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Interaction picture and smatrix 
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#1
Dec2012, 10:38 AM

P: 229

Consider the Smatrix:
<fU(t,t)i> When going into the interaction picture, this becomes: [tex]<f_IU_I(t,t)i_I> [/tex] where the propagator is the interaction picture propagator, and the states are interaction states. Can you say that: [tex]<f_IU_I(t,t)i_I>= <fU_I(t,t)i>[/tex] ? It seems that the two sides of the above equation differ by phase factors that don't really matter when calculating probabilities. However, when making calculations, don't the states i> and f> need to be in the Schrodinger picture? 


#2
Dec2012, 05:40 PM

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P: 5,451

OK, you start with the expression
[tex]\langle f_S\,U_S\,i\rangle_S = (\langle f_S\,\Omega^\dagger)\;(\Omega\, U_S\,\Omega^\dagger)\;(\Omega\,i\rangle_S) = \langle f_I\,U_I\,i\rangle_I[/tex] U denotes the time evolution operator and Ω is the unitary operator [tex]\Omega_{S \to I}(t) = e^{iH_0t}[/tex] [tex]\Omega\,\Omega^\dagger = 1[/tex] which transformes states and operators from the Schrödinger to the interactionpicture. Now let's look at your expression [tex]\langle f_S\,U_I\,i\rangle_S[/tex] We can rewrite this as [tex]\langle f_S\,U_I\,i\rangle_S = \langle f _S\;\Omega\, U_S\, \Omega^\dagger\; i\rangle_S[/tex] Why should the two expressions only differ by a phase factor? 1) U depends on the full Hamiltonian [tex]H = H_0 + V[/tex] whereas Ω depends on H_{0}; so we get [tex]\Omega\, U_S\, \Omega^\dagger = e^{iH_0t} e^{i(H_0 + V)t} e^{iH_0t}[/tex] Using a Taylor series expansion for U we get a sum of terms [tex]e^{iH_0t} (H_0 + V)^n e^{iH_0t}[/tex] which can be reduced to individual terms [tex]e^{iH_0t} (H_0 + V) e^{iH_0t} = H_0 + e^{iH_0t} V e^{iH_0t} [/tex] But H_{0} and V do not commute in general. So your expression which mixes Schrödinger states and interaction operators is meaningless. 


#3
Dec2012, 07:58 PM

P: 229

The reason I'm asking this is because I have a textbook that talks about the Smatrix as: S=<fUi> then it somehow replaces U with the interaction propagator U_{I} and then writes: S=<fU_{I}i> and I'm wondering don't the states have to change to when you change to the interaction picture? The idea is that i>=(creation operators) 0> but are the creation operators in the interaction picture? 


#4
Dec2112, 02:32 AM

Sci Advisor
P: 5,451

Interaction picture and smatrix
I see your point.
What you need is the formal derivation of the Smatrix including a formal proof of the existence of the limit t → ∞, and of the applicability of replacing interacting states with free states (or "switching off" the interaction). This is nontrivial, even many textbooks don't care about the proof. I don't have a good refernce at hand, but you may google for "Moeller operators"; perhaps this helps. 


