Joel Broida
UCSD Fall 2009
Phys 130B
QM II
Homework Set 5 Solutions
1. From E = p2 /2m + V we have p2 = 2m(E − V ) so that
Hrel = −
p4
(E − V )2
1
=
−
=−
(E 2 − 2EV + V 2 )
3
2
2
8m c
2mc
2mc2
and hence
1
(E 2 − 2EhV i + hV 2 i) .
2mc2
For the harmonic oscillator we have
1
En = ~ω n +
2
(1)
Erel = hHrel i = −
and
1
1 2
kx = mω 2 x2 .
2
2
To evaluate hV i and hV 2 i where the expectation values are taken with respect to the harmonic oscillator states, we use the creation and annihilation
operators
∓ip + mωx
a± = √
2~mω
to write
r
~
(a+ + a− ) .
x=
2mω
Since the orthonormal harmonic oscillator states |ni satisfy
√
√
and
a− |ni = n |n − 1i
a+ |ni = n + 1 |n + 1i
V =
we have
hV i =
1
~k
khx2 i =
hn|(a2+ + a+ a− + a− a+ + a2− )|ni
2
4mω
=
~k
hn|a+ a− + a− a+ |ni
4mω
~k
~k
[n + (n + 1)] =
4mω
2mω
~ω
1
=
n+
.
2
2
=
1
1
n+
2
Note that the only nonzero terms in the expectation value are those with an
equal number of creation and annihilation operators. With this in mind, we
then also have
hV 2 i =
k2 4
~2 k 2
hn|(a2+ + a+ a− + a− a+ + a2− )2 |ni
hx i =
4
16m2 ω 2
=
=
=
~2 k 2
hn|(a2+ a2− + a+ a− a+ a− + a+ a− a− a+
16m2 ω 2
+ a− a+ a+ a− + a− a+ a− a+ + a2− a2+ )|ni
~2 k 2
[n(n − 1) + n2 + n(n + 1)
16m2 ω 2
+ n(n + 1) + (n + 1)2 + (n + 1)(n + 2)]
~ω
4
2
[6n2 + 6n + 3]
If we note that 2hV i = E, it is then easy to substitute the result for hV 2 i to
obtain
3~2 ω 2
(1)
Erel = −
[2n2 + 2n + 1] .
32mc2
2. From equation 2.59 in the notes, we
(0) En n
(1)
Efs = − 2 α2
−
n
j + 1/2
have
13.6 eV 2
n
3
3
=−
α
−
4
n4
j + 1/2 4
where
e2
1
≈
.
~c
137.036
(a) The n = 2 level has l = 0, 1 so the possible j values are j = 1/2 and
j = 3/2, 1/2. Thus the n = 2 level will split into two levels. We have
2
3
13.6 eV 2
(1)
= −5.66 × 10−5 eV
α
Efs (j = 1/2) = −
−
24
1/2 + 1/2 4
2
3
13.6 eV 2
(1)
= −1.13 × 10−5 eV
α
−
Efs (j = 3/2) = −
24
3/2 + 1/2 4
α=
(b) The n = 3 level has l = 0, 1, 2 so the possible j values are j = 1/2,
j = 3/2, 1/2 and j = 5/2, 3/2. Thus the n = 3 level will split into three
levels. The fine structure corrections are
13.6 eV 2
3
3
(1)
Efs (j = 1/2) = −
= −2.01 × 10−5 eV
α
−
34
1/2 + 1/2 4
3
3
13.6 eV 2
(1)
= −0.67 × 10−5 eV
α
−
Efs (j = 3/2) = −
34
3/2 + 1/2 4
2
(1)
Efs (j = 5/2) = −
13.6 eV 2
3
3
= −0.22 × 10−5 eV
α
−
34
5/2 + 1/2 4
(c) The energy level diagram showing all n = 3 to n = 2 transitions looks like
this (very much not to scale):
n=3
j = 5/2
j = 3/2
j = 1/2
n=2
j = 3/2
j = 1/2
1 2 3 4 5 6
(d) The difference in energy levels for a transition is given by
(0)
(1)
(0)
(1)
(0)
(0)
(0)
(0)
(1)
(1)
(E3 + E3,fs ) − (E2 + E2,fs ) = (E3 − E2 ) + (E3,fs − E2,fs )
= (E3 − E2 ) + ∆E
where
(0)
=−
E3
so that
13.6
= −1.51 eV
32
and
(0)
(0)
E3 − E2
(0)
E2
=−
13.6
= −3.4 eV
22
= 1.89 eV .
