g (E ) dE d E d Ω = 2E d E dE dΩ E dΩ2 . (10. (10.10) 2π h ¯3 Hence show from Fermi’s golden rule that the cross-section for elastic scattering of such particles by a weak potential V ( V (x) from momentum ¯ momentum ¯hk into the solid angle d d 2 Ω around momentum ¯ momentum ¯hk′ is
2
2
m2 dσ = d 2 Ω d3 x ei(k−k )·x V ( V (x) . (10. (10.11) 4 2 (2π (2π) h ¯ Explain in what sense the potential has to be ‘weak’ for this Born Born approximation to the scattering cross-sectio cross-section n to be valid. valid. π/L, where where nx is an integer, and similarly for ky , kz . So eac each h stat statee Soln: We have kx = 2nx π/L, occupies occupies volume volume (2π/L (2π/L))3 in k in k-space. -space. So the number of states in the volume element k element k 2 dk d2 Ω is
2
g (E )dE )dE d Ω =
′
L 2π
3
k 2 dk d2 Ω
Using k Using k 2 = 2mE/ 2 mE/¯h ¯ 2 to eliminate eliminate k we obtain the required expression. In Fermi’s formula we must replace g replace g((E ) dE by g by g((E ) dE d E d2 Ω because this is the density of states that will make our detector ping if d2 Ω is its angular resolution. Then the probability per unit time of pinging is 3
2π L dk 2 ˙ = 2π g (E )d P )d2 Ω out V in 2 = k2 d Ω out V in 2 h ¯ h ¯ 2π dE The matrix element is 1 out V in = 3 d3 x e−ik ·x V ( V (x)eik·x L ˙ = dσ Now the cross section dσ dσ is defined by P d σ incoming flux = (v/L (v/L 3 )dσ )dσ = (¯ hk/mL hk/mL3 )dσ )dσ. Putting everything together, we find
| | | |
| |
| | | |
′
×
hk ¯hk 1 d σ = mL3 L6
2
3
d xe
′
−ik ·x
ik·x
V ( V (x)e
⇒
3
2π h ¯
L 2π
k2
dk 2 d Ω dE
mk dk/d k/ dE dσ = (2π (2π)2 h ¯2
3
d xe
′
−ik ·x
ik·x
V ( V (x)e
2
.
Eliminating k Eliminating k with ¯h2 k dk = m = md dE we obtain the desired expression. The Born approximation is valid providing the unperturbed wavefunction is a reasonable approximation proximation to the true wavefunc wavefunction tion throughout throughout the scattering scattering potential. potential. That is, we must be able to neglect “shadowing” by the scattering potential.
31 Assume that a LiH molecule comprises a Li + ion electrostatically bound to an H − ion, and that in the molecule’s molecule’s ground state the kinetic energies energies of the ions can b e neglected. Let the centres centres of the two ions be separated by a distance b distance b and calculate the resulting electrostatic binding energy under the assumption assumption that they attract like point charges. charges. Given Given that the ionisation ionisation energy of Li is 0.40 and using the result of Problem 11.6, show that the molecule has less energy than that of well separated hydrogen and lithium atoms for b b < 4. 4 .4a0 . Does this calculation suggest that LiH is a stable molecule? Is it safe to neglect the kinetic energies of the ions within the molecule? separated,, the energy required required to strip an electron from the Li Soln: When the LI and H are well separated − and park it on the H is E = (0. (0.4 + 1 0.955) = 0.445 . Now we recover recover some some of this energy by letting the Li+ and H− fall towards towards each other. When they have reached reached distance distance b the energy released is e2 a0 =2 4πǫ 0 b b This energy equals our original outlay when b = b = (2/ (2/0.445)a 445)a0 = 4.49 49a a0, which establishes the required proposition. In LiH the Li-H separation will be < 2a 2 a0 , because only at a radius of this order will the electron clouds of the two