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Why application of probability in QM is fundamentally different than application of probability in other areas?

+ 6 like - 0 dislike
4093 views

Well the title is pretty much to the point. So i just re-state it here:

Why application of probability in QM is fundamentally different than application of probability in other areas?

Quantum Mechanics applies probability according to the same probability theory that other areas of physics/engineering/etc.. etc.. apply probability.

Why is there a difference?

"Naively" one would assume these cases:

  1. Either it is not the same probability (theory?)

  2. Or it is a matter of interpretation (of the formalism?)

  3. ??

UPDATE:

The answers already posted have some differences and some added references to other formulations. Personally i will have to read them through and update here.

Feel free to add more answers if you like.

This post imported from StackExchange Physics at 2014-06-03 16:28 (UCT), posted by SE-user Nikos M.
asked Jun 3, 2014 in Theoretical Physics by Nikos M. (80 points) [ no revision ]
retagged Jul 16, 2014 by dimension10
Most voted comments show all comments
I would definitely say that 'normal' probability theory is perfectly capable of encompassing QM. The physical meaning that we choose to attach to things in this context may be strange and counterintuitive, but that's a separate issue from the mathematics.

This post imported from StackExchange Physics at 2014-06-03 16:28 (UCT), posted by SE-user Danu
i say it is (just) the interpretation. BUT if one follows the (older and current) debates on the nature of QM, it is as though the application is quite different. Maybe i am wrong

This post imported from StackExchange Physics at 2014-06-03 16:28 (UCT), posted by SE-user Nikos M.
The question is not clear to me. Would you provide some example which the probability is used different than QM?

This post imported from StackExchange Physics at 2014-06-03 16:28 (UCT), posted by SE-user user26143
@ People I do not understand the close vote, all five of you. To me the question is clear and the answer by V.Moretti enlightening. I had not realized that the run of the mill probability had to be extended to include the difference in the mathematical formulation of QM

This post imported from StackExchange Physics at 2014-06-03 16:28 (UCT), posted by SE-user anna v
@anna v Indeed I voted against closing the thread. This is, in my opinion, a fundamental question of mathematics AND physics. I voted for reopening it.

This post imported from StackExchange Physics at 2014-06-03 16:28 (UCT), posted by SE-user V. Moretti
Most recent comments show all comments
I'm not an expert, but I was under the impression that the frequentist approach is used in QM. We have a large number of tests of a system prepared in the same quantum state and we observe how many times the measured result is such and such. That gives us the probability.

This post imported from StackExchange Physics at 2014-06-03 16:28 (UCT), posted by SE-user mpv
In QM the 'or' is very different. The classical rule $P(A or B)=P(A)+P(B)$ for independent events $A$ or $B$ does not generically apply in QM (as e.g. very well illustrated by the famous two slits experiments and the interference pattern). The amplitudes sum up, but then you square it and a new mixed term responsible for the interference appears.

This post imported from StackExchange Physics at 2014-06-03 16:28 (UCT), posted by SE-user TwoBs

5 Answers

+ 8 like - 0 dislike

The theory of probability used in QM is intrinsically different from the one commonly used for the following reason: The space of events is non-commutative (more properly non-Boolean) and this fact deeply affects the conditional probability theory. The probability that A happens if B happened is computed differently in classical probability theory and in quantum theory, when A and B are quantistically incompatible events. In both cases probability is a measure on a lattice, but, in the classical case, the lattice is a Boolean one (a $\sigma$-algebra), in the quantum case it is not.

To be clearer, classical probability is a map $\mu: \Sigma(X) \to [0,1]$ such that $\Sigma(X)$ is a class of subsets of the set $X$ including $\emptyset$, closed with respect to the complement and the countable union, and such that $\mu(X)=1$ and: $$\mu(\cup_{n\in \mathbb N}E_n) = \sum_n \mu(E_n)\quad \mbox{if $E_k \in \Sigma(X)$ with $E_p\cap E_q= \emptyset$ for $p\neq q$.}$$ The elements of $\Sigma(X)$ are the events whose probability is $\mu$. In this view, for instance, if $E,F \in \Sigma(X)$, $E\cap F$ is logically interpreted as the event "$E$ and $B$". Similarly $E\cup F$ corresponds to "$E$ or $F$" and $X\setminus F$ has the meaning of "not $F$" and so on. The probability of $P$ when $Q$ is given verifies $$\mu(P|Q) = \frac{\mu(P \cap Q)}{\mu(Q)}\:.\tag{1}$$

If you instead consider a quantum system, there are "events", i.e. elementary "yes/no" propositions experimentally testable, that cannot by joined by logical operators.

