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I want to use Matt Strassler's definition of the word "particle" as a specific example:

Matt Strassler writes:

(1) "...all the elementary “particles” (i.e. quanta) of nature are quanta of waves in fields..."

My first question is, in this sentence, is the word "quanta" short for "quanta of waves in fields?"

Sentence (1) simply defines the word "particle" to be another name for "quanta".

I looked up quanta and quanta means

a wave that for a given frequency, have minimum energy and amplitude.

So, a quantum is a wave. Is this correct?

From this analysis, I conclude that, in particle physics, a particle is a wave, because a particle is defined as a quantum.

In mathematical terms:

$quantum \equiv wave$

or

$particle \equiv wave \equiv quantum$

So, in particle physics, the word particle, has the properties of a wave, and it has no other properties. A particle, in particle physics, for instance, is not a spherical object with finite radius.

This is confirmed by Matt Strassler, who writes

(2) "...the word “particle” in particle physics has the same meaning as the word “quantum”..."

Sentence (2) is incomplete, because the word "particle" has several other meanings besides being another name for quanta. This is very confusing to non-physicists because the word "particle" has been defined many times and physicists pick and choose a meaning that saves the context.

My questions are:

1) The word "quantum" is a perfectly good word meaning a wave with minimum amplitude; why are physicists insisting on calling a quantum a particle?

2) Is there a unique definition of the word particle in physics not dependent on context (not casuistic) and that all physicists can agree upon?

If you cannot answer any of these, can you clarify if "particle" is another name for "quantum".

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3  
Matt Strassler is talking about "elementary particles". They are certainly not spherical objects with finite radius. But they don't behave exactly like waves, either. Explaining what they do behave like to non-physicists is quite difficult, because they don't behave like anything that we have good intuition for. And you can certainly have quanta which are not elementary particles (e.g., phonons). –  Peter Shor Dec 1 '12 at 15:06
    
Related: physics.stackexchange.com/q/35781/2451 and links therein. –  Qmechanic Dec 1 '12 at 15:27
    
@PeterShor: Strassler draws "particle" as a quantum which looks exactly like a wave as in this slide called "Quantum" physics.rutgers.edu/~strassler/QuestHiggs.pdf. What is the reason to believe that this quantum drawn as a wave is not a wave? –  Zeynel Dec 1 '12 at 15:30
1  
If you are interested, Susskind discussing what we mean by particles at around 25 min in youtube.com/watch?v=-BleG7PBwEA –  DJBunk Dec 1 '12 at 18:19
    
@DJBunk: Thanks for the link. Here's the transcript of the relevant part: docs.google.com/document/d/… It all boils down to this sentence at 31:53: "a particle or not that depends on the excitation spectrum of the energy levels above the ground state. If they are well-separated for some reason, then the answer is, it will behave like a particle." –  Zeynel Dec 2 '12 at 12:43

6 Answers 6

My questions are:

1) The word "quantum" is a perfectly good word meaning a wave with minimum amplitude; why are physicists insisting on calling a quantum a particle?

The word "quantum" in physics is used as a "definite quantity of a variable" in contrast to "a continuous quantity of that variable". There are quanta of energy, for example the energy levels of an atom which can be probed by a quantized photon energy

atomic spectral lines

Neutrons and protons can be considered as quanta building up a nucleus.

The location of atoms and molecules in a crystal can be considered as space quanta, because only specific sites can be accessed.

So the definition of a "quantum" as a "wave" is wrong. A particle may be a quantum if it builds up a specific unit, as for example the quarks can be considered as quanta which build up hadrons ( bricks to be crude).

2) Is there a unique definition of the word particle in physics not dependent on context (not casuistic) and that all physicists can agree upon?

Yes, a particle is an entity whose position can be defined by an (x,y,z,t) and its four momentum by ( p_x,p_y,p_z,E) as with macroscopic particles. That is why it is called a particle. BUT talking of elementary particles quantum mechanics reigns and the Heisenberg Uncertainty Principle holds.

