The LSZ reduction formula based on S-5 In order to describe scattering experiments we need to construct appropriate initial and final states and calculate scattering amplitude. Summary of free theory: one particle state: vacuum state is annihilated by all a s: then, one particle state has normalization: normalization is Lorentz invariant! see e.g. Peskin & Schroeder, p. 23
Let s define a time-independent operator: that creates a particle localized in the momentum space near wave packet with width σ and localized in the position space near the origin. (go back to position space by Fourier transform) is a state that evolves with time (in the Schrödinger picture), wave packet propagates and spreads out and so the particle is localized far from the origin in at. for in the past. In the interacting theory is a state describing two particles widely separated is not time independent
A guess for a suitable initial state: Similarly, let s consider a final state: we can normalize the wave packets so that where again and The scattering amplitude is then:
A useful formula: Integration by parts, surface term = 0, particle is localized, (wave packet needed). E.g. is 0 in free theory, but not in interacting one!
Thus we have: or its hermitian conjugate: The scattering amplitude: we put in time ordering (without changing anything) is then given as (generalized to n i- and n f-particles):
Lehmann-Symanzik-Zimmermann formula (LSZ) Note, initial and final states now have delta-function normalization, multiparticle generalization of. We expressed scattering amplitudes in terms of correlation functions! Now we need to learn how to calculate correlation functions in interacting quantum field theory.
Comments: we assumed that creation operators of free field theory would work comparably in the interacting theory... acting on ground state: we want, so that we can always shift the field by a constant so that is a Lorentz invariant number is a single particle state otherwise it would create a linear combination of the ground state and a single particle state
one particle state: is a Lorentz invariant number we want, since this is what it is in free field theory, correctly normalized one particle state. creates a we can always rescale (renormalize) the field by a constant so that.
multiparticle states: is a Lorentz invariant number in general, creates some multiparticle states. One can show that the overlap between a one-particle wave packet and a multiparticle wave packet goes to zero as time goes to infinity. see the discussion in Srednicki, p. 40-41 By waiting long enough we can make the multiparticle contribution to the scattering amplitude as small as we want.
Summary: Scattering amplitudes can be expressed in terms of correlation functions of fields of an interacting quantum field theory: provided that the fields obey: Lehmann-Symanzik-Zimmermann formula (LSZ) these conditions might not be consistent with the original form of lagrangian!
Consider for example: After shifting and rescaling we will have instead:
Path integral in quantum mechanics based on S-6 Consider nonrelativistic quantum mechanics of one particle in one dimension with the hamiltonian: P and Q obey: Probability amplitude for the particle to start at q at time t and end up at position q at time t is where and are eigenstates of the position operator Q. In the Heisenberg picture: and we can define instantaneous eigenstates: Probability amplitude is then:
to evaluate the transition amplitude: let s divide the time interval T = t - t into N+1 equal pieces = insert N complete sets of position eigenstates
let s look at one piece first: Campbell-Baker-Hausdorf formula complete set of momentum states
to evaluate the transition amplitude: let s divide the time interval T = t - t into N+1 equal pieces = insert N complete sets of position eigenstates - we find: important for general form of hamiltonians with terms containing both P and Q in our case, it doesn t make any difference
taking the limit we get: should be understood as integration over all paths in phase space that start at and end at (with arbitrary initial and final momenta) In simple cases when hamiltonian is at most quadratic in momenta, the integral over p is gaussian and can be easily calculated: lagrangian prefactor can be absorbed into the definition of measure
taking the limit we get: should be understood as integration over all paths in phase space that start at and end at (with arbitrary initial and final momenta) In simple cases when hamiltonian is at most quadratic in momenta: where is calculated by finding the stationary point of the p-integral by solving: for p and plugging the solution back to
What is it good for? Consider e.g.: the result can be simply written using path integral as: Similarly: time ordering is crucial! Time-ordered products appeared in LSZ formula for scattering amplitudes!
