The Quantum Mechanics Solver 10 ppsx

9.3 Solutions 85
9.2.2. Given the definitions of ω
0
and Ω
e
, one has
ω
2
0
+ ω
2
e
=
9
4

4π
0
¯h
Mq
e

2


αc
3

2
B
2

0
+ f
2
(n) E
2
0

,
where α is the fine structure constant and c the velocity of light. The experi-
mental line
(αc/3)
2
B
2
0
+ f
2
(34)E
2
0
goes through the points E
2
0
=0,B
2
0
 87 × 10
−4
T
2

and B
2
0
=0,E
2
0

4 × 10
6
V
2
m
−2
. This gives f(34) = 34.
9.2.3. Indeed, the very simple result found by Pauli was f(n)=n.
10
Energy Loss of Ions in Matter
When a charged particle travels through condensed matter, it loses its kinetic
energy gradually by transferring it to the electrons of the medium. In this
chapter we evaluate the energy loss of the particle as a function of its mass and
its charge, by studying the modifications that the state of an atom undergoes
when a charged particle passes in its vicinity. We show how this process can
be used to identify the products of a nuclear reaction.
The electric potential created by the moving particle appears as a time-
dependent perturbation in the atom’s Hamiltonian. In order to simplify the
problem, we shall consider the case of an atom with a single external electron.
The nucleus and the internal electrons will be treated globally as a core of
charge +q, infinitely massive and, therefore, fixed in space. We also assume
that the incident particle of charge Z
1

q is heavy and non-relativistic, and that
its kinetic energy is large enough so that in good approximation its motion
can be considered linear and uniform, of constant velocity v,whenitinteracts
with an atom.
Here q denotes the unit charge and we set e
2
= q
2
/(4πε
0
). We consider the
x, y plane defined by the trajectory of the particle and the center of gravity
of the atom, which is chosen to be the origin, as shown on Fig. 10.1.
Let R(t) be the position of the particle at time t and r =(x, y, z) the
coordinates of the electron of the atom. The impact parameter is b and the
notation is specified in Fig. 10.1. The time at which the particle passes nearest
to the atom, i.e. x = b, y = 0 is denoted t =0.WewriteE
n
and |n for the
energy levels and corresponding eigenstates of the atom in the absence of an
external perturbation.
10.1 Energy Absorbed by One Atom
10.1.1. Write the expression for the time-dependent perturbing potential
ˆ
V (t) due to the presence of the charged particle.
88 10 Energy Loss of Ions in Matter
10.1.2. We assume that the impact parameter b is much larger than the
typical atomic size, i.e. b r,sothat|R(t)||r| for all t. Replace
ˆ
V (t)

by its first order expansion in |r|/|R| and express the result in terms of the
coordinates x and y of the electron, and of b, v and t.
10.1.3. Initially, at time t = −∞, the atom is in a state |i of energy E
i
.
Using first order time-dependent perturbation theory, write the probability
amplitude γ
if
to find the atom in the final state |f of energy E
f
after the
charged particle has passed (t =+∞). We set ω
fi
=(E
f
−E
i
)/¯h and we only
consider the case E
f
= E
i
.
10.1.4. The calculation of γ
if
involves the Bessel function K
0
(z). One has

∞

0
cos ωt
(β
2
+ t
2
)
1/2
dt = K
0
(ωβ)

∞
0
t
sin ωt
(β
2
+ t
2
)
3/2
dt = ωK
0
(ωβ) .
Express γ
if
in terms of K
0
and its derivative.

The asymptotic behavior of K
0
is K
0
(z) −ln z for z  1, and K
0
(z) 

2π/z e
−z
for z  1. Under what condition on the parameters ω
fi
, b and v
is the transition probability P
if
= |γ
if
|
2
large?
Show that, in that case, one obtains
P
if


2Z
1
e
2
¯hbv


2
|f|ˆx|i|
2
.
10.1.5. Give the physical interpretation of the condition derived above. Show
that, given the parameters of the atom, the crucial parameter is the effective
interaction time, and give a simple explanation of this effect.
10.2 Energy Loss in Matter
We assume in the following that the Hamiltonian of the atom is of the form
Fig. 10.1. Definition of the coordinates.
10.2 Energy Loss in Matter 89
ˆ
H
0
=
ˆ
p
2
2m
+ V (
ˆ
r) .
10.2.1. Thomas–Reiche–Kuhn Sum Rule.
(a) Calculate the commutator [ˆx,
ˆ
H
0
].
(b) Deduce from this commutator a relation between the matrix elements

i|ˆx|f and i|ˆp|f,where|i and |f are eigenstates of
ˆ
H
0
.
(c) Applying a closure relation to [ˆx, ˆp]=i¯h,showthat:
2m
¯h
2

f
(E
f
− E
i
)|f|ˆx|i|
2
=1
for all eigenstates |i of H
0
.
10.2.2. Using the Thomas–Reiche–Kuhn sum rule, calculate the expectation
value δE of the energy loss of the incident particle when it interacts with the
atom.
Let E be the energy of the particle before the interaction. Which parame-
ters does the product EδEdepend on?
10.2.3. Experimental Application. We are now interested in incident par-
ticles which are fully ionized atoms (Z
1
= Z,whereZ is the atomic number),

whose masses are, to a good approximation, proportional to the mass num-
ber A = Z + N (where N is the number of neutrons of the isotope). When
these ions traverse condensed matter, they interact with many atoms of the
medium, and their energy loss implies some averaging over the random impact
parameter b. The previous result then takes the form
EδE= kZ
2
A,
where the constant k depends on the nature of the medium.
Semiconductor detectors used for the identification of the nuclei in nu-
clear reactions are based on this result. In the following example, the ions to
be identified are the final state products of a reaction induced by 113 MeV
nitrogen ions impinging on a target of silver atoms.
In Fig. 10.2 each point represents an event, i.e. the energy E and energy loss
δE of an ion when it crosses a silicon detector. The reference point corresponds
to the isotope A =12ofcarbon
12
6
C (we use the notation
A
Z
Nforanucleus
charge Z and mass number A) which loses δE = 30 MeV at an energy E =
50 MeV.
(a) Calculate the constant k and the theoretical prediction for the energy
loss at 60 and 70 MeV. Put the corresponding points on the figure.
(b) Assuming the reaction could produce the following isotopes:
– boron, Z =5,A =10, 11, 12
–carbon,Z =6,A =11, 12, 13, 14
– nitrogen Z =7,A =13, 14, 15, 16,

90 10 Energy Loss of Ions in Matter
10
20
30
40
50
50
60 70 80 90 100
40
30
20
10
E(MeV)
14
7
N(113MeV)→Ag
12
6
C
δE(MeV)
Fig. 10.2. Energy loss δE versus energy E through a silicon detector, of the final
products of a reaction corresponding to 113 MeV nitrogen ions impinging on a target
of silver atoms
what nuclei are effectively produced in the reaction? Justify your answers
by putting the points corresponding to E =50MeVandE =70MeV
on the figure.
10.3 Solutions
Section 10.1: Energy Absorbed by One Atom
10.1.1. The interaction potential between the particle and the atom is the
sum of the Coulomb interactions between the particle and the core, and those

between the particle and the outer electron:
ˆ
V (t)=
Z
1
e
2
R(t)
−
Z
1
e
2
|R(t) −
ˆ
r|
.
10.1.2. For |R||r|,wehave
1
|R − r|
=

R
2
− 2R · r + r
2

−1/2

1

R
+
r ·R
R
3
.
10.3 Solutions 91
Therefore
ˆ
V (t) −
Z
1
e
2
R
3
(t)
ˆ
r ·R(t) .
Since R(t)=(b, vt, 0),we obtain
ˆ
V (t) −
Z
1
e
2
(b
2
+ v
2

t
2
)
3/2
(ˆxb +ˆyvt) .
10.1.3. To first order in
ˆ
V , the probability amplitude is
γ
if
=
1
i¯h

+∞
−∞
e
iω
fi
t
f|
ˆ
V (t)|idt.
Inserting the value found above for
ˆ
V (t), we find
γ
if
= −
1

i¯h

+∞
−∞
Z
1
e
2
e
iω
fi
t
(b
2
+ v
2
t
2
)
3/2
(bf|ˆx|i + vtf|ˆy|i)dt.
10.1.4. One has

∞
0
cos ωt dt
(β
2
+ t
2

)
3/2
= −
1
β
d
dβ
K
0
(ωβ)=−
ω
β
K

0
(ωβ) .
Setting β = b/v, the amplitude γ
if
is
γ
if
=i
2Z
1
e
2
ω
fi
¯hv
2

(K
0
(ω
fi
b/v) f |ˆy|i−K

0
(ω
fi
b/v) f |ˆx|i) .
The probability P
if
= |γ
if
|
2
is large if K
0
or K

0
are also large. This happens
for ω
fi
b/v  1. In this limit, K
0
(z) ∼−ln z and K

0
(z) ∼−1/z,andwe

obtain
γ
if
=i
2Z
1
e
2
¯hvb

f|ˆx|i−f|ˆy|i
ω
fi
b
v
ln
ω
fi
b
v

.
Since |f|ˆx|i|  |f|ˆy|i|, one can neglect the second term (x ln x  1for
x  1) and we obtain, for ω
fi
b/v  1,
P
if
= |γ
if

|
2


2Z
1
e
2
¯hbv

2
|f|ˆx|i|
2
.
10.1.5. The time τ = b/v is the characteristic time during which the inter-
action is important, as we can see on the above formulas. For t  τ ,the
interaction is negligible.
The condition ω
fi
τ  1 means that the interaction time τ must be much
smaller than the Bohr period ∼ 1/ω
fi
of the atom. The perturbation
ˆ
V (t)
must have a large Fourier component at ω = ω
fi
if we want the probability
P
if

