Let me preface this by saying that I don't have an issue with this:

$$ \langle\Omega|T\phi_H\cdots\phi_H|\Omega\rangle = \frac{\langle 0|T\phi_I\cdots\phi_IS|0\rangle}{\langle 0|S|0 \rangle}, $$

what I want to know is why we want to calculate $\langle\Omega|T\phi_H\cdots\phi_H|\Omega\rangle$ at all.

Say you have an initial state $|i\rangle$ in the interaction picture of a real scalar field theory which is an eigenstate of the free Hamiltonian but not of the interacting Hamiltonian. However, just because this isn't an eigenstate of the Hamiltonian doesn't mean it's unphysical, it's just a superposition state:

$$ |i\rangle = |\Omega\rangle\langle\Omega|i\rangle + \sum_{n=1}^{\infty}|n\rangle\langle n|i\rangle $$

where $|n\rangle$ are the eigenstates of the full interacting Hamiltonian (whatever they might be). You can expand $|i\rangle$ in terms of a product of fields acting on $|0\rangle$ (the vacuum of the free theory):

$$ |i\rangle = \prod_{a=1}^{b}\int\left(d^3y^{(a)}\ 2E_{q^{(a)}}\ e^{iq^{(a)} \cdot y^{(a)}}\phi[t;y^{(a)}]\right)|0\rangle $$

and there's still nothing wrong with this statement. If you're really having consternations you can expand $|0\rangle$ in terms of $|\Omega\rangle$ and $|n\rangle$ and get everything in terms of the interacting Hamiltonian eigenbasis (if you do this and then evolve forward in time you end up with the first equation). You can do a similar construction for $|f\rangle$, the final state.

Then what you want to calculate is the probability that $|i\rangle$ has evolved into $|f\rangle$ after some time $t_1 - t_0$ has passed:

$$ P = \langle f|U_{int}(t_1,t_0)|i\rangle $$

and then using the above expansion for $|i\rangle$ and $|f\rangle$ you can get:

$$ P = \\ \prod_{a=1}^{b}\int\left(d^3y^{(a)}\ 2E_{q^{(a)}}\ e^{iq^{(a)} \cdot y^{(a)}}\right)\prod_{a=1}^{b'}\int\left(d^3z^{(a)}\ 2E_{r^{(a)}}\ e^{ir^{(a)} \cdot z^{(a)}}\right) \\ \langle 0|\prod_{a=1}^{b'}\left(\phi[t;z^{(a)}]\right) U_{int}(t_1,t_0) \prod_{a=1}^{b}\left(\phi[t;y^{(a)}]\right)|0 \rangle $$

and the last term looks like

$$ \langle 0|T\phi_I\cdots\phi_I U_{int}(t_1,t_0)|0\rangle $$

with no division by $\langle 0|S|0 \rangle$ or even $t_{0,1} \to \pm \infty,\ U_{int}(t_1,t_0) \to S$ necessary.

My question is what is wrong with the way we constructed $P$ that requires taking the infinite limit in time and dividing out the vacuum bubbles ($\langle 0|S|0 \rangle$)? As far as I can tell none of this is necessary; we choose $|i\rangle$ and there's nothing stopping us from choosing it to be a free field eigenstate, and we choose $|f\rangle$ similarly. The only pitfalls I can see are if $|i\rangle$ is orthogonal to the entire interacting eigenbasis (which would make it incomplete as a basis) or that $P=0$ because evolution under the interacting Hamiltonian takes $|i\rangle$ far away from the non-interacting states.


1 Answer 1

  1. There is no notion of creation/annihilation operators in an interacting theory. This means that, in general, the equation $$ |i\rangle = \prod_{a=1}^{b}\int\left(d^3y^{(a)}\ 2E_{q^{(a)}}\ e^{iq^{(a)} \cdot y^{(a)}}\phi[t;y^{(a)}]\right)|0\rangle $$ is meaningless. More precisely, you can always define $|i\rangle$ by this expression, but this has nothing to do with the bunches of particles we scatter off each other in accelerators. It is a valid in state, but it does not correspond to sharp wave packets that represent particles.

    As it should be more or less clear by the standard derivation of the LSZ formula, you must assume an adiabatic switching off of the interactions, such that the fields become free in the asymptotic past and future. Only when the fields are free you get a well-defined creation operator, so that you can prepare the in state by acting with them on the vacuum. Without the switching off, and without the asymptotic limit, you can formally calculate stuff with your $|i\rangle$ state, but it will have nothing to do with that you measure in an experiment.

  2. You forgot to normalise your states. To make this clear, note that if you have no particles in neither the in nor the out state (that is, a vacuum to vacuum transition), your amplitude reads $$ P_{0\to 0}=\langle 0|U(-\infty,+\infty)|0\rangle $$ which is in general different from one. This is clearly wrong: the amplitude to end up with the vacuum, if you begin with the vacuum, is one. To make this true, you must divide your formula by the factor $\langle 0|U(-\infty,+\infty)|0\rangle$, which makes it properly normalised. In this case, you get $P_{0\to0}=1$, as one would expect.


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