High School Mapping between rules of classical vs QM probability?

[etotheipi]

Apologies in advance for my ignorance, I don't really have a reference to consult and Google hasn't been too helpful! In standard probability theory there are a few common useful formulae, e.g. for two events $S$ and $T$ $$P(S\cup T) = P(S) + P(T) - P(S\cap T)$$ $$P(S \cap T) = P(S) \times P(T | S)$$ I was reading Binney's notes and he says for two mutually exclusive events $S$ and $T$ (let's say with complex probability amplitudes $A(S)$ and $A(T)$) the probability of $S$ or $T$ happening goes as $$P(S \cup T) = |A(S \cup T)|^2 = |A(S) + A(T)|^2 = \dots$$It appears that the first two formulas I quoted work right so long as you change out $P$ for $A$ and work with amplitudes instead. So for instance our second rule might become $$A(S\cap T) = A(S) \times A(T | S)$$ Are these just special cases, or is it always valid to apply the rules of classical probability in terms of amplitudes here instead? Or are these formulae even useful? Thanks!

 

[Strilanc]

Multi-event amplitude expressions like $A(S \cup T)$ don't have a well defined meaning. This is one of the annoying things about amplitudes. There can be an amplitude for $S$, and an amplitude for $T$, but in general there's no well behaved amplitude you can then assign to $S \cup T$. The issue is that correctly calculating the interference effects of processes that contribute amplitude into the $S \cup T$ system requires knowing specifically whether the amplitude is going into $S$ or into $T$.

This does has its upsides. If you think you have a single-variable quantum system $X$, but actually have a multi-variable system $X = Y \cup Z$, you can notice interference effects playing out incorrectly. Quantum mechanical systems betray unknown degrees of freedom in a way that probabilistic systems do not. See https://www.scottaaronson.com/blog/?p=3327:

Quantum mechanics says that, no, you can verify that two particles are perfectly identical by doing an experiment where you swap them and see what happens. If the particles are identical in all respects, then you’ll see quantum interference between the swapped and un-swapped states. If they aren’t, you won’t.

 

[etotheipi]

In that case, does the construction $$P(S \cup T) = |A(S \cup T)|^2 = |A(S) + A(T)|^2 = p(S) + p(T) + 2\Re{(A(S)A^*(T))}$$

with interference term $2\Re{A(S)A^*(T)}$ only apply to certain specific examples (I think Binney was referencing the two-slit interference example) but does not hold in the general case? I wonder if works here only because in this cases both events (left slit/right slit) have equal weighting?

Thanks for your reply, I'll have a read through that article also!

 

[Strilanc]

I would say that particular derivation is wrong because it says $A(S \cup T)$. I would instead write it as (for disjoint and atomic $S$ and $T$):

$P(S \cup T) = P(S) + P(T) = |A(S)|^2 + |A(T)|^2$

whereas for non-atomic possibly-overlapping S and T:

$$P(S \cup T) = \sum_{x \in S \cup T} P(x) = \sum_{x \in S \cup T} |A(x)|^2$$

where each $x$ is an atomic disjoint case.

It may be that in the specific case of the double slit experiment that you can get away with the equation you mentioned, that there's some kind of constraint on the distribution of amplitude that allows reasoning about regions as if they had an amplitude for a photon being contained within, but in general I would be very cautious about using that reasoning.

 

[etotheipi]

Your construction does make more sense to me, since it's more similar to how disjoint probabilities work in normal statistics. I'll tread with caution in that case!

Just for reference, the derivation in question is covered in page 5 of this set of notes (out of about 302... hmm looks like I've still got a little way to go )

 

[Strilanc]

Ah I see, in the notes they are talking about ROUTES, not OUTCOMES. I was speaking in terms of outcomes. In other words, they are talking about the entries in a unitary matrix whereas I was talking about the entries in a unit vector (the wave function). Confusingly, both types of entry are called amplitudes.

(This ambiguity is also present in statistics. The entries in a stochastic matrix, a thing which acts on probability distribution, are called probabilities. The weights in a probability distribution are also called probabilities.)

If I was trying to be explicit I would call the entries in a unitary matrix something like "amplitude flows". It is safe to group amplitude flows into composite objects with an associated total amplitude flow into a single outcome, and to then add together the amplitude flows of disjoint composite objects with the same outcome. The fact that this works is one of the key ideas in Feynman's path integration formulation of quantum mechanics.

 

[etotheipi]

Yes that is confusing! I didn't even know there was a difference between routes/outcomes so I just treated them as being synonymous...

Needless to say there are a lot more subtleties to probability in QM than I had naïvely suspected . I'll try and keep that distinction in mind and hopefully when I get a little bit further into the material things will become a little clearer...

Thanks for the clarification!