In quantum mechanics, what can be measured as a particle in one experiment as here we see an antiproton hitting a proton at rest and generating a number of pions,

antiproton proton annihilation

where we can measure the four momentum of each particle in the picture and the track it left in space, can be a probability wave in another experiment.

For completeness here is the image used in a similar question:

two slit electron

This two slit experiment with an electron beam starts with individual electrons, (particles) and detectors are counting the location of their impact. In a) it looks random and it is only after enough statistics is accumulated that the interference pattern is observed, on the distribution in the x y plane of individual hits. A materialization of the probability distribution. In this particular experiment the slit through which the individual electron particle) passed could be identified, non destructively. The interference pattern is there. The boundary condition of two slits is what shows the probability wave nature of the particle.

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Very good answer, I am still reading it, but this picture of a quantum in the slide called "QUANTA" here physics.rutgers.edu/~strassler/QuestHiggs.pdf looks exactly like a wave. Do you say that this quantum is not a wave although it is drawn as a wave? –  Zeynel Dec 1 '12 at 15:10
    
The problem I see with this simplified lecture is that the crucial point that "the wave is a probability wave" is ignored. Any particle can be represented by a probability wave not a wave like an acoustic wave. It is not something that changes in the medium, as air when it vibrates. It is a mathematical function that can be computed by the boundary conditions which says how probable it is to find a particle hit at a specific ( x,y,z,t). NOT that the particle is distributed in space, as an acoustic wave is distributed in air. –  anna v Dec 1 '12 at 15:26
    
This is even more confusing. So waves that Strassler draws as "quanta" which are the pictures of "ripples in the Higgs field" are actually probability waves and not real ripples in a field? physics.rutgers.edu/~strassler/QuestHiggs.pdf –  Zeynel Dec 1 '12 at 15:56
    
Yes, they are probability waves that can be computed by the field theoretical models, i.e. Feynman diagrams. Not ripples in a field in space. The drawing is misleading because it makes one think that the particle is distributed as a wave in space. see the electron two slit experiment in my other answer that I linked above. Individual electrons go through it seems randomly, but the probability function appears when the statistics are large. –  anna v Dec 1 '12 at 16:01
    
In your picture about an antiproton hitting a proton, we have a picture of particles not probabilities. So, how are we supposed to picture the particles that leave these trails? Are they spherical objects? Are they waves? –  Zeynel Dec 2 '12 at 13:24

There are two definitions of particle in physics, both commonly used. I prefer the following terms:

  1. Asymptotic particle: The fundamental incoming S-matrix states.
  2. Field particle: an excitation of a fundamental field in a physical Lagrangian.

The two definitions are different in the case of composite particles, which are particles only in the sense of 1 (except for that they are also type 2 particles of effective theories) and for particles like quarks and gluons, which are only particles in the second sense but not the first.

The first definition is Wigner's, and it dominated high energy physics until 1974. Once quarks were recognized to be real and confined, the second definition took over. The two definitions are identical in free field theories, and they overlap in interacting field theories with nonconfined particles. But for strong interactions, the spectrum particles are hadrons, while the fundamental fields are quarks and gluons, they are completely different, and require different tools to analyze.

The spectrum particles in the strong interactions are described by S-matrix theory. They are approximately like strings in string theory. The fundamental particles are described by field theoretic perturbation theory. Much of physics of the last 30 years is devoted to reconciling the two different kinds of descriptions.

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In his article http://profmattstrassler.com/articles-and-posts/particle-physics-basics/fields-and-their-particles-with-math/5-waves-quantum/, Strassler tries to present an explanation of particles in quantum field theory (QFT) in the most elementary terms possible. The result is that he draws a picture and uses terminology that is not commonly used, though it gives some insight into what QFT is about.

So let me write a narrative in slightly more abstract terms, but in the standard notation and terminology, and then relate it to what Strassler writes.

