# Calculation of the cross section

Why, when we calculate the total cross section, we make the average other initial states and the sum over final states?

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## 2 Answers

The summing over final states and the averaging over initial states is a good observation that I always emphasize as the origin of the arrow of time. As soon as one considers mathematical logic, this asymmetry has to arise.

Why are we summing over final states? Because "we don't care" about which of them occurs (and no one knows). We're calculating the probability - or cross section (which is the same thing, up to unproblematic coefficients) - of getting a final state $F_1$ or $F_2$. Both of them are OK so I use "or".

But the probability $$P(F_1 \mbox{or} F_2) = P(F_1)+P(F_2)$$ is simply the sum if the two final states are mutually excluding - e.g. orthogonal.

Things are very different for the initial states. Now, it's not quite true that "we don't care" whether the state was $I_1$ or $I_2$. Instead, "we don't know" which of them it was (but we know that one of them was the right one). We have to make probabilistic assumptions about the initial state. The most "balanced" one is that each $I_1$ and $I_2$ appear with the 50 percent probability. There could also be an asymmetric choice of the "priors" - leading to a weighted average - but it's important that $$P(I_1)+P(I_2)=1$$ The probabilities of the evolution include the calculate probabilities of transitions from $I_k$ to $F_l$ - given by the Feynman diagrams etc. - but for each $I_k$, we must also multiply this by the probability that $I_k$ occurred in the first place. This reduces the calculation by a factor of two. The total and averaged probability is $$P(I\to F) = 1/N_{initial} \sum_{ij} P(I_i \to F_j)$$ The very asymmetry here - the extra factor $1/N_{initial}$ without its "final" counterpart - is the reason why low-entropy states are favored as initial states while the high-entropy ones are favored as final states. In particular, you may compute the ratio of the probabilities from I to F, and from F to I, and they will differ by the factor of $$N_{final}/N_{initial} = exp(S_f-S_i)$$ or the exponential of the entropy difference between the final and initial state. This number's being very different from one is the reason why the transitions to lower-entropy states can't occur in practice.

Many people - including professional physicists - remain extremely confused about these points when they suggest that the arrow of time remains mysterious. It doesn't. The second law of thermodynamics may be proved and it all boils down to your question and the right answer.

The only assumptions I made are those about the addition of probabilities of assumptions and their effects - and these logical rules are fundamentally asymmetric when it comes to the role of the assumptions and their consequences. This logical arrow of time can't be removed from any reasoning about a world that depends on time - time only copies the logical relationship of implication. And this logical arrow of time is the source of the thermodynamic arrow of time as well.

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thank you very much for your answer, very complete and relevant – Andrea Amoretti Jan 14 '11 at 15:36
I disagree with "we don't know". Everything is dictated by a particular experimental situation. We can determine what fraction of the incident (projectile) flux $j$ represents this or that particular polarization state so we take the corresponding weight into account according to our measurements, if and only if both polarizations contribute. As well, summation over final states is not always necessary, especially if we can experimentally distinguish different final states. This all may be necessary in some calculations if all different initial and final states contribute (inclusive picture). – Vladimir Kalitvianski Jan 18 '11 at 23:32
I was thinking about this question but decided to look it up before asking. As usual, Lubos and his great answers. Many thanks for your answer. I learned a lot from you in this forum. – stupidity May 31 '12 at 13:42

We do all the "cross section business" because we want to predict results of experiments.
Let's take for example some particle with two polarizations states: "+" and "-". You know that experimentalists will collide 1 000 000 pairs of particles, with polarisation of initial particles being unknown. Best thing you can do is to hope that in experiment polarization would be evenly distributed. So that you will have around 250000 of "++" collisions, 250000 of "+-", and the same for "-+" and "--". And in order to predict the result you will need to do your calculations for each polarization configuration and average over them.

If in experiment the polarization is fixed, then you don't need to do the averaging. And if you know the polarization density matrix or some spectrum for the initial state, then you got to average according to them.

Next. You know that after collision they will detect some resulting particles. Let's say that your calculations predict 1000 of "+" polarized particles and 500 of "-" polarization. But it turns out that the detector do not detect polarization of final particles. Best thing you can do -- sum over final polarizations and predict 1500 for the final result.

I made an example with polarization, while it can be extended to any properties/degrees of freedom in the experiment. If you do not know some property in initial state -- you should average over them. If you do not know some property in final state -- sum over them.

This works also for total/differential cross section. If the detector can tell where does particle fly -- then you can work with differential cross section. But if the detector just tells you that "there is a particle" without any info on the direction -- then you will need to integrate the differential cross section (sum over final states) to get total cross section.

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Ok, but, for exaple, if you have completely umpolarised incident beam but you now for some reason the composition (the density matrix), also in this case you do simply the average other inital states or the weight of each initial state is in accordance with the density matrix? – Andrea Amoretti Jan 10 '11 at 13:21
Of course if you know the polarisation density matrix or some spectrum for the initial state, then you got to average according to them. (Inserted in main answer.) – Kostya Jan 10 '11 at 13:26
ok thank you for the answer – Andrea Amoretti Jan 10 '11 at 13:37
Good answer. And while we're at it, we might as well point out that this is precisely the reason why there is second law of thermodynamics. We average over initial states (assuming a priori Bayesian uniform density) but have to sum over final states. This induces the time asymmetry into the reversible microscopic dynamics. – Marek Jan 10 '11 at 17:08
@Marek: Can you and/or Lubosh tell more about this "irreversibility", please? If we speak of polarization states, I do not feel the difference between averaging and summation. – Vladimir Kalitvianski Jan 19 '11 at 9:40
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