LSZ reduction coeffs etc (Srednicki)

[LAHLH]

In free field theory one particle states can be created as: |k\rangle =a^{\dag}(\vec{k})|0\rangle. Just as we can expand the field operator in terms of the creation and annihilation operators it is possible to obtain an expression for the creation operator in terms of the field, it turns out to be: a^{\dag}(\vec{k})=-i\int\,\mathrm{d}^3x\,e^{ikx}\overleftrightarrow{\partial_0}\phi(x).

Next if we define a time independent operator as a^{\dag}_1\equiv \int\,\mathrm{d}^3k\,f_1(\vec{k})a^{\dag}(\vec{k}), where f_1(\vec{k})\propto \exp{\left[-(\vec{k}-\vec{k_1})^2/4\sigma^2\right]}. Then it is clear to me that this will create a particle localized in momentum space near \vec{k_1}, but should it be obvious that this particle is also localized in position space near the origin? Perhaps |k\rangle =a^{\dag}(\vec{k})|0\rangle states are just automatically at the origin, but I don't see where position enters this exactly, how would one create a particle located somewhere else in spacetime for example?

Srednicki then considers the state a^{\dag}_1 |0\rangle and says that if we time evolve it in the Schroedinger picture (since I guess states are time dependent in the Schroedinger as oppose to Heisenberg say where states are independent of time and operators inherit all time dependence) then we the wave packet will propagate and spread out ( I think he means propagate in space, and spread out in momentum space?). So he concludes the state is far from the origin at t goes to +/-infinity. Fair enough. Then he considers the state a^{\dag}_1 a^{\dag}_2|0\rangle. Now he states that \vec{k_1}\neq\vec{k_2} so my understanding is that both particle would be created at the space-time origin, but localized around different momenta in momentum space. Thus in the far past, my guess would be that both particles would have spread out in momentum space, maybe even overlapped in momentum space, and stayed together in position space but just propagated somewhere else. I don't understand then why they are necessarily widely separated in the far past as he states

 

[LAHLH]

So that was the initial set up and was for free field theory. Next we want to look at interacting field theory, however Srednicki says that a^{\dag}(\vec(k)) is no longer time independent. Is this because of the formula 5.10? , i.e.because the field no longer satisfies the KG equation? Is there a better way to see this, since he states it as if it's obvious and before getting to (5.10) (which is for a^{\dag}_1 rather than directly for a^{\dag}(\vec{k})anyway.)

Why did he have to introduce this a^{\dag}_1 anyway, would the argument he goes on to present not have followed if he had just defined his intitial state as say |i\rangle =\lim_{t\to\infty} a(\vec{k_1})a(\vec{k_2})|0\rangle why did he need to create states which we localized around some momentum rather than just a specific momentum as usual? (I'm thinking Heisenberg forbids having some state at specific momentum, so that's why he had to go to all the trouble, but I've never see this done in other places as far as I can recall)

He later says the wave packets no longer play a key role so take the f profile to be the delta function, but I fail to see how the argument would have been different without packets anyway?

 

[LAHLH]

Finally Srednicki goes on to say the derivation of the LSZ has relied on us supposing that the creation ops of free field theory will work equally well in interacting theory.

1)What is it that means we have used the creation ops of free theory here? just the fact that we used a^{\dag}(\vec{k})=-i\int\,\mathrm{d}^3x\,e^{ikx}\overleftrightarrow{\partial_0}\phi(x) and this is only valid for free theory? meaning our a dagger must be the free field theory creation op? Is there not an interacting field theory creation op we could just use instead to create our initial and final states?
2) Is the central issue that because we have used these free field theory creation ops to create our initial and final states we now may have issues if we're not careful, since they may not produce single particle states but may produce mixtures of single state/ground state or single state/multiparticle state?
3) If \langle 0|\phi(0)|0\rangle \neq 0 how exactly does this mean that a^{\dag}_1(\pm\infty) doesn't create a single particle state as desired? I originally thought about this by expanding phi in terms of creation/annihilation ops and it seemed to make sense. But \phi(0) here isn't the free field so we can't expand it like that, and what does \phi(0) and to do with t\to\infty anyhow? Maybe it's just because by (5.17) this is equal to \langle 0|\phi(x)|0\rangle and as t goes to infinity we hope this becomes a free field that can be expanded in terms of creation/annilihation ops, it's value being \langle 0|a^{\daq}(\infty)|\rangle

 

[Oudeis Eimi]

In free field theory one particle states can be created as: |k\rangle =a^{\dag}(\vec{k})|0\rangle. Just as we can expand the field operator in terms of the creation and annihilation operators it is possible to obtain an expression for the creation operator in terms of the field, it turns out to be: a^{\dag}(\vec{k})=-i\int\,\mathrm{d}^3x\,e^{ikx}\overleftrightarrow{\partial_0}\phi(x).

Next if we define a time independent operator as a^{\dag}_1\equiv \int\,\mathrm{d}^3k\,f_1(\vec{k})a^{\dag}(\vec{k}), where f_1(\vec{k})\propto \exp{\left[-(\vec{k}-\vec{k_1})^2/4\sigma^2\right]}. Then it is clear to me that this will create a particle localized in momentum space near \vec{k_1}, but should it be obvious that this particle is also localized in position space near the origin? Perhaps |k\rangle =a^{\dag}(\vec{k})|0\rangle states are just automatically at the origin, but I don't see where position enters this exactly, how would one create a particle located somewhere else in spacetime for example?
[/B]

f1 is a Gaussian in momentum space, so its Fourier transform gives you another Gaussian
in position space; the lack of any x dependence in f1 causes its FT to be centered at
the origin.

