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In QFT we usually want to calculate objects of the form $\langle f|\hat S|i\rangle$ which yields the probability amplitude for the process $i \to f$. We can expand the scattering operator $\hat S$ in terms of the famous Dyson series. The first term in this series is simply $1$. This term only contributes something if $i = f$. In this case, $$\langle i|\hat S |i\rangle = \langle i|1 + \ldots |i\rangle = \langle i |i\rangle + \ldots$$ Depending on which states we use as initial and final states this first term in the expansion either yields $$ \langle i |i\rangle = (2\pi)^3 \delta(0)$$ or $$\langle i |i\rangle = 1.$$ This is, of course, only a first order approximation but still seems to suggest that this transition has a probability of $100\%$ (or "$\infty$" if we use non-normalizable initial and final states).

Moreover, for example, in $\phi^4$ theory the first correction term for $2 \to 2$ scattering reads $$ A^{(1)} = - i \lambda .$$ The total amplitude therefore reads $$ A \approx A^{(0)} + A^{(1)} = 1 - i \lambda . $$ This implies that, for example, for $\lambda= 0.1$, we find $$ P = | A^{(0)} + A^{(1)}|^2 = |1 - i 0.1|^2= 1.0049876 ... \, . $$

The textbooks I consulted seem to gloss over this first term in the expansion. Therefore my question is, how can we make sense of the perturbation series if the first order approximation already seems to use up $100\%$ of the probability?

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    $\begingroup$ If the states are not normalized, you probably want to study $\frac{\langle i \vert S \vert i\rangle}{\langle i \vert i \rangle}$ instead $\endgroup$
    – KF Gauss
    Commented Oct 25, 2019 at 15:32
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    $\begingroup$ I disagree that this is a 'first order approximation' - it is a zeroth order approximation. What number would you like it to be? $\endgroup$
    – jacob1729
    Commented Oct 25, 2019 at 16:21

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The reason is in fact in the first line of the question: $\langle f|\hat S|i\rangle$ is the probability amplitude rather than probability. The probability is the absolute value squared: $$|\langle i|\hat S|i\rangle|^2=|\langle i|i\rangle+\dots|^2=|1+\dots|^2,$$ which can (and must) be smaller than 1. The first term must contribute exactly $1$ for the probabilities to sum up to $1$ (if the series converge). A very clear and visual discussion of this effect can be found in Figure 69 on p.108 of R.Feynman, QED: The strange theory of light and matter. Princeton University Press, 2006.

EDIT. But this explanation removes only a part of the problem. For instance, if $\hat S$ is the identity operator, the paradox is still there. A more deep reason is that 'probability' does not make any sense until a probability space, i.e., the set of possible outcomes of an experiment, is introduced. In quantum mechanics this set is an orthonormal basis in a Hilbert space. Then the inner product with the basis elements is interpreted as the transition probability (or probability density). Thus it is in general just not valid to take $f=i$ for the object $\langle f|\hat S|i\rangle$ to be still interpreted as a probability amplitude. Taking $f=i$ is possible, only if $i$ belongs to the implicit basis of final states. Say, in the case when $\hat S$ is the identity operator, the latter implies that all the other $\langle f|i\rangle=0$, and the probabilities sum up to $1$.

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  • $\begingroup$ Awesome thanks! I will check the reference you mentioned. $\endgroup$
    – jak
    Commented Oct 25, 2019 at 17:32
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    $\begingroup$ Here's what I've come up with: in a first order approximation we predict that in 100% of all cases no scattering happens at all. But if we include higher order terms, this prediction gets refined and we find, for example, that only in 92% of all cases no scattering happens. $\endgroup$
    – jak
    Commented Oct 26, 2019 at 10:36

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