An example is $P=$"the $x$ component of this electron is $1/2$" and $Q=$"the $y$ component of this electron is $1/2$". There is no experimental device able to assign a truth value to $P$ and $Q$ simultaneously, so that elementary propositions as "$P$ and $Q$" make no sense. Pairs of propositions like $P$ and $Q$ above are physically incompatible.

In quantum theories (the most elementary version due to von Neumann), the events of a physical system are represented by the orthogonal projectors of a separable Hilbert space $H$. The set ${\cal P}(H)$ of those operators replaces the classical $\Sigma(X)$.

In general, the meaning of $P\in {\cal P}(H)$ is something like "the value of the observable $Z$ belongs to the subset $I \subset \mathbb R$" for some observable $Z$ and some set $I$. There is a procedure to integrate such class of projectors to construct a self-adjoint operator, and this is the physical meaning of the spectral theorem.

If $P, Q \in {\cal P}(H)$, there are two possibilities: $P$ and $Q$ commute or they do not.

Von Neumann's fundamental axiom states that commutability is the mathematically corresponding of physical compatibility.

When $P$ and $Q$ commutes $PQ$ and $P+Q-PQ$ still are orthogonal projectors, that is elements of ${\cal P}(H)$.

In this situation, $PQ$ corresponds to "$P$ and $Q$", whereas $P+Q-PA$ corresponds to "$P$ or $Q$" and so on, and classical formalism holds true this way. As a matter of fact, a maximal set of pairwise commuting projectors has formal properties identical to those of classical logic: is a Boolean $\sigma$-algebra.

In this picture, a quantum state is a map assigning the probability $\mu(P)$ that $P$ is experimentally verified for every $P\in {\cal P}(H)$. It has to satisfy: $\mu(I)=1$ and $$\mu(\sum_{n\in \mathbb N}P_n) = \sum_n \mu(P_n)\quad \mbox{if $P_k \in {\cal P}(H)$ with $P_p P_q= P_qP_q =0$ for $p\neq q$.}$$

Celebrated Gleason's Theorem, establishes that, if $dim(H)>2$, the measures $\mu$ are all of the form $\mu(P)= tr(\rho_\mu P)$ for some mixed state $\rho_\mu$ (a positive trace-class operator with unit trace), biunivocally determined by $\mu$. In the convex set of states, the extremal elements are the standard pure states. They are determined, up to a phase, by unit vectors $\psi \in H$, so that, with some trivial computation (completing $\psi_\mu$ to an orthonormal basis of $H$ and using that basis to compute the trace), $$\mu(P) = \langle \psi_\mu | P \psi_\mu \rangle = ||P \psi_\mu||^2\:.$$

(Nowadays, there is a generalized version of this picture where the set ${\cal P}(H)$ is replaced by the class of bounded positive operators in $H$ (the so-called "effects") and Gleason's theorem is replaced by Busch's theorem with a very similar statement.)

Quantum probability is therefore given by the map, for a given generally mixed state $\rho$, $${\cal P}(H) \ni P \mapsto \mu(P) =tr(\rho_\mu P) $$

It is clear that, as soon as one deals with physically incompatible propositions, (1) cannot hold just because there is nothing like $P \cap Q$ in the set of physically sensible quantum propositions. All that is due to the fact that the space of events ${\cal P}(H)$ is now noncommutative.

