My confusion about quantum theory is twofold:

  1. I lack an adequate understanding of how the mathematics of quantum theory is supposed to correspond to phenomena in the physical world
  2. I still have an incomplete picture in my mind of how cause and effect relationships occur at the quantum level of reality.

This is why phenomena such as "entanglement" make absolutely no sense to me. So, in an attempt to come to some understanding of all of this, I would like to know that if what we conceptualize as a "field" is merely an interaction among particles, and particles themselves are actually fluctuations in "fields", then which comes first, particles or fields?

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    $\begingroup$ I would say quantum fields are more fundamental. Numbers of particles, what kind (i.e. which field they belong to), where and when they are and their state of motion, are merely ways of describing states of fields. $\endgroup$ Commented Jul 1, 2014 at 3:46
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    $\begingroup$ Firstly, fields are more fundamental. More helpfully, perhaps something that can help lead your intuition and learning about how the mathematics fit into the physical world is that the mathematics best describes what information can be known about physical phenomenon and how they interact, not necessarily the physical thing itself. The wavefunction is an incredibly useful tool for encapsulating what we know about what something is and how it behaves, but it's not 'real' physical phenomena in the way I think you're conceptualizing it. $\endgroup$
    – user28754
    Commented Jul 1, 2014 at 4:02
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    $\begingroup$ I think fields are more fundamental. You can have fields without particles, but you cannot have particles without fields. For example you have the Higgs field in your apartment, but I'm sure there is no Higgs boson at your premises. $\endgroup$
    – mpv
    Commented Jul 1, 2014 at 8:58
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    $\begingroup$ A related article: There are no particles, there are only fields, Am. J. Phys. 81, 211 (2013) and the comment on it and other citations $\endgroup$
    – Tarek
    Commented Jul 1, 2014 at 11:14
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    $\begingroup$ @trysis Well I'm pretty sure an "electron" is not a couple of quarks bound together... $\endgroup$
    – Lurco
    Commented Jul 2, 2014 at 14:20

6 Answers 6


This is a tricky question because it asks about the meaning of words. People use the word "particle" to refer to various, not always well defined, notions in physics.

In the end, I think the simplest and more correct single way to categorize the terms is to interpret "particle" as "excitation of a field". For example, if someone says

There are two electrons in this box"

I would mentally translate that to

The electron field in this box has two units of excitation.

This is all much easier to think about if you're familiar with the so-called "second quantization".$^{[1]}$

Second quantization

Consider a one-dimensional infinite wall potential (i.e. "particle in a box"). The system has a set of discrete energy levels, which we can index as

$$\left\{ A, B, C, D, \ldots \right\}$$

If we have only one particle, we can denote its state as e.g. $|\Psi \rangle_1 = |B\rangle + |D\rangle$.$^{[2]}$ This is the so-called first quantization. If we have two particles, the situation is significantly more complex because, as you have probably learned, quantum particles are indistinguishable. You probably learned that you have to symmetrize (bosons) or antisymmetrize (fermions) the state vector to account for the fact that the particles are indistinguishable. For example, if you say that particle #1 is in state $|\Psi\rangle_1$ as written above, and particle #2 is in state $|\Psi\rangle_2=|C\rangle$, then the total system state is (assuming boson particles):

\begin{align} \left \lvert \Phi \right \rangle &= (|B\rangle_1 + |D\rangle_1)|C\rangle_2 + |C\rangle_1 (|B\rangle_2 + |D\rangle_2) \\ &= |B\rangle_1 |C\rangle_2 + |D\rangle_1 |C\rangle_2 + |C\rangle_1 |B\rangle_2 + |C\rangle_1 |D\rangle_2 \, . \end{align}

This notation is horrible. In symmetrization/antisymmetrization you are basically saying:

"My notation contains information that it shouldn't, namely the independent states of particles which are actually indistinguishable, so let me add more terms to my notation to effectively remove the unwanted information."

This should seem really awkward and undesirable, and it is.

Let us consider an analogy for why the symmetrized state is such a bad representation. Consider a violin string with a set of vibrational modes. If we want to specify the state of the string, we enumerate the modes and specify the amplitude of each one, i.e. we write a Fourier series

$$\text{string displacement}(x) = \sum_{\text{mode }n=0}^{\infty}c_n \,\,\text{[shape of mode }n](x).$$

The vibrational modes are like the quantum eigenstates, and the amplitudes $c_n$ are like the number of particles in each state. With this analogy, the first quantization notation, in which we index over the particles and specify each one's state, is like indexing over units of amplitude and specifying each one's mode. That's obviously backwards. In particular, you now see why particles are indistinguishable. If a particle is just a unit of excitation of a quantum state, then just like units of amplitude of a vibrating string, it doesn't make any sense to say that the particle has identity. Units of excitation have no identity because they're just mathematical constructs to keep track of how excited a particular mode is.

