Kern- und Teilchenphysik I Lecture 2: Fermi s golden rule

Transcription

1 Kern- und Teilchenphysik I Lecture 2: Fermi s golden rule (adapted from the Handout of Prof. Mark Thomson) Prof. Nico Serra Dr. Patrick Owen, Mr. Davide Lancierini

2 Lifetime The probability that a particle survive a time t + dt is given by P (t + dt) = P (t)(1 dt), where is the decay probability per unit of time. We are basically saying that the probability that a particle decay at each time does not depend on the history of the particle. P (t + dt) = P (t)(1 dt)! P (t + dt) P (t) (t) =P = (t) dt( therefore dt) we have dp dt = P! P (t) = e t In order to find the coe cient we can just impose P (0) = 1, i.e. the particle t (by definition) exists at t=0, so we get P (t) =e We can derive the Probability density Function (PdF) such that R 1 P(t)dt =1 0 which implies A R 1 e t dt =1! A 1 e t 1 0 = A 1 =1! P(t) = 1 t e 0 Finally if we compute the expectation value of t we get hti = = R 1 0 P(t)dt = 1 In a general frame P (t) =e t and P(t) = e t 2

3 Lifetime The decay width measures the probability per time units Its units are GeV Physical meaning: E t ~ - The mass of the particle we quote is the position of the peak - Particles with very short lifetime have a large width, particles with a long lifetime have a narrow width - Except for particles decaying strongly we cannot usually resolve the width 3

4 Breit-Wigner Let s take a non stable particle (t) = (0)e iet /2t! (t) 2 / e t An unstable particle can be describe by having an imaginary part ( /2) in the energy If the particle is unstable it means it is not an eigenstate of the Hamiltonian (indeed it has a small imaginary part of the energy). Let s write this state as a function of the energy eigenstates: (E) = R (0)e imt+ 2 t e iet / 1 E M+(i 2 ) 2 / 1 (E M) 2 +( 2 ) 2 More rigorous arguments lead to the same result 4

5 Cross Section σ = no of interactions per unit time per target incident flux Flux = number of incident particles/ unit area/unit time The cross section, σ, can be thought of as the effective crosssectional area of the target particles for the interaction to occur. In general this has nothing to do with the physical size of the target although there are exceptions, e.g. neutron absorption here σ is the projective area of nucleus dσ dω = Differential Cross section no of particles per sec/per target into dω incident flux or generally dσ d... e e θ p with 5

6 Cross Section In time δt a particle of type a traverses region containing particles of type b A v a v b A σ «Interaction probability obtained from effective cross-sectional area occupied by the particles of type b Interaction Probability = Rate per particle of type a = n b v σ Consider volume V, total reaction rate = = As anticipated: Rate = Flux x Number of targets x cross section 6

7 Luminosity We define the luminosity such that: W = dp dt = L [L] =[ 1 L 2 T ] We often see the integrated luminosity as L = Z L = Z L! P = Z Ldt! W = L Wdt = L W W = N b a! L = N b a = #particles At 7

8 Mean free path Consider to shoot particles at an infinitesimal slab Mean Free Path: ` = 1 n b If the scattering centers are nucleons ` = 1 N A b where N A is the Avogadro number 8

9 Units for cross-section The cross-section is often measured in barn (b) 1b = (10fm) (10fm) = 100fm 2 = m 2 = cm 2 E.g. Total pp cross section at 7TeV 90mb compute number of interactions in 1 year running at the luminosity of L = cm 2 s 1 9

10 Units for cross-section The cross-section is often measured in barn (b) 1b = (10fm) (10fm) = 100fm 2 = m 2 = cm 2 E.g. Total pp cross section at 7TeV 90mb compute number of interactions in 1 year running at the luminosity of L = cm 2 s 1 L = R Ldt = cm 2 s s = = cm 2 = b =2fb 1 = mb 1 N = L = mb mb =

11 Decay rate and Cross Section In particle physics we are mainly concerned with particle interactions and decays, i.e. transitions between states these are the experimental observables of particle physics Calculate transition rates from Fermi s Golden Rule is number of transitions per unit time from initial state to final state not Lorentz Invariant! is Transition Matrix Element is density of final states «Rates depend on MATRIX ELEMENT and DENSITY OF STATES is the perturbing Hamiltonian the ME contains the fundamental particle physics just kinematics 11

12 Decay rate and Cross Section «Aiming towards a proper calculation of decay and scattering processes e + e µ + µ e q e q Need relativistic calculations of particle decay rates and cross sections: Need relativistic treatment of spin-half particles: Dirac Equation Need relativistic calculation of interaction Matrix Element: Interaction by particle exchange and Feynman rules + and a few mathematical tricks along, e.g. the Dirac Delta Function 12