ions generate sufficient repulsion to balance the electrostatic attraction we have been calculating. calculating. At this separation separation the energy will be decidedly decidedly less than that of isolated isolated Li and H, so yes the molecule will be stable. In its ground state the molecule will have zero angular momentum, so there is no rotational kinetic kinetic energy to worry worry about. However However the length of the Li-H bond will oscillate oscillate around its equilibrium librium value, roughly as a harmonic harmonic oscillator, oscillator, so there will be zero-point zero-point energy. energy. However However,, this energy will suffice only to extend the bond length by a fraction of its equilibrium value, so it does not endanger the stability of the molecule. 11.7∗
R
−
R
R
R
∼
11.8∗
Two Two spin-on spin-onee gyros are in a box. box. Expres Expresss the states states j, j, m in which the box has definite angular momentum as linear combinations of the states 1, m 1, m′ in which the individual gyros have have definite angular angular momentum. momentum. Hence Hence show that 1 0, 0 = ( 1, 1 1 , 1 1, 0 1, 0 + 1, 1 1, 1 ). (11. (11.4) 3 By considering the symmetries of your expressions, explain why the ground state of carbon has l has l = = 1 rather than l than l = 2 or o r 0. 0 . What is the total spin angular momentum of a C atom? Soln: We have that J − 2, 2 = 2 2, 1 , J − 2, 1 = 6 2, 0 , J − 1, 1 = 2 1, 0 , J − 1, 0 = 2 1, 1 . We start from 2, 2 = 1 , 1 1 , 1 and apply J apply J − to both sides, obtaining 1 2 2, 1 = 2( 1, 0 1, 1 + 1, 1 1, 0 ) 2, 1 = ( 1, 0 1, 1 + 1, 1 1, 0 ) 2 Applying J − again we find 1 2, 0 = ( 1, 1 1, 1 + 2 1, 0 1, 0 + 1, 1 1, 1 ) 6 Next we identify 1, 1 as the linear combination of 1, 1 1, 0 and 1, 0 1, 1 that’s orthogonal to 2, 1 . It clearly is 1 1, 1 = ( 1, 0 1, 1 1, 1 1, 0 ) 2 We obtain 1, 0 by applying J applying J − to this 1 1, 0 = ( 1, 1 1, 1 1, 1 1, 1 ) 2 and applying J applying J − again we have 1 1, 1 = ( 1, 1 1, 0 1, 0 1, 1 ) 2
| |
| | |
| √ | − | − | | | | − |
| | √ | | √ | | | |
| √ | −
| √ | | | | ⇒ | √ | | | | | √ | − | | | | | − | | | | | |
|
| √ | | − | |
|
| √ | − | − | | −
| − √ | − | − | | −
32 Finally we have that 0, 0 is the linear combination of 1, 1 1, 1 , 1, 1 1, 1 and 1, 0 1, 0 that’s orthogonal to both 2, 0 and 1, 0 . By inspection it’s the given expression. The kets for j = 2 and j = 0 are symmetric under interchange of the m values of the gyros, while that for j = 1 is antisymmetric under interchange. Carbon has two valence electrons both in an l an l = 1 state, so each electron maps to a gyro and the box to the atom. When the atom is in the 1, 1 state, for example, from the above the part of the wavefunction that described the locations of the two valence electrons is 1 ( x1 1, 0 x2 1, 1 x 1 , x 2 1, 1 = x 1 1, 1 x 2 1 , 0 ) 2 This function is antisymmetric in its arguments so vanishes when x1 = x2 . Hence Hence in this state state of the atom, the electrons do a good job of keeping out of each other’s way and we can expect the electron-electron repulsion to make this state (and the other two l = 1 states) lower-lying than the l = 2 or l or l = 0 states, which lead to wavefunctions that are symmetric functions of x x 1 and x 2 . Since the wavefunction has to be antisymmetric overall, for the l the l = = 1 state it must be symmetric in the spins of the electrons, so the total spin has to be 1.
