Chapter 4. Quantum Mechanics. Contents

Transcription

1 Chapter 4 Quantum Mechanics Contents Fourier Analysis in QM: Momentum space wave functions Completeness and orthogonality Initial condition problems for quantum wells Klein-Gordon equation Relativistic Schrödinger equation Negative energy solutions Perturbation Theory in QM: Time independent perturbation theory Time dependent perturbation theory Fermi s Golden Rule Learning Outcomes Know the properties of coordinate and momentum space wave functions and be able to calculate one from the other in simple problems. Know the meaning of completeness and orthogonality and their implications for initial condition problems. 89

2 Be able to compute the time evolution of an initial wave function in square well problems. Be able to compute the Klein-Gordon equation starting from the relativistic relation for energy and momentum. Understand the meaning of negative energy solutions to the Klein- Gordon equation. Understand the derivations of energy shifts and transition rates in perturbation theory. Be able to calculate energy shifts and transition rates in simple perturbed problems. Reference Books PHYS2003 Quantum Physics notes. Introduction to Quantum Mechanics, Griffiths Quarks and Leptons, Halzen and Martin Quantum Physics, Eisberg and Resnick 90

3 4.1 Introduction In this part of the course we will study a number of techniques in Quantum Mechanics (QM). To begin with (chapter 4.2) we will review parts of your second year course but bearing in mind what you have learnt about Fourier analysis. Since QM is a theory of waves it is not surprising that Fourier techniques are very important. In chapter 4.3 we will move on to investigate the relativistic Schrödinger equation, more commonly known as the Klein- Gordon equation. Here we will encounter negative energy solutions and we will see the standard interpretation due to Stückelberg and Feynman. Finally, in chapter 4.4 we will work through the major results in QM perturbation theory, a technique for solving problems that are close to previously solved problems. It is most useful in scattering experiments where the deviation of a particle as the result of an interaction is small relative to its free motion. 4.2 Quantum Mechanics Review In QM the behaviour of a particle is controlled by a wave equation. A free particle is associated with a wave ψ = e i(kx wt) (4.1) where the wave number k and angular frequency w are related to the momentum and energy of the particle p = h λ k = p (4.2) E = hν w = E (4.3) The properties of the particle can therefore be obtained from the wave by acting on it with operators Eψ = i t ψ (4.4) pψ = i x ψ (4.5) The free wave equation is an eigenfunction of these operators with the values of E and p being the eigenvalues. For a classical particle we require that energy is conserved so E = p2 2m + V (4.6) 91

4 which, using the operators, we can rewrite as a wave equation i t ψ = 2 2m This is the Schrödinger equation which is central to QM. 2 x 2 ψ + Vψ (4.7) Time Independent Schrödinger Equation In problems where V is independent of time there are always solutions to the Schrödinger equation of the form ψ(x, t) =u(x)e iet/ (4.8) where u(x) satisfies (simply substitute this solution into the full Schrödinger equation) the time independent Schrödinger equation 2 2m 2 x 2 u(x)+v (x)u(x) =Eu(x) (4.9) Interpretation The amplitude of the wave function ψ (x, t)ψ(x, t) (which in the time independent case is just u (x)u(x)) is associated with the probability of finding a particle at x. Remembering that x is continuous the precise statement is u (x)u(x)dx = prob of finding particle between x and x +dx (4.10) Graphically this looks like the probability of finding the particle in the dx spatial slice is just the area under the curve u u in that slice. Since the particle must be somewhere with probability one we must have u (x)u(x)dx = 1 (4.11) Formally we find observable properties of the particles using the operators p = x = u (x) u (x) xu(x)dx (4.12) ( i ) x u(x)dx (4.13) 92