13 Decay rate and Cross Section Consider the two-body decay i θ 1 Want to calculate the decay rate in first order perturbation theory using plane-wave descriptions of the particles (Born approximation): 2 as where N is the normalisation and For decay rate calculation need to know: Wave-function normalisation Transition matrix element from perturbation theory Expression for the density of states All in a Lorentz Invariant form «First consider wave-function normalisation Previously (e.g. part II) have used a non-relativistic formulation Non-relativistic: normalised to one particle in a cube of side 13

14 Decay rate and Cross Section Apply boundary conditions ( ): Wave-function vanishing at box boundaries quantised particle momenta: Volume of single state in momentum space: a a a Normalising to one particle/unit volume gives number of states in element: Therefore density of states in Golden rule: Integrating over an elemental shell in momentum-space gives 1 with E 2 = p 2 + m 2 2E de = 2p dp dp de = E p = 1 14

15 Decay rate and Cross Section In the relativistic formulation of decay rates and cross sections we will make use of the Dirac δ function: infinitely narrow spike of unit area a Any function with the above properties can represent e.g. (an infinitesimally narrow Gaussian) In relativistic quantum mechanics delta functions prove extremely useful for integrals over phase space, e.g. in the decay and express energy and momentum conservation 15

16 Decay rate and Cross Section «We will soon need an expression for the delta function of a function Start from the definition of a delta function Now express in terms of and then change variables where x From properties of the delta function (i.e. here only non-zero at ) Rearranging and expressing the RHS as a delta function x (1) 16

17 Fermi s Golden Rule Rewrite the expression for density of states using a delta-function Note : integrating over all final state energies but energy conservation now taken into account explicitly by delta function Hence the golden rule becomes: since the integral is over all allowed final states of any energy For dn in a two-body decay, only need to consider one particle : mom. conservation fixes the other i 1 θ However, can include momentum conservation explicitly by integrating over the momenta of both particles and using another δ-fn 2 Energy cons. Mom. cons. Density of states 17

18 Lorentz Invariant PH In non-relativistic QM normalise to one particle/unit volume: When considering relativistic effects, volume contracts by a a a a a/γ Particle density therefore increases by «Conclude that a relativistic invariant wave-function normalisation needs to be proportional to E particles per unit volume Usual convention: Previously used Normalise to 2E particles/unit volume normalised to 1 particle per unit volume Hence is normalised to per unit volume Define Lorentz Invariant Matrix Element,, in terms of the wave-functions normalised to particles per unit volume a 18

19 Lorentz Invariant PH 19

20 Fermi s Golden Rule For the two body decay «Now expressing in terms of gives Note: uses relativistically normalised wave-functions. It is Lorentz Invariant is the Lorentz Invariant Phase Space for each final state particle the factor of arises from the wave-function normalisation This form of is simply a rearrangement of the original equation but the integral is now frame independent (i.e. L.I.) is inversely proportional to E i, the energy of the decaying particle. This is exactly what one would expect from time dilation (E i = γm). Energy and momentum conservation in the delta functions (prove this in Question 2) 20

21 Fermi s Golden Rule - Let s consider the total Hamiltonian H= H0 + H H 0 ki = E k ki k are the eigenstates of the unperturbed Hamiltonian (H 0 ) i d (t) dt =[H 0 + H 0 (t)] h k ji = kj - We can still write a generic wave function as a function of the eigenstates of the unperturbed Hamiltonian - This time the coefficient will depend on time, since the eigenstates of H0 are not stationary (x, t) = X k c k (t) k (x)e ie kt 21

22 Fermi s Golden Rule - Let s consider the total Hamiltonian H= H0 + H H 0 ki = E k ki k are the eigenstates of the unperturbed Hamiltonian (H 0 ) i d (t) dt =[H 0 + H 0 (t)] - We can still write a generic wave function as a function of the eigenstates of the unperturbed Hamiltonian - This time the coefficient will depend on time, since the eigenstates of H0 are not stationary (x, t) = X k c k (t) k (x)e ie kt h k ji = kj Coefficients depend on time Eigenstate of H0 at time t 22

23 Fermi s Golden Rule Let s now put (x, t) into the Schrödinger equation: dck (t) i X dt k ke ie kt ie k c k (t) k e ie kt = X c k (t)h 0 k e iekt + X k k c k (t)h 0 ke ie kt X k E k c k (t) k e ie kt i X k dc k (t) dt ke ie kt = X k c k (t)h 0 ke ie kt Let s now assume that ii = i (e.g. excited state of Hydrogen atom) and that H 0 is constant for t>0. If H 0 is su ciently small c i (t) ' 1 and c k6=i (t) ' 0 i P dc k (t) k dt ke iekt ' H 0 ie ieit, where c k6=i ' 0 was used on the left side 23

