Particle or Wave? Duality Explained

In summary, quantum objects, such as electrons, can exhibit properties of both particles and waves, depending on how they are measured. However, this is not a dualistic concept and the wave-particle duality is outdated. In modern quantum mechanics, the wave function represents the probability distribution for the position of a particle at a given time. This eliminates the need for concepts like "quantum weirdness" and "wave-particle duality." Additionally, the wave function can take on different values depending on the type of particle it represents.
  • #1
rahaverhma
69
1
How anything can be a particle or wave at the same time.
 
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  • #2
Quantum scale objects (electrons say), can be one or the other depending on how you measure them.
I think it's best to just call them quantum objects.
 
  • #3
Yeah but at a particular time object or a Quantum object behave like a particle or wave but not simultaneously both
 
  • #5
Feymann has a good view on this.They are acting just like "Quantum Mechanical Way".This a problem which turning physics to english as he claimed.I agree with Feymann
 
  • #6
rahaverhma said:
How anything can be a particle or wave at the same time.

Wave-particle duality is an outdated concept, so don't bother too much with that.
 
  • #7
I just want to know that waves we talk about of particles, are they electromagnetic waves ??
,and when someone is talking about wavelike property of macroscopical objects are they talking about our oscillation up and down ie. Our wave function otherwise how can someone think of us as a wave.
 
  • #8
rahaverhma said:
I just want to know that waves we talk about of particles, are they electromagnetic waves ??
No, the wave we're talking about are not electromagnetic waves.
How anything can be a particle or wave at the same time.
It can't. A quantum object is neither a particle nor a wave, although it has some properties that we usually associate with waves and some properties that we usually associate with classical particles.
 
  • #9
Which waves are these?
 
  • #11
I mean what type of wave are they actually are?
 
  • #12
As several posters have noted before, there is no wave-particle dualism in "modern quantum mechanics" (the today still valid version of QM has been developed independently in 1925/26 by Heisenberg+Jordan+Born, Dirac, and Schrödinger).

In non-relativistic quantum mechanics one associates a wave function with the "state" of a single particle like an electron, ##\psi(t,\vec{x})##. It takes values in ##\mathbb{C}## (for scalar particles, spin ##s=0##) or ##\mathbb{C}^{2s+1}## (for particles with spin ##s \in \{1/2,1,\ldots \}##). The meaning of the wave function is that
$$P(t,\vec{x})=|\psi(t,\vec{x})|^2$$
is the probability distribution for the position ##\vec{x}## of the particle at time ##t## (Born's Rule).

All the "quantum weirdness" like "wave-particle duality" and similar ideas of "old quantum theory" vanish, as soon as you accept this probabilistic meaning of the wave function (or more generally any kind of quantum state).
 
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  • #13
... "probability waves" ...
 
  • #14
As I understand, in modern quantum mechanics, the wave-particle duality was "outdated." There is no duality, there are only quantum objects, or better to say: vectors in the Hilbert space of infinite dimensions. What we perceive as a "wave" is only a "name" just to give a name to something, that "in essence" is a mathematical function, like sin and cos.
And what is sin and cos? Trigonometric functions ...
 
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  • #15
Karolus said:
What we perceive as a "wave" is only a "name" just to give a name to something, that "in essence" is a mathematical function, like sin and cos.
And what is sin and cos? Trigonometric functions ...
Partially true, but I think there is more to it than that, so I also partially disagree, but the full analysis would have to go on a separate thread. Roughly and briefly three points here:
1. Structure of wave as (amplitude function)×f(x-vt, y, z) [f being sin or cos etc. type function, or combination etc.], or F(x-vt, y, z) in general etc., or includes different types of modulation ...
2. Thus not limited only to sin, cos ... , unless perform Fourier alalysis (expansion).
3. Besides waves there is also wave-packets, very useful and common even in QM, and these are not sin, cos only (nothing but ... , sometimes, unless Fourier ...).

Thus it's a big issue.
 
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  • #16
vanhees71 said:
As several posters have noted before, there is no wave-particle dualism in "modern quantum mechanics" (the today still valid version of QM has been developed independently in 1925/26 by Heisenberg+Jordan+Born, Dirac, and Schrödinger).

In non-relativistic quantum mechanics one associates a wave function with the "state" of a single particle like an electron, ##\psi(t,\vec{x})##. It takes values in ##\mathbb{C}## (for scalar particles, spin ##s=0##) or ##\mathbb{C}^{2s+1}## (for particles with spin ##s \in \{1/2,1,\ldots \}##). The meaning of the wave function is that
$$P(t,\vec{x})=|\psi(t,\vec{x})|^2$$
is the probability distribution for the position ##\vec{x}## of the particle at time ##t## (Born's Rule).