Functional derivatives: Dirac delta function continuous generalization of they are defined to satisfy all the usual rules of derivatives (product rule,...) Consider modifying hamiltonian to: Then we have: And we find, e.g.:
more examples: after we bring down as many qs and ps as we want we can set and return to the original hamiltonian:
Finally, we want both initial and final states to be ground states and take the limits and : looks complicated, we will use the following trick instead: is the ground-state wave function eigenstate of H corresponding eigenvalue is wave function of n-th state let s replace with and take the limit : every state except the ground state is multiplied by a vanishing factor!
thus we have: Similarly, for the replacement picks up the ground state as the final state in the limit. we can integrate over q and q which leads to a constant factor that can be absorbed into the normalization of the path integral. Thus, with the replacement the boundary conditions and we have: we don t have to care about
Adding perturbations: we can simply write (suppressing the ): Finally, if perturbing hamiltonian depends only on q, and we want to calculate only time-ordered products of Qs, and if H is no more than quadratic in P and if the term quadratic in P does not involve Q, then the equation above can be written as:
Path integral for harmonic oscillator Consider a harmonic oscillator: based on S-7 ground state to ground state transition amplitude is: external force equivalent to thus going to lagrangian formulation (integrating over p) we get:
using Fourier-transformed variables: and setting for simplicity, we get and thus: E = E
it is convenient to change integration variables: then we get: a shift by a constant and the transition amplitude is:
But since, if there is no external force, a system in its ground state remain in its ground state. thus we have: or, in terms of time-dependent variables: using inverse Fourier transform where:
Comment: is a Green s function for the equation of motion of the harmonic oscillator: you can evaluate it explicitly, treating the integral as a contour integral in the complex E-plane and using the residue theorem. Make sure you are careful about closing the contour in the correct half-plane for t > t and t < t and that you pick up the correct pole. you should find:
Let s calculate correlation functions of Q operators: for harmonic oscillator we find: For odd number of Qs there is always one f(t) left-over and the result is 0!
For even number of Qs we pair up Qs in all possible ways: in general: We can now easily generalized these results to a free field theory...
Path integral for free field theory Hamiltonian density of a free field theory: based on S-8 similar to the hamiltonian of the harmonic oscillator dictionary between QM and QFT: classical field operator field classical source we repeat everything we did for the path integral in QM but now for fields; we divide space and time into small segments; take a field in each segment to be constant; the differences between fields in neighboring segments become derivatives; use the trick: multiplying by is equivalent to replacing with which we often don t write explicitly;... eventually we can integrate over momenta and
obtain path integral (functional integral) for our free field theory: path in the space of field configurations Comments: lagrangian is manifestly Lorentz invariant and all the symmetries of a lagrangian are preserved by path integral lagrangian seems to be more fundamental specification of a quantum field theory
to evaluate we can closely follow the procedure we did for the harmonic oscillator: Fourier transform:
change of integration variables: a shift by a constant
But
Thus we have: where we used inverse Fourier transform to go back to position-functions is the Feynman propagator, a Green s function for the Klein-Gordon equation: integral over zero s component can be calculated explicitly by completing the contour and using the residue theorem, the three momentum integral can be calculated in terms of Bessel functions
Now we can calculate correlation functions: we find: For odd number of there is always one J left-over and the result is 0!
For even number of we pair up in all possible ways: in general: Wick s theorem
Path integral for interacting field Let s consider an interacting phi-cubed QFT: based on S-9 with fields satisfying: we want to evaluate the path integral for this theory:
it can be also written as: epsilon trick leads to additional factor; to get the correct normalization we require: and for the path integral of the free field theory we have found:
assumes thus in the case of: the perturbing lagrangian is: counterterm lagrangian in the limit we expect and we will find and
Let s look at Z( J ) (ignoring counterterms for now). Define: exponentials defined by series expansion: let s look at a term with particular values of P (propagators) and V (vertices): number of surviving sources, (after taking all derivatives) E (for external) is E = 2P - 3V 3V derivatives can act on 2P sources in (2P)! / (2P-3V)! different ways e.g. for V = 2, P = 3 there is 6! different terms
V = 2, E = 0 ( P = 3 ): ¹ ¹ ² ² ¹ ¹ ¹ ¹ ¹ ¹ ² ² ² ² ² ² ³ ³ ³ ³ ³ ³ 3! 3! 2 2 2 2! 6 6 3! 2 2 2 x 1 x 2 = 1 12 dx 1 dx 2 (iz g g) 2 1 i (x 1 x 2 ) 1 i (x 1 x 2 ) 1 i (x 1 x 2 ) symmetry factor
V = 2, E = 0 ( P = 3 ): ¹ ¹ ² ² ¹ ¹ ¹ ¹ ¹ ¹ ² ² ² ² ² ² ³ ³ ³ ³ ³ ³ 3! 3! 3! 2 x 1 x 2 2! 6 6 3! 2 2 2 = 1 8 dx 1 dx 2 (iz g g) 2 1 i (x 1 x 1 ) 1 i (x 1 x 2 ) 1 i (x 1 x 1 ) symmetry factor
Feynman diagrams: a line segment stands for a propagator vertex joining three line segments stands for a filled circle at one end of a line segment stands for a source e.g. for V = 1, E = 1 What about those symmetry factors? symmetry factors are related to symmetries of Feynman diagrams...
