Confusion (6) from Weinberg's QFT.(in and out state)

1. Aug 10, 2011

kof9595995

In page 110(section 3.1), after giving the relation between in and out states and free particle states,
$\Psi_{\alpha}^{\pm}=\Omega(\mp\infty)\Phi_{\alpha}$.........(3.1.13)
Then he says in page 111:
But I don't see why it's necessary.

2. Aug 10, 2011

meopemuk

Energy eigenstate has trivial time evolution (multiplication by a phase factor). So, there is no meaningful scattering in energy eigenstates. You need to have a smooth superposition of energy eigenstate in order to see a non-trivial time dependence of the wave function and a non-trivial scattering (i.e., the difference between wave function shapes in the remote past and the remote future).

Eugene.

3. Aug 11, 2011

vanhees71

Another explanation of the same fact: Scattering "states" are not true states since by definition they belong to the continuous part of the spectrum of the Hamiltonian. They are to be understood as generalized functions. A true state has to be square integrable and thus normlizable to 1.

In physical terms, what you have to consider to understand scattering theory correctly are wave packets that have a macroscopically sharp momentum and position. Of course, you can never beat the Heisenberg uncertainty relation $\Delta x \Delta p \geq \hbar/2$. That's what I mean by "macroscopically sharp". A good discussion on this quite subtle issue can be found in Messiah's textbook on quantum mechanics and also in Peskin and Schroeder's QFT book.

4. Aug 15, 2011

kof9595995

Emm, it seems there's a difference between your explanations. For meopemuk "meaningful" means "non-trivial", but for vanhees it means "not ill-defined", so which is the one Weinberg refers to?

5. Aug 15, 2011

The_Duck

I've just been reading the same part of Weinberg. I think it is not too hard to see that this relation is simply wrong if you try to apply it to energy eigenstates. Equation 3.1.13 is an abbreviated form of equation 3.1.12. Equation 3.1.12 and the equation below it say that if you take a wave packet of free-particle states and evolve it to t = -infinity using the free hamiltonian, and the analogous wave packet of in states, and evolve that to t = -infinity using the interacting hamiltonian, then these two things become equal. Suppose you try to do away with the wave packets. Then you are are taking an eigenstate of the free hamiltonian and evolving it to t = -infinity with the free hamiltonian, and taking an eigenstate of the interacting hamiltonian and evolving it to t = -infinity with the interacting hamiltonian, and hoping that these things become equal. But in this case time evolution is just multiplication by a phase, so you are hoping that the free eigenstate and the interacting eigenstate differ only by a phase, i.e. that they are the same state, which is clearly wrong.

6. Aug 16, 2011

kof9595995

That makes sense, but I'm more confused: If 3.1.12 is true for every smooth superposition, how come it's not true for a single component? What is this kind of mathematical phenomenon?

7. Aug 16, 2011

meopemuk

kof9595995,

In scattering we consider states which have specific time dependence: the wave packet propagates freely in the remote past; at time around t=0 it collides with the scattering center; and in the remote future it moves away from the scattering center propagating freely again. If the time evolution of a state does not look like the one described above, then this process cannot be called "scattering" and we cannot apply the idea of S-matrix to it.

Energy eigenstates (= stationary states) have trivial time evolution. In particular, the probability density of such states does not change with time. So, clearly, such eigenstates do not look like scattering states at all, and scattering theory does not apply to them. S-matrix theory can be applied only to linear combinations of stationary states. Not all linear combinations of energy eigenstates can be called "scattering states", but only those whose time evolution satisfies the definition of a scattering process in my first paragraph.

Eugene.

8. Aug 16, 2011

kof9595995

Emm, are you saying that for energy eigenstates S-matrix formalism is just trivial? Or it's simply wrong to apply it to energy eigenstates?

9. Aug 16, 2011

kof9595995

And another question,$\Phi_{\alpha}$ and $\Psi_{\alpha}$ are defined as eigenstates of $H_0$ and$H$ respectively. Yet $\Phi_{\alpha}$ and $\Psi_{\alpha}$ has to be related by (3.1.12), but can this always be done mathematically?
I mean in a sense $H$ and $H_0$ are independent since $V$ can be anything, so $\Phi_{\alpha}$ and $\Psi_{\alpha}$ should be independent, then how do we guarantee that (3.1.12) can always be satisfied?

10. Aug 16, 2011

meopemuk

I think it is more correct to say that the S-matrix formalism is not applicable to energy eigenstates. There is no scattering in stationary states. So, scattering theory doesn't work there.

Eugene.

11. Aug 17, 2011

kof9595995

Then what is the mathematical reason for this invalidity? Is it related to post 5?

12. Aug 17, 2011

meopemuk

You post 5 is about Lorentz invariance of the S-matrix, so it is not related to this post.

The reason for inapplicability of the S-matrix to stationary states is physical rather than mathematical. Scattering theory is a mathematical formalism, which is designed to describe a very specific physical process - the scattering. In the simplest scattering event you have two particles that move toward each other from a large distance. While the distance between the particles is large the interaction between them is negligible and their dynamics is well described by the non-interacting Hamiltonian. Then at t=0 the two particles collide, the interaction between them becomes strong, their wave functions change rather abruptly during a short time interval. After that, the particles move away from each other, their separation grows, and interaction becomes insignificant again.

The S-matrix formalism is just a simplified method for the description of such scattering states. If the state you are dealing with does not have the time evolution described above, then you are not allowed to apply the scattering theory to such a state. Of course, formally, you can still multiply this state vector by the S-matrix. But the result of such multiplication will be physically meaningless.

Eugene.

13. Aug 19, 2011

kof9595995

Actually I meant post #5 by The_Duck
I think there shall be some mathematical difficulty there, or else energy eigenstates should just somehow appear as the trivial part of the S-matrix. I think what The Duck pointed out has a lot to do with it, I just can't see the whole picture clearly.

14. Aug 19, 2011

strangerep

This part in Weinberg should be read with the caveats near eqns(3.1.8) in mind. He's relating the free eigenstates $\Phi_\alpha$ to the asymptotic in/out states $\Psi_\alpha^\pm$. The latter are of course not the same thing as eigenstates of the full Hamiltonian.

IOW, Weinberg is assuming that the free eigenstates span the asymptotic state space. This is true if the interaction part of the Hamiltonian vanishes sufficiently quickly at large times and large separations. However, this is not always so (e.g., in QED), and special techniques must be employed instead in such cases.

The following papers contain some useful insights into the details of this problem.

R. Horan, M. Lavelle, D. McMullan,
"Asymptotic Dynamics in QFT",
Arxiv preprint hep-th/9909044.

R. Horan, M. Lavelle, D. McMullan,
"Asymptotic Dynamics in QFT -- When does the coupling switch off?",
Arxiv preprint hep-th/0002206.