We have
∆E1 = (−2.01 − (−1.13)) × 10−5 eV = −0.88 × 10−5 eV
∆E2 = (−0.67 − (−1.13)) × 10−5 eV = 0.46 × 10−5 eV
∆E3 = (−0.22 − (−1.13)) × 10−5 eV = 0.91 × 10−5 eV
∆E4 = (−2.01 − (−5.66)) × 10−5 eV = 3.65 × 10−5 eV
∆E5 = (−0.67 − (−5.66)) × 10−5 eV = 4.99 × 10−5 eV
∆E6 = (−0.22 − (−5.66)) × 10−5 eV = 5.44 × 10−5 eV
Note that the 1/2 → 3/2 transition has an energy spacing that is less
than the spacing between the unperturbed n = 3 and n = 2 levels.
3
(e) Using ∆Ei = hνi and h = 4.136 × 10−15 eV-sec, we find the following
frequencies (in Hz):
ν1 = 2.13 × 109
ν2 = 1.11 × 109
ν4 = 8.82 × 109
ν5 = 12.06 × 109
ν3 = 2.20 × 109
ν6 = 13.15 × 109
Therefore the spacings of adjacent transitions are as follows. (Note that
the spacings within each n level are much closer than the spacing between
n = 3 and n = 2.)
ν2 − ν1 = (∆E2 − ∆E1 )/h = 3.24 × 109 Hz
ν3 − ν2 = (∆E3 − ∆E2 )/h = 1.09 × 109 Hz
ν4 − ν3 = (∆E4 − ∆E3 )/h = 6.62 × 109 Hz
ν5 − ν4 = (∆E5 − ∆E4 )/h = 3.24 × 109 Hz
ν6 − ν5 = (∆E6 − ∆E5 )/h = 1.09 × 109 Hz
3. For the n = 3 level of hydrogen we have the possible values l = 0, 1, 2 and the
(0)
unperturbed energy E3 = −13.6/32 = −1.51 eV.
(a) In the strong-field regime we have the first-order energy correction due to
the magnetic field (equation (2.70))
(1)
Enlml ms = µB B(ml + 2ms ) := β(ml + 2ms )
and this is in addition to the fine structure energy (equation (2.72))
l(l + 1) − ml ms
1
A
α2
(1)
−
eV := −1.51α2 eV
Efs = −1.51
3
l(l + 1)(l + 1/2)
4
3
where for l = 0 we take the term in square brackets to be equal to 1. The
total energy of each state is then given by (in eV)
(0)
(1)
(1)
E3 = E3 + Efs + Enlml ms
A 2
= −1.51 1 + α + β(ml + 2ms ) .