5 u * u dx x Momentum Space Wave Functions In the above discussion we have described the particle by its wave function at a particular point in space and then shown how to calculate it s momentum with an operator. Alternatively we could write a wave function that describes the probability of the particle having momentum in some dp interval directly and then calculating the position becomes more complicated. In fact it is possible to set up this momentum space wave function such that φ (p) φ(p) dp = prob. of particle having momentum p to p +dp (4.14) φ (p) φ(p)dp = 1 (4.15) with the properties of the particle being given by the operator relations φ (p) φ (p) pφ(p)dp = p (4.16) ( i ) p φ(p)dp = x (4.17) The relationship between ψ(x) and φ(p) is given by a Fourier Transform φ(p) = 1 2π ψ(x)eipx/ dx (4.18) 93

6 or inversely ψ(x) = 1 2π φ(p)e ipx/ dp (4.19) We can demonstrate that the Fourier Transform indeed has the correct properties by checking the consistency of the three operator equations above. Firstly consider [ ] [ ] φ (p) φ(p) dp = 1 2π dp dx e ipx ψ (x ) dx e ipx ψ(x ) = dx dx 1 2π ψ (x )ψ(x ) dpe ip(x x ) (4.20) We recognise the dp integral as the Fourier expansion of a delta function δ(x x 0 )= 1 2π e iω(x x 0) dω (4.21) So with ω = p/ and dω = dp/ φ (p) φ(p)dp = dx dx δ(x x )ψ (x )ψ(x ) = dx ψ (x )ψ(x ) (4.22) = 1 The equations are consistent. Secondly we can check the relation for the expectation value of the particles position ( ) φ (p) i φ(p)dp p 94

7 [ ]( = 1 2π dp dx e ipx ψ (x ) i ( ix )) [ dx ] e ipx ψ(x ) = dx dx 1 2π ψ (x ) x ψ(x ) dpe ip(x x ) = dx dx δ(x x )ψ (x ) x ψ(x ) = dx ψ (x ) x ψ(x ) = x (4.23) Finally we check the expectation value for momentum [ ]( ) [ ] φ (p) pφ(p)dp = 1 2π dp dx e ipx ψ (x ) i x dx e ipx ψ(x ) = dx dx δ(x x )ψ (x ) = dx ψ (x ) ( i x ) ψ(x ) ( i x ) ψ(x ) = p (4.24) Everything is nicely consistent. 95

8 4.2.4 Square Well Example A simple, interesting example of a QM system is the square potential well. We assume that the particle can not penetrate the infinite barriers V= V=0 V= x=0 x=a Since the potential is time independent the solution takes the form ψ(x, t) =u(x)e iet/ (4.25) and we must solve the time independent Schrödinger equation 2 d 2 u(x)+v (x)u(x) =EU(x) (4.26) 2m dx2 Of course in the region of interest the potential is just V = 0. The solutions to this equation take the form u(x) =A sin kx + B cos kx (4.27) The integration constants are fixed by the boundary conditions of ψ vanishing at x =0,a so with n integers 1, 2, 3... u n (x) =A sin nπx a Substituting this solution into the Schrödinger equation we find E n = 2 2m ( nπ a (4.28) ) 2 (4.29) 96

9 Finally to find the constant A we can require ψ(x, t) is correctly normalized ψ ψdx = 1 = a 0 A2 sin 2 nπx a dx (4.30) The full solution is therefore = A 2 a 2 ψ n (x, t) = 2 nπx sin a a e iet/ (4.31) 97

10 4.2.5 Completeness The square well problem shares features with a typical Fourier analysis problem. The solutions in each case are of the form sin nπx. Thus we can make a a Fourier expansion of any initial condition for the wave form and then determine the evolution. For example if we take an initial wave function form at t =0 k x=0 x=a then we can write u(x, t = 0) = c n u n (x) (4.32) where the c n are the Fourier coefficients. Taking into account the normalization of u n (x)) they are given by c n = 8k a n 2 π 2 2 sin nπ 2. (4.33) We now know the time evolution since we know that each individual term evolves as n=1 u n (x, 0) e ient/ u n (x, 0). (4.34) Resuming the Fourier series at time t gives the evolution of the initial condition (to a precision determined by how many terms you resum). 98