All the "quantum weirdness" like "wave-particle duality" and similar ideas of "old quantum theory" vanish, as soon as you accept this probabilistic meaning of the wave function (or more generally any kind of quantum state).

Any reason for separating the scalar particles out of the ##\mathbb{C}## formula, please? Is this OK:

In non-relativistic quantum mechanics one associates a wave function with the "state" of a single particle like an electron, ##\psi(t,\vec{x})##. It takes values in ##\mathbb{C}^{2s+1}## (for particles with spin ##s \in \{0,1/2,1,\ldots \}##?

Thanks.
 
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  • #17
I remember there were three kinds of interpretations of a measurement on a particle.

1. realist position, which implies that the particle was there before measurement. Because you didn't do measurement, you did not know it was there. This implies a loophole in quantum mechanics that there is hidden variable to determine position of the particle. This is what Einstein believed in.

2. orthodox position, the particle is not anywhere, but observation created a form of measurement (forced wavefunction of the particle to collapse into one determinate state) The method of observation deeply affects the form of measurement you get.

3. agnostic position, refuse to answer such question because it is nutty to ask for position of a particle before measurement.

Bell's inequality proved that the statistic of realist position and orthodox position are different. This eliminated agnostic position. And the result is that orthodox position is correct (This is why Einstein was wrong).

Now back to your question. If you measure position of a particle, you get position of a particle, which is a particle. Now if you measure momentum of the particle, which is related to wavelength by de broglie formula, you get wavelength, which is a wave. Because position and momentum do not commute each other, you cannot get both information accurately. When position wavefunction collapse, you get measurement of a particle, but the momentum measurement is lost. Vice versa, this shows that a free particle has position and momentum, which is particle and wave property. Measurement method will change their property, which is different from classical mechanics. In classical mechanics, properties are static, particle is particle, wave is wave.
 
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  • #18
sunmaggot said:
I remember there were three kinds of interpretations of a measurement on a particle

These descriptions seem oversimplified to me. They also seem to assume a collapse interpretation of QM; there are also no collapse interpretations.
 
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  • #19
PeterDonis said:
These descriptions seem oversimplified to me. They also seem to assume a collapse interpretation of QM; there are also no collapse interpretations.

maybe I used wrong words. Three opinions maybe better?
 
  • #20
sunmaggot said:
I remember there were three kinds of interpretations of a measurement on a particle.

1. realist position, which implies that the particle was there before measurement. Because you didn't do measurement, you did not know it was there. This implies a loophole in quantum mechanics that there is hidden variable to determine position of the particle. This is what Einstein believed in.

2. orthodox position, the particle is not anywhere, but observation created a form of measurement (forced wavefunction of the particle to collapse into one determinate state) The method of observation deeply affects the form of measurement you get.

3. agnostic position, refuse to answer such question because it is nutty to ask for position of a particle before measurement.

Bell's inequality proved that the statistic of realist position and orthodox position are different. This eliminated agnostic position. And the result is that orthodox position is correct (This is why Einstein was wrong).

Now back to your question. If you measure position of a particle, you get position of a particle, which is a particle. Now if you measure momentum of the particle, which is related to wavelength by de broglie formula, you get wavelength, which is a wave. Because position and momentum do not commute each other, you cannot get both information accurately. When position wavefunction collapse, you get measurement of a particle, but the momentum measurement is lost. Vice versa, this shows that a free particle has position and momentum, which is particle and wave property. Measurement method will change their property, which is different from classical mechanics. In classical mechanics, properties are static, particle is particle, wave is wave.

I agree, then the "wave particle duality", or whatever you want to call it, is not outdated
 
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  • #21
sunmaggot said:
maybe I used wrong words. Three opinions maybe better?
But, as PeterDonis implies, there are more ...
[Three are the most basic and common ones]
 
  • #22
Stavros Kiri said:
But, as PeterDonis implies, there are more ...
[Three are the most basic and common ones]
these three are from griffith's book, so maybe there are more.
 
  • #23
We call photons quantum objects, that can behave like a wave or like a particle. How can an object behave like a wave? An object is supposed to have its own identity. Can it keep its identity when it behaves like a wave?
 
  • #24
calinvass said:
. How can an object behave like a wave? An object is supposed to have its own identity.

