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This is a follow up on this question, the answer of which points towards Quantum Mechanics.
As stated I am not a phycisist so please forgive my ignorance.
I will try to understand the issue by going in small steps (questions).
Wiki says:

In quantum physics, a quantum vacuum fluctuation (or quantum fluctuation or vacuum fluctuation) is the temporary change in the amount of energy in a point in space,1 arising from Werner Heisenberg's uncertainty principle

My question here is the following (actually has 2 parts):
1) The statement the temporary change in the amount of energy is what is meant when I read (and in the answer in my previous post) that things pop out of "nothing"?
2) Reviewing the uncertainty principle in wiki the concept (as I am capable of understaning it) is that we can not know both the position and the direction of a particle/object at the same time.How can from this conclude that things can come up out of "nothing ? It is not clear to me.
If we can not know where a particle currently is since we are looking into another property of it i.e. momentum then if we start looking into its position then does that mean that it appears out of "nothing"?
If someone could help me understand this in lamens terms it would be much appreciated.

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Jim
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2 Answers2

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Taking the first part of your question first: how do we get the energy-time uncertainty from the position-momentum uncertainty? This turns out to be surprisingly difficult to do rigorously. Even Heisenberg was only able to give an approximate derivation of it (based on a property called Compton wavelength). I found some of rigorous derivations here, here and here, but these are utterly impenetrable for the beginner. Even the Compton wavelength argument is a bit involved, so what I'm going to give is a justification based on dimensional analysis. This doesn't prove the energy-time uncertainty relation, but it shows it is plausible.

The Heisenberg uncertainty principle relates position and momentum. Position has units of distance, e.g. metres, and from basic mechanics ditsance is velocity times time:

$$x = vt$$

Momentum has units of mass times velocity:

$$p = mv$$

So if you multiply together position and momentum (as the Heisenberg UP does) you get:

$$x \times p = vt \times mv = t \times mv^2$$

I've rearranged the right hand side slightly because kinetic energy is $1/2mv^2$, so the right hand side looks like time times energy i.e.

$$x \times p = t \times E$$

NB this doesn't prove that $\Delta x \Delta p = \Delta t \Delta E$ but it shows that it's plausible.

Now onto the second part of your question (assuming I've convinced you that the energy-time UP follows from the position-momentum UP).

First let's ask is the energy-time UP real. Yes it is, and we can observe it fairly easily. You've probably heard that if you excite an atom it will emit light as it returns to it's ground state, and the frequency of the light emitted depends on the energy difference between the excited and ground states. This creates the atomic spectrum, which is routinely used for identifying atoms. Helium was first identified in the atmosphere of the Sun using this technique. Anyhow, the lines in the atomic spectrum don't have a precise frequency. If you measure them carefully you'll find they span a range of frequencies. Part of the broadening is from mundane sources like the doppler shift, but part arises from the E-t uncertainty principle.

So the E-t uncertainty principle is real, and it means we can't be certain about the energy of an atom unless we watch it for an infinite time. But exactly the same argument means that if we take some patch of vacumm we can't be certain about it's energy unless we watch it for an infinite time. That means the energy of the vacuum must fluctuate i.e. energy must spring into existance from nothing.

You may still be a bit unconvinced, but we can actually measure this spontaneous creation in the vacuum using the Casimir effect, so we know it really happens.

Hopefully by now you're convinced about (temporary) creation from nothing, and I guess your next question is precisely what happens when a virtual particle is created. Sadly I can't give you an answer for this. We have mathematical models for the process, like Quantum Field Theory, but whether this is what eally happens, or even if "what really happens" is a meaningful question, I don't know.

Response to comment: this links up with my answer to your other question, What is meant by "Nothing" in Physics/Quantum Physics?, so I thought I'd expand this answer rather trying to put everything in comments.

Anyhow, you ask a fair question. I've taken the position that the vacuum is effectively nothing plus the vacuum fluctuations, and you're asking me how I know it's not something plus the vacuum fluctuations. Actually this is sort of where we came in with your first question in the series.

My answer is that we can do experiments on the vacuum to see what's there. For example we can measure the vacuum fluctuations using the Casimir effect, and we get the answer our theory predicts. We can shine light through the vacuum to see if there's anything there, and we can weigh the vacuum (i.e. see if the vacuum has any gravitational attraction). In all cases we get the results our experiment predicts, and that's why I say the vacuum is effectively nothing plus the vacuum fluctuations.

You could argue that there is something present that we haven't worked out how to detect yet, but without any theoretical backing for this it's like saying there are fairies at the end of the garden that we haven't worked out how to detect yet!