Symmetry factors: we can rearrange three derivatives without changing diagram we can rearrange three vertices we can rearrange two sources we can rearrange propagators this in general results in overcounting of the number of terms that give the same result; this happens when some rearrangement of derivatives gives the same match up to sources as some rearrangement of sources; this is always connected to some symmetry property of the diagram; factor by which we overcounted is the symmetry factor
the endpoints of each propagator can be swapped and the effect is duplicated by swapping the two vertices propagators can be rearranged in 3! ways, and all these rearrangements can be duplicated by exchanging the derivatives at the vertices
All these diagrams are connected, but Z( J ) contains also diagrams that are products of several connected diagrams: e.g. for V = 4, E = 0 ( P = 6 ) in addition to connected diagrams we also have : and also: and also:
All these diagrams are connected, but Z( J ) contains also diagrams that are products of several connected diagrams: e.g. for V = 4, E = 0 ( P = 6 ) in addition to connected diagrams we also have : A general diagram D can be written as: the number of given C in D additional symmetry factor not already accounted for by symmetry factors of connected diagrams; it is nontrivial only if D contains identical C s: particular connected diagram
Now is given by summing all diagrams D: any D can be labeled by a set of n s thus we have found that is given by the exponential of the sum of connected diagrams. imposing the normalization means we can omit vacuum diagrams (those with no sources), thus we have: vacuum diagrams are omitted from the sum
If there were no counterterms we would be done: in that case, the vacuum expectation value of the field is: only diagrams with one source contribute: and we find: (the source is removed by the derivative) we used since we know which is not zero, as required for the LSZ; so we need counterterm
Including term in the interaction lagrangian results in a new type of vertex on which a line segment ends e.g. corresponding Feynman rule is: at the lowest order of g only contributes: in order to satisfy we have to choose: Note, must be purely imaginary so that Y is real; and, in addition, the integral over k is ultraviolet divergent.
to make sense out of it, we introduce an ultraviolet cutoff and in order to keep Lorentz-transformation properties of the propagator we make the replacement: the integral is now convergent: we will do this type of calculations later... and indeed, is purely imaginary. after choosing Y so that we can take the limit Y becomes infinite... we repeat the procedure at every order in g
e.g. at we have to sum up: and add to Y whatever term is needed to maintain... this way we can determine the value of Y order by order in powers of g. Adjusting Y so that means that the sum of all connected diagrams with a single source is zero! In addition, the same infinite set of diagrams with source replaced by ANY subdiagram is zero as well. Rule: ignore any diagram that, when a single line is cut, fall into two parts, one of which has no sources. = tadpoles
all that is left with up to 4 sources and 4 vertices is:
finally, let s take a look at the other two counterterms: we get it results in a new vertex at which two lines meet, the corresponding vertex factor or the Feynman rule is Summary: we have calculated in theory and expressed it as we used integration by parts for every diagram with a propagator there is additional one with this vertex where W is the sum of all connected diagrams with no tadpoles and at least two sources!
Scattering amplitudes and the Feynman rules based on S-10 We have found Z( J ) for the phi-cubed theory and now we can calculate vacuum expectation values of the time ordered products of any number of fields. Let s define exact propagator: short notation: thus we find: W contains diagrams with at least two sources +...