11 This is an example of a general rule in QM called completeness: any wave function may be expanded as a series of the eigenfunction solutions of the Schrödinger equation relevant to that problem. In other words in any problem we may write φ(x) = n c n u n (x), (4.35) for any function φ(x), where Hu n = E n u n. (4.36) We won t prove this here but if it weren t true we d be in all sorts of trouble! Imagine we had found all the solutions of the Schrödinger equation and then wrote down an initial condition that couldn t be rewritten in terms of those solutions... we d have no idea how to evolve that initial condition - which is silly! Completeness therefore has to be true for the theory to make sense. 99

12 4.2.6 Orthogonality It is also important in these initial condition problems that there is a unique way of writing u(x, 0) = n c n u n (x) (4.37) If it were not unique then a given initial condition would have more than one expansion which would evolve differently. Again the evolution would be undetermined and the theory not make sense. Each u n (x) therefore contains unique information. Orthogonality is a mathematical statement of this fact u n(x)u m (x)dx = δ nm (4.38) where δ nm = 1 if m = n and δ nm = 0 if m n. You can think of this expression as similar to a dot product between the coordinate axes vectors (î, ĵ, ˆk) - the axes contain the separate information about the three directions in the space and the dot product is zero between any two orthogonal directions. Proof: The u n are eigenfunctions of the Hamiltonian H satisfying Hu n = E n u n so consider u i Hu j dx (4.39) We can act with H to either the left or right in which case we will find E j u i u j dx = E i u i u j dx (4.40) which can only be true for i j if the wave functions are orthogonal and both sides are zero. When i = j the integral over the wave function squared is just the usual probability of finding the particle in all space and is set equal to one. 100

13 4.3 Klein-Gordon equation The Schrödinger equation Going back to the Schrödinger equation, it is useful to remember that a way of deriving it for a free particle of mass m is to start with the classical relation between energy and momentum, and write E and p in terms of differential operators, E = p2 2m, (4.41) E i t, p i. (4.42) The resulting equation is understood to act on a wave function ψ(x, t), and so it reads i ψ t + 2 2m 2 ψ =0. (4.43) As already discussed last year ψ 2 is the probability density, with ψ 2 d 3 x being the probability of finding the particle in the volume d 3 x. From now on we will use the notation ρ ψ 2. The main application within this Quantum Mechanics section of the concepts developed is to the study of scattering processes, where the particles concerned are in motion. It makes therefore sense to define the density flux of a beam of particles, j. This flux obeys a continuity equation, namely the rate of decrease of the number of particles in a given volume must equal the flux of particles out of that volume. We can write that as ρdv = j. ˆn ds, (4.44) t V where ˆn is a unit vector along the outward normal to the element ds of the surface S enclosing volume V. Using Gauss s theorem j. ˆn ds =. jdv, (4.45) S and putting these two last equations together we arrive at ρ t V S +. j =0. (4.46) To determine j we use the Schrödinger, equation eq. (4.43), given that ρ t ψ = ψ t + ψ ψ, (4.47) t 101

14 and we can replace the partial derivatives by the corresponding 2 terms. Then ρ t i 2m (ψ 2 ψ ψ 2 ψ ) = 0. (4.48) This last equation allows us to identify the probability flux density as j = i 2m (ψ ψ ψ ψ ). (4.49) For example in the case of the free particle of energy E and momentum p, we know that a solution to the Schrödinger equation is given by which means that and ψ = Ne ip.x iet/, (4.50) ρ = N 2, (4.51) j = p m N 2. (4.52) The Relativistic Schrödinger Equation It should be clear to us that the Schrödinger equation, eq. (4.43), does not respect Lorentz covariance, i.e. space and time coordinates behave in completely different ways. The way to obtain a relativistic version of the Schrödinger equation is to take as starting point the relativistic relation between energy and momentum E 2 = p 2 + m 2. (4.53) We can apply to this equation the same substitutions of eq. (4.42), and act on a wave function φ to obtain 2 φ t φ = m 2 φ. (4.54) This is the Klein-Gordon equation, which can also de denoted as the relativistic Schrödinger equation. If we multiply it by iφ, and its conjugate equation by iφ the result is t [ i ( φ φ t φ φ t )] + [ i(φ φ φ φ )] = 0 (4.55) By comparison with the non relativistic continuity equation, eq. (4.46) we can identify the probability density as ( ρ = i φ φ ) t φ φ, (4.56) t 102