3
4
Tabulating these results, we obtain the following table:
state
|n l ml ms i
ml + 2ms
A
total energy E3 (eV)
1
3/4
−1
3/4
−1.51(1 + α2 /4) + β
l = 1 |3 1 1 1/2i
2
1/4
l = 1 |3 1 1 −1/2i
0
7/12
l = 1 |3 1 0 1/2i
1
5/12
l = 1 |3 1 0 −1/2i
−1
5/12
0
7/12
l = 1 |3 1 −1 −1/2i
−2
1/4
l = 2 |3 2 2 1/2i
3
1/12
l = 2 |3 2 2 −1/2i
1
l = 2 |3 2 1 1/2i
2
13/60 −1.51(1 + 13α2 /180) + β
l = 2 |3 2 1 −1/2i
0
l = 2 |3 2 0 1/2i
1
l = 2 |3 2 0 −1/2i
−1
l = 2 |3 2 −1 1/2i
0
l = 0 |3 0 0 1/2i
l = 0 |3 0 0 −1/2i
l = 1 |3 1 −1 1/2i
l = 2 |3 2 −1 −1/2i
−2
l = 2 |3 2 −2 −1/2i
−3
l = 2 |3 2 −2 1/2i
−1
7/60
−1.51(1 + α2 /4) + β
−1.51(1 + α2 /12) + 2β
−1.51(1 + 7α2 /36)
−1.51(1 + 5α2 /36) + β
−1.51(1 + 5α2 /36) − β
−1.51(1 + 7α2 /36)
−1.51(1 + α2 /12) − 2β
−1.51(1 + α2 /36) + 3β
−1.51(1 + 7α2 /180) + 2β
11/60 −1.51(1 + 11α2 /180)
3/20
3/20
−1.51(1 + α2 /20) + β
−1.51(1 + α2 /20) − β
11/60 −1.51(1 + 11α2 /180)
7/60
−1.51(1 + 7α2 /180) − 2β
13/60 −1.51(1 + 13α2 /180) − β
1/12
−1.51(1 + α2 /36) − 3β
(b) For the weak-field regime we have (equation (2.73))
av
Emag
= βmj gJ
where
gJ = 1 +
=
j(j + 1) + s(s + 1) − l(l + 1)
2j(j + 1)
3 3/4 − l(l + 1)
+
2
2j(j + 1)
and also (equation (2.59))
(1)
Efs = −1.51
α2
A
1
1
eV := −1.51α2 eV .
−
3 j + 1/2 4
3
5
Thus the total energy is given by
(0)
(1)
av
E3 = E3 + Efs + Emag
A 2
= −1.51 1 + α + βmj gJ eV .
3
Since the possible values of j are l + s, l + s − 1, . . . , |l − s|, and mj =
j, j − 1, . . . , −j, we have the following table.
state
|3 l j mj i
gJ
A
total energy E3 (eV)
l = 0; j = 1/2 |3 0 1/2 1/2i
2
3/4
2
3/4
−1.51(1 + α2 /4) + β
l = 1; j = 3/2 |3 1 3/2 3/2i
4/3
1/4
4/3
1/4
l = 1; j = 3/2 |3 1 3/2 −1/2i 4/3
1/4
l = 1; j = 3/2 |3 1 3/2 −3/2i 4/3
1/4
l = 1; j = 1/2 |3 1 1/2 1/2i
2/3
3/4
l = 1; j = 1/2 |3 1 1/2 −1/2i 2/3
3/4
l = 2; j = 5/2 |3 2 5/2 5/2i
6/5 1/12 −1.51(1 + α2 /36) + 3β
l = 2; j = 5/2 |3 2 5/2 1/2i
6/5 1/12 −1.51(1 + α2 /36) + 3β/5
l = 0; j = 1/2 |3 0 1/2 −1/2i
l = 1; j = 3/2 |3 1 3/2 1/2i
−1.51(1 + α2 /4) − β
−1.51(1 + α2 /12) + 2β
−1.51(1 + α2 /12) + 2β/3
−1.51(1 + α2 /12) − 2β/3
−1.51(1 + α2 /12) − 2β
−1.51(1 + α2 /4) + β/3
−1.51(1 + α2 /4) − β/3
6/5 1/12 −1.51(1 + α2 /36) + 9β/5
l = 2; j = 5/2 |3 2 5/2 3/2i
l = 2; j = 5/2 |3 2 5/2 −1/2i 6/5 1/12 −1.51(1 + α2 /36) − 3β/5
l = 2; j = 5/2 |3 2 5/2 −3/2i 6/5 1/12 −1.51(1 + α2 /36) − 9β/5
