<f|S|i>

[JonathanApps]

In Quantum Field Theory we calculate the matrix elements which give the probability that an initial state |i> will turn into a final state |f>, e.g. |i> could be an electron and a positron, and |f> could be two photons.

 

I don't really understand why this works the way it's done. The interaction process is meant to be a collision: the initial particles start off far away from each other so they don't interact, then come together briefly and react to produce the final result; the resultant particles move apart again. It's a scattering process.

 

So why do we look at the matrix elements for states which have definite momentum? These are spread all over space so there's no "collision" as it were: they always have the potential to interact, for all t. A typical inital state that gets considered is something like a^{\dagger}_p b^{\dagger}_q |0>, i.e. 2 particles with definite momenta p and q.

 

What we surely need is packet states that represent particles with a reasonably well-defined position, moving towards each other. These would have a range of momenta. I can't understand why the definite momentum states work, and correctly predict experimental results in real particle accelerators.

 

Sorry I'll try latex:

 

[math]

 

a^{\dagger}_{p}b^{\dagger}_{q}|0>

 

[/math]

What I tried was

 

"[math]

a^{\dagger}_{p}b^{\dagger}_{q}|0>

[\math]"

 

Didn't work.

Edited October 22, 2014 by JonathanApps

[ajb]

You could use a density matrix in the interaction picture, never looked in to details here.

[Sensei]

[math]a^{\dagger}_{p}b^{\dagger}_{q}|0>[/math]

 

You need /math not \math...

Click Preview Post, until you have everything fixed.

 

[JonathanApps]

[math]

a^{\dagger}_{p}b^{\dagger}_{q}|0>

[/math]

[elfmotat]

Klauber addresses this point in section 8.9 of his book (which I highly recommend by the way). Particles are indeed typically wave-packets, which are a bit harder to deal with than momentum eigenstates. The way most textbooks deal with this is to make an approximation.

 

Interactions only occur where the interaction part of the Hamiltonian is nonzero. For example, take it to be the QED interaction: [math]\mathcal{H}_{int} = -e \bar{\psi} A^{\mu} \gamma_{\mu} \psi[/math]. In order for the interaction to be significant, the overlap of [math]\psi (x)[/math], [math]\bar{\psi} (x)[/math], and [math]A^{\mu} (x)[/math] has to be significant. Typically this overlap only occurs in a localized region of space, so an approximation can be made: we take the fields to be evenly distributed sinusoids (momentum eigenstates) inside some box of volume V, while outside of the box the fields are all zero.

 

He also provides the following (very helpful) illustration:

 

[JonathanApps]

Thanks; that makes some sense.

 

But the interaction should typically take place over a time of about Lm/(p-q), where L is the length of the box and m is electron mass (there might be a gamma factor in there!)

We do the integration from t = -infinity to infinity. That seems to be giving the particles a lot more opportunity to interact than they have.....?

Edited October 22, 2014 by JonathanApps

[elfmotat]

Thanks; that makes some sense.

 

But the interaction should typically take place over a time of about Lm/(p-q), where L is the length of the box and m is electron mass (there might be a gamma factor in there!)

We do the integration from t = -infinity to infinity. That seems to be giving the particles a lot more opportunity to interact than they have.....?

The creation and annihilation operators become functions of time when interactions are allowed, and we assume that in the far past and the far future there are no interactions taking place.

 

The reason the Feynman rules work is due to the LSZ reduction formula. A lot of intro texts don't ever mention LSZ, or even wave-packets in general, which can cause confusion. See for example here: http://isites.harvard.edu/fs/docs/icb.topic474176.files/LSZ.pdf .

 

Also, you can make nice pretty bras and kets with \langle and \rangle. For example, "\langle p | x \rangle" renders as: [math]\langle p | x \rangle[/math].

Edited October 23, 2014 by elfmotat

[ajb]

The creation and annihilation operators become functions of time when interactions are allowed, and we assume that in the far past and the far future there are no interactions taking place.

Exactly, such set-ups are known as asymptotically free.

 

Just to throw a spanner into the works, Haag's theorem tells us that the interaction picture is non-sence. If a state is asymptotically free then it is always free. The interaction picture is not mathematically well defined. Yet it, via the LSZ reduction, gives calculations of S-matrices that agree very well with nature.