15 while the flux density is given by j = i(φ φ φ φ ). (4.57) In the particular case of a free particle of energy E and momentum p the solution of the Klein-Gordon equation is φ = Ne ip.x iet, (4.58) in, from now on, units of = 1, c = 1. Using this form for φ in eqs (4.56,4.57) we obtain ρ = i( 2iE) N 2 =2E N 2, (4.59) j = i(2ip) N 2 =2p N 2. (4.60) Note that the expression for the probability density contains a factor of E, the energy of the particle. It is convenient to express all this in four vector notation, in particular we will make use of the D Alembertian operator, µ µ, (4.61) which is the contraction of the four vector derivative µ =( / t, ) with itself. Then the Klein-Gordon equation reads ( + m 2 )φ =0, (4.62) and the probability density and flux can be grouped in a four vector, with the continuity equation reading The free particle solution is expressed as j µ =(ρ, j) =i(φ µ φ φ µ φ ), (4.63) µ j µ =0. (4.64) φ = Ne ip.x (4.65) where x.p = x µ p µ. Plugging this expression into eq. (4.63) we obtain j µ =2p µ N 2. (4.66) The next thing to do is to study the possible values of the energy for this free particle that obeys the Klein-Gordon equation. To do that we substitute the solution (4.65) into (4.62) and the result are the eigenvalues of the energy E = ± p 2 + m 2, (4.67) 103

16 that is, negative energy solutions are possible! Moreover these E<0 solutions are associated to a negative probability density, as it is clear from eq. (4.60). In other words E < 0 solutions with ρ< 0. (4.68) This is a problem that we have to address if we want to continue our study of relativistic quantum mechanics Interpretation of negative energy states It seems from looking at the Klein-Gordon equation that the problem of negative energy states arises, from a mathematical point of view, from the fact that this equation is quadratic in both space and time, which itself arises from the relativistic energy-momentum relation, eq. (4.53). One, therefore, has to take the square root of the resulting expression for the energy 2 and that implies having two possible signs in the solution. In 1927 Dirac derived a relativistic wave equation which was linear in space and time (i.e. proportional to µ rather than ). In that way the probability density was well defined and positive, but he could not avoid having negative energy solutions. Moreover his equation only described spin 1/2 particles. This is known today as the Dirac equation which describes spin 1/2 fermions and antifermions. In 1934 Pauli and Weisskopf revisited the Klein-Gordon equation and introduced a slight modification. They added a factor of e (the electron charge) to the expression of the current, eq. (4.63), which now reads j µ = ie(φ µ φ φ µ φ ). (4.69) Now j 0 = ρ can be interpreted as a charge density, not a probability density, and this explains the fact that it can take negative values (for negatively charged particles). The prescription for dealing with negative energy states is due to Stückelberg (1941) and Feynman (1948). Their idea, commonly accepted today, is that a negative energy solution describes both a particle propagating backward in time and a positive energy antiparticle propagating forward in time. This can be easily seen with an example: consider an electron of energy E, momentum p and charge e. Its corresponding four vector current will be j µ (e )= 2e N 2 (E,p). (4.70) If we now consider a positron, the antiparticle of the electron, with the same energy and momentum, its corresponding four current will read j µ (e + ) = 2e N 2 (E,p), (4.71) 104