4-point function: we have dropped terms that contain Let s define connected correlation functions: does not correspond to any interaction; when plugged to LSZ, no scattering happens and plug these into LSZ formula.
at the lowest order in g only one diagram contributes: S = 8 derivatives remove sources in 4! possible ways, and label external legs in 3 distinct ways: each diagram occurs 8 times, which nicely cancels the symmetry factor.
General result for tree diagrams (no closed loops): each diagram with a distinct endpoint labeling has an overall symmetry factor 1. Let s finish the calculation of y z putting together factors for all pieces of Feynman diagrams we get:
For two incoming and two outgoing particles the LSZ formula is: and we have just written terms of propagators. The LSZ formula highly simplifies due to: in We find:
four-momentum is conserved in scattering process Let s define: scattering matrix element From this calculation we can deduce a set of rules for computing.
Feynman rules to calculate : for each incoming and outgoing particle draw an external line and label it with four-momentum and an arrow specifying the momentum flow draw all topologically inequivalent diagrams for internal lines draw arrows arbitrarily but label them with momenta so that momentum is conserved in each vertex assign factors: 1 for each external line for each internal line with momentum k for each vertex sum over all the diagrams and get
Additional rules for diagrams with loops: a diagram with L loops will have L internal momenta that are not fixed; integrate over all these momenta with measure divide by a symmetry factor include diagrams with counterterm vertex that connects two propagators, each with the same momentum k; the value of the vertex is now we are going to use to calculate cross section...
Cross sections and decay rates based on S-11 Particle physics experiments typically measure cross sections and decay rates. Kinematics of a scattering process: two convenient frames: center-of-mass, or CM frame: particles can have different mass fixed target, or FT frame (lab frame):
CM frame (we choose to be in z-direction): there is only one free initial parameter. However it is convenient to define: which is Lorentz invariant; in the CM frame it is equal to then we find: center-of-mass energy squared
finally instead of, it is convenient to define a Lorentz scalar:
Mandelstam variables: they satisfy: the scattering matrix element in can be simply written as:
Fixed target frame: in this case from we have: comparing it with the result in the CM frame, we find
Formula for the differential scattering cross section: we assume the experiment is taking place in a big box of volume V, and lasts for a large time T (we should be thinking about colliding wave packets but we will simplify the discussion somewhat, for more precise treatment see e.g. Peskin and Schroeder) probability for 1,2 1,2...,n is: norm of a single particle state is: thus we have: probability per unit time
this is the probability per unit time to scatter into a set of outgoing particles with precise momenta. we should sum over each momenta due to the box we have: in a small range; vector with integer entries in the limit of large L we have: thus we should consider:
finally to get the cross section we should divide by the incident flux: cross section x incident flux = Probability per unit time = the number of particles per unit volume that are striking the target particle times their speed (easy to evaluate in the FT frame): we have one particle in V with speed and so the incident flux is thus in the CM frame (using and ) we find: where we defined the n -body Lorentz-invariant phase-space measure:
Two outgoing particles: Lorentz invariant, we can compute it in any frame, it is convenient to work in the CM frame:
can be evaluated using differential solid angle in our case f(x)=0 for thus we have:
or, in a frame independent form: in general, depends on both s and and so the formula is more complicated than in the CM frame
Total cross section: the number of identical outgoing particles of type i For two outgoing particles we have: symmetry factor dlips treats outgoing particles as an ordered list of momenta or, equivalently: correspond to and.
Let s get back to the scattering process in theory we considered: In the CM frame: all masses equal we obtain as a complicated function of s and. In the nonrelativistic limit, or : differential cross section almost isotropic.
Let s get back to the scattering process in theory we considered: In the CM frame: all masses equal we obtain as a complicated function of s and. In the extreme relativistic limit, or : diff. cross section sharply peaked in the forward and backward directions.
Let s get back to the scattering process in theory we considered: integrating over t (for fixed s) we can calculate the cross section: correspond to and. we get:
In the nonrelativistic limit, or : In the extreme relativistic limit, or :
Formula for the differential decay rate: we assume that the LSZ formula is valid for a single particle that can decay following the derivation of : with the only difference being: identifying with gives: In the CM frame ; in other frames, the relative factor accounts for relativistic time dilation of the decay rate. Finally, a total decay rate: