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, http://www.gsi.de (Courtesy James Rich) 10.4 Comments 95 http://www-aix.gsi.de/ bio/home.html http://www.sgsmp.ch/protsr-f.htm This promising sector of medical applications in rapidly developing at present. . 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. 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. 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