A better way to specify a quantum state is to list each possible state and say how excited it is. In quantum mechanics, excitations come in discrete units $^{[3]}$, so we could specify a state like this:

$$|n_A\rangle_A |n_B\rangle_B |n_C\rangle_C |n_D\rangle_D$$

where $n_i$ is an integer. In this notation, the state $|\Psi\rangle_1$ from before is written

$$|\Psi\rangle_1 = |0\rangle_A |1\rangle_B |0\rangle_C |0\rangle_D + |0\rangle_A |0\rangle_B |0\rangle_C |1\rangle_D.$$

For compactness this would often be written $|\Psi\rangle_1=|0100\rangle + |0001\rangle$. The more complex two particle state would be

$$\left \lvert \Phi \right \rangle = |0\rangle_A |1\rangle_B |1\rangle_C |0\rangle_D + |0\rangle_A |0\rangle_B |1\rangle_C |1\rangle_D$$

or, more compactly,

$$\left \lvert \Phi \right \rangle = |0110\rangle + |0011\rangle \, .$$

This is the so-called second quantization notation. Note that it has less terms than the first quantized version. This is because it doesn't need to undo information that it's not supposed to have.

Back to fields vs. particles

The second quantized notation is far better because it naturally accounts for the "indistinguishable" particles. But, what we really learned, is that particles are actually units of excitation of quantum states. In the field theory language, we'd say that the particle is a unit of excitation of the various modes of the field. I won't say that either fields or particles are more fundamental because one has little meaning without the other, but now that we understand what "particle" really means, the whole situation is hopefully much clearer to you.

P.S. I do hope you'll ask for clarification as needed.

[1] The term "second quantization" is stupid, so don't try to interpret it.

[2] We ignore normalization.

[3] Hence the term "quantum".

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    $\begingroup$ @Benjohn: What's more fundamental, a violin string or the shape that it's in? $\endgroup$
    – DanielSank
    Commented Jul 2, 2014 at 0:11
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    $\begingroup$ @DanielSank Maybe "fundamental" isn't a good word: it implies merit or importance, which is not what I'm trying to understand. Perhaps the question isn't well understood because it seems too obvious to need an answer… What I'm trying to understand is, is it like a violin string: there is a string (a field), and vibrational modes of it are particles? Or is it like a swarm of flying birds: individual particles that can be construed as a single (changing) thing made from parts? Or something else? Or isn't it known? Or doesn't it make sense to ask? Sorry – I'm not being wilfully silly :-) $\endgroup$
    – Benjohn
    Commented Jul 2, 2014 at 11:49
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    $\begingroup$ @DanielSank Why do you make analogy where $c_n$ is like number of particles, if it in fact could be negative or even fractional? How could number of particles in a particular state be negative or fractional? $\endgroup$
    – Ruslan
    Commented Oct 1, 2014 at 10:41
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    $\begingroup$ @Ruslan: I should have said $|c_n|$ is like the number of particles. Does that answer your question regarding the phase and negativity? $\endgroup$
    – DanielSank
    Commented Oct 1, 2014 at 17:04
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    $\begingroup$ @Ruslan: Certainly it is more like average particle number. Anyway the point is just that in classical descriptions of things with excited modes you don't consider the units of excitation to have identity, and that it's the same story in quantum. The dimensions don't match because I wrote the state of the classical string in a particular basis. I could have written a more abstract equation $|\text{state of string}\rangle = \sum_{\text{mode }s} c_s |s\rangle$. $\endgroup$
    – DanielSank
    Commented Oct 1, 2014 at 19:08

What follows is an answer from an experimental particle physicist, i.e. one who has more knowledge of theoretical physics than the average educated person, but not in a position to teach it :). I can use theoretical results and study data and validate or falsify a theory.

I would like to know that if what we conceptualize as a "field" is merely an interaction among particles (bosons and fermions in the case of Quantum fields),

The concept of "field" in physics is general and mathematical.

A field is a physical quantity that has a value for each point in space and time. ....A field can be classified as a scalar field, a vector field, a spinor field or a tensor field according to whether the value of the field at each point is a scalar, a vector, a spinor or a tensor, respectively........a field can be either a classical field or a quantum field, depending on whether it is characterized by numbers or quantum operators respectively.

So the quantum mechanical field has operators sitting there at every spacetime point, which when acting will give a measurement value for that field. It does not carry interactions unless put in by hand/experiment.

and particles (themselves) are actually fluctuations in "fields",

Particles are described as excitations of an all permeating quantum field.

then which comes first in the hierarchy of cause and effect relationships, particles or "fields"?

Well, if you did not have the field there would not be a possibility for the particle to manifest itself , in the same way if you do not have space there is noplace to measure a classical electric field. It is like an underlying coordinate system. A field is more than a cause, it is a framework where cause and effect (interactions) can be described.

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    $\begingroup$ This is a good answer. OP, you can take the one I posted as an expansion of the idea of particles as excitations of fields. $\endgroup$
    – DanielSank
    Commented Jul 1, 2014 at 4:34
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    $\begingroup$ @DanielSank Thanks, your answer is good for somebody who has some grounding in formal physics courses. $\endgroup$
    – anna v
    Commented Jul 1, 2014 at 5:28
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    $\begingroup$ Indeed. It's good when multiple people post answers targeted toward different background levels. Much more useful for future readers :) $\endgroup$
    – DanielSank
    Commented Jul 1, 2014 at 5:33
  • $\begingroup$ What about the interacting fields? Do they not interact without supervision? $\endgroup$ Commented Nov 8, 2020 at 15:55
  • $\begingroup$ In quantum field theory it is not the fields that interact, but the particles as represented in Feynman diagrams. which are decide when one has to calculate the crossection for a specific interaction $\endgroup$
    – anna v
    Commented Nov 8, 2020 at 16:13

I would like to add to DanielSank's fantastic answer, as I've just had a thought on another way to state his brilliant passage:

Consider a violin string which has a set of vibrational modes. If you want to specify the state of the string, you enumerate the modes and specify the amplitude of each one, eg with a Fourier series

$$\text{string displacement}(x) = \sum_{\text{mode }n=0}^{\infty}c_n \,\,\text{[shape of mode }n](x).$$

The vibrational modes are like the quantum eigenstates, and the amplitudes $c_n$ are like the number of particles in each state. With that analogy, the first quantization notation, where you index over the particles and specify each one's state, is like indexing over units of amplitude and specifying each one's mode. That's obviously backwards. In particular, you now see why particles are indistinguishable. If a particle is just a unit of excitation of a quantum state, then just like units of amplitude of a vibrating string, it doesn't make any sense to say that the particle has identity. All units of excitation are the same because they're just mathematical constructs to keep track of how excited a particular mode is.

A better way to specify a quantum state is to list each possible state and say how excited it is .....

A companion analogy, actually mathematically exact, is the analogy between the quantum harmonic oscillator base states and the integers as described by the Peano axioms. When we make this analogy, it makes Daniel's great explanation even clearer. Consider a mode oscillator for the second quantized EM field. When a mode is in its second excited state, we say that the field has had two photons added to it. But the Fock state $|2\rangle$ of the field is just that: a state. In other words, there is no further information needed to fully describe this mode oscillator, and it is meaningless to try to tell which photon was which in their addition to reach this state: whichever "order" we add them in, the system state change wrought by their addition is it precisely the same in both cases.

Therefore, to try to tell the difference between these two photons is like trying to say that when we add two ones together to get $1+1=2$, the two ones have separate identities. Is the one on the left of $+$ distinguishable from the one on the right? Indeed, if we make the analogy below, the error in trying to tell the difference between the two 1s is precisely the same error in asserting that the "particles" added to the mode oscillator have distinct identities.

$$\begin{array}{|c|c|c|}\hline\\\text{Quantum HO}&&\text{Peano}\\\text{Object}&&\text{Object}\\\\\hline\\ \text{Number States}&\leftrightarrow&\mathbb{N}\\|0\rangle&\leftrightarrow&0\\\text{Raising operator }a^\dagger&\leftrightarrow&\text{Successor function }S\\\text{Lowering operator }a&\leftrightarrow&\text{Inverse Successor }S^{-1}\\\text{(}a\,|0\rangle=\text{null ket)}&&\text{(}S^{-1}\,0\text{ undefined)}\\\\\hline\end{array}$$

So, in the light of both Daniel's answer and mine, one would have to say wholeheartedly that fields are the fundamental entities, physics is interested in excitations of fields and how the fields change state when they interact, and "particles", as Daniel says, are a terrible description of what's going on.