To obtain a Gaussian centered at x0 you'd add an x0-dependent term to f1 (its exact
form eludes me at the moment, but you can easily derive it by applying a translation
to the FT of f1 and then taking its inverse FT).

Srednicki then considers the state a^{\dag}_1 |0\rangle and says that if we time evolve it in the Schroedinger picture (since I guess states are time dependent in the Schroedinger as oppose to Heisenberg say where states are independent of time and operators inherit all time dependence) then we the wave packet will propagate and spread out ( I think he means propagate in space, and spread out in momentum space?).

I don't have (nor have read) the book, but I bet he really means position space in both
cases. A free Gaussian wavepacket both translates *and* spreads out in position space.
IOW, the dynamics generated by the wave equation is dispersive. This is already shown
in many (most?) elementary, non-relativistic QM texts.

So he concludes the state is far from the origin at t goes to +/-infinity. Fair enough. Then he considers the state a^{\dag}_1 a^{\dag}_2|0\rangle. Now he states that \vec{k_1}\neq\vec{k_2} so my understanding is that both particle would be created at the space-time origin, but localized around different momenta in momentum space. Thus in the far past, my guess would be that both particles would have spread out in momentum space, maybe even overlapped in momentum space, and stayed together in position space but just propagated somewhere else. I don't understand then why they are necessarily widely separated in the far past as he states

 

[LAHLH]

I don't have (nor have read) the book, but I bet he really means position space in both
cases. A free Gaussian wavepacket both translates *and* spreads out in position space.
IOW, the dynamics generated by the wave equation is dispersive. This is already shown
in many (most?) elementary, non-relativistic QM texts.

Well the book is available free online for reference, but if we start with the two creation operators creating particles both at the origin of spacetime, but centred at different places in momentum space, then what dictates that these two particles must end up far away from each other in position space as t goes to infinity? they were created at the same place in position space...What is determining the evolution of position here?

 

[Oudeis Eimi]

Well the book is available free online for reference,

Oh! This is a good thing to know.

but if we start with the two creation operators creating particles both at the origin of spacetime, but centred at different places in momentum space, then what dictates that these two particles must end up far away from each other in position space as t goes to infinity? they were created at the same place in position space...What is determining the evolution of position here?

Well, isn't it obvious? "Different places in momentum space" implies in general different
speeds and directions, right? In particular, in the centre of momentum frame, they'll
be going in opposite directions.

 

[LAHLH]

Oh! This is a good thing to know.


Well, isn't it obvious? "Different places in momentum space" implies in general different
speeds and directions, right? In particular, in the centre of momentum frame, they'll
be going in opposite directions.

Yeah, I had had that thought already. But it seems like a very classical mechanics explanation I suppose (despite it being true), what I'm getting at is, isn't there some operator governing the spatial evolution of these states or something?

 

[Oudeis Eimi]

Yeah, I had had that thought already. But it seems like a very classical mechanics explanation I suppose (despite it being true), what I'm getting at is, isn't there some operator governing the spatial evolution of these states or something?

Of course there is, the hamiltonian.

In the Schröedinger picture, just solve Schröedinger's equation with the free hamiltonian.
You'll obtain a couple Gaussians whose 'peaks' in position space propagate with constant
speed and whose width varies as

\sigma(t)=\sigma_0\sqrt{1+(at/m)^2},

with a some constant (don't remember the details offhand).

------EDIT
The above expression for sigma(t) is true in a non-relativistic scenario. I don't know the exact form of sigma(t) for the relativistic case.

 

[LAHLH]

Of course there is, the hamiltonian.

In the Schröedinger picture, just solve Schröedinger's equation with the free hamiltonian.
You'll obtain a couple Gaussians whose 'peaks' in position space propagate with constant
speed and whose width varies as

\sigma(t)=\sigma_0\sqrt{1+(at/m)^2},

with a some constant (don't remember the details offhand).

------EDIT
The above expression for sigma(t) is true in a non-relativistic scenario. I don't know the exact form of sigma(t) for the relativistic case.

This would be the case for regular QM I guess, but what has the Schroedinger equation got to do with Klein-Gordon fields? The field operator is governed by the KG equation, but then the states in the Schroedinger picture are vectors in some Fock space evolving according to...erm..really the Schroedinger equation? (maybe I need to go back to basic qft but I just can't see why now, I'm used to just seeing the Heisenberg pic and the states being time independent so not having to worry how they evolve)

 

[naima]

I found the http://demonstrations.wolfram.com/EvolutionOfAGaussianWavePacket/" of a free particle like in Srednicki's textbook .
It is peaked on p_0 t.

 

[Oudeis Eimi]

This would be the case for regular QM I guess, but what has the Schroedinger equation got to do with Klein-Gordon fields? The field operator is governed by the KG equation, but then the states in the Schroedinger picture are vectors in some Fock space evolving according to...erm..really the Schroedinger equation? (maybe I need to go back to basic qft but I just can't see why now, I'm used to just seeing the Heisenberg pic and the states being time independent so not having to worry how they evolve)

In QFT, the Schrödinger equation is still valid. One generally works in the interaction picture, so it has the form

i \frac{d|\psi(t)>}{dt} = \hat{H_I}|\psi(t)>

with \hat{H_I} the interaction hamiltonian. For a free field this reduces to the Heisenberg picture and the states stay fixed.

In principle, one could work in the Schrödinger picture, but the square root in the KG hamiltonian would make things difficult. Therefore, it's simpler to just work in the Heisenberg picture and time-evolve the fields.

For the setup described in your initial post, the expected value of the density operator evaluated with the state a_1^\dagger a_2^\dagger |0> should give you two 'bumps' in the field moving in different directions.