7
$\begingroup$

It's heard quite often that fundamental particles (photons, quarks, etc) act as both particles and waves.

Now, I'm looking at it from a Quantum Field perspective. Is this localized energy ripple what the wave is? And is the fact that it is localized make it a particle?

Quantum Field

As shown in the GIF above, there is a red lattice, which is a quantum field. This is an oscillating ripple. Is this the wave nature of the particle? Is that what causes the result of Thomas Young's double slit experiment? Also, there is a green layer representing the particle's position, representing the disturbance in the quantum field as a particle. Is this the explanation for the photoelectric effect?

Ultimately, is this the explanation for the wave-particle duality?

Improve this question
$\endgroup$
5
  • $\begingroup$ It fits pretty nicely. Please read about second quantization and Fock space. $\endgroup$
    – Prof. Legolasov
    Commented Jul 6, 2017 at 3:29
  • 1
    $\begingroup$ This isn't a direct answer but might be of interest $\endgroup$
    – John Rennie
    Commented Jul 6, 2017 at 5:35
  • $\begingroup$ @SolenodonParadoxus Hmm... The answers I got say otherwise, don't they? $\endgroup$
    – user143777
    Commented Jul 6, 2017 at 12:07
  • 1
    $\begingroup$ One thing that is very helpful is to know that the term "wave particle duality" causes all sorts of problems because people describe it as "a wave and a particle at the same time," which is not correct. What is more correct is to say that the entity's state is not perfectly described as a particle, and not perfectly described as a wave. It always has some character of both. This confusion is then compounded by the fact that in QM, we talk about the "wavefunction," which is not the same wave as they are talking about in wave/particle duality, despite having "wave-" in it. $\endgroup$
    – Cort Ammon
    Commented Jul 6, 2017 at 16:52
  • $\begingroup$ @AdityaRadhakrishnan the 1-particle subspace of a free QFT is just a quantum-mechanical system. Thus my comment. $\endgroup$
    – Prof. Legolasov
    Commented Jul 7, 2017 at 1:48

5 Answers 5

Reset to default
19
$\begingroup$

No, this is not at all how quantum field theory works.

  1. A "quantum field" does not have a definite value at any time, it is an operator in the quantum theory, not something that has a fixed numerical value, therefore representing it as a lattice as you have done does not reflect the quantum nature of the field. This is the classical picture of the field, just like a point particle is the classical picture of the electron, not its quantum picture.

  2. The quantum field and the particle states are different things - the field is an operator and the particle is a state in the quantum theory. You can use (parts of) the quantum field operator to create particles, but the notion of particle is much more elusive than it being a simple ripple in a classical field. For more on this see this answer of mine on real particles and this question and its answers on virtual particles.

  3. The "wave-particle duality" is, in any case, a somewhat vague notion that has no real formal counterpart in modern quantum mechanics. Quantum objects are just that, quantum objects. They have aspects of waves (e.g. they can "interfere", they can obey wave-like equations, they "spread") and they have aspects of particles (e.g. they can (but not must be) localized at "points", they have mass) but they are neither. And I'm sure you can find quantum behaviour that you'll not be able to attribute to either a wavy or a particle nature, such as Bell experiments about entanglement (which cannot be explained classically, and hence any attempt to explain them with a particle or wave picture must necessarily fail).

Improve this answer
$\endgroup$
1
  • $\begingroup$ With respect of 3. The "wave-particle duality", it's also worth noting that both are mathematical abstractions, ideals, that don't match reality anyway (neither infinitely long (waves) nor infinitely intense (particles)), so the classical reality is somewhere between - a Plurality of basis functions. In particular Wavelets, which are rather recent (in the scheme of things). The Friendly Guide to Wavelets", Kaiser G, 1994, 0-8176-3711-7, Ch 9. Electro Magnetic wavelets, is interesting in that area. $\endgroup$
    – Philip Oakley
    Commented Aug 10, 2017 at 15:22
4
$\begingroup$

Here is an answer by an experimentalist:

Quantum field theory is a necessary mathematical tool in order to be able to calculate interactions between elementary particles , as defined presently in the standard model of particle physics.

This introduction in the wiki article is sufficient for a broad picture.

In theoretical physics, quantum field theory (QFT) is the theoretical framework for constructing quantum mechanical models of subatomic particles in particle physics and quasiparticles in condensed matter physics. QFT treats particles as excited states of the underlying physical field, so these are called field quanta.

For the QFT of particle physics, each particle/quantum-mechanical-entity in the table defines a field, a photon field, an electron field etc which covers all the phase space. This field is the ground state of the corresponding free particle solution of the appropriate equation, the Dirac for the electron, the quantized Maxwell for for the photon etc . These fields acquire a physical meaning when operated with the number operator, which is a differential operator and creates or annihilates a particle at that (x,y,z,t) where the ground state/field is defined.

So in your diagram, if one of the axis is space, the other is time , and when going up to the green level one electron is created with a creation operator, and going down it is destroyed with a destruction operator, and moves to the next time position. A consistent wave packet in (x,t) would move in the x direction like a bump. See the answer here for particles as wavepackets where one has to enter in the mathematics.

