Understanding the Interaction of Scalar Fields in Yukawa Theory

[Rick89]

Hi, I am having this problem learning interacting fields. In scalar Yukawa theory, two scalar fields (one real call meson phi, one complex call nucleon psi) (interaction term psi-dagger*psi*phi) why can't we write at first order in the interaction an amplitude from a single meson state to vac. (I know it does not make sense, but I don't understand why coming from the formalism) like:
<0| c*c-dagger*a|meson>
where c destroys anti-nucleon, and a destroys meson. Isn't this a term in psi-dagger*psi*phi, when expanded in normal modes? Thanx alot

 

[haushofer]

Well, which vacuum would you take?

 

[Rick89]

well, I suppose the vacuum of the free theory, without interaction would do...is it important?
what I'm not understanding is probably trivial, just I don't see why we compute decay of a meson using b-dagger*c*dagger*a (that is destroy a meson and create a pair of nucleon-antinucleon) but cannot have c*c-dagger*a
Thanx

 

[Rick89]

maybe I see what you mean, is it because of amputated diagrams and all that story?

 

[Rick89]

anyone with more detail? I still don't understand how to treat such terms.. Thanx

 

[weejee]

The process you are talking about (meson -> nothing) is not allowed because it violates energy conservation. However, according to your theory, (meson -> nucleon + antinucleon) is possible, just as (photon -> electron + position).

 

[Rick89]

I know what you're saying, but then why is the matrix element different from zero, there must be an error, where? Thanx

 

[haushofer]

well, I suppose the vacuum of the free theory, without interaction would do...is it important?

Yes, because in general the vacuum of the free theory is NOT the same as the vacuum of the interacting theory.

Take as a simple example a QM harmonic oscillator. You know what the ground state is of that theory. Now add a small perturbation. Is the ground state still the same?

 

[Dickfore]

<0| c*c-dagger*a|meson>

This matrix element does not describe a single meson state going to a vacuum state. If you recall LSZ reduction formula, you will find that the transition amplitude for the process you are discussing is:

<br /> \langle f | i \rangle = i \, \int{d^{4}x \, e^{-i k x} \,(\partial^{2} + m^{2}) \, \langle 0|T \varphi^{\dagger}(x) |0 \rangle}<br />

If you calculate this at tree level in Yukawa theory, you should get 0.

 

[Rick89]

to haushofer: I know that, it's true. But does it matter for what I asked?
to Dickfore: I am not sure I understand exactly where your formula comes from, but I have a feeling my thing is in a step in evaluating it. It's quite simple to see where my matrix element comes from, I'm considering (following an approximate treatment you can find on page 55 of these notes:http://www.damtp.cam.ac.uk/user/tong/qft/qft.pdf) initial and final states as eigenstates of the free theory as he does. and taking the first term only in Dyson's formula for the interaction S matrix.

 

[Rick89]

if you look at that page (it's the number written, not the page of the pdf I meant) you will see that in meson decay he's doing that. just what about the term I was mentioning? Thanx

 

[weejee]

The matrix element <0| c*c-dagger*a|meson> itself is nonzero, but we have the overall delta function that mandates the conservation of energy and momentum, as you can see from equation 3.30 (page 56) in your note. Since the initial and final energies are different, we have zero transition probability.

 

[Rick89]

Hi, thanks for your reply, I'm not sure about this (I'm not doing that course yet, I'm only reading it for myself) , check what I say please. The delta fn there comes from the spacetime integral of the exponentials. In my case,once acted with a already as in the notes, we commute the c past the c-dagger, the two exponentials cancel(is it right?) the delta kills one momentum integral and what about the other? Does it diverge? There is an energy integrated in the denominator, I doubt it converges? I suppose the spacetime integral of the exp. coming from acting with a gives the delta that forces the four momentum of the meson to zero, so is that telling us the process is impossible? Thanx for any explanation!

 

[weejee]

As you've said there is no a priori reason not to think about the process, meson -> vacuum.
Why don't you just reproduce the calculation that is done in the lecture note with |f>=|0>? You will find that the matrix element is nonzero, but you will also find that you still have the overall delta function as in equation (3.30), and that is what makes the transition probability equal to zero. It is essentially the same as Fermi's golden rule in usual time-dependent perturbation theory of quantum mechanics.
I'm sorry if I sound like saying the same thing over again, but I can't really find how to articulate it more clearly.
Hope it helped. Good luck in studying QFT!

 

[Rick89]

yes, thanx, but that's what I was doing. As I said the delta function is there, it comes from the integral in d^4x of the exp(-i p.x) since there are no other terms with x (the two exponentials of the two psi's cancel) it gives delta(p), imposing the constraints (conservation laws) you're talking about since the whole four momentum of the meson can only be 0 if the meson is not there. But when you do the integrals in d^3k (coming from the mode expansions of psi and psi-dagger) one integral is killed by a delta fn, but the other one seems to me to diverge... Is that normal? Thanx, I might be doing sth stupid...