l = 2; j = 5/2 |3 2 5/2 −5/2i 6/5 1/12 −1.51(1 + α2 /36) − 3β
l = 2; j = 3/2 |3 2 3/2 3/2i
4/5
1/4
l = 2; j = 3/2 |3 2 3/2 1/2i
4/5
1/4
l = 2; j = 3/2 |3 2 3/2 −1/2i 4/5
1/4
l = 2; j = 3/2 |3 2 3/2 −3/2i 4/5
1/4
−1.51(1 + α2 /12) + 6β/5
−1.51(1 + α2 /12) + 2β/5
−1.51(1 + α2 /12) − 2β/5
−1.51(1 + α2 /12) − 6β/5
(c) Now for the intermediate-field regime. Note that there are 18 degenerate
unperturbed states. (The hydrogen atom energies En are n2 -fold degenerate, and this becomes 2n2 if we include spin. Alternatively, the l = 0
level has one state, the l = 1 level has 2l + 1 = 3 states, and the l = 2 level
has 2l + 1 = 5 states, for a total of 9. Multiplying this by 2 for spin and
we have a total of 18 states.) As in the notes, we choose to work in the
|j mi basis because both Hrel and Hso are already diagonal in this basis,
and hence so is Hfs . This means that we can immediately use equation
6
(2.59) to write
(1)
Efs
(0) 2 E α
3
3
1
=−
eV
−
34
j + 1/2 4
4
13.6α2
− 1 eV
=−
108
j + 1/2
4
= −3γ
− 1 eV
j + 1/2
where it is convenient to define
γ=
13.6 2
α .
324
Thus we have
(1)
Efs = −9γ
for j = 1/2
= −3γ
for j = 3/2
= −γ
for j = 5/2 .
(1)
Efs
(1)
Efs
The magnetic Hamiltonian is
Hmag =
eB
µB B
β
(Lz + 2Sz ) =
(Lz + 2Sz ) = (Lz + 2Sz ) .
2me c
~
~
To evaluate the matrix elements of Hmag , we need to write our 18 basis
|j mi states in terms of |ml ms i states using Clebsch-Gordan coefficients.
For l = 0 and l = 1 the |j mi states are just the 8 states constructed in
the notes, Section 2.5.3, where we worked out the intermediate-field case
for n = 2. Therefore the β terms in the matrix of H ′ = Hfs + Hmag are
the same as in the notes, and the Hfs contributions are −9γ for j = 1/2
and −3γ for j = 3/2 (as compared with −5ξ and −ξ in the notes). Thus
this part of H ′ consists of the blocks
"
(−9γ + β);
(−9γ − β);
−3γ + 23 β
−
−
√
2
3 β
√
2
3 β
−9γ + 31 β
#
(−3γ + 2β);
;
"
−3γ − 23 β
−
√
2
3 β
(−3γ − 2β)
−
√
2
3 β
−9γ − 31 β
#
.
So now we need only construct the remaining 10 states corresponding to
l = 2, or j = 5/2, 3/2. Since these are the result of adding a spin 2 state to
a spin 1/2 state, we use the table of Clebsch-Gordan coefficients to write
the various |j mi states as linear combination of |m1 m2 i = |ml ms i states.
