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What causes it and how does it occur? If you do post some mathematics, please explain what each term means too please.

Qmechanic
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Matthew
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    "What causes it"... you must be new here. – Mitchell Porter Nov 12 '11 at 00:46
  • ok then, just answer the "how does it occur" part. and yes I am, how could you tell :) – Matthew Nov 12 '11 at 01:58
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    Hi Matt, see here for a very resonable introduction. Good luck on your quest for quantum answers, you'll need it! – MoonKnight Nov 12 '11 at 02:19
  • yeah, that wikipedia page is definitely in layman's terms – Matthew Nov 12 '11 at 05:29
  • how can the conservation of energy "appear" to be violated without it actually being violated? what does that even mean? – Matthew Nov 12 '11 at 05:33
  • on the simple wikipedia page for quantum fluctuations, it said that photons have no energy value, and thus the conservation of energy principle isn't violated. I know that photons are made up of positrons and electrons, so does that mean that the first particles in the early universe were electrons? – Matthew Nov 12 '11 at 05:34
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    Matthew, the general answer to all your questions about quantum mechanics is that in quantum theory, everything is fundamentally random, and the order of the world we see is how things average out. It's how these "averages" change over time that looks deterministic. Everything fluctuates in all possible ways, but some ways are more probable than others, which is why e.g. a rock sits there being a rock, rather than disintegrating immediately. – Mitchell Porter Nov 15 '11 at 08:37
  • @MitchellPorter I think that's worth being an answer. – Volker Siegel Jul 19 '14 at 07:18

2 Answers2

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Quantum fluctuations are a popular buzzword for the statistical triviality that the variance (the spread of values) of a random variable A (in context of quantum physics, this could be the position of a particle or the amount of energy that it has) with zero mean is typically not zero - except that A is now an operator. Some people, therefore, think that this deserves a much more mysterious name.

Taken from the section ''Does the vacuum fluctuate?'' in Chapter A8: Virtual particles and vacuum fluctuations of A theoretical physics FAQ

Max
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Arnold Neumaier
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    That's not bad and interesting in itself but can not be the full story. "Fluctuation" implies some temporal variation. Whereas the variance you are talking about is just the spread a distribution has around its mean value. – Raskolnikov Mar 02 '12 at 19:56
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    @Raskolnikov: There are two uses of the term; a technical one that means precisely what I wrote, and popular one that associates with the quantum fluctuation weird stuff with a temporal behavior that is not observable, and thus figures only in the minds of those who enjoy quantum mysticism. (For example, the vacuum is temporally completely inert, but the vacuum fluctuations of most fields are nonzero.) This is explained in more depth in other articles of the chapter cited. – Arnold Neumaier Mar 02 '12 at 20:06
  • You're right. I guess I haven't been doing QM for too long. I've lost acquaintance with the terminology. – Raskolnikov Mar 02 '12 at 20:12
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Fluctuations in the mean are also called fluctuations. It gives a notion about how reliable the mean value is (the second moment of the distribution). Any quantity that we are uncertain about will have that uncertainty encoded in a probability distribution, Quantum mechanics is no different in that respect then any other theory of inference, it is only different in that it claims that the uncertainty is intrinsic whereas other theories of inference simply assume that the data is observable in principle but not in practice.

We use the term `quantum fluctuation' therefore to impose the idea of fluctuations on physical variables that we classically thought of as being exact and obtainable such as position and momentum.

An interesting and quick calculation in scalar free field theory gives an interesting example of `quantum fluctuations' $$\langle \phi(x)\rangle_0=0$$

$$Var(\phi)_0=\langle\phi(x)^2\rangle_0-\langle \phi(x)\rangle_0=\langle\phi(x)^2\rangle_0 =\int \frac{d^3k}{(2\pi)^3}\frac{1}{\sqrt{\vec k^2+m^2}}\rightarrow\infty$$

The average value of the field is vanishing, but when we ask the extent to which this result can be trusted, it cannot, our ignorance is infinite.

SalmonProtocol
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  • @SalmonProtocol- How is the non-zero variance, related to virtual particles in the Feynman diagram? Or in other words, if the variance were zero, wouldn't the vacuum bubble diagrams arise? – SRS Aug 14 '16 at 14:15