Undergrad Can the energy of a particle ensemble in QFT be bounded over time?

[asimov42]

Peter and LeandroMd0, thanks to you both for putting up with all of these questions. I can certainly understand the approximation, and I know the central limit theorem well. So this, in fact, makes total sense to me.

The reason I asked the last question is because, in another posting (where I tried to clear things up a bit but asked essentially the same question), mfb replied that:

All this is irrelevant for practical measurements. You simply do not care about things with 10^-1000 probability, although the mathematics requires them to be there. Removing these odd things artificially would need new physical laws, and there is no evidence for such a change.

This quote seems to imply that we absolutely need to use functions with long (infinite) tails to represent reality, or somehow the math breaks and ruins physics. So again, I'm in a slightly confused state - are we dealing with approximations (makes 100% total sense)? And why would recognizing that we're dealing with approximations ruin physics (require new laws) as mbf suggests?

p.s. I'll have to send you both beers via mail for all your help with this

 

[PeterDonis]

This quote seems to imply that we absolutely need to use functions with long (infinite) tails to represent reality, or somehow the math breaks and ruins physics.

I think what @mfb was trying to say was that, mathematically, the functions with infinite tails are what our current mathematical formulations of physical laws give us. Is that because those functions with infinite tails actually "represent reality"? Or is it just because we haven't figured out yet how to write mathematical formulations of physical laws that make correct predictions but don't have the infinite tails? I don't think we know for sure; but @mfb also made the point that we have no way of testing the difference in practice, since the predicted probability, from the math that has functions with infinite tails, for measuring, say, the energy of an electron in the table in front of you to be large enough to send it flying out of the table, is so low that we would not expect to see such an event for a time much longer than the lifetime of the universe.

To me, treating the functions with infinite tails as approximations allows us to not have to worry about which of the above possibilities is actually true, because treating them as approximations means only using the functions in regimes where we know we can test their predictions.

 

[asimov42]

Whew ok - thanks again all. To be clear on the QM side of things - take the electron in my desk which I (surprisingly) measure with my apparatus to have an energy much larger than than a highly energetic cosmic ray. Now, I've made a measurement which had a low probability - and the energy of the particle was not precisely defined until I measured it... that energy had to come from somewhere, as @LeandroMdO noted. Until I make the measurement, the energy imparted by all of the entangled particles is also unknown - so the measurement causes wavefunction collapse...

There's no question here - just interesting to think that the energy imparted by the whole chain can only be determined after a measurement...

 

[asimov42]

The answer, of course, is that the heights aren't really normally distributed, but that approximation is often good enough for government work. We choose the normal distribution because of the central limit theorem. If a series of random processes add or remove energy from a particle, sampling from a fixed but unknown distribution, you'd expect the distribution of particle momenta to approach a normal distribution. It is, however, only an approximation, just like the normal distribution is an approximation to the real distribution describing the heights of school children.

Thanks @LeandroMdO - so you would say that the continuous momentum function is simply an approximation? But certainly provides accurate predictions over any range we can measure...

 

[vanhees71]

vahhees71, can you comment on my post about the momenta being bounded? Obviously measuring an infinite momentum is not possible, but this doesn't say anything about measuring, say, an electron with an enormous momentum, which the Gaussian wave function clearly indicates is possible... (there's a clear distinction here between infinite and enormous)

Perhaps, I misunderstood your question. I still do not understand your problem properly obviously. You just construct a single-particle state with a square-integrable wave function in momentum space by (for simplicity I consider a Klein-Gordon field)
$$|\phi \rangle=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{p}}{(2 \pi)^3} \frac{1}{\sqrt{2 E_{\vec{p}}}} \hat{a}^{\dagger}(\vec{p}) |\Omega \rangle.$$
Here, $|\Omega \rangle$ is the free-particle vacuum, $\hat{a}^{\dagger}(|\vec{p}|)$ the creation operator of the relativistically covariantly normalized plane-wave mode, i.e., fulfilling the commutator relation
$$[\hat{a}(\vec{p}),\hat{a}^{\dagger}(\vec{p}')]=(2 \pi)^3 2 E_{\vec{p}} \delta^{(3)}(\vec{p}-\vec{p}'),$$
and $E(\vec{p})=\sqrt{m^2+\vec{p}^2}$.

Then $|\phi \rangle$ defines a free-particle wave packet normalized to 1:
$$\langle \phi|\phi \rangle=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{p}}{(2 \pi)^3} |\phi(\vec{p})|^2=1$$

 

[LeandroMdO]

Thanks @LeandroMdO - so you would say that the continuous momentum function is simply an approximation? But certainly provides accurate predictions over any range we can measure...

It's not the continuity that is an approximation, but the choice of a wave packet with a Gaussian form. In reality it doesn't have to be Gaussian. We use Gaussians for convenience of study and because the central limit theorem ensures it's a good approximation for many relevant situations.

 

[asimov42]

Sorry @LeandroMdO - I should have said continuous and non-zero over the whole real line? A continuous function could have a value of zero, of course.

 

[PeterDonis]

You just construct a single-particle state with a square-integrable wave function in momentum space

This is a free particle wave packet, so it would not describe, for example, an electron bound in an atom. The OP is talking about making a measurement of momentum on an electron in such a bound state. Scattering theory, which is basically what the wave packet you wrote down applies to, cannot be used in such a case.