Just like the waves of a single oscillator (think of a vibrating string) can be represented as a superposition of harmonic waves with given frequencies $\omega$ (Fourier analysis), so the waves of an extended medium (such as water) can be represented as a superposition of plane waves with given 4-momentum vectors $p$. The vectors $p$ label the different modes of the superposition $p$ is composed of $p_0=\pm E/c=\pm \omega \hbar/c$ where $E$ is the energy, $\omega$ the angular frequency (and $\hbar/c$ is a constant proportionality factor) and a 3-vector $\mathbb p$, the spatial momentum, telling the direction and speed of the motion of the wave. They are related by the so-called dispersion relation $p_0^2= \mathbb p^2+(mc)^2$, where $m$ is the mass of the (free) field. If there is just a single mode (rather than a superposition, which is the usual case) then an observer moving with the speed of the wave will see the 4-vector with zero spacial momentum (rest frame); so one can consider it to have just one component determined by the frequency.

Now a quantum field is not an ordinary field but an operator-valued field. You may think of an operator as a random variable, which takes different values - called eigenvalues - upon each measurement (if it can be measured) according to a probability distribution characteristic of the state of the field. (Though the measurement of fields is a complex subject, so this is quite simplified.)

Doing the decomposition into plane waves with a quantum field, each amplitude becomes an operator $a(p)$ or $a(\omega)$ itself, called a creation or annihilation operator depending on the sign of $p_0$; the creation operators have an additional * in the formula. These operators have arbitrary complex eigenvalues (just as in a classical field, the amplitudes are arbitrary comlex numbers). [Thus there is no smallest amplitude.]

However, due to the peculiarities of eigenvalues (which one can see already when the operators are just 2 by 2 matrices), the eigenvalues of a product are generally not equal to the products of the eigenvalues of the factors. In the case of a single quantum oscillator, it turns out that the product $a^*a$ has very few eigenvalues only, namely only the nonnegative integers. If you would measure $a^*a$ (which is not actually possible), you would therefore get one of the numbers $n=0,1,2,...$, counting the excitation level (or the ''number of quanta'', though this terminology is rarely used). One therefore calls $a^*a$ the number operator. Level 0 is the ground state, and higher and ligher levels have more and more energy.

Now we take all modes into account, allowing for the full field in place of just one plane wave mode. Then the $a$ depend on the 4-momentum $p$, and to get a number operator one must integrate the operators $a^*(p)a(p)$ over all possible 3-momenta. Then again, the eigenvalues of the number operator are nonnegative integers $n$, this time interpreted as the number of particles detected. The standard way of talking about this situation is by saying that, in QFT, ''a particle is an elementary excitation of a quantum field''.

Identifying an elementary excitations with a quantum, a single particle becomes a single field quantum. Moreover, if one simplify the noncommutative aspects of operators by treating the amplitude $a$ as a kind of square root of the number $n$, one might say that a quantum is ''a wave with minimal aplitude'', though this is quite misleading if you attach more than metaphorical meaning to it. This gives Strassler's imagery.

Note that in a general state of a quantum field, it is meaningless to talk about the number of particles it contains, just as it is meaningless to talk about the position of a quantum oscillator. What one can talk about is about the mean position of the oscillator and the mean number of particles, which is usually enough to draw macroscopic conclusion (in quantum statistical mechanics).

Particle detection events, on the other hand, concern very special (asymptotic) states, where interaction with recording equipment singles out a particular measured state with a definite observable particle content. It is this particle content that is represented by the traces in bubble chambers and photographs from CERN such as those mentioned in the answer by anna_v.

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A particle is the name for a quantum system for which $P_\mu P^\mu$ is a positive multiple of the identity (the mass squared), the energy is positive, and the angular momentum takes only certain discrete values (spin).

EDIT: I will add below a few extra comments because the original question was about thee relation of particles and waves.