#5
Dec2112, 04:22 AM

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Thanks
P: 2,501

Also one should be warned that not always you can use naive "freeparticle" states as the asymptotic states. The most simple example is the nonrelativistic Coulombscattering cross section. This describes the scattering of a particle on a classical Coulomb potential (roughly the scattering of a lowenergy electron (forgetting about its spin and magnetic moment) on the Coulomb potential of an atomic nucleus. The differential cross section is
[tex]\frac{\mathrm{d} \sigma}{\mathrm{d} \Omega} \propto \frac{1}{\sin^4(\vartheta/2)}.[/tex] The total cross section is divergent, because you have to integrate this expression (multiplied by [itex]\sin \vartheta[/itex]) over [itex]\vartheta[/itex] from [itex]\vartheta=0[/itex] to [itex]\pi[/itex]. The integral is divergent on both boundaries of the integral. This result is exact within nonrelativistic quantum theory. So it's not an artifact of perturbation theory. Within nonrelativistic quantum theory you can "cure" the problem by the fact that usually there is no pure Coulomb field in reality, but other charges around the atomic nucleus screen it to a Yukawalike potential [itex]1/r \rightarrow \exp(r/r_0)/r[/itex], where [itex]r_0[/itex] is the socalled Debye radius. With this potential, which falls off exponentially for [itex]r \rightarrow \infty[/itex] you get a perfectly finite total cross section. Using the Born approximation (1storder perturbation theory) you get the same result for the Coulomb potential, the reason being that for the Coulomb potential the exact scattering amplitude deviates from the Born approximation only by a (nontrivial) phase, which doesn't enter into the expression for the cross section, because that's obtained by the modulus of the amplitude squared, but the nontrivial phase factor points into the direction what's wrong with the naive expression for the cross section: You miss the phase factor, because you use (hidden or in good textbooks explicitly) the naive planewave initial and final states for the incoming and outgoing electron after regularizing the expression by using a Yukawalike potential and letting the Debye radius [itex]r_0 \rightarrow 0[/itex] after calculating the entire expression for the matrix element and the cross section. A more detailed analysis shows that the true asymptotic state, within nonrelativistic quantum theory, is not a plane wave due to the longrange nature of the Coulomb potential but a socalled "distorted wave", which in this case can be found by solving the scattering problem exactly, which is possible for the Coulomb potential due to the famous additional symmetry (Lenz vector) of the Coulomb problem. In the asymptotic limit the corresponding wave function is given by the plane wave with the above mentioned additional phase factor. A very intuitive picture, aiming directly at the asymptotic perturbative value of the matrix element without the necessity to first calculate the exact solution for the timeindependent Schrödinger equation, occurs by doing the analysis of the asymptotic in the interaction picture, i.e., solving the timedependent scattering problem, is given in the beginning of the paper P.P. Kulish and L.D. Faddeev. Asymptotic conditions and infrared divergences in quantum electrodynamics. Theor. Math. Phys., 4:745, 1970. http://dx.doi.org/10.1007/BF01066485 This paper also gives a treatment of the problem with the infinite Coulombscattering cross section within QED. The divergence of the Coulombscattering problem is still not solved, despite invoking the argument with the screening of the pure Coulomb potential by other charges around it. The divergence of the Coulomb cross section comes from the longrange nature of the Coulomb potential, i.e., the fact that the photon is massless. The usual perturbative treatment is to introduce a fictitious photon mass, which makes all particles massive in QED (and does not spoil renormalizability, because QED is an Abelian Gauge theory) and then consider the fact that the accelerated electron radiates off photons and that thus the final asymptotic state is never a single electron at the same energy as the incoming electron but at best an electron with a slightly smaller energy plus a bunch of soft photons. Thus, even if you start with a singleelectron zerophoton initial asymptotic state, in the final state you have to consider all processes with a single electron and any number of soft photons with a total energy less than the detector's energy resolution, because this defines the true physical meaning of a process measured as "elastic" by the detetector. After this softphoton resummation the result for the (perturbative) cross section becomes finite in the limit of vanishing photon mass. This famous result by Bloch and Nordsieck can be found in a very concise derivation in Weinberg's Quantum Theory of Fields Vol. 1. A more complete analysis of the asymptotic states in QED is given in the above mentioned paper: The final state to use is not a singleelectron momentum eigenstate but a singleelectron distorted wave plus a coherent photon state, which describes the irradiation of (soft) photons. A full treatment of these problems are given in M.S. Swanson. Reduction formulas for quantum electrodynamics. Phys. Rev. D, 25:20862102, 1982. 10.1103/PhysRevD.25.2086 and references therein. 


#6
Dec2112, 07:12 AM

P: 1,020




#7
Dec2112, 08:21 AM

P: 229

Thanks everyone. Equation 4.89 of Peskin and Schroeder discusses it a little. As you were all saying, the initial and final states are interacting planewave states, not free ones. But if you write everything in terms of noninteracting plane wave states, it all simplifies to <fU_{I}i>, where i> and f> are noninteracting plane wave states.
I also vaguely recall the distorted waves. The idea is that asymptotically you don't have a free planewave. Instead you have to expand in the basis of the full Hamiltonian, and the full Hamiltonian has asymptotic solutions of free plane wave + spherical wave. Which makes sense, since after scattering you should get a spherical wave and not a plane wave. But what's nice is that the Fourier coefficients of the asymptotic solutions are the same as the Fourier coefficients of the free planewaves, so if you want to find the amplitude between two plane wave states, you can use their Fourier coefficients in the noninteracting theory as the coefficients for the asymptotic states, and sandwich the states between the interaction propagator. Or something like that. The book I had glossed over the whole asymptotic thing. But it's a particle physics book (for experimentalists) rather than a QFT book, so that's probably why. Thanks. 


#8
Dec2112, 12:52 PM

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P: 2,501

Fortunately you don't need to calculate the asymptotic states with the full Hamiltonian, which would mean to solve the interacting QFT (e.g. by solving the nonlinear operator equations in the Heisenberg picture), and that's not possible in practice. At least nobody has found a solution for the full interacting theory (except in toy models in 1+1 dimensions, etc.).
What you need to find is only the correct asymptotic solution. The most simple example is given in the paper by Kulish and Faddeev, which I've cited in my previous post. You use a decomposition of the Hamiltonian into a free+corrections for the asymptotic dynamics and the "rest Hamiltonian". The result are distorted waves for the asymptotic states, and the perturbation theory works in the same way as for shortrange potentials for the "rest Hamiltonian". 


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