In order to determine the hyperfine constant A of positronium, it is of interest to study the energy and the lifetime of the level corresponding to the state |ψ+, defined in question 3.2,
Trang 18.3.1. (a) Taking into account the result of question 2.2 and setting ω =
−γB, write the action of ˆ HZ on the basis states{|σ1, σ2}.
(b) Write in terms of A and ¯ hω the matrix representation of
ˆ
H = ˆ HSS+ ˆHA+ ˆHZ (8.4)
in the basis{|S, m} of the total spin of the two particles.
(c) Give the numerical value of ¯hω in eV for a field B = 1 T Is it easy
experimentally to be in a strong field regime, i.e ¯hω A?
8.3.2 Calculate the energy eigenvalues in the presence of the field B; express
the corresponding eigenstates in the basis {|S, m} of the total spin The largest eigenvalue will be written E+ and the corresponding eigenstate|ψ+ For convenience, one can introduce the quantity x = 8¯ hω/(7A), and the angle
θ defined by sin 2θ = x/ √
1 + x2, cos 2θ = 1/ √
1 + x2.
8.3.3 Draw qualitatively the variations of the energy levels in terms of B.
Are there any remaining degeneracies?
8.4 Decay of Positronium
We recall that when a system A is unstable and decays: A → B + · · · , the probability for this system to decay during the interval [t, t+dt] if it is prepared
at t = 0, is dp = λe −λt dt, where the decay rate λ is related to the lifetime
τ of the system by τ = 1/λ If the decay can proceed via different channels, e.g A → B + · · · and A → C + · · · , with respective decay rates λ1 and λ2,
the total decay rate is the sum of the partial rates, and the lifetime of A is
τ = 1/(λ1+ λ2).
In all what follows, we place ourselves in the rest frame of the positronium
8.4.1 In a two-photon decay, or annihilation, of positronium, what are the
energies of the two outgoing photons, and what are their relative directions?
8.4.2 One can show that the annihilation rate of positronium into photons
in an orbital state|n, l, m is proportional to the probability for the electron
and positron to be at the same point, i.e to|ψ nlm(0)|2 In what orbital states
is the annihilation possible?
8.4.3 In quantum field theory, one can show that, owing to charge
conjuga-tion invariance,
(a) a singlet state, S = 0, can only decay into an even number of photons:
2, 4, · · ·
(b) a triplet state, S = 1, can only decay into an odd number of photons:
3, 5, · · ·
In the orbital ground state ψ100, split by spin–spin interactions as calculated
in Sect 2, the lifetime of the singlet state is τ ∼ 1.25 × 10 −10 s, and the
Trang 2lifetime of either of the three triplet states is τ3∼ 1.4 × 10 −7 s Quantum field
theory predicts:
λ2= 1
τ2
= 4πα2c ¯h
mc
2
|ψ100(0)|2, λ3= 1
τ3
= 4
9π (π
2− 9) α λ2 .
Compare theory and experiment
8.4.4 In order to determine the hyperfine constant A of positronium, it is
of interest to study the energy and the lifetime of the level corresponding to the state |ψ+, defined in question 3.2, as a function of the field B.
¿From now on, we assume that the field is weak, i.e |x| = |8¯hω/(7A)| 1,
and we shall make the corresponding approximations
(a) What are, as a function of x, the probabilities pS ans pT of finding the state|ψ+ in the singlet and triplet states respectively?
(b) Use the result to calculate the decay rates λ+2 and λ+3 of the state |ψ+ into two and three photons respectively, in terms of the parameter x, and
of the rates λ2 and λ3 introduced in question 4.4
(c) What is the lifetime τ+(B) of the state |ψ+? Explain qualitatively its dependence on the applied field B, and calculate τ+(B) for B = 0.4 T
1 One measures, as a function of B, the ratio R = τ+(B)/τ+(0) of the life-time of the|ψ+ state with and without a magnetic field The dependence
on B of R is given in Fig 8.1, with the corresponding error bars.
(i) What estimate does one obtain for the hyperfine constant, A, using the value of the magnetic field for which the ratio R has decreased by
a factor two?
(ii) How do theory and experiment compare?
Fig 8.1 Variation of the ratio R defined in the text as a function of the applied
magnetic field B
Trang 38.5 Solutions
Section 8.1: Positronium Orbital States
In positronium, we have, by scaling:
8.1.1 A reduced mass µ = m/2.
8.1.2 A center of mass Hamiltonian ˆH = ˆ p2/2µ − q2/4π0r.
8.1.3 The energy levels En = −(1/2)µc2α2/n2 = −(1/4)mc2α2/n2 The
degeneracy is n2 for each level, as in the hydrogen atom; the bound state energies are half of those of hydrogen
8.1.4 The Bohr radius is a0 = ¯h/(µcα) = 2¯ h/(mcα) = 2aH
0 ∼ 1.06 ˚A The diameter of positronium isr = 3a0/2 = 3aH
0, and, since the proton is fixed, the diameter of the hydrogen is 2rH= 3aH
0 Therefore the two systems have
the same size
8.1.5 The ground state wave function is ψ100(r) = e −r/a0/
πa3, and we have|ψ100(0)|2= (mcα/(2¯ h))3/π.