17 given that its charge is +e. We can shift signs twice within this equation to obtain j µ (e + )= 2e N 2 ( E, p), (4.72) which can be read as the current for an electron of energy E and momentum p. Therefore the emission of a positron of energy E is equivalent to the absorption of an electron of energy E. This is easily seen if we focus on the energy dependence of the free particle solution of the Klein-Gordon equation, i.e. e ie.t =e i( E).( t). (4.73) Here we see explicitly that the antiparticle going forward in time is equivalent to the particle of negative energy going backward in time. This procedure is also known as the single particle wave function formalism, as it interprets any scattering process involving relativistic and quantum systems in terms of just the particles, without involving the antiparticles. It can be applied not only to single particle states but to a many particle system, as we can see in the particular example of the double scattering of an electron due to a potential. Figure (a) represents what we would naively think it would happen, i.e. as the electron propagates in the region it interacts twice with the potential and t 1 and t 2 >t 1. There is, however a second possibility, shown in (b): at t 2 the electron scatters backward in time. This would be a negative energy electron or, in other words, a positive energy positron (e + ). In order to explain its origin we must follow the whole sequence of events. At t 1 an electron-positron pair is created out of the vacuum. The positron in that pair propagates and annihilates with the electron that had entered the region of space affected by the potential. The remaining electron of the pair is the one that exits this region. Given that we have no way of distinguishing between them, both possibilities (with identical initial (A) and final (B) states) have to be considered in the calculation. To note for the future is the fact that the quantum vacuum is indeed highly populated, and that, therefore, calculations which do not take this fact into account will not reproduce the experimental results. After this descriptive picture of how to understand scattering processes it is time to do some calculations, and the tool to use is called perturbation theory. 4.4 Perturbation Theory Perturbation theory is a technique for solving problems where a system we understand is tweaked by a small change. For example we know how to 105

18 Figure 4.1: Different time orderings of a scattering process. 106

19 find the solutions of the Schrödinger equation for a particle in a square well potential, or a simple harmonic potential - perturbation theory can tell us the solutions (approximately) if these potentials are modified a little. Let s begin by studying the case where the modification is time independent Time Independent Perturbation Theory Consider a time-independent problem in QM we can already solve. means we have found solutions to the Schrödinger equation This where H 0 ψ(x) =E 0 ψ(x) (4.74) H 0 = 2 + V (x) (4.75) 2m x2 We assume that the different solutions φ i have different energies E 0i. Now imagine perturbing the problem by changing the potential by a small amount. Thus 2 H = H 0 + H p (4.76) Since it s a small change most likely the wave function solutions haven t changed much and we can write the new solutions as The Schrödinger equation now becomes ψ i = φ i + δφ i (4.77) (H 0 + H p )(φ i + δφ i ) = (E 0i + E p )(φ i + δφ i ) (4.78) H p, δφ i and E p are all small so we can expand this equation zeroth-order: first order: H 0 φ i = E 0i φ i H 0 δφ i + H p φ i = E 0i δφ i + E p φ i We ve dropped terms that are the square of a small quantity. Now we use the completeness of the set of states φ i to write δφ i = n i c n φ n (4.79) 107

20 (Note: the δφ i is the amount that φ i shifts away from being φ i so we don t include φ i in the sum) The first order expression is H 0 c n φ n + H p φ i = E 0i c n φ n + E p φ i (4.80) n i In the first term we can act with H 0 on φ n and get a factor of E 0n. Now multiply on the left by φ j and integrate over all space φ j E 0n c n φ n dx + n i φ jh p φ i dx = E 0i n i φ j c n φ n dx + E p n i φ jφ i dx (4.81) Using the orthogonality of the wave functions ( φ jφ i dx = δ ij ) we find E 0n c n δ nj + φ jh p φ i dx = E 0i c n δ jn + E p δ ij (4.82) n i and performing the sums c j E 0j + n i φ jh p φ i dx = E 0i c j + E p δ ij (4.83) Now set i = j so the first term on each side cancel and E p = φ i H p φ i dx (4.84) or if j i so the δ ij are zero φ j H p φ i dx c j = (E 0i E 0j ) We have obtained the lowest order perturbation theory results (4.85) E i = E 0i + φ i H pφ i dx ψ i = φ i + n i φ n H p φ i dx (E 0i E 0n ) φ n (4.86) Of course these are not exact - we threw away some small terms. We can get a better approximation by taking the above answers and allowing small corrections on top 108