The wave nature is evident in the calculated probability distributions for a given reaction, and these show wave behavior because the underlying ground levels on which calculations and expansions are carried out are wavefunction solutions of quantum mechanical equations. QFT is a method of organizing the perturbative expansion which approximates the exact solutions of a scattering problem, but it is based on the postulates of quantum mechanics.

So the wave nature does not come because of QFT structures,but because the QFT mathematical structures are built on wavefunctions. So a similar plot would illustrate the wavepacket nature of particle representations in QFT, but the particle/wave duality comes from the nature of the wavefunctions describing the ground state, on which the quantum field creation and annihilation operators work.

Improve this answer
$\endgroup$
1
$\begingroup$

No. Wave-particle duality is due to the fact that, in quantum mechanics, not all particle properties commute. This ultimately means properties of particles can be described using a wavefunction (an expansion of the quantum state in some basis of properties). This function maps properties to probability amplitudes, and permits interference patterns and other strange wave-like phenomena loosely termed "wave-particle duality".

Quantum field theory has an analogous formalism. The properties of fields do not all commute, and so field properties can be described using a wavefunctional.

The commutation relations of quantum theory are a nomological ingredient of quantum theory and are not explained by some deeper physical principle.

Improve this answer
$\endgroup$
6
  • $\begingroup$ "This function maps properties to probability amplitudes..." not really. The wave function maps a state $|\psi\rangle\mapsto \langle x|\psi\rangle\in\mathbb{C}$. $\endgroup$
    – gented
    Commented Jul 6, 2017 at 12:16
  • $\begingroup$ The wavefunction maps properties to probability amplitudes. In quantum mechanics, a property $P$ is associated with the Hilbert subspace $\mathcal{P}$, and the wavefunction $\psi$ maps the property $P$ to probability amplitude $\langle P | \psi \rangle$. I.e. $P \mapsto \langle P | \psi \rangle \in \mathbb{C}$ where $|P\rangle$ is a state that lies in $\mathcal{P}$ $\endgroup$
    – DJames
    Commented Jul 6, 2017 at 18:58
  • 1
    $\begingroup$ Probability amplitudes are not restricted to positive real numbers. See equation 3.1 in the Feynman lectures here. Also, the position basis is just one option for writing down the wavefunction. You can also, for example, use a momentum basis. See here for example $\endgroup$
    – DJames
    Commented Jul 6, 2017 at 23:06
  • 1
    $\begingroup$ I believe there is a lot of misunderstandings in your terminology: 1) "Probability" amplitudes must be positives, amplitudes (without the pre-noun "probability") are the scalar product (it's stated even in your link if your read carefully). 2) The wave function is by definition defined against the $|x\rangle$ basis and $\langle p|\psi\rangle$ can be of course therefrom derived (it is the Fourier transform). If you use non-standard terminology or convention you should at least pre-specify it, as standard QM definitions are otherwise. $\endgroup$
    – gented
    Commented Jul 6, 2017 at 23:26
  • 2
    $\begingroup$ I'm using standard terminology. "Probability amplitude" refers to the scalar products in question.. And physicists regularly talk about wavefunctions with coordinates other than position (e.g. Spin-coordinates of molecular orbitals) $\endgroup$
    – DJames
    Commented Jul 7, 2017 at 14:21
0
$\begingroup$

According to QFT there's no wave, and there are no particles at all. There are only and only quantized fields and they can unify the two contradictory sides of the coin, the wave-particle duality enigma or paradox is completely solved within the framework of QFT. The problem was firstly posed by two theorems by Einstein in his famous Photoelectric paper.

1-Computing the number of microstates of the radiation field using Planck's entropy that was compatible only with the existence of a photon gas(particles)

2-Energy fluctuations that consisted of a wave-like contribution and a corpuscular(particle-like) contribution.

And the first rigorous solution was presented by Jordan who invented the idea of Quantum Fields.

Improve this answer
$\endgroup$
-3
$\begingroup$

First, 'wave' is a form (descriptor) and particle is an entity/object, so you and everyone else are comparing incomparables. That said, a particle does have a form which we can describe as"field-form" [something extending in {occupying} space (volume)]. Just as a wave extends in time (wave cycles reside/occupy time) so a particle extends in [occupies] space. For electrons in motion, why cannot the rest mass have the 'field-form' character and the electron's kinetic energy have the [oscillating] waveform character? Which character dominates depends upon rest mass quantity versus kinetic energy quantity.

As for asking QFT to determine the actual, physical presence of waveform or field-form ['particle'] for moving masses, here is a quote from Stanford Encyclopedia of Philosophy to consider: "one reason why the ontological interpretation of QFT is so difficult is the fact that it is exceptionally unclear which parts of the formalism should be taken to represent anything physical in the first place."

Short answer, QFT doesn't care about physical entities; it postulates entities [fields] it can manipulate mathematically to satisfy experimental results [the ether theory can do the same thing; it too can account for space contraction and time dilation].

Suggest you google 'einstein' and 'method' since E had similar critiques of the duality.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.