Note that we group these so that linear combinations of the same |ml ms i
7
states are kept together. This will put the matrix of H ′ = Hfs + Hmag
into block diagonal form.
| 25
5
2i
= |2 12 i
| 25 − 25 i = | −2 − 12 i
q
q
= 15 |2 − 21 i + 45 |1 21 i
| 25 32 i
q
q
| 32 32 i
= 45 |2 − 21 i − 15 |1 21 i
q
q
= 25 |1 − 21 i + 35 |0 21 i
| 25 12 i
q
q
| 32 12 i
= 35 |1 − 21 i − 25 |0 21 i
q
q
| 25 − 21 i = 35 |0 − 21 i + 25 | −1 12 i
q
q
| 23 − 21 i = 25 |0 − 21 i − 35 | −1 12 i
q
q
| 25 − 23 i = 45 | −1 − 12 i + 51 | −2 21 i
q
q
| 32 − 23 i = 15 | −1 − 12 i − 54 | −2 21 i
We already showed what the matrix elements of Hfs are for these states,
so we need only evaluate the matrix elements of Hmag . Each of the above
pair of equations yields a 2 × 2 matrix since each pair is orthogonal to any
other pair. For example, we have
!
r r 5 3E
1E
1
4 1E
= β(Lz + 2Sz )
+
Hmag 2 −
1
2 2
5
2
5
2
!
r r 1E
1
4 1E
+2
2 −
1
5
2
5
2
=β
and hence
5 3E
D5 3
Hmag 2 2
2 2
!
! r
r r D
r D
1 1 1E
1
4
1 4 1E
2 − +
1 +2
=β
2 −
1
5
2
5
2
5
2
5
2
=β
1
4
+2·
5
5
=
9
β
5
(1)
so that together with Efs = −γ (since this is the diagonal element for
j = 5/2) we would have the H ′ matrix element −γ + (9/5)β. And the
other nonvanishing matrix element for the | 52 32 i state is
5 3E
D3 3
2
2
2
=− β
−2·
=β
Hmag 2 2
2 2
5
5
5
8
but this is off-diagonal so there is no additional term due to Hfs .
Continuing in this manner, we find the following blocks corresponding to
these 10 states:
(−γ + 3β); (−γ − 3β)
√
"
#
#
"
−γ + 95 β
− 52 β
− 56 β
−γ + 35 β
;
√
− 52 β
−3γ + 56 β
−3γ + 25 β
− 56 β
√
"
#
#
"
−γ − 53 β
−γ − 95 β
− 56 β
− 52 β
.
;
√
−3γ − 65 β
− 52 β
−3γ − 52 β
− 56 β
Thus we see that the matrix of H ′ is the direct sum of six 1 × 1 blocks and
six 2 × 2 blocks. The 1 × 1 blocks are already the eigenvalues, so we need
to find the eigenvalues of the six 2 × 2 blocks. There are three distinct
characteristic equations (and another three that result from letting β →
−β). These are
2
1
2
−3γ + β − λ
−9γ + β − λ − β 2 = 0
(1a)
3
3
9
9
6
4
−γ + β − λ
−3γ + β − λ − β 2 = 0
(1b)
5
5
25
3
2
6
−γ + β − λ
−3γ + β − λ − β 2 = 0 .
(1c)
5
5
25
Multiplying these out we have the three equations
λ2 + λ(12γ − β) + γ(27γ − 7β) = 0
33
2
2
2
=0
λ + λ(4γ − 3β) + 3γ − γβ + 2β
5
11
2
λ + λ(4γ − β) + γ 3γ − β = 0 .
5
Using the quadratic formula, these have the roots
r
1
1
(a)
λ± = −6γ + β ± 9γ 2 + γβ + β 2
2
4
r
3
3
1
(b)
λ± = −2γ + β ± γ 2 + γβ + β 2
2
5
4
r
1
1
1
(c)
λ± = −2γ + β ± γ 2 + γβ + β 2
2
5
4
9
(2a)
(2b)
(2c)
(3a)
(3b)
(3c)
Combining our results, we have the following nine energy levels:
(0)
E1 = E3 − 9γ + β
(0)
E2 = E3 − 3γ + 2β
(0)
E3 = E3 − γ + 3β
E4 =
(0)
E3
− 6γ +
(0)
E5 = E3 − 6γ +
(0)
E6 = E3 − 2γ +
(0)
E7 = E3 − 2γ +
(0)
E8 = E3 − 2γ +
(0)
E9 = E3 − 2γ +
r
1
1
β + 9γ 2 + γβ + β 2
2
4
r
1
1
β − 9γ 2 + γβ + β 2
2
4
r
3
3
1
β + γ 2 + γβ + β 2
2
5
4
r
3
3
1
β − γ 2 + γβ + β 2
2
5
4
r
1
1
1
β + γ 2 + γβ + β 2
2
5
4
r
1
1
1
β − γ 2 + γβ + β 2
2
5
4
plus another nine levels that result from letting β → −β.