| | |
| − | | | −
| |
|
| √ | | − | |
11.9∗
Suppose we have three spin-one gyros in a box. Express the state 0, 0 of the box in which it has no angular momentum as a linear combination of the states 1, m 1, m′ 1, m′′ in which the individual individual gyros have have well-defined well-defined angular momenta. momenta. Hint: Hint: start with just two gyros in the box, giving states j, j, m of the box, and argue that only for a single value of j will it be possible to get 0, 0 by adding the third gyro; use results from Problem 11.8. Explain the relevance of your result to the fact that the ground state of nitrogen has l = 0. Deduce the value of the total electron spin of an N atom. Soln: Since when when we add add gyros with with spins j spins j 1 and j and j 2 the resulting j resulting j satisfies j j1 j2 j j1 + j + j2 , we will be able to construct the state 0, 0 on adding the third gyro to the box, only if the box has j = j = 1 before adding the last gyro. From Problem 11.8 we have that 1 0, 0 = ( 1, 1 1 , 1 1, 0 1, 0 + 1, 1 1, 1 ), 3 where we can consider the first ket in each product is for the combination of 2 gyros and the second ket is for the third gyro. We use Problem 11.8 again to express the kets of the 2-gyro box as linear combinations of the kets of individual gyros: 1 1 1 0, 0 = 1, 1 1, 0 1, 0 1, 1 1, 1 1, 1 1, 1 1, 1 1 , 1 1, 0 3 2 2 1 + 1, 0 1, 1 1, 1 1 , 0 1, 1 , 2 1 = 1, 1 1 , 0 1, 1 1, 0 1, 1 1, 1 1, 1 1, 1 1 , 0 6
|
| | |
| |
|
| − | ≤ ≤
|
| √ | − | − | | | | −
|
√ √ | − | − | | − | − √ | − | − | √ | | − | | | − √ | − | | − | | − | − | − | | | | − | | | | − | | | + 1 , 1 1 , 1 1, 0 + 1, 0 1, 1 1, 0
| − |
1, 1 1, 0 1, 0
This state is totally antisymmetric under exchange of the m the m values of the gyros. When we interpret the gyros as electrons and move to the position representation we find that the wavefunction of the valence electrons is a totally antisymmetric function of their coordinates, x1 , x2 , x3 . Hence the electrons do an excellent job of keeping out of each other’s way, and this will be the ground state. To be b e totally antisymmetric antisymmetric overall, overall, the state must be symmetric symmetric in the spin labels of the electrons, so the spin states will be + + + and the states obtained from this by application application of J J − . Thus Thus the total spin spin will be s = s = 32 .
| | |
11.10∗
Consider a system made of three spin-half particles with individual spin states . Write down a linear combination of states such as + + (with two spins up and one down) that is symmetric under any exchange of spin eigenvalues . Write down three other other totally symmetric symmetric states and say what total spin your states correspond to. Show that it is not possible to construct a linear combination of products of which is totally antisymmetric. What consequences do these results have for the structure of atoms such as nitrogen that have three valence electrons?
|± |±
| | | |− ±
|±