21 (H 0 + H p )(ψ i + δψ i ) = (E i + E p)(ψ i + δψ i ) (4.87) We then repeat everything above so for example E = E i + ψi H p ψ i dx (4.88) or substituting in terms of the unperturbed results E = E 0i + φ i H p φ i dx + n i φ nh p φ i dx 2 (E i E n ) (4.89) By repeatedly doing this we can make the result arbitrarily good. The first order shift in the energy is easy to calculate since it only requires knowledge of the unperturbed φ i in question. The first order shift in the wave function, and hence second order shift in the energy, requires a knowledge of all the φ n to calculate the shifts for one φ i Example: Perturbed Square Well Consider the square well problem we solved before. Now imagine that the potential is perturbed by a term!*$+,-.'/0"1022!!"#"!"""""$""""""" ""%"&"$"&"'() 109

22 We can calculate the shift in the energy of the solutions u n as follows E n = u n(x) Vu n (x)dx = 2α a a/2 x sin 2 nπx 0 a dx (4.90) = αa 8 αa 2n 2 π 2 sin 2 (nπ/2), where we have integrated by parts with u = x, v = sin 2 (nπx/a), therefore u = 1 and v = x/2 a sin(nπx/a) cos(nπx/a)/(2nπ) Non Relativistic, Time Dependent Perturbation Theory Let us now proceed one step further and work in three spatial dimensions. Our starting point will be the known solutions to the free Schrödinger equation H 0 φ n = E n φ n, (4.91) where φ mφ n d 3 x = δ mn. (4.92) Here H 0 is a time-independent Hamiltonian, and we are going to add a potential, V (x,t), so that the equation to solve (in units og = 1) is (H 0 + V (x,t))ψ = i d ψ. (4.93) dt That is, have to solve the time dependent Schrödinger equation, as the potential we are considering depends on both space and time. Once again we are going to make use of the completeness of the solutions of the original Hamiltonian, H 0 to write ψ = n c n (t)φ n (x)e ient. (4.94) The coefficients c n (t) are therefore carrying the response to the perturbation introduced through the potential. If we replace this last ansatz in the Schrödinger equation eq. (4.93), and use the time independent equation, eq. (4.91), we obtain i n dc n dt φ n(x)e ient = n V (x,t)c n (t)φ n (x)e ient. (4.95) 110

23 In order to isolate the differential equation for a single coefficient, c f, we multiply both sides of the previous equation by φ f and integrate over space, using the orthonormality condition, eq. (4.92). Then dc f dt = i c n (t) φ fvφ n d 3 xe i(e f E n)t. (4.96) n The key point throughout this derivation is the fact that the potential is small, and as such we can assume that, before it acts, the particle is in an eigenstate φ i of the unperturbed Hamiltonian, This implies that c i ( T/2) = 1, c n ( T/2) = 0,n i. (4.97) dc f dt = i which can be integrated to give c f (t) = i t T/2 φ fvφ i d 3 xe i(e f E i )t, (4.98) dt φ fvφ i d 3 xe i(e f E i )t, (4.99) which, at time t = T/2 when the interaction has ceased, can be written as T/2 T fi c f (T/2) = i dt d 3 x[φ f (x)e ie f t ] V (x,t)[φ i (x)e ieit ]. T/2 (4.100) This last equation can be written in a covariant way, i.e. using four vectors, although you must remember that this expression is non relativistic (has been derived making use of the Schrödinger equation!). Finally we get T fi = i d 4 xφ f (x)v (x)φ i(x). (4.101) Here we must understand that x x µ. The immediate physical intuition would tells us to associate T fi 2 with the probability of the particle being scattered from an initial state i into a final state f. However there is a conceptual problem associated to this reading, which is easily seen in the case of a time independent interaction, i.e. when V ((x,t)=v (x). Then T fi = iv fi dte (E f E i )t = 2πiV fi δ(e f E i ), (4.102) 111

