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Interaction picture and s-matrix

by geoduck
Tags: interaction, picture, smatrix
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geoduck
#1
Dec20-12, 10:38 AM
P: 229
Consider the S-matrix:

<f|U(t,-t)|i>

When going into the interaction picture, this becomes:

[tex]<f_I|U_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_I|U_I(t,-t)|i_I>= <f|U_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?
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tom.stoer
#2
Dec20-12, 05:40 PM
Sci Advisor
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 interaction-picture.

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 H0; 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 H0 and V do not commute in general.

So your expression which mixes Schrödinger states and interaction operators is meaningless.
geoduck
#3
Dec20-12, 07:58 PM
P: 229
Quote Quote by tom.stoer View Post

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?
But the Schrodinger states |i> and |f> are eigenstates of the free Hamiltonian Ho? So that Ω on the RHS, when it operates on the states, just creates a phase factor?

The reason I'm asking this is because I have a textbook that talks about the S-matrix as:

S=<f|U|i>

then it somehow replaces U with the interaction propagator UI and then writes:

S=<f|UI|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?

tom.stoer
#4
Dec21-12, 02:32 AM
Sci Advisor
P: 5,451
Interaction picture and s-matrix

I see your point.

What you need is the formal derivation of the S-matrix 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 non-trivial, 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.
vanhees71
#5
Dec21-12, 04:22 AM
Sci Advisor
Thanks
P: 2,501
Also one should be warned that not always you can use naive "free-particle" states as the asymptotic states. The most simple example is the non-relativistic Coulomb-scattering cross section. This describes the scattering of a particle on a classical Coulomb potential (roughly the scattering of a low-energy 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 non-relativistic quantum theory. So it's not an artifact of perturbation theory. Within non-relativistic 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 Yukawa-like 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 (1st-order 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 (non-trivial) 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 non-trivial 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 plane-wave initial and final states for the incoming and outgoing electron after regularizing the expression by using a Yukawa-like 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 non-relativistic quantum theory, is not a plane wave due to the long-range 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 time-independent Schrödinger equation, occurs by doing the analysis of the asymptotic in the interaction picture, i.e., solving the time-dependent 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 Coulomb-scattering cross section within QED. The divergence of the Coulomb-scattering 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 long-range 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 single-electron zero-photon 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 soft-photon 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 single-electron momentum eigenstate but a single-electron 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, 25:2086-2102, 1982.
10.1103/PhysRevD.25.2086

and references therein.
andrien
#6
Dec21-12, 07:12 AM
P: 1,020
Quote Quote by geoduck View Post
But the Schrodinger states |i> and |f> are eigenstates of the free Hamiltonian Ho? So that Ω on the RHS, when it operates on the states, just creates a phase factor?

The reason I'm asking this is because I have a textbook that talks about the S-matrix as:

S=<f|U|i>

then it somehow replaces U with the interaction propagator UI and then writes:

S=<f|UI|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?
is not s matrix defined for interaction representation i.e. only interaction part is concerned.
geoduck
#7
Dec21-12, 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 plane-wave states, not free ones. But if you write everything in terms of non-interacting plane wave states, it all simplifies to <f|UI|i>, where |i> and |f> are non-interacting plane wave states.

I also vaguely recall the distorted waves. The idea is that asymptotically you don't have a free plane-wave. 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 plane-waves, so if you want to find the amplitude between two plane wave states, you can use their Fourier coefficients in the non-interacting 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.
vanhees71
#8
Dec21-12, 12:52 PM
Sci Advisor
Thanks
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 non-linear 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 short-range potentials for the "rest Hamiltonian".


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