Section 8.2: Hyperfine Splitting
8.2.1 In the orbital ground state, the degeneracy is 4, corresponding to the
number of independent spin states
8.2.2 Since the masses are equal, but the charges are opposite, we have
γ1= q/m, γ2=−q/m, γ = q/m.
8.2.3 As usual, we can express the spin–spin operator in terms of the total
spin S as S1· S2= [S2− S2
1− S2
2]/2 Hence, the orbital ground state is split
into:
(a) the triplet states: | + +, (| + − + | − +)/ √2,| − −, with the energy
shift:
ET= A/4 ,
(b) the singlet state: ( | + − − | − +)/ √2, with the energy shift:
ES=−3A/4
8.2.4 (a) There is a mass factor of ∼ 1/2000, a factor of ∼ 2.8 for the
gyromagnetic ratio of the proton, and a factor of 8 due to the value of the wave function at the origin Altogether, this results in a factor of∼ 22/2000 ∼ 1% for the ratio of hyperfine splittings H/(e+e −).
(b) The numerical value of A is
12π
q¯ h mc
2mcα
¯
h
3
= 1
3mc
2α4∼ 4.84 × 10 −4 eV.
Trang 4(c) This corresponds to a transition frequency of ν = A/h 117 GHz This
prediction is not in agreement with the experimental result (∼ 200 GHz).
8.2.5 (a) Taking into account ˆHA, the triplet state energy is A while the
singlet state energy is−3A/4 The splitting is δE = 7A/4 = 8.47 × 10 −4 eV.
(b) The corresponding frequency is ν = δE/h ∼ 205 GHz, in agreement with
experiment
Section 8.3: Zeeman Effect in the Ground State
8.3.1.
(a) The Zeeman Hamiltonian is ˆHZ= ω( ˆ S 1z − ˆ S 2z), therefore, we have
ˆ
HZ| + + = ˆ HZ| − − = 0
ˆ
HZ| + − = ¯hω| + −
ˆ
HZ| − + = −¯hω| − +
In terms of total spin states, this results in
ˆ
HZ|1, 1 = ˆ HZ|1, −1 = 0
ˆ
HZ|1, 0 = ¯hω|0, 0
ˆ
HZ|0, 0 = ¯hω|1, 0
(b) Hence the matrix representation in the coupled basis:
ˆ
HZ=
⎛
⎜
⎝
0 0 0 ¯hω
0 0 ¯hω 0
⎞
⎟
⎠ ,
where the elements are ordered according to:|1, 1, |1, −1, |1, 0, |0, 0.
8.3.2 Similarly, one has the matrix representation of the full spin
Hamil-tonian:
ˆ
H =
⎛
⎜
⎝
0 0 ¯hω −3A/4
⎞
⎟
⎠
In a field of 1 T, |¯hω| = q¯hB/m = 2µBB 1.16 × 10 −4 eV The strong field
regime corresponds to |¯hω| A , i.e B 4 T, which is difficult to reach.
Trang 58.3.3 (a) Two eigenstates are obvious:|1, 1 and |1, −1, which correspond
to the same degenerate eigenvalue A of the energy The two others are obtained
by diagonalizing a 2× 2 matrix:
|ψ+ = cos θ |1, 0 + sin θ |0, 0 ,
|ψ − = − sin θ |1, 0 + cos θ |0, 0 ,
corresponding to the energies
E ±=A8 ± 7A
8
2
+ (¯hω)2
1/2
= A 8
1± 71 + x2
.
(b) The triplet states | + + and | − − remain degenerate, as shown in
Fig 8.2
Fig 8.2 Variation of the hyperfine energy levels with applied magnetic field
Section 8.4: Decay of Positronium
8.4.1 In a two-photon decay, the outgoing photons have opposite momenta,
their energies are both mc2= 511 keV
8.4.2 The wave function vanishes at the origin, except for s-waves
(|ψ nlm(0)|2 = 0 if l = 0), owing to the centrifugal barrier Therefore the decay can only occur when the positronium is in an s-state.
8.4.3 The given formulas correspond to λ2= mc2α5/(2¯ h) which yields τ2=
1.24 × 10 −10 s and τ3= 1.38 × 10 −7 s, in agreement with experiment.
