# What are quantum fields?

1. Jul 6, 2013

### eightsquare

I did some research on Wikipedia, and I quote from the write up on Real Particles:
"The term(Virtual Particles) is somewhat loose and vaguely defined, in that it refers to the view that the world is made up of "real particles": it is not; rather, "real particles" are better understood to be excitations of the underlying quantum fields. Virtual particles are also excitations of the underlying fields, but are "temporary" in the sense that they appear in calculations of interactions, but never as asymptotic states or indices to the scattering matrix."

What I wanted to ask was, what is the meaning of "excitation of the underlying fields"? What quantum field are they talking about? I always thought that in quantum mechanics fields were basically made up of particles. But that brings me back in a loop.

2. Jul 6, 2013

### eightsquare

Why do particles move when force is applied?

According to Wikipedia, a particle is nothing but an excitation of the underlying quantum field. So how exactly are forces defined in quantum mechanics and why does the 'excitation' tend to move when we apply a force to it?

3. Jul 6, 2013

### wotanub

Welcome to PF, 82. I'm not sure what level of physics you have, I think a "mid-level" understanding of quantum mechanics is needed to answer this question. At least the quantum harmonic oscillator and some classical field theory.

I know about this in terms of very basic QFT. Do some research on quantizing the EM field. One can write down the Hamiltonian for an EM field, and find it has a form like the harmonic oscillator. The eigenstates of the corresponding number operator are called Fock states. When the raising operator is applied to the Fock state it increases the quanta of the field by one. The quanta is called a photon and the raising operator is re-termed the creation operator since it creates a qunata. Similarly, the lowering operator is called the annihilation operator. I imagine the approach is similar for whatever kind of field you would like to quantize.

Here's a good lecture video that explains it very clearly.
And the Wikipedia article. http://en.wikipedia.org/wiki/Quantization_of_the_electromagnetic_field

In QM, we really don't discuss forces per se, we prefer to talk about energy. Particles experience force because they have potential energy, and they move because they have momentum, and the force can change momentum, just like in classical mechanics. You can always write down the force, equations of motion, or what have you from the Hamiltonian (energy), but equations of motion really take a back seat in quantum mechanics.

Last edited by a moderator: Sep 25, 2014
4. Jul 6, 2013

5. Jul 7, 2013

### mattt

Quantum Fields are some mathematical objects that appear in some mathematical structures used to produce predictions of the outcomes of some experiments.

6. Jul 7, 2013

### The_Duck

You can think of a field as the surface of a pond and "excitations" as traveling waves on the surface. Forces appear when different parts of the water have different properties (different "potential energy"), which can cause the waves to bend (toward regions of lower potential energy) instead of propagating in straight lines.

7. Jul 8, 2013

### No-where-man

This is very interesting. But is entire universe (well at least this visible observable part of the universe) made of one quantum field or more quantum fields that are directly connected and full of interactions?

8. Jul 8, 2013

### eightsquare

@No-where-Man- I'm pretty sure more than one field. Every elementary particle is considered to be a quanta of their respective fields, and as there are many elementary particles, I'd say there are many fields full of interactions.

To everyone- The vacuum fluctuation model of the beginning of the universe imply that these fields exist even in a vacuum, even when there is no real particle affecting the space around it. Maybe this is a fallout of the Heisenberg uncertainty principle, but does it mean that even the existence of fields is uncertain on the most fundamental level and that even nothing can have fluctuations due to uncertainty?

9. Jul 9, 2013

### mpv_plate

Yes. The fields cannot have definite zero value and definite zero conjugate momentum everywhere all the time because of the canonical commutation relations. Even if there are no particles in the fields.

10. Jul 9, 2013

### No-where-man

I'm still confused with the answer, I read a lot about Heiseneberg's uncertainity principle, and I'm still not sure if I understand it.
Why wouldn't fields have 0 value and definite zero conjugate momentum everywhere all the time (whatever that means), just because you cannot 100% know both position and momentum at the same time?
I don't understand why would this be a case?

I'm not questioning Heisenberg's uncertainty principle at all, I just don't understand it.