24 where V fi d 3 xφ f(x)v (x)φ i (x), (4.103) and we have taken the limit t and we have used the Fourier representation of the delta function, i.e. δ(x x 0 )= 1 2π e iω(x x 0) dω. (4.104) The presence of the delta function in eq. (4.102) tells us about the conservation of energy in the scattering process, that is the energies of initial and final states are the same. But then, by the uncertainty principle, it would be needed an infinite amount of time to conduct the experiment, i.e. to see a transition from initial to final state. This is clearly not realistic, and the meaningful quantity is the transition probability per unit time, W = lim 2π T fi 2 T T. (4.105) Going back to eq. (4.102), we can now square it and compute W. Note that we will compute one of the time integrals straightaway, while leaving the other in terms of T, W = lim 2π V fi 2 T T δ(e f E i ) = lim 2π V fi 2 T T δ(e f E i ) T/2 T/2 T/2 T/2 dte i(e f E i )t dt =2π V fi 2 δ(e f E i ).(4.106) In the second line we have applied the delta function to make E i = E f in the exponent. Then the time integral is equal to T which cancels that of the denominator. Finally, to link this calculation with experiment, we must assume that the final state is unknown and, therefore, there is a variety of possible final states (we can still assume that the initial state is the one we have prepared). If ρ(e f ) is the density of final states, then ρ(e f )de f is the number of states with energy between E f and E f +de f. Integrating over this density we obtain W fi =2π ρ(e f )de f V fi 2 δ(e f E i ), (4.107) which, integrating over de f and using the delta function leads to W fi =2π V fi 2 ρ(e i ). (4.108) 112

25 This is Fermi s Golden Rule, which tells us about the probability of a transition i f per unit time. It is remarkably accurate when compared to atomic physics experiments Feynman Rules On can try to improve the calculation that led to Fermi s Golden Rule, by iterating the procedure. That is, we take the solution for the c n (t) coefficients given by eq. (4.99) and we substitute in eq. (4.96). We are therefore no longer assuming that the initial conditions c i = 1. c n =0,n i persist throughout the process, but rather input the solutions obtained for c f (t) for the c n,n i. Then, eq. (4.96) becomes dc f dt = iv fi e i(e f E i )t [ t + ( i) 2 V ni n i T/2 dt e i(en E i)t ] V fn e i(e f E n)t, (4.109) where the first line is the first order result we obtained before. The new term is going to generate a correction to T fi as follows T fi = 2πiV fi δ(e f E i ) (4.110) t V fn V ni dte i(e f E n)t dt e i(en E i)t, n i where again the first line represents the first order result we already obtained. In the second line we must compute the integral in t first, for which it is necessary to add a small quantity ɛ to the exponent, to ensure that the lower limit is finite. Then we would take ɛ 0 at the end of the calculation t dt e i(en E i iɛ)t = i ei(en E i iɛ)t E i E n + iɛ. (4.111) Plugging this back in eq. (4.111), and computing the delta function integral we obtain the second order correction, T fi = 2πiV fi δ(e f E i ) 2πi n i Which is equivalent to making the replacement V fi V fi + n i V fn V ni E i E n + iɛ δ(e f E i ), (4.112) V fn 1 E i E n + iɛ V ni +... (4.113) 113