What about the limiting cases of a very strong field (i.e., γ ≪ β) and
very weak field (i.e., β ≪ γ)? In the strong-field case, we ignore√terms of
order γ 2 , and expand the square root to first order in γ/β using 1 + x ≈
1 + x/2. For example,
r
r
1/2
1
1
β
β
γ
9γ 2 + γβ + β 2 ≈ γβ + β 2 =
≈ +γ.
1+4
4
4
2
β
2
Continuing in this manner yields the strong-field limits
(0)
E4 ≈ E3 − 5γ + β
(0)
E5 ≈ E3 − 7γ
7
(0)
E6 ≈ E3 − γ + 2β
5
13
(0)
E7 ≈ E3 − γ + β
5
9
(0)
E8 ≈ E3 − γ + β
5
11
(0)
E9 ≈ E3 − γ
5
Since 324 = 9(36), we note that
γ=
13.6 2
α2
α = 1.51
324
36
10
and hence it is not hard to see that these limits reproduce the table in
part (a). Similarly, in the weak-field limit β ≪ γ we ignore terms of order
β 2 , expand the square root to first order in β/γ, and find that
2
(0)
E4 ≈ E3 − 3γ + β
3
1
(0)
E5 ≈ E3 − 9γ + β
3
9
(0)
E6 ≈ E3 − γ + β
5
6
(0)
E7 ≈ E3 − 3γ + β
5
3
(0)
E8 ≈ E3 − γ + β
5
2
(0)
E9 ≈ E3 − 3γ + β
5
Again, we see that these reproduce the results of part (b).
In the figure below we plot E1 – E9 . The three points on the E-axis
(1)
represent the three possible values of Efs that we would have if B = β =
0.
E
25
20
15
10
5
Β
2
4
6
8
10
-5
The slope of the curves for E1 – E3 are constant, but the slope of the
curves for E4 – E9 varies as β increases. For example, at small β the
slope of E4 is 2/3, but at large β it is 1. Similarly, the slope of E5 starts
out as 1/3, but then decreases to 0. Also note that this is only half of the
energies. There are another nine curves that result from letting β → −β,
and hence they have a negative (downwards) slope.
11
4. As Rin Example 3.1 of the notes, the potential energy is given by
−e E · dr so that
2
2
H ′ (t) = −eE e−t /τ z .
R
F · dr =
The transition probability is given by
Z
2
1 ∞ ′
Pif (t) = 2 Hf i (t)eiωf i t dt
~
−∞
where the hydrogen atom wave functions ψnlm are ψi = ψ100 and ψf = ψ2lm .
Since the matrix element is proportional to h2 l m|z|1 0 0i, we see that parity
considerations tell us that the only nonzero term is when l = 1. This is because z is an odd function, and the parity of the spherical harmonics goes like
(−1)l . Since Y00 is even, we must have a transition to an odd state, and hence
l = 1 is the only possibility for n = 2. Thus the potential final states are ψ210 ,
ψ211 and ψ21 −1 .