[JonathanApps]

The creation and annihilation operators become functions of time when interactions are allowed, and we assume that in the far past and the far future there are no interactions taking place.

 

 

 

Cheers. But so then the commutation relations

 

[math]

[a^{\dagger}_{p},a_{q}]=\delta(p-q)

[/math]

 

won't be true for all time, and then we mess up the field theory, don't we?

 

I've heard of switching H_{I} on gradually, and then off again, by multiplying it with something f(t) that starts at 0, goes to 1 for a bit, then ends at 0, but again I don't see how you can do this without messing things up. We would screw up the formula for the S matrix in terms of H_{I} because it would now be f(t)H_{I}......

PS perhaps the answer to the above is in the LSZ link. It looks good - I'll have a look at it. Cheers.

[elfmotat]

 

Cheers. But so then the commutation relations

 

[math]

[a^{\dagger}_{p},a_{q}]=\delta(p-q)

[/math]

 

won't be true for all time, and then we mess up the field theory, don't we?

 

I've heard of switching H_{I} on gradually, and then off again, by multiplying it with something f(t) that starts at 0, goes to 1 for a bit, then ends at 0, but again I don't see how you can do this without messing things up. We would screw up the formula for the S matrix in terms of H_{I} because it would now be f(t)H_{I}......

PS perhaps the answer to the above is in the LSZ link. It looks good - I'll have a look at it. Cheers.

 

If you solve for the creation operator in terms of the field in free theories, you get:

 

[math]a^{\dagger} (\mathbf{k}) = -i \int d^3 x \, e^{-ik \cdot x} \, \overleftrightarrow{\partial_0} \, \varphi (x)[/math]

 

where [math]f \overleftrightarrow{\partial_0} g = f \partial_0 g - g \partial_0 f[/math]. In interacting theories this becomes the definition of creation (and annihilation) operators. But with this definition, in interacting theories they obviously become time-dependent. So we write:

 

[math]a^{\dagger} (\mathbf{k},t) = -i \int d^3 x \, e^{-ik \cdot x} \, \overleftrightarrow{\partial_0} \, \varphi (x)[/math].

 

When doing scattering we assume that in the limit [math]t \to \pm \infty[/math] these operators become the operators of the free theory.

[JonathanApps]

Those 2 lines look the same - k.x is a 4D scalar product, right?

 

I thought in the IP the time-dependence of the field operators was only due to the non-interacting Hamiltonian, as for no interaction. So the time-dependence of the "a"s should be unchanged....?

Are you in the Interaction Picture here?

[elfmotat]

Those 2 lines look the same - k.x is a 4D scalar product, right?

 

I thought in the IP the time-dependence of the field operators was only due to the non-interacting Hamiltonian, as for no interaction. So the time-dependence of the "a"s should be unchanged....?

 

Yes, [math]k \cdot x = k_\mu x^\mu[/math].

 

The two lines look the same, but the solutions of [math]\varphi (x)[/math] are not the same because the field equations contain additional terms in interacting theories.

[JonathanApps]

OK, sure, but not in the Interaction Picture, surely? The field equations are the same; it's the quantum state that's time-dependent.

Edited October 23, 2014 by JonathanApps

[elfmotat]

OK, sure, but not in the Interaction Picture, surely? The field equations are the same; it's the quantum state that's time-dependent.

 

Right. I'm working in Heisenberg picture. The LSZ reduction formula is what justifies our use of the interaction picture, which is technically ill-defined.

[JonathanApps]

Ah, thanks. You had me very worried for half an hour. I'll look at the LSZ stuff.

 

It seems basic QFT courses don't go anywhere near justifying their claims in this area. Not very impressive.

 

Thanks for all your help.

[timo]

Just to be sure: You are aware that S is a linear operator so that from the matrix elements in any basis you can calculate the <i|S|f> for any states |i> and |f>, even if they are not elements of your basis, right? Irrespective of this, I agree that many QFT courses are lousy exercises of tedious calculations.

[JonathanApps]

Yes; it's just that |i> and |f> seem unrealistic when the QFT courses decide they're going to be definite momentum.