Trang 68.4.4 (a) For a given value of the applied field, with the positronium
pre-pared in the state |ψ+, the probabilities of finding the system in the singlet and triplet states are respectively pS = sin2θ ∼ x2/4 and pT = cos2θ ∼
1− x2/4.
(b) The rate for |ψ+ to decay into two photons is the product of the
prob-ability of finding |ψ+ in the singlet state with the singlet state decay rate:
λ+2 = pSλ2∼ x2λ2/4 = x2/(4τ2)
Similarly, one has
λ+3 = pTλ3∼ (1 − x2/4)λ3= (1− x2/4)/τ3 .
(c) The lifetime of the|ψ+ state is
λ+2 + λ+3 =
τ3
1− x42 + x42τ3
τ2
1 + 16¯h2ω2
49A2
τ3
τ2 .
As the field B increases, the state |ψ+, which is purely triplet for B = 0,
acquires a greater and greater singlet component Therefore its lifetime
de-creases as B inde-creases For B = 0.4 T, one has τ+= 0.23 τ3= 3.2 × 10 −8 s.
(d) Experimentally, one has R ∼ 0.5, i.e x2τ3/4τ2 1 for B ∼ 0.22 T Therefore x 6 × 10 −2 and, since A = 8¯ hω/7x and ¯ hω = 2.3 × 10 −5 eV, the
result is A ∼ 4.4 × 10 −4 eV, in good agreement with theoretical expectations.
Section 8.5: References
S DeBenedetti and H.C Corben, Positronium, Ann Rev Nucl Sci., 4, 191
(1954)
Stephan Berko and Hugh N Pendleton, Positronium, Ann Rev Nucl Sci.,
30, 543 (1980).
A.P Mills and S Chu, Precision Measurements in Positronium, in Quan-tum Electrodynamics, ed by T Kinoshita (World Scientific, Singapore 1990)
pp 774-821
Trang 7The Hydrogen Atom in Crossed Fields
We study the modification of the energy spectrum of a hydrogen atom placed
in crossed static electric and magnetic fields in perturbation theory We thus recover a result first derived by Pauli
In his famous 1925 paper on the hydrogen atom, W Pauli made use of the particular symmetry of the Coulomb problem In addition to the hydrogen spectrum, he was able to calculate the splitting of the levels in an electric field (Stark effect) or in a magnetic field (Zeeman effect) Pauli also noticed that he could obtain a simple and compact formula for the level splitting
in a superposition of a magnetic field B0 and an electric field E0 both sta-tic and uniform, and perpendicular to each other In this case, he found that a
level with principal quantum number n is split into 2n − 1 sublevels
En+ δEn(k) with
δE (k)
n = ¯hk (ω2
0+ ω2
where k is an integer ranging from −(n−1) to n−1, ω0and ωeare respectively
proportional to B0 and E0, and ωe can be written as
ωe=3
2Ωef (n) with Ωe=
4π0¯h
M qe E0 , where M and qe are the mass and charge of the electron, and where f (n) depends on n only.
It is only in 1983 that Pauli’s result was verified experimentally Our
pur-pose, here, is to prove (9.1) in the special case n = 2, to calculate ω0 and ωe
in that case, and, by examining the experimental result for n = 34, to guess what was the very simple formula found by Pauli for f (n).
Trang 89.1 The Hydrogen Atom in Crossed Electric
and Magnetic Fields
We consider the n = 2 level of the hydrogen atom We neglect all spin effects.
We assume that B0is along the z axis and E0 along the x axis We use first
order perturbation theory
9.1.1 What are the energy levels and the corresponding eigenstates in the
presence of B0only? Check that (9.1) is valid in this case and give the value
of ω0?
9.1.2 In the presence ofE0 only, the perturbing Hamiltonian is the electric dipole term ˆH E=− ˆ D.E0=−qer.Eˆ 0 Write the matrix representing ˆH E in
the n = 2 subspace under consideration.
We recall that:
(a) ∞
0 r3R 2s (r) R 2p (r) dr = 3 √
3 a1 where R 2s and R 2pare the radial wave
functions for the level n = 2, l = 0 and n = 2, l = 1 respectively, and where a1= ¯h2/(M e2) is the Bohr radius (e2= q2
e/4π0)
(b) In spherical coordinates (θ polar angle and φ azimuthal angle), the l = 0 and l = 1 spherical harmonics are
Y00(θ, φ) = √1
4π , Y
±1
1 (θ, φ) = ∓
3
8π sin θ e
±iφ ,
Y0(θ, φ) =
3
9.1.3 Calculate the energies of the levels originating from the n = 2 level in the presence of the crossed fields E0 and B0 Show that one recovers (9.1)
with ωe= (3/2)f (2)Ωe, and give the value of f (2).