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    $\begingroup$ Hey, this is the first time I really read this answer carefully and it's interesting! Thanks! $\endgroup$
    – DanielSank
    Commented Sep 6, 2015 at 0:56
  • $\begingroup$ It might be worth clarifying that $0 = \text{null ket} \neq |0\rangle$, as explained here: physics.stackexchange.com/a/141098/35699. $\endgroup$
    – user76284
    Commented May 17, 2018 at 14:03

When I studied quantum mechanics, my professor advised that I avoid the question "which is more fundamental?" and replace it with "which is more useful?". The problem is that our brains are programmed to think classically, so many concepts in QM have no classical analogue. For that reason, we usually discuss them mathematically in order to avoid ambiguity. One the one hand, we could say that fields are more fundamental and that particles are just excitations of the underlying fields. This accounts for some of the odd behaviours of particles (e.g. why particles of the same type are indistinguishable). However, when we do experiments, we tend to observe discrete objects rather than continuous fields. Ultimately, if I had to give an answer, I'd say that symmetries and commutation relations are fundamental, and representing them as fields or particles or whatever you want is just a matter of preference.

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    $\begingroup$ "which is more useful?" Nice. When I realized this was the right way to think about mathematical objects, physics became a lot easier. $\endgroup$
    – DanielSank
    Commented Jul 1, 2014 at 4:35
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    $\begingroup$ Definitely. We tend to perceive reality this way automatically without even thinking about it. For example, which is more fundamental: the guy standing there about to hit me in the head with a stick or a bundle of organic molecules arranged so that their large scale motions will produce a net force on my head? Now, which one of these interpretations is more useful? $\endgroup$ Commented Jul 1, 2014 at 4:49
  • $\begingroup$ Heh. While I might argue that the bundle of organic molecules is more fundamental in the scientific sense, I think I understand your meaning. $\endgroup$
    – DanielSank
    Commented Jul 1, 2014 at 4:54
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    $\begingroup$ Also, I should add that in Haag's axiomatization of local QFT, neither fields nor particles are necessary. The fundamental object is a way of assigning algebras of observables to open sets of space-time in a causal and Lorentz invariant way (an isotonic co-presheaf of C* algebras for the mathematically minded.) I like axiomatizations like this because you can use whatever kind of ontology you want as long as it satisfies the axioms. Either way you get the same observables and the same predictions, which is what science is really about. (Unfortunately AQFT is in its early days.) $\endgroup$ Commented Jul 2, 2014 at 19:45
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    $\begingroup$ I understood some of those words. $\endgroup$
    – DanielSank
    Commented Jul 2, 2014 at 20:18

There have been answers that fields are more useful, because they enable you to more quickly calculate the predictions of quantum field theory. This is true, but useful is not the same as fundamental.

We may take either one of two approaches to quantum field theory. We may start by defining the commutation relation for the field operators (e.g. by second quantisation). This approach makes the field operators fundamental in a mathematical sense. It says nothing about what field operators mean physically. This is the usual modern approach.

Alternatively we may start with relativistic quantum mechanics for single particles (typically electrons), we construct Fock space for multiparticle states, and we define field operators from creation and annihilation operators for the purpose of constructing operators to describe interactions between particles. This was the original approach, pre-war, for quantum electrodynamics.

Both approaches lead to essentially the same theory, but the approach from relativistic quantum mechanics has considerably more mathematical overhead. Field theorists usually think this does not offer anything useful. However, in questions of what is fundamental, I think should be looking at which is a better picture of physical reality, not simply which is the more useful mathematical structure.

In either case, we should recognise that quantum field theory is, strictly speaking, not mathematically valid. It is a pseudo-mathematical theory, which allows us to make correct predictions, but its fundamental elements, the quantum fields, have no mathematical definition. They are often described as operator valued distributions, but they cannot be defined in distribution theory. The reason is that products of field operators, as used in perturbation theory, contain powers of delta functions. It is mathematically proven that, not only is the square of the delta function not defined, it cannot be consistently defined. In the sense in which mathematicians use the word "exist", quantum field theory on a continuum does not exist. The same problem existed in the pre-war approach based on relativistic quantum mechanics, leading Dirac to despair "this Schrodinger equation has no solutions".