26 in the first order expression. This equation is nothing but the perturbation series for the amplitude, in this case showing the first and second order in V. They can be put in direct correspondence with a set of Feynman diagrams corresponding to a scattering process between points a and b in space time, as shown in fig In fig. 4.3 we see the process in the absence of scattering, t t b b t c c t a a x Figure 4.2: Space-time diagram to show the location of the initial (a) and final (b) states, as well as intermediate point c. plus the first and second order contributions to the transition amplitude. These diagrams define what we call Feynman rules for scattering processes, that are summarised as follows: for every interaction vertex (such as c) we have a factor of V ni for the propagation of each intermediate state (line cd in the figure) we have a factor of 1/(E i E n ) (also called the propagator) at every vertex integration over space and time must be understood A couple of remarks are in order: note that the intermediate states are virtual, that is they do not conserve energy. There is no delta function accompanying them, and also if the energies were the same the propagator would blow up. There is an overall delta of conservation of initial and final energy, but it does not apply to the intermediate states Initial Response to a Perturbation One example we can make progress with is studying the initial response of a system to the imposition of a perturbation at, say, the time t a = 0. Thus 114

27 area of space with potential b b d b a a c a c no scatters one scatter two scatters Figure 4.3: Zeroeth order plus first and second order contributions to the a b transition. V p = 0 for t<0 V p = V p (x) for t 0 (4.114) Suppose the system begins in the state φ i what is the transition amplitude to a different state φ m at time t b = T? To compute this, we go back to equation (4.99), modifying the time limits to fit this particular problem. Then T c m (T ) = i dt 0 φ mv p φ i d 3 xe i(em E i)t = i(v p ) mi 1 i(e m E i ) (ei(em E i)t 1). (4.115) The probability of a transition from an initial state i to a final state m will then be the square of the previous quantity P mi = c m (T ) 2 =(V p ) mi(v p ) mi 1 (E m E i ) 2 ei(em E i)t 1 2, (4.116) where, as usual, (V p ) mn = φ mv p φ n d 3 x. To compute the modulus square of the previous equation we do the following e i(em E i)t 1 2 = (e i(em E i)t 1)(e i(em E i)t 1) (4.117) = e i(em E i)t e i(em E i)t = 2[1 cos((e m E i )T )]. 115

28 Using the trigonometric formula for the cosine of the double angle, cos α = cos 2 (α/2) sin 2 (α/2) we have ( ) e i(em Ei)T 1 2 = 4 sin 2 (Em E i )T, (4.118) 2 so that the probability reads P mi = c m (T ) 2 =(V p ) mi(v p ) mi sin 2 ( (Em E i )T 2 ) ( Em E i 2 ) 2. (4.119) Plotting the factor sin 2 x/x 2 at fixed T gives a curve of the form It s maximum is at x = 0, that is E m = E i and it falls to zero when Em E i = π. Essentially 2 T only transitions between states in this energy range occur. As the maximum of the function is at T 2, and the width of the main peak is 1/T it is easy to deduce what happens in the limit of very large T : the peak narrows and the function tends to a delta function (which is multiplied by T ). We therefore revert to the previous situation where the interaction was always switched on, i.e. we must divide by T in order to get the transition rate, namely 1 lim T T ( ) 2 sin(xt) = πδ(x). (4.120) Example - Perturbed Square Well II x Consider the perturbed square well we had in the previous example. The probability of a transition from the ground state to an excited state as a result of imposing the perturbation is 116

29 is [ ] (En E P n1 (t) (V p ) n1 2 sin2 1 ) t 2 (E n E 1 ) 2 /4 (4.121) The energies of the level are already known so the only thing to calculate (V p ) n1 = ψ n(x)v p ψ 1 (x)dx = 2α a = α a a/2 x sin πx 0 a a/2 0 x [ cos (n+1)πx a nπx sin dx a Integrating by parts one finds, for example, cos (n 1)πx a ]. (4.122) (V p ) 13 = aα 2π 2 (4.123) 117