11. Jul 9, 2013

### The_Duck

The uncertainty principle is more general than the one between position and momentum of particles. In field theory, we have an uncertainty relation between the value of a field at a point, and its time rate of change at that point. (Partly to make an analogy to the more familiar position-momentum uncertainty relation, we sometimes call the time rate of change of the field its "conjugate momentum." This is jargon: "momentum" here has nothing to do with the momentum of particles or anything like that).

The position-momentum relation for particles implies that if you confine a particle to some region, its momentum becomes uncertain. We might poetically think of this uncertainty as a result of constant unobserved fluctuations in the position and momentum of the particle. The uncertainty relation described above for fields has a similar result. If a field, for example the electromagnetic field, is "naturally" zero in vacuum, it cannot be exactly zero all the time because of the uncertainty relation between the value of the field and its time rate of change: if we know that the field has a value close to zero, then there is some uncertainty in its time rate of change. Again we can poetically think of this as a result of constant unobserved "vacuum flucuations" of the field, which consist of the value of the field fluctuating in some region around zero.

12. Jul 9, 2013

### hilbert2

A question about this... The generalized uncertainty principle relates the combined uncertainty of two observables to their commutator (two commuting observables can be measured accurately at the same time, a noncommuting observable pair can not). Do the field strength and its conjugate momentum of a fermionic field obey a commutation or an anticommutation relation? With the creation and annihilation operators of fermion fields there seem to be anticommutation rules.

13. Jul 9, 2013

### No-where-man

But wouldn't be possible to actually detect energy value of the field, if you know at least one of properties (either position or momentum)?
I mean can the non-zero energy be detected?
http://www.sciencedaily.com/releases/2013/03/130303154958.htm

Thanks.

14. Jul 9, 2013

### mpv_plate

You can actually know both of them simultaneously (technical remark: quentum fields are not observable, so I'm talking here figuratively), but due to the commutation relations there is a non-zero probability that the values of the fields or their momenta will differ from zero when you next "measure" them.

Heisenberg uncertainty is not saying that you cannot know the non-commuting quantities simultaneously. The uncertainty is only defining the statistical features of repeatedly measured values. But during one measurement you can measure both non-commuting quantities to arbitrary precision simultaneously.

Some details about the vacuum state of quantum fields are here. Go to page 55, paragraph 3.4.3 Zero Point (Vacuum) Energy.

15. Jul 9, 2013

### No-where-man

Ok, thanks, I thought about zero-point energy but I actually thought this was a crackpot hypothesis not made by true physics, I was wrong:
Can you just comment this link I found:
http://www.sciencedaily.com/releases/2013/03/130303154958.htm

Last edited: Jul 9, 2013
16. Jul 9, 2013

### The_Duck

An anticommutation relation. Also my statement above that the conjugate momentum is the time rate of change of the field isn't true for fermionic fields.

Yes. One way of thinking about the Casimir effect is in terms of the "zero-point energy" due to these "vacuum fluctuations."

There's a caveat, though: we can't detect the *absolute* energy density due to vacuum fluctuations. In the Casimir effect, for instance, we can only detect the *change* in vacuum energy density produced by the presence of a pair of conducting plates.

17. Jul 9, 2013

### Maui

Their experiment highlights a property of quantum systems - that their behavior is dependent on what can be known. You still need a 'strong' measurement for any real reading, the weak measurements are only revealing what the system is doing mostly(this would be classically impossible so interpret as you deem fit). The bottom line is this - they are not really getting aroung the Heisenberg uncertainty principle. They are getting more data than 'usual'(about the first property) but it's as ambiguous as Heisenberg postulated close to a century ago.

Last edited: Jul 9, 2013
18. Jul 10, 2013

### eightsquare

Why are there only a finite number of fields? And do we know why they interact with each other the way they do?

19. Jul 10, 2013

### No-where-man

I will bet that there is a truly infinite number of fields.

20. Jul 10, 2013

### ZapperZ

Staff Emeritus
Where do you get such a thing?

Zz.