24 Fermi s Golden Rule i P k 6 dc k (t) dt ke iekt ' H 0 ie ieit, ' We want to compute the transition probability to the final state fi, soimultiply by hf = h f (e.g. transition to the ground state of the Hydrogen atom) dc f (t) dt = ihf H 0 iie i(e f E i )t 24

25 Fermi s Golden Rule 25

26 Fermi s Golden Rule 26

27 Decay rate calculation «Because the integral is Lorentz invariant (i.e. frame independent) it can be evaluated in any frame we choose. The C.o.M. frame is most convenient In the C.o.M. frame and Integrating over using the δ-function: i 1 θ now Writing since the δ-function imposes 2 For convenience, here is written as 27

28 Decay Rate Calculation Which can be written in the form (2) where and Note: imposes energy conservation. determines the C.o.M momenta of the two decay products i.e. for «Eq. (2) can be integrated using the property of δ function derived earlier (eq. (1)) 2 i θ 1 where is the value for which All that remains is to evaluate 28

29 Decay Rate Calculation giving: But from, i.e. energy conservation: In the particle s rest frame (3) can be obtained from VALID FOR ALL TWO-BODY DECAYS! 29

30 Cross section calculation Consider scattering process Start from Fermi s Golden Rule: Now where is the transition matrix for a normalisation of 1/unit volume For 1 target particle per unit volume the parts are not Lorentz Invariant 30

31 Cross section calculation To obtain a Lorentz Invariant form use wave-functions normalised to per unit volume particles Again define L.I. Matrix element The integral is now written in a Lorentz invariant form The quantity can be written in terms of a four-vector scalar product and is therefore also Lorentz Invariant (the Lorentz Inv. Flux) Consequently cross section is a Lorentz Invariant quantity Two special cases of Lorentz Invariant Flux: Centre-of-Mass Frame Target (particle 2) at rest (see appendix I) Prof. M.A. Thomson Michaelmas

32 Cross section calculation We will now apply above Lorentz Invariant formula for the interaction cross section to the most common cases used in the course. First consider 2 2 scattering in C.o.M. frame Start from Here «The integral is exactly the same integral that appeared in the particle decay calculation but with replaced by 32

33 Appendix In case you want to have fun with other scattering processes you can look at the appendix

34 Cross section calculation In the case of elastic scattering For calculating the total cross-section (which is Lorentz Invariant) the result on the previous page (eq. (4)) is sufficient. However, it is not so useful for calculating the differential cross section in a rest frame other than the C.o.M: because the angles in refer to the C.o.M frame For the last calculation in this section, we need to find a L.I. expression for «Start by expressing in terms of Mandelstam t i.e. the square of the four-momentum transfer 34

35 Cross section calculation Want to express where in terms of Lorentz Invariant s In C.o.M. frame: x z giving therefore hence Finally, integrating over (assuming no dependence of ) gives: 35

36 LI differential cross section All quantities in the expression for are Lorentz Invariant and therefore, it applies to any rest frame. It should be noted that is a constant, fixed by energy/momentum conservation As an example of how to use the invariant expression we will consider elastic scattering in the laboratory frame in the limit where we can neglect the mass of the incoming particle E 1 m 2 e.g. electron or neutrino scattering In this limit 36

37 2->2 Body scattering in the Lab The other commonly occurring case is scattering from a fixed target in the Laboratory Frame (e.g. electron-proton scattering) First take the case of elastic scattering at high energy where the mass of the incoming particles can be neglected: 1 2 θ 3 e.g. 1 e e 3 X X 2 4 Wish to express the cross section in terms of scattering angle of the e 4 therefore Integrating over The rest is some rather tedious algebra. start from four-momenta so here But from (E,p) conservation and, therefore, can also express t in terms of particles 2 and 4 37

38 2->2 Body scattering in the Lab Note E 1 is a constant (the energy of the incoming particle) so Equating the two expressions for t gives so using gives Particle 1 massless In limit 38

39 2->2 Body scattering in the Lab In this equation, E 3 is a function of θ : giving General form for 2 2 Body Scattering in Lab. Frame «The calculation of the differential cross section for the case where m 1 can not be neglected is longer and contains no more physics (see appendix II). It gives: Again there is only one independent variable, θ, which can be seen from conservation of energy i.e. is a function of 39