S-matrix in Weinberg's book

In summary, the S-matrix is defined as the inner product of the in- and out-states, and in the context of Lorentz invariance, the same transformation rule as that for free particle states applies to both in- and out-states. However, this rule may not apply to all states, as the in- and out-states may only appear non-interacting to observers at temporal infinity. This is because they are able to be eigenstates of the full Hamiltonian, as shown by Equation (3.1.12). This approach is also used in non-relativistic QM scattering theory.
  • #1
diraq
14
0
Hi All,

The S-matrix is defined as the inner product of the in- and out-states, as in Eq. (3.2.1) in Weinberg's QFT vol 1:
[itex]S_{\beta\alpha}=(\Psi_\beta^-,\Psi_\alpha^+)[/itex]

When talking about the Lorentz invariance of S-matrix, the Lorentz transformation induced unitary operator [itex]U(\Lambda,a)[/itex] is applied both on the in- and out-states, and the transformation rule is the same as that for free particle states, i.e., Eq. (3.1.1).

However, since [itex]\Psi_\alpha^\pm[/itex] are the eigenstates of the full Hamiltonian with a non-zero interaction term, how can [itex]\Psi_\alpha^\pm[/itex] be transformed according to the same rule for the states composed of free particles? In the paragraph under Eq. (3.1.5), Weinberg explicitly indicates that the rule of Eq. (3.1.1) can only be applied to non-interacting particle states.

I appreciate any help from you to eliminate my miss/non-understanding. Thanks.
 
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  • #2
I think actually that 3.1.1 doesn't apply to the in and out states although the notation seems to indicate this. If you read the paragraph under 3.1.7 Weinberg says:

"On the other hand, the transformation rule (3.1.1) does apply in scattering processes at [itex] t\rightarrow \pm \infty[/itex]."​

In other words in general it does not apply to In and Out states. In and Out states appear non-interacting only to observers at [itex]t=\pm\infty[/itex]
 
  • #3
Thank you very much.

I am still confused. In- and out-states are non-interacting but according to Weinberg they are still the eigenstates of the full Hamiltonian.

It looks like that Weinberg took a different perspective. The asymptotic hypothesis that will turn off the interaction at [itex]t\rightarrow\pm\infty[/itex] is not included here. The Hamiltonian is is time dependent in asymptotic hypothesis, but Weinberg treats it as time-independent--only the state vector evolves with time.

bobloblaw said:
I think actually that 3.1.1 doesn't apply to the in and out states although the notation seems to indicate this. If you read the paragraph under 3.1.7 Weinberg says:

"On the other hand, the transformation rule (3.1.1) does apply in scattering processes at [itex] t\rightarrow \pm \infty[/itex]."​

In other words in general it does not apply to In and Out states. In and Out states appear non-interacting only to observers at [itex]t=\pm\infty[/itex]
 
  • #4
I am still confused. In- and out-states are non-interacting but according to Weinberg they are still the eigenstates of the full Hamiltonian

In and Out states only appear non-interecting to observers sitting at temporal infinity. To all other observers they appear to be "interacting" in the sense that they no longer have definite particle content. That is how they are able to be eigenstates of the full Hamiltonian. Equation (3.1.12) makes this more precise. The point is that we can find states that satisfy (3.1.12) despite the Hamiltonian being time independent, ie we can find states that look like they are non-interacting to observers at infinity.

If you look at eq. (3.1.7) you can see that at any finite t the In and Out states have components along every multiparticle state but at [itex]t\rightarrow\pm\infty[/itex] [itex]\Psi^\mp \rightarrow\Phi[/itex]. Also as [itex]V\rightarrow 0[/itex] we also get [itex]\Psi^\mp \rightarrow\Phi[/itex]. So in this sense [itex]t\rightarrow\pm\infty[/itex] is equivalent to [itex]V\rightarrow 0[/itex].

This method of doing things appears also in scattering in non-relativistic QM. For example it is discussed in Sakurai's book in Chapter 7: Scattering theory.
 
  • #5
Thank you very much. I think I have got the idea from your help.
 
  • #6
sweet! glad I could help.
 

1. What is the S-matrix in Weinberg's book?

The S-matrix, or scattering matrix, is a mathematical tool used in quantum field theory to calculate the probability of particles interacting and scattering off of each other. It is a key concept in Weinberg's book The Quantum Theory of Fields and is used to describe the behavior of particles in high-energy collisions.

2. How is the S-matrix calculated?

The S-matrix is calculated using Feynman diagrams, which are graphical representations of particle interactions. These diagrams allow for the calculation of the amplitude of a particle scattering process, which is then used to determine the probability of the process occurring.

3. What is the importance of the S-matrix in quantum field theory?

The S-matrix is important because it allows for the calculation of scattering processes in quantum field theory, which is crucial for understanding the behavior of particles at high energies. It also provides a way to test the predictions of quantum field theory through experimental results.

4. Can the S-matrix be used for all particle interactions?

The S-matrix is most commonly used for interactions between elementary particles, such as those described in the Standard Model of particle physics. However, it can also be applied to other types of interactions, such as nuclear or atomic interactions.

5. Are there any limitations to the S-matrix approach?

While the S-matrix approach is a powerful tool for calculating particle interactions, it does have limitations. For example, it does not take into account quantum effects such as tunneling, and it is not applicable in certain extreme scenarios such as black holes. Additionally, the S-matrix approach is not always the most efficient method for calculating particle interactions, and alternative approaches may be used in certain cases.

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