ADDENDUM. Actually, it is possible to extend the notion of logical operators $AND$ and $OR$ even for ${\cal P}(H)$ and that was the program of von Neumann and Birkhoff (the quantum logic). In fact just the lattice structure of ${\cal P}(H)$ permits it, or better is it. The point is that the physical interpretation of this extension of $AND$ and $OR$ is not clear. The resulting lattice is however non-Boolean. In other words, for instance, these extended $AND$ and $OR$ are not distributive as the standard $AND$ and $OR$ are (this reveals their quantum nature). However, the found structure is well known: A $\sigma$-complete, orthomodular, separable, atomic, irreducible and verifying the covering property, lattice. At the end of 1900 it was definitely proved, by Solèr, a conjecture due to von Neumann stating that there are only three possibilities for practically realizing such lattices: The lattice of orthogonal projectors on a separable complex Hilbert space, the lattice of orthogonal projectors on a separable real Hilbert space, the lattice of orthogonal projectors on a separable quaternionic Hilbert space. Requiring the existence time reversal symmetry introduces a complex structure on real Hilbert spaces giving rise to a complex Hilbert space. Conversely, still nowadays, it is not obvious if it is possible to rule out quaternionic Hilbert spaces to describe quantum physics.

This post imported from StackExchange Physics at 2014-06-03 16:29 (UCT), posted by SE-user V. Moretti
answered Jun 3, 2014 by Valter Moretti (2,025 points) [ no revision ]
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This is very interesting. Could you give a good reference?

This post imported from StackExchange Physics at 2014-06-03 16:29 (UCT), posted by SE-user nephente
Well, barring the old, but always wonderful, von Neumann's book about mathematical foundations of Quantum Mechanics, I think the book by Varadarajan on foundations of QM deals with these topics, too and there are some other books by Jauch (also Piron) relying on similar ideas for instance. (It is a bit embarrassing, but I could also suggest my book "Spectral theory and Quantum Mechanics" Springer 2013.) However, all that is more related to mathematics than physics.

This post imported from StackExchange Physics at 2014-06-03 16:29 (UCT), posted by SE-user V. Moretti
@V.Moretti Thank you for making me aware of your own book! I just had a look at the table of contents and it seems to cover many topics i always wanted to learn about in quantum theory.

This post imported from StackExchange Physics at 2014-06-03 16:29 (UCT), posted by SE-user nephente
@nephente Take into account that it is a first edition, so containing several misprints end errors of various kind, there is an (under construction) errata corrige on my webpage...

This post imported from StackExchange Physics at 2014-06-03 16:29 (UCT), posted by SE-user V. Moretti

I wonder how this answer is related to the simple fact that one usually adds probabilities (amplitudes square), whereas in QM one adds amplitudes. 

Most recent comments show all comments
Yes the author is the same and the book is that you pointed out.

This post imported from StackExchange Physics at 2014-06-03 16:29 (UCT), posted by SE-user V. Moretti
@WetSavannaAnimalakaRodVance Thank you for providing the title of the Varadarajan book!

This post imported from StackExchange Physics at 2014-06-03 16:29 (UCT), posted by SE-user nephente
+ 2 like - 0 dislike

The application of probability in areas other than quantum mechanics is a clever way to model situations that are complex enough so that the exact analysis is non feasible, or at least highly tedious.

On the other hand in QM nature is inherently probabilistic. When you make an observation the quantum state your system is in has a probability for each possible outcome. It is no more a trick to make calculations. It is a feature of nature. That is the difference.

This post imported from StackExchange Physics at 2014-06-03 16:28 (UCT), posted by SE-user silvrfück
answered Jun 3, 2014 by Dmitry hand me the Kalashnikov (720 points) [ no revision ]
+1, thanks, will wait for other answers also

This post imported from StackExchange Physics at 2014-06-03 16:28 (UCT), posted by SE-user Nikos M.
should i assume that your answer selects the 2nd option? (from the options given in question)

This post imported from StackExchange Physics at 2014-06-03 16:28 (UCT), posted by SE-user Nikos M.
regarding to your first point. 1)It is the same probability theory. 2)Yeah, it is a feature we assume (Max Born was the one who came with the probabilistic interpretation I think to recall)

This post imported from StackExchange Physics at 2014-06-03 16:28 (UCT), posted by SE-user silvrfück
@silvrfück: as I have written in the comments to the question: it is not the same probability theory since for mutually exclusive independent events the probabilities do not simply sum up.