You are using an inadequate definition of "object". Or perhaps "object" is simply an inadequate term to use to describe quantum entities like a photon. "Identity" in the sense you are using that concept simply doesn't apply to such entities.
 
  • #25
N88 said:
Any reason for separating the scalar particles out of the ##\mathbb{C}## formula, please? Is this OK:

In non-relativistic quantum mechanics one associates a wave function with the "state" of a single particle like an electron, ##\psi(t,\vec{x})##. It takes values in ##\mathbb{C}^{2s+1}## (for particles with spin ##s \in \{0,1/2,1,\ldots \}##?

Thanks.
That's of course true.
 
  • #26
calinvass said:
We call photons quantum objects, that can behave like a wave or like a particle. How can an object behave like a wave? An object is supposed to have its own identity. Can it keep its identity when it behaves like a wave?
Again: Wave-particle dualism is an outdated concept for nearly 100 years now, and if there is anything that is no particle, then it's a photon. It is not even possible to define a position observable for it!
 
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  • #27
vanhees71 said:
Again: Wave-particle dualism is an outdated concept for nearly 100 years now, and if there is anything that is no particle, then it's a photon. It is not even possible to define a position observable for it!
Then that goes into "particles &_fields (or field particles)", which is a different but relevant kind of dualism question ...

Also
vanhees71 said:
... photon. It is not even possible to define a position observable for it!

that's because in the 2nd quantization the wave function becomes a field operator and gets quantized (into states, e.g. photon states) ...
 
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  • #28
calinvass said:
We call photons quantum objects, that can behave like a wave or like a particle. How can an object behave like a wave? An object is supposed to have its own identity. Can it keep its identity when it behaves like a wave?
Photons are field particles, and fields can be waves (e.g. Electomagnetic field and EM waves). Furthermore, [in QED] photons (or photon states) are the results of the 2nd Quantization (of the EM Lagrangian). [See also previous post, but please maintain separate.]

Added note: other field particles (or gauge bosons) in the Standard Model are the 8 gluons (mediators or quanta of the Strong interaction), the two W particles (W+, W-) and the one Z particle (Z0), i.e. the three Electroweek mediators of the Weindberg-Salam theory, [and, of course, the photon, ... makes a total of 12 field particles]. Then of course there is the Higgs boson(s), that give rise to mass.
Beyond the standard model, the graviton, hasn't been discovered yet.
[Field particles (or gauge bosons) are usually massless, unless they acquire mass with the Higgs mechanism (e.g. Spontaneous Symmetry Breaking) ...]
 
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  • #29
weirdoguy said:
Wave-particle duality is an outdated concept, so don't bother too much with that.
Its not a concept it's a proven theory and it's not outdated we still use it.
 
  • #30
I've not seen a single empirical evidence for wave-particle duality. So how can you say, it's a "proven theory"? Since 1925, fortunately we do not need to use such shaky and self-contradictory concepts anymore!
 
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  • #31
But using photons (or other field particles) as an example to resolve the wave-particle issue is not a good idea, because these are (or can be considered as) virtual particles (see https://en.m.wikipedia.org/wiki/Virtual_particle)

It would be better if one considered as an example the Schrodinger equation and 1st quantization, to resolve the issue ...
 
  • #32
I don't know why people keep saying wave-particle duality is outdated. The energy and momentum of a particle are associated with the frequency and wavelength of a wave. You have to introduce wave concepts like frequency into the description of particles.

The equation E = hf and the corresponding one for momentum are the foundation of quantum theory, as fundamental as t2 - x2 = s2 is to relativity. They will never become outdated.
 
  • #33
The fact that the vision particle wave, as has been definitively surpassed in 1925, (why?), In 1927 de Broglie introduced the concept of "wave and particle", as illustrated by John Bell in his "speakable and unspeakable in quantum mechanics" , where in chapter 20 "six possible worlds in quantum mechanics", discusses the problem of the various interpretations of quantum mechanics, including the wave-particle. The book is about 1960, and here, what need had John Bell to dedicate a book if, since 1925 these problems were already solved?
 
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  • #34
Ddddx said:
I don't know why people keep saying wave-particle duality is outdated.

One of the reasons might be that they are experts in the field and know what they are talking about!

For example, if you look at one of the most popular undergraduate textbooks on QM by Griffiths, the wave-particle duality gets one mention on page 420 as a "historical footnote".
 
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  • #35
I don't know what Griffiths wants to attack in that footnote, but he doesn't seem to be questioning the validity of the fundamental equations like E = hf.
 

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