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John Rennie
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    Thank you for the answer.I am not sure I get this:That means the energy of the vacuum must fluctuate i.e. energy must spring into existance from nothing Why MUST it be concluded that it sprung from "nothing" and not from a really low energy level (not-zero) which is steady at that level and is a starting point? – Jim Jun 29 '12 at 11:49
  • @Jim: I've updated my answer to address your comment. – John Rennie Jun 29 '12 at 14:21
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    "I've taken the position that the vacuum is effectively nothing plus the vacuum fluctuations". But only "something" can fluctuate. If the vacuum fluctuates, it is not "nothing", it is some thing. – Alfred Centauri Jun 29 '12 at 15:00
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    @JohnRennie:Thanks for the update!I will bother you with one last question.You note that we assume that the vacuum is effectively nothing, because this assumption is consistent with experimental results that are based on theories which already assume vacuum as effectively nothing.But you seem to consider absurd the counter argument that there could be something there that we just haven't detected yet.But assuming that something is there (beyond us yet) leads to a model that is intuitive.But the current is counter-intuitive(or not?). Isn't it more reasonable to assume that reality is intuitive? – Jim Jun 29 '12 at 15:03
  • @JohnRennie:Please forgive my simplicity in my question.It is due to the missing background – Jim Jun 29 '12 at 15:04
  • Intuition is based on experience. If you have no experience of QM then it probably seems unintuitive. Once you've learned the maths behind it QM seems not only intuitive but actually beautiful because everything fits together so well. The same is true of most modern physics e.g. general relativity. – John Rennie Jun 29 '12 at 15:10
  • I think that it is counter-intuitive to someone like me, due to the way it uses terms like "nothing" etc which have a different meaning from what a beginner understands. – Jim Jun 29 '12 at 15:48
  • @Jim: if I may elaborate science started useing the word nothing to help beginners understand fundamental concepts the problem with this practice is for someone like you who has a proper sense of intuition this immediately raises "red flags" so to speak. – Argus Jul 01 '12 at 08:07
  • @JohnRennie, I don't get it. You say " ... if we take some patch of vacuum, we can't be certain about it's energy unless we watch it for an infinite time. That means the energy of the vacuum must fluctuate ..." Why does the lack of knowledge about the energy mean that it must fluctuate? Why can't it also, just as easily, be non-fluctuating? Seems like a lack of knowledge shouldn't be used to say something must be true. – dcgeorge Aug 09 '13 at 16:04
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I'll try and explain this intuitively. I've no doubt that you'll get some answers that are more difficult to understand at your level (and I'll probably up vote them).

The uncertainty principle relates position and momentum. This actually applies to a single dimension at a time; one can, for example, measure the x-coordinate of position and the y-coordinate of momentum (theoretically, if not practically) to any accuracy, simultaneously.

So let's make this explicit. We'll let the position be $z$ and the momentum be $p_z$.

But according to Einstein's relativity, space and time are interchangeable (to some degree). So $z$ is equivalent to time $t$. So if there's an uncertainty relationship between $z$ and $p_z$ there must also be an uncertainty relationship between $t$ and, well, what would that other thing be?

Hmmmm. Position $z$ is not conserved and has a direction. Momentum $p_z$ is conserved and has a direction. And time $t$ does not have a direction. (We say that position and momentum are vectors, time is a scalar.) So in looking for the thing that is the counterpart to time, we want something that (a) doesn't have a direction, and (b) is conserved. Energy turns out to be the answer.

Carl Brannen
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  • 1)to any accuracy, simultaneously. do you mean to any accuracy, NOT simultaneously? 2)Ok, you replace the equivalent forms to associate t and energy as counterparts. But how does this show that things pop out of "nothing"? I lost you there – Jim Jun 29 '12 at 07:23
  • No, I mean to any accuracy, simultaneously. The concept that there might be limitations that (for example) relate momentum in the z direction with position in the x direction, is an active research area in theoretical quantum mechanics (no practical applications yet, AFAIK). Look for stuff like "noncommutative quantum mechanics" on arXiv.org
    • – Carl Brannen Jun 30 '12 at 03:38
  • If energy can pop out of nothing, and the relationship is Delta E x Delta T >= h-bar/4 (I could easily have forgotten a constant factor here), then for small enough Delta T, you can have Delta E as large as you like. Once Delta E is larger than the mass of an electron and a positron, it's possible for that Delta E to create an electron and positron pair. (The reason for using electron is that it's the lightest and so easiest charged particle. The positron weighs the same amount. The reason for the pair is that their charges are opposite and so charge is conserved.)
    • – Carl Brannen Jun 30 '12 at 03:43