This post imported from StackExchange Physics at 2014-06-03 16:29 (UCT), posted by SE-user TwoBs
@TwoBs never heard that. Could you please make your point clearer. I would like to learn this

This post imported from StackExchange Physics at 2014-06-03 16:29 (UCT), posted by SE-user silvrfück
@silvrfück: the answer given by alanf just a few minutes ago restates this point and provide also several link. But as I said: in QM, the amplitudes for mutually exclusive events are summed, and then squared to get the probability. This is clearly not the same that summing up the exclusive probabilities of each event separately

This post imported from StackExchange Physics at 2014-06-03 16:29 (UCT), posted by SE-user TwoBs
+ 2 like - 0 dislike

The probabilities in QM are given by the square amplitudes of the relevant terms in the wavefunction, or by by the expectation value of the relevant projector or POVM. However, it is not the case that those numbers always act in a way that is consistent with the calculus of probability.

For example, if there are two mutually exclusive ways for an event to happen then the calculus of probability would say that the probability for that event is the sum of the probabilities of it happening in each of those ways. But in single photon interference experiments this doesn't seem to work. There are two routes through the interferometer, the photon cannot be detected on both routes at once, so they are mutually exclusive, right? So then to get the probability of the photon emerging from a particular port on the other end you should just add the probability of it going along each route. But that calculation gives the wrong answer: you can get any probability you like by changing the path lengths see:

http://arxiv.org/abs/math/9911150.

So then you have the problem of explaining under what circumstances the calculus of probabilty applies.

You ask about frequentist approaches to quantum probability. There are some such approaches, e.g. - Hugh Everett's 1957 paper and his PhD. thesis:

http://www-tc.pbs.org/wgbh/nova/manyworlds/pdf/dissertation.pdf.

I think these arguments don't work because the frequency approach itself doesn't work. Why would the relative frequency over an infinite number of samples have anything to do with what is observed in a laboratory? And if there is some explanation, then why are we bothering with this relative frequency stuff rather than using the actual explanation? The best explanation of why it is applicable is the decision theoretic approach:

http://arxiv.org/abs/quant-ph/9906015

http://arxiv.org/abs/0906.2718.

The best attempt at explaining the circumstances under which it holds is given by the requirements that quantum mechanics imposes on the circumstances under which information can be copied:

http://arxiv.org/abs/1212.3245.

This post imported from StackExchange Physics at 2014-06-03 16:29 (UCT), posted by SE-user alanf
answered Jun 3, 2014 by alanf (20 points) [ no revision ]
thanks for the papers, havent seen this approach, will read them through

This post imported from StackExchange Physics at 2014-06-03 16:29 (UCT), posted by SE-user Nikos M.
+ 1 like - 0 dislike

Maybe you will find the essay Quantum Theory From Five Reasonable Axioms by Lucien Hardy interesting. In the abstract it says:

In this paper it is shown that quantum theory can be derived from five very reasonable axioms. The first four of these axioms are obviously consistent with both quantum theory and classical probability theory. Axiom 5 (which requires that there exist continuous reversible transformations between pure states) rules out classical probability theory.

This post imported from StackExchange Physics at 2014-06-03 16:29 (UCT), posted by SE-user doetoe
answered Jun 3, 2014 by doetoe (110 points) [ no revision ]
good one, will check it out, very interesting i actually learn sth more

This post imported from StackExchange Physics at 2014-06-03 16:29 (UCT), posted by SE-user Nikos M.
+ 1 like - 1 dislike

All probabilities arise from quantum mechanics. As pointed out here, there are no known examples of classical probabilities that don't have a quantum mechanical origin. Here we're not nitpicking about some small quantum effects, whether you consider coin throws, betting on the digits of pi etc., the probabilities can always be shown to be purely of quantum mechanical origin, and therefore the treatment according to quantum mechanics applies. Classical probability theory is thus not fundamental, it should be derived as an appropriate approximation from quantum mechanics.

This post imported from StackExchange Physics at 2014-06-03 16:29 (UCT), posted by SE-user Count Iblis
answered Jun 3, 2014 by Count Iblis (60 points) [ no revision ]

Betting on the digits of pi (or Chaitin's Omega number) has a quantum mechanical origin? I think the authors of the paper you cite have not really grasped the meaning of the term **algorithmically** random. See http://en.wikipedia.org/wiki/Chaitin's_constant

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