9.2 Pauli’s Result
The first experimental verification of Pauli’s result was performed in 1983.1
In Fig 9.1, the points correspond to a sub-level with a given value of k arising from the n = 34 level of an hydrogen-like atom All points correspond to the same energy of this level, but to different values of the static fields E0 and
B0
Knowing that ωe is a function of the principal quantum number n of the form: ωe = (3/2)f (n)Ωe, and that ω0 and Ωe are the constants introduced above, answer the following questions:
1 Fig 9.1 was obtained by F Biraben, D Delande, J.-C Gay, and F Penent, with
rubidium atoms prepared in a Rydberg state, i.e with an electron placed in a
strongly excited level (see J.-C Gay, in Atoms in unusual situations, J.-P Briand
ed., p 107, Plenum, New York, 1986)
Trang 9Fig 9.1 Values of the electric and magnetic fields giving rise to the same sub-level
energy of the n = 34 level of a hydrogen-like atom
9.2.1 Does the experimental data agree with (9.1)?
9.2.2 Write the quantity ω2+ ω2
e in the form λ
γB2+ f2(n)E2 , give the
value of the constant γ, and calculate f (34).
9.2.3 Guess Pauli’s result concerning f (n).
9.3 Solutions
Section 9.1: The Hydrogen Atom in Crossed Electric
and Magnetic Fields
9.1.1 Consider a state|n, l, m The orbital magnetic moment of the
elec-tron is ˆµorb = γ0L, with γˆ 0 = qe/(2M ) The magnetic Hamiltonian is
ˆ
H = −ˆµorb.B = −(qe/2M ) ˆ L z B0
At first order perturbation theory, the energy levels originating from the n = 2 subspace (angular momentum l = 0 or l = 1) are m¯ hω0 with m = −1, 0, +1, and ω0=−qeB0/(2M ) (ω0> 0 for B0> 0) The corresponding states are
|2s and |2p, m = 0 δE = 0
|2p, m = −1 δE = −¯hω0
|2p, m = +1 δE = +¯ hω0 .
9.1.2 The Hamiltonian is ˆH E=−qex Eˆ 0 We have to calculate the 16 matrix elements2, l , m |ˆx|2, l, m The integral to be evaluated is
Trang 102, l , m |ˆx|2, l, m =
Y l m (θ, φ)
∗
sin θ cos φ Y l m (θ, φ) d2Ω
×
∞
0
r3 (R 2,l (r)) ∗ R
2,l (r) dr
The angular integral vanishes if l = l We need only consider the terms l = 0,
l = 1 (and the hermitian conjugate l = 1, l = 0), i.e.
3√
3 a1
1
√ 4π
2π
3 (−Y1
1(θ, φ) + Y −1
1 (θ, φ)) Y1m (θ, φ) d2Ω
where we have incorporated the radial integral given in the text One therefore
obtains 3 a1(δ m,−1 − δ m,1 )/ √
2 The only non-vanishing matrix elements are
2s| ˆ H |2p, m = ±1 and their hermitian conjugates.
Setting Ωe = 4π0¯hE0/(M qe) = qeE0a1/¯ h, we obtain the matrix
ˆ
H E= 3¯√ hΩe
2
⎛
⎜
⎝
⎞
⎟
⎠
where the rows (columns) are ordered as 2p, m = 1, 0, −1; 2s.
9.1.3 We want to find the eigenvalues of the matrix
¯
h
⎛
⎜
⎝
2
2
3Ωe/ √
2 0 −3Ωe/ √
⎞
⎟
⎠
There is an obvious eigenvalue λ = 0 since the |2p, m = 0 and |2s states
do not mix in the presence the electric field The three other eigenvalues are easily obtained as the solutions of:
λ(¯ h2ω20− λ2) + 9 ¯h2Ω2eλ = 0 , i.e λ = 0 and λ = ±¯hω2+ 9Ω2
e The shifts of the energy levels are therefore: δE = 0 twice degenerate, and
δE = ±¯hω2+ 9Ω2
e If we adopt the prescription given in the text, we obtain
ωe= 3Ωe=⇒ f(2) = 2
Section 9.2: Pauli’s Result
9.2.1 We remark that the experimental points are aligned on a straight line
aB2
0+ bE2
0 = constant which is in agreement with (9.1), i.e a constant value
of ω2+ ω2 corresponds to a constant value of each energy level
...There is an obvious eigenvalue λ = since the |2p, m = 0 and |2s states
do not mix in the presence the electric field The three other eigenvalues are easily obtained as the solutions... give the
value of the constant γ, and calculate f (34).
9. 2.3 Guess Pauli’s result concerning f (n).
9. 3 Solutions
Section 9. 1:... have incorporated the radial integral given in the text One therefore
obtains a1(δ m,−1 − δ m,1 )/ √
2 The only non-vanishing