33 There are just three three of these product product states to consider because there there are just three places to put the single minus sign. The sum of these states is obviously totally symmetric: 1 ψ = + + + + + + + + 3 Three other totally symmetric state are clearly + + + and what you get from this ket and the one given by everywhere interchanging + and . These four kets are the kets 32 , m . A totally antisymmetric state would have to be constructed from the same three basis kets used above, so we write it as ψ ′ = a + + +b+ + +c + + On swapping the spins of the first and the third particles, particles, the first and third kets would would interchange, interchange, and this would have to generate a change of sign. So a = c and b = 0. Similarly, by swapping the spins on the first and second particles, we can show that a = 0. Hence ψ = 0, and we have shown that no nonzero ket has the required symmetry. States that satisfy the exchange principle can be constructed by multiplying a spatial wavefunction function that is totally antisymmetric antisymmetric in its arguments arguments by a totally totally symmetric symmetric spin function. Such states have have maximum total spin. In contrast to the situation situation with helium, conforming states states cannot be analogously constructed by multiplying a symmetric wavefunction by an antisymmetric spin function. Soln:
| √ | | |− | |−| |−| | | | | − |
|
| | |− | |−| |−| | − |
11.11∗
In this problem we use the variational principle to estimate the energies of the singlet and triplet states 1s2s of helium by refining the working of Appendix K. The idea is to use as the trial wavefunction symmetrised products of the 1s and 2s hydrogenic wavefunctions (Table 8.1) with the scale length aZ replaced by a by a 1 in the 1s wavefunction and by a different length a2 in the 2s wavefu wavefunction nction.. Explain Explain physically physically why with this choice of wavefunc wavefunction tion we expect H to be minimised with a with a 1 0.5a0 but a but a 2 distinctly larger. Using the scaling properties of the expectation values of the kinetic-energy and potential-energy operators, operators, show that a20 4a 4 a0 a20 a 0 H = + + 2a 2a0 (D(a1 , a2 ) E (a1 , a2 )) , a21 a1 4a22 a2 where D D and E are E are the direct and exchange integrals. Show that the direct integral can be written ∞ 2 1 D = dx x2 e−2x 8 (8 + 6y 6y + 2y 2 y 2 + y 3 )e−y , a2 0 4y where x x r1 /a1 and y y = r = r 1 /a2. Hence show that with α 1 + 2a 2 a2 /a1 we have 1 a22 4 6 6 12 D = 1 + 3 + 4 + 5 . 2 2 a1 a1 α α α α Show that with y with y = r = r 1 /a2 and ρ ρ = αr = αr2 /2a2 the exchange integral is 2 0∗ 0 E = = d3 x1 Ψ10 (x1 )Ψ20 (x1 ) (a1 a2 )3/2
∼
−
−
3
αy/ 2
× − − 1 r1
2a2 α
b
0
dρ (ρ2
a
b
dρ (ρ
a
2
dρ (ρ
3
−ρ
/α)e − ρ /α)e
ρ3 /α)e /α)e−ρ =
≡
−
√
and
±
≡
Using
−
R
+
2
∞
2a2 α
αy/2
dρ (ρ
2
/α)e − ρ /α)e
−[{(1 − α3 )(2 + 2ρ 2ρ + ρ2 ) − α1 ρ3 }e−ρ ]ba
ρ2 /α)e /α)e−ρ =
−ρ
−[{(1 − α2 )(1 + ρ) − α1 ρ2}e−ρ]ba
.
34 show that
∞
− − −{ − × { −
2 E = = (a1 a2 )3 1 r1
0
dr1 r12 1
r1 2a2
e−αr1 /2a2
3
2a2 α
3 ) α
2(1
3 )(2 α
(1
2a2 α
+
+ αy + αy + 14 α2 y 2 )
− 18 α2y3}e−αy/2
2
(1
− α2 )(1 + 12 αy) αy) − 14 αy 2 }e−αy/2
50 66 8a22 10 + , 3 α5 a1 α α2 Using the above results, show numerically that the minimum of H occurs near a a 1 = 0.5a0 and a2 = 0.8a0 in both the singlet and triplet cases. Show that for the triplet the minimum is 60 60..11eV and for the singlet it is 57 57..0 eV. eV. Compare these results with the experimental values and the values obtained in Appendix K. Soln: We’d expect the 2s electron to see a smaller nuclear charge than the 1s electron and therefore to have a longer scale length since the latter scales inversely with the nuclear charge. The 1s orbit taken on its own has K has K = (a0 /a1 )2 because the kinetic energy is for hydrogen and it is proport proportion ional al to the invers inversee square square of the wavefu wavefunct nction ion’s ’s scale scale length. length. The 