The wave functions are given by ψnlm = Rnl Ylm , and the relevant functions
are
−5/2
a0
re−r/2a0
R21 = √
24
Y00
Y11
−3/2 −r/a0
R10 = 2a0
1
= √
4π
r
=−
Y10
3 iφ
e sin θ
8π
Y1−1
e
=
r
3
cos θ
4π
=
r
3 −iφ
e
sin θ
8π
Since z = r cos θ and the volume element depends on sin θ dθ, the fact that
π
Z π
Z π
sin3 θ 2
2
cos θ sin θ dθ =
sin θ d sin θ =
=0
3 0
0
0
tells us that the only nonzero matrix element Hf′ i ∼ h2 1 m| cos θ|1 0 0i is
h2 1 0| cos θ|1 0 0i. So we have
Z
2
2
Hf′ i = −eE e−t /τ
R21 Y10 [r cos θ]R10 Y00 r2 sin θ drdθdφ
r
Z
1 −4
1
a
r4 e−3r/2a0 cos2 θ drd cos θdφ
= −eE e
2π 8 0
r
Z
1 −4 ∞ 4 −3r/2a0
−t2 /τ 2 2
= −eE e
a0
dr .
r e
3 8
0
−t2 /τ 2
The integral was done in HW4, Problem 3, and is
5
Z ∞
2a0
4 −3r/2a0
.
dr = 4!
r e
3
0
12
This leaves us with
Hf′ i
= −eE e
−t2 /τ 2
r
1
8
2 9 a0
35
and therefore
Pif (t → ∞) =
eE
~
2 215 a20
310
Z
∞
e
−t2 /τ 2 iω21 t
e
−∞
2
dt .
But this integral was evaluated in Example 3.1 of the notes and is equal to
√
2
2
πτ 2 e−ω21 τ /4 . Thus we finally arrive at
Pif (t → ∞) =
eE
~
2 215 a20
310
2
πτ 2 e−ω21 τ
2
/2
.
5. (a) For a spin one-half particle at rest, the Hamiltonian is simply
H = −µ · B = −
gq
γ~
S · B = −γS · B = − σ · B
2mc
2
where the Pauli matrices are given by
"
#
"
#
0 1
0 −i
σx =
σy =
1 0
i 0
σz =
"
1
0
0 −1
#
.
For the given field, this becomes
H=−
=−
=−
γ~
(Brf cos ωt σx − Brf sin ωt σy + B0 σz )
2
#
"
B0
Brf cos ωt + iBrf sin ωt
γ~
2
γ~
2
"
Brf cos ωt − iBrf sin ωt
#
B0
Brf eiωt
.
Brf e−iωt
−B0
−B0
(b) The time-dependent Schrödinger equation is
i~
which becomes
"
i~
ȧ(t)
ḃ(t)
#
γ~
=−
2
∂
χ(t) = Hχ(t)
∂t
"
B0
Brf eiωt
Brf e−iωt
−B0
13
#"
a(t)
b(t)
#
.
Equating components on both sides of this equation we have
ȧ =
iγ
iγ
B0 a + Brf eiωt b
2
2
ḃ =
iγ
iγ
Brf e−iωt a − B0 b
2
2
Defining
ω0 = γB0
and
Ω = γBrf
these equations become
ȧ(t) =
ḃ(t) =
i
Ωeiωt b + ω0 a
2
i
Ωe−iωt a − ω0 b .
2
(4a)
(4b)
(c) Taking the derivative of (4a) and substituting (4b) for ḃ we have
−2i ä = iωΩeiωt b + Ωeiωt ḃ + ω0 ȧ
= iωΩeiωt b + Ωeiωt
i
Ωe−iωt a − ω0 b + ω0 ȧ
2
i
i
= iωΩeiωt b + Ω2 a − ω0 Ωeiωt b + ω0 ȧ
2
2
i
ω0 + ω0 ȧ + Ω2 a .