to be significant (the shorter in time the perturbation, the larger the
92 10 Energy Loss of Ions in Matter
spread of its Fourier transform in frequency). In the opposite limiting case,
where the perturbation is infinitely slow, the atom is not excited.
This observation provides an alternative way to evaluate the integrals of
question 1.3. The only values of t which contribute significantly are those for
which t is not too large, compared to τ (say |t|10 τ ). If ω
fi
τ  1, one
can replace e
iω
fi
t
by 1 in these integrals; the second integral is then zero for
symmetry reasons and the first one is easily evaluated, and gives the desired
result.
Section 10.2: Energy Loss in Matter
10.2.1. Thomas–Reiche–Kuhn Sum Rule.
(a) We find [ˆx,
ˆ
H
0
]=i¯hˆp/m.
(b) Taking the matrix element of this commutator between two eigenstates
|i and |f of
ˆ
H
0
, we obtain:
i¯h

m
f | ˆp | i = f | [ˆx,
ˆ
H
0
] | i =(E
i
− E
f
)f | ˆx | i .
(c) We now take the matrix element of [ˆx, ˆp]=i¯h between i| and |i and
we use the closure relation:
i¯h =

f
i | ˆx | ff | ˆp | i−

f
i | ˆp | ff | ˆx | i
=
m
i¯h

f
(E
i
− E
f
) |f | ˆx | i|
2

−
m
i¯h

f
(E
f
− E
i
) |i | ˆx | f |
2
=
2m
i¯h

f
(E
i
− E
f
)|f | ˆx | i|
2
,
which proves the Thomas–Reiche–Kuhn sum rule.
10.2.2. The expectation value δE of the energy transferred to the atom is
δE =

f
(E
f

− E
i
) P
if
=

2Z
1
e
2
¯hbv

2

f
(E
f
− E
i
) |f | ˆx | i|
2
.
Making use of the Thomas–Reiche–Kuhn sum rule, we obtain
δE =
2Z
2
1
e
4
mb

2
v
2
,
where m is the electron mass. If the ion has mass M, its kinetic energy is
E = Mv
2
/2, and we therefore obtain a very simple expression:
10.3 Solutions 93
EδE=
M
m

Z
1
e
2
b

2
,
where we see that the product EδEdoes not depend on the energy of the
incident particle, but is proportional to its mass and to the square of its charge.
10.2.3. With the
12
6
C point, one obtains k =3.47. We have put the calculated
points of the various isotopes on Fig. 10.3.
We make the following observations:
(a) For boron, the isotopes

10
Band
11
B are produced, but not
12
B.
(b) For carb on,
12
C is produced more abundantly than
13
C,
14
Cand
11
C.
(c) For nitrogen, there is an abundant production of
14
N, a smaller produc-
tion of
15
N, but practically no
13
Nor
16
N.
10
20
30
40
50

50
60 70 80 90 100
40
30
20
10
E(MeV)
14
7
N(113MeV)→Ag
12
6
C
δE(MeV)
10
11
11
12
13
14
14
15
7
N
6
C
5
B
Fig. 10.3. Interpretation of the data of Fig. 10.2
94 10 Energy Loss of Ions in Matter

10.4 Comments
Ionization of matter has numerous applications, for instance in developing
detectors for particle and nuclear physics, or in defining protection regulations
against radioactivity. In order to calculate the energy loss of an ion in matter,
one must integrate the above results over the impact parameter. In practice,
taking everything into account, one ends up with the following formula, due
to Hans Bethe and Felix Bloch, for the rate of energy loss per unit length:
−
dE
dx
=
4πK
2
Z
2
e
4
N
m
e
c
2
β
2
(ln

2m
e
c
2

β
2
I(1 − β
2
)

− β
2
) (10.1)
where β = v/c, K is a constant, N is the number density of atoms in the
medium and I is the mean excitation energy of the medium (I ∼ 11.5eV).
The cases of protons or heavy ions is of great interest and, in comparatively
recent years, it has allowed a major improvement in the medical treatment of
tumors in the eyes (proton therapy) and in the brain (ion therapy) . Owing
to the factor 1/β
2
, or equivalently 1/v
2
, in (10.1), practically all the energy
is deposited in a very localized region near the stopping point. Figure (10.4)
shows the comparison between the effect of ion beams and photons. One can
see the enormous advantage, from the medical point of view, of heavy ion
beams. These permit to attack and destroy tumors in a very accurate and
localized manner, as opposed to γ rays which produce damages all around the
point of interest.
Pioneering work on brain tumor therapy has been developed in Darmstadt
at the Heavy ion accelerator facility. Information can be found on the sites
Fig. 10.4. Energy loss of ions (left) and survival rate of cells (right) as a function
of the penetration depth. The dashed curve corresponds to the same quantities for
photons. We can see the considerable medical advantage to use heavy ion beams.

Document from the data of Heavy ion therapy at GSI, Darmstadt,
(Courtesy James Rich)
10.4 Comments 95
bio/home.html
/>This promising sector of medical applications in rapidly developing at
present.