1s potent potential ial energy is W = 4(a 4(a0 /a1 ) because in hydrogen it is 2 , and it’s proportional to the nuclear charge charge and to the inverse inverse of the wavefu wavefunctio nction’s n’s scale length. Similarly Similarly,, the 2s orbit taken on its 1 1 2 own has K = 4 (a0 /a2 ) and W = (a0 /a2) , both just 4 of the 1s values from the 1/n 1 /n2 in the Rydberg formula. formula. The electron-elect electron-electron ron energies energies are ( D E )2a )2a0 because = e2 /8πǫ0 a0 . The required expression for H now follows. When the scale length a length a Z is relabelled a relabelled a 1 where it relates to the 1s electron and is relabelled a 2 where it relates to the 2s electron, equation (K.2) remains valid with ρ redefined to ρ to ρ r2 /a2 and a nd x x replaced by y by y r1 /a2 . With these definitions the first line of equation (K.2) remains valid and the second line becomes becomes ∞ 2 1 D = dx x2 e−2x 8 (8 + 6y 6y + 2y 2 y 2 + y 3 )e−y a2 0 4y (11. (11.5) ∞ ∞ 1 x −2x x 2 2 3 4 −(2x+y) = 8 dx x e dx 2 (8y (8y + 6y 6 y + 2y 2y + y )e 2a2 y y 0 0 =
−
−
R
−
R
R
− R
R
−
R
±
R
R
≡
≡
Now x/y Now x/y = a = a 2 /a1 and
− − − −
∞ n −αy 0 dy y e
= α −(n+1) n! so with α
2 a2 /a1 we have ≡ 1 + 2a
1 a2 a 32 8 6 2 1 D = 2 + 2! + 3! + 4! 3 2a2 a1 a1 α2 α3 α4 α5 1 a22 4 6 6 12 = 1 + 3+ 4+ 5 2 2 a1 a1 α α α α which agrees with equation (K.2) when a 1 = a = a 2 = a = a Z as it should. Equation (K.3) for the exchange integral becomes 1 0∗ 0 E = = d3 x1 Ψ10 (x1 )Ψ20 (x1 ) / 3 2 2(a 2(a1 a2 )
√
×
dr2 dθ2
r22 (1
×
−
(11. (11.6)
2
(11. (11.7)
− r2/2a2)sin θ2e−αr /2a |r12 + r22 − 2r1r2 cos θ2|
After integrating over θ over θ as in Box 11.1, we have 2 0∗ 0 E = d3 x1 Ψ10 (x1 )Ψ20 (x1 ) (a1 a2 )3/2 r1 r2 r2 dr2 2 1 e−αr2 /2a2 + r 2 a 1 2 0
√
2
∞
r1
.
−
dr2 r2 1
r2 2a2
e
−αr2 /2a2
36
Estimatess of the energy in electron volts of the 1s2s triplet triplet excited state of helium. helium. The estimates estimates Figure Figure 11.4 Estimate are obtained by taking the expectation of the Hamiltonian using anti-symmetrised products of 1s and 2s hydrogenic wavefunctions that have scale lengths a lengths a 1 and a and a 2 , respectively. respectively.
our final result is 8a2 16 E = = 2 23 1 α3 1 α a1 α3 8a2 = 5 23 16 1 α3 1 α a1 8a2 50 66 = 5 23 10 + 2 α a1 α α
− − − − − − − 2 α
2 α
(1
4(1
−
3 4 1 α ) α α2 (1
−
2 4 2α ) + ( α
−
2 1) 3 (1 α
−
3 1 2α ) + 2
6 (1 α4
−
4 2α )
− α3 )(1 − α1 ) + ( α4 − 1)(2 − α3 ) + α3 (1 − α2 )
,
which when a when a 1 = a 2 = a = a Z agrees with equation (K.4) as it should. Figure 11.4 shows H for the triplet state as a function of a1 and a2 . The The surfac surfacee has its minimum 60 60..11eV at a1 = 0.50 50a a0, a2 = 0.82 82a a0 . As expected expected,, this this minim minimum um is deeper deeper than our estimate 57 57..8 eV from perturb perturbati ation on theory theory,, and it occurs occurs when when a2 is significantly greater than 0.5a0. It is closer to the experimental value, 59 59..2 eV, than the estimate estimate from perturb p erturbation ation theory. theory. A variational value is guaranteed to be larger than the experimental value only for the ground state, and our variational value for the first excited state lies below rather than above the experimental value. The variational estimate of the singlet 1s2s state’s energy is 57 57..0 eV, which which lies between between the values from experiment ( 58 58..4 eV) and perturbation perturbation theory theory ( 55 55..4 eV). eV).
− −
−
−
−
−