A field $\phi(x)$ associated to a certain particle is an operator (or a collection of operators, one for every spacetime point) that sends the vacuum state $|0\rangle$ into a particle localized in space and time, $\phi(x)|0\rangle=|\phi(x)\rangle \approx |x\rangle$. It turns out that for consistency of the theory (which is a QFT) these operators $\phi(x)$ satisfy certain Lorentz invariant wave equations. This in particular mean that $|\phi(x)\rangle$ and the various amplitudes $\langle A|\phi(x)\rangle$ obey those wave equations.

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Then any quantum field would be a particle. Usually one wants an irreducible representation of the Poincare group, not just any. –  Arnold Neumaier Dec 2 '12 at 14:44
    
yes, I agree, one wants irreducible representations. I didn't add this information explicitly because I thought it was quite clear from my answer where I've mentioned definite mass and spin. –  argopulos Dec 2 '12 at 15:58
    
Where did you mention definite spin? Moreover, one can have multiple copies of rerpresentations with smae mass and same spin. –  Arnold Neumaier Dec 2 '12 at 16:22

Assuming that "particle" refers to elementary particles, then yes. An elementary particle is a quantum of energy.

Basically the confusion arises by people trying to simplify what is really a subtle distinction, and adding to that there has historically been some confusing terminology.

Note - Particle in context of my post will always refer to an elementary particle

No doubt you have heard of the uncertainty principle. What that says is that the position and momentum of a particle cannot be simultaneously known to some arbitrary precision. There is an uncertainty associated with the particle's position and momentum, which is fundamental in nature. It is inherently 'fuzzy' in some sense, with it's position not perfectly defined.

But how exactly can we say where a particle is in the first place? Or tell how fast it's moving? That is given by the particle's wavefunction. Now even though the name suggests that it is a wave, what the wavefunction actually describes is the probability that the particle is at a given location at a give time. For a 'free particle' (which is a particle with no external forces acting on it), the wavefunction is defined by a 'wavepacket'. This wikipedia page has a nice diagram explaining that.

The wavepacket has a reasonably well defined position and momentum at any given point, therefore we can pinpoint where it is quite well. This is what people refer to as particles - the quantized wavepacket, which is actually just a wavefunction that describes a probability. Therefore what all of the above is saying is that when you can predict reasonably accurately where something is, and how fast it's going, we can call that a particle. Which isn't a bad definition at all. :)

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Contrary to what Matt Strassler writes in his blog, there no waves in particle physics and this is why the discipline is named "particle physics" and not "wave physics".

Intuitively an elementary particle is the most basic piece of matter known. Each elementary particle is characterized by a set of physical properties: mass, charge, and spin. E.g. a photon is an elementary particle with zero mass, zero charge and spin $1$. Everything around us is made of lots of those elementary pieces. A good introduction to those themes is given in the CERN site:

http://public.web.cern.ch/public/en/science/Glossary-en.php#P

http://public.web.cern.ch/public/en/science/standardmodel-en.html

A discussion of why a particle is not a wave neither has the properties of a wave is given here

http://statintquant.net/siq/siqse3.html#x42-60003

Let us see the rest of questions.

No a quantum is not a wave. The term quantum usually refers to a small --of the order of Planck size-- quantity of a physical property. A quantum of energy is a small quantity of energy: e.g., the small difference of energy between two levels of energy in a hydrogen atom. E.g. a chronon is the hypothetical quantum of time.

Elementary particles are quantum particles and quantum particles are not spherical objects with finite radius. In fact technically elementary particles are considered pointlike objects

About your specific questions:

1) As stated "quantum" does not mean "a wave with minimum amplitude".

2) Yes, technically an elementary particle is defined as an irreducible element of the Poincaré group. In more laymen terms this means that any elementary particle is unambiguously identified and characterized by its mass, charge, and spin.

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well, you should add, depending on the particle, color charge, parity, R-parity etc, if we go into supersymmetry. –  anna v Dec 2 '12 at 13:50
    
@AnnaV: I restricted my answer to the known physics --The Standard Model-- and avoided speculative beliefs –  juanrga Dec 3 '12 at 11:27

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