= iΩeiωt b ω −
2
2
But from (4a) we can write
iΩeiωt b = 2ȧ − iω0 a
so that
i
ω0 + ω0 ȧ + Ω2 a
−2i ä = (2ȧ − iω0 a) ω −
2
2
i 2
= 2ω ȧ +
Ω − ω0 (2ω − ω0 ) a
2
i 2
= 2ω ȧ +
Ω − ω 2 + (ω − ω0 )2 a .
2
Rearranging, this becomes
1 2
Ω − ω 2 + (ω − ω0 )2 a = 0 .
4
This is now just a second-order differential equation with constant coefficients, and its solutions are of the form ert where r is a root of the
polynomial
1 2
r2 − iωr +
Ω − ω 2 + (ω − ω0 )2 = 0 .
4
ä − iω ȧ +
14
From the quadratic formula, the roots of this are
r=
iω
iω iω ′
ip 2
Ω + (ω − ω0 )2 :=
±
±
2
2
2
2
and hence the general solution for a can be written in the form
ω′t
ω′t
iωt/2
+ β sin
.
a(t) = e
α cos
2
2
Clearly we have α = a(0) := a0 . To find β, we take the derivative of a(t)
to write
iω iωt/2
ω ′ iωt/2
ω′t
ω′t
ω′t
ω′t
ȧ(t) =
− e
.
α cos
α sin
e
+ β sin
− β cos
2
2
2
2
2
2
Then (using α = a0 )
ω′
iω
a0 + β
2
2
2
iω
β = ′ ȧ(0) − a0 .
ω
2
ȧ(0) =
or
Using (4a) to evaluate ȧ(0) and letting b(0) = b0 , this becomes
2 i
i
iω
β= ′
(Ωb0 + ω0 a0 ) − a0 = ′ [Ωb0 + (ω0 − ω)a0 ] .
ω 2
2
ω
Thus the solution for a(t) is
ω′t
ω′t
i
a(t) = a0 cos
eiωt/2 .
+ ′ [Ωb0 + (ω0 − ω)a0 ] sin
2
ω
2
To solve for b(t) you could repeat all this work starting from (4b). However, it is a lot easier to realize that equations (4a) and (4b) are symmetric
under the exchange a ↔ b along with ω → −ω and ω0 → −ω0 . This means
that we can immediately write down the solution for b(t) as
ω′t
i
ω′t
e−iωt/2 .
+ ′ [Ωa0 + (ω − ω0 )b0 ] sin
b(t) = b0 cos
2
ω
2
(d) If the particle is initially in the spin up state, then a0 = 1 and b0 = 0.
Therefore
i
ω ′ t −iωt/2
b(t) = ′ Ω sin
e
ω
2
so the probability of being in the spin down state is
2
Ω
ω′t
2
Pdown (t) = |b(t)| =
sin2
.
′
ω
2
15
(e) The resonance curve satisfies the equation
P (ω) =
Ω2
.
(ω − ω0 )2 + Ω2
To plot this in Mathematica so the whole curve would show, I simply let
Ω = 2 and ω0 = 2. Then schematically, the resonance curve looks like
this.
1.0
0.8
0.6
0.4
0.2
-10
5
-5
10
No matter what Ω is, the maximum value of P (ω) is 1. Therefore, the
width at half-maximum comes from solving
1
Ω2
=
2
(ω − ω0 )2 + Ω2
which implies that (ω − ω0 )2 = Ω2 or
ω = ω0 ± Ω .
Thus the width at half-maximum is
∆ω = 2Ω .
(f) For the proton we have
gp e
ω0 = γB0 =
B0
2mp c
=
5.586(4.803 × 10−10 statC)
104 gauss
2(1.673 × 10−24 g)(2.998 × 1010 cm/s)
= 2.674 × 108 rad/s
so that
ω0
= 4.257 × 107 Hz .
2π
By definition, Ω = γBrf . Then
ν0 =
∆ω = 2Ω = 2(2.674 × 104 )10−2 = 534.8 rad/s
or
∆ν =
534.8
= 85.12 Hz .
2π
16