To take this any further one needs to study the mathematical foundations of quantum mechanics. As treated by Dirac, and shown by von Neumann, quantum mechanics is a probabilistic model of measurement results. It is distinguished from classical probability theory in one important respect. Whereas in classical probability theory outcomes are determined by unknown quantities, in quantum probability, outcomes are actually indeterminate.

There is an immediate implication for our understanding of a physical particle. A classical particle always has a definite, but possibly unknown, position, but, in the general case, a quantum particle does not have a position. We can only give a probability for the position where a quantum particle may be found.

To understand this better, we should reflect that (as observed by Newton himself), even in the macroscopic world, the empirical property of position only exists as a relative quantity. You cannot say where something is unless you say where it is relative to other matter. Newton inferred the existence of absolute space from the working of his equations, but he also remarked that only relative quantities can be observed. You cannot say where something is in absolute space.

Position is a relationship resulting from the interactions of an object with its environment. Macroscopic matter is in continuous interaction with its environment, and always has position, but a quantum particle may have too few interactions to generate the property of position. It may not even be possible for the particle to have a position. We can measure the position of an electron, but we can only measure where a photon was annihilated.

Position can exist in the results of measurement (including measurement by eye). Following von Neumann, we can set up a probabilistic theory of the results of measurement. Since we are talking of measurement results, we should modify von Neumann's treatment by taking into account that all measurements have finite range and resolution. As a theory of measurement results, Hilbert space should strictly be finite dimensional, not infinite dimensional as is normal for treatments of quantum mechanics.

I have developed this approach in my books and in Mathematical Implications of Relationism, which contains in appendices key arguments from The Hilbert space of conditional clauses and A Construction of Full QED Using Finite Dimensional Hilbert Space.

The implication is that the particle model, as advocated by Dirac and Feynman, is indeed the more fundamental representation of reality. As Dyson wrote

“In Feynman’s theory the graph corresponding to a particular matrix element is regarded, not merely as an aid to calculation, but as a picture of the physical process which gives rise to that matrix element”.

As I wrote in Mathematical Implications of Relationism

In a particle interpretation, Feynman diagrams also give a pictorial representation of the fundamental structure of matter. We cannot say what the precise configuration of particle interactions in any given instance, but we represent each possible configuration as a graph and sum over the possibilities, using the interpretation of a sum as logical disjunction. Only the topology of lines and vertices in a graph is relevant. The paper on which a graph is drawn has no meaning (a graph isomorphism is an edge-preserving bijection which defines equivalence classes of graphs with identical structure and meaning. Isomorphic graphs can be regarded as identically the same). Thus, spacetime structure does not appear in Feynman diagrams, except in so far as that energy-momentum has four components. Thus, Feynman diagrams describe the fundamental structure of a particulate model in which only particles exist and in which other properties, including spacetime geometry, emerge on large scales from interactions between particles.

A major justification for the particle interpretation is the locality, or microcausality, condition obeyed by field operators. This can be interpreted as meaning that interactions take place at a point, albeit a point for which the property of position cannot, in general, be defined.

  • $\begingroup$ but a quantum particle may have too few interactions to generate the property of position. Aren't quantum particles constantly interacting? $\endgroup$ Commented Nov 8, 2020 at 16:55
  • $\begingroup$ @ArmanArmenpress, It depends. Photons only interact when they are created and annihilated. The double slits experiment, for any particle, is the same as for photons. The particle is not interacting. Electrons in an atomic orbital, are interacting, but not in such a way as to generate the property of exact position. $\endgroup$ Commented Nov 8, 2020 at 17:12
  • $\begingroup$ Look at the path integral for a flying photon. It interacts with other quantum fields all the time in every possible way. $\endgroup$ Commented Nov 8, 2020 at 17:16
  • $\begingroup$ @ArmanArmenpress. Not in qed, or in any established physics. $\endgroup$ Commented Nov 8, 2020 at 17:22
  • $\begingroup$ One-loop correction to the photon propagator. researchgate.net/figure/… $\endgroup$ Commented Nov 8, 2020 at 17:30

In my opinion, at this moment of time fields are more fundamental than particles because are more useful. It is easier to describe e.g. processes of creation and anihilation of pairs electron-positron using the notion of field. Vector potential A plays dual role, it represents photons in these processes and describes an interaction in Dirac equation which creates and anihilates these particles. In Young-Mills theories of weak and strong interactions some massive particles behave just like photon. Keeping in mind only particle image it would be impossible to understand these 'particle' processes. However, in the future someone maybe will discover a new more fundamental notion! One must not forget about wave-particle dualism. In my opinion, it is still alive.


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