I Particle energy in QFT..,

asimov42

Ok, so, in general, even in the relativistic case, one may measure an arbitrarily large momentum with some (very tiny) non-zero probability. Assuming we use a wave function / wave packet formulation that allows for a solution.

Here's what really bothers me: take QFT, and run a scattering experiment where at t = -∞ the particles going in are asymptotic free states with well-defined energies. Likewise, at t = +∞ one ends up with particles again with well defined energies, and energy is conserved. So there is no possibility here for arbitrarily large momentums - i.e., exactly zero probability that at any point the energy can be larger than the input energy, the energy is bounded.

Now, in some sense (and please correct me if I'm wrong), the entire universe is one large scattering experiment. Then in no case should there ever be a situation in which the momentum wave function for a particle (or whatever this translates to in terms of field excitations in QFT) can have an arbitrarily large momentum (e.g., one should never use a Gaussian with non-zero tails if one truly want real answers... granted the Gaussian makes a great approximation if you basically ignore the tails, which is the case if I understand correctly).

Am I correct with the above?

LeandroMdO

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)
I think what's missing from your understanding here is a bit of physics.

A wave packet in a realistic setting is not just some distribution handed from god. For simplicity, consider a classical electromagnetic wave: what does it mean to say that the wave has a Fourier component with an extremely high momentum? Well, it means that the source of the wave radiated that amount at the corresponding frequency. Energy, momentum, etc, are conserved, and they all come from the source, so if you find a Fourier component that carries quite a bit of energy, well, that's the reason.

In quantum mechanics the situation is a little stranger because of the question of measurement, but fundamentally it is the same: when you measure the momentum of some particle, you're also making an indirect (weak) measurement of whatever it is the particle entangled itself with in the past. So, for example, if an electron is accelerated by a pair of capacitor plates, the final state is entangled because of conservation of energy and momentum: a state in which the electron moves up has the plates moving down a bit (a very small amount if the plates are macroscopic, but in principle it's there). If, perchance, you measure the electron's momentum and get a very high amount, you're also indirectly measuring the momentum of the plates that accelerated it, getting a correspondingly huge amount, and so on through the entire history of interactions that led to this electron finding its way into your detector.

In short, if you measure a huge momentum in a wave packet, something put it there.

asimov42

I think what's missing from your understanding here is a bit of physics.

... If, perchance, you measure the electron's momentum and get a very high amount, you're also indirectly measuring the momentum of the plates that accelerated it, getting a correspondingly huge amount, and so on through the entire history of interactions that led to this electron finding its way into your detector.

In short, if you measure a huge momentum in a wave packet, something put it there.
LeandroMdO thanks - yes, this is exactly what I would expect from the physics of the situation, specifically the entanglement, and makes perfect sense - it's precisely the answer I expected.

But the math still indicates (non-zero tails in momentum space over the whole real line), at least from what I understand, that you might measure an enormous momentum value - the question is why is this a possibility? It is of course possible that whatever the particle was entangled with in the past imparted this enormous momentum but... Sigh, I know I'm beating a dead horse, and I apologize - but how can the momentum space wave function being non-zero over ±∞ possibly represent physical reality where there is (presumably) only a finite amount of energy.

LeandroMdO

LeandroMdO thanks - yes, this is exactly what I would expect from the physics of the situation, specifically the entanglement, and makes perfect sense - it's precisely the answer I expected.

But the math still indicates (non-zero tails in momentum space over the whole real line), at least from what I understand, that you might measure an enormous momentum value - the question is why is this a possibility? It is of course possible that whatever the particle was entangled with in the past imparted this enormous momentum but... Sigh, I know I'm beating a dead horse, and I apologize - but how can the momentum space wave function being non-zero over ±∞ possibly represent physical reality where there is (presumably) only a finite amount of energy.
Let's just put it this way. Take, for example, heights of children age 10. We often assume a normal distribution, say, with mean 1 m and standard deviation 0.1 m. Given this, what's the probability that a child will have a height
1. smaller than -0.5 m and 2. larger than 3 m?

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.

PeterDonis

Mentor
the math still indicates (non-zero tails in momentum space over the whole real line), at least from what I understand, that you might measure an enormous momentum value
As I've said: this is because the math is only an approximation, and you are trying to use it in a regime where the approximation breaks down. The correct answer is to not ask the math that question. It isn't to keep wondering why the math says something that doesn't make sense. Any approximation will tell you things that don't make sense if you apply it in a regime where it breaks down.

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

Mentor
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.

• mfb

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

Gold Member
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.

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PeterDonis

Mentor
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.

"Particle energy in QFT..,"

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