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Light is an electromagnetic wave and it possess both properties of a particle and wave.

enter image description here

Looking at the image I can imagine how it behaves like a wave.
However, I'm not sure how it can behave like a particle.
If it behaves like a particle shouldn't it also look like one?

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That's a deep question. The idea of particle-wave duality is behind the theory of quantum mechanics. In quantum mechanics the state of a system, be it a free particle for example, is given in term of a wave-function $\psi$ given by Schrodinger's equation $$i\hbar\frac{\partial\psi}{\partial t} =\hat{H}\psi$$ Whenever you measure something on a state, it collapses on a given eigenvector and the result of that measurement is given in term of the eigenvalue corresponding to that eigenvector.

The simple example is the one of a free particle, be it a photon, in one dimension for which the hamiltonian is just $$\hat{H} = \frac{\hat{p}^2}{2m} $$ so that Schrodinger's equation reads

$$i\hbar\frac{\partial\psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x^2}$$

for which a solution is the following wavefunction

$$\psi(x,t) = e^{\frac{i}{\hbar}\left(px-\frac{p^2}{2m}t\right)} $$

The only measurable quantity for a free particle is it's momentum. Without going into the details, if you measure the momentum from this wavefunction you'll get $p$. So, by doing an experiment, what you'll measure is a particle with momentum $p$.

The theory for photons is actually more complicated than this since you absolutely need a relativistic formulation of quantum mechanics, but the idea behind it is just about the same.

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  • $\begingroup$ just to get the notation clear, is there a difference between $\hat H$ and $H$ for the Hamiltonian? $\endgroup$ – Alessio Popovic Mar 2 '20 at 11:16
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    $\begingroup$ @AlessioPopovic Yes, there is! Whenever you see a hat on some quantity it means operator. $\hat{H}$ is the hamiltonian operator while $H$ is the hamiltonian. Whenever you measure something, in quantum mechanics, you do it with some operator that acts on the state. The hamiltonian operator, for example, gives back the energy $$\hat{H}|\psi\rangle = E|\psi\rangle $$ $\endgroup$ – Davide Morgante Mar 2 '20 at 11:19
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In this picture it shows a wavelength $\lambda$, which is certainly a wavelike property. Fine.

It also has a frequency $f$ - not explicitly drawn, but easy to deduce from $\lambda$ and $c$.

It also has an energy $E$. You have to work a bit harder for that, calculating the Poynting vector $E \times H$, integrating over the two cycles in the picture, and making some assumption about the area, which isn't shown as that would make the picture too complicated but must be there in reality.

It also has a momentum which is just $E/c$. This is also known as radiation pressure.

Those are all wavelike properties, and there in the picture.

But the energy $E$ is not arbitrary. It's restricted to integer multiples of $h f$, where $h$ is Planck's constant $6.626\, 10^{-34}Js$. The energy comes in fixed size chunks (or quanta) - and chunkiness in momentum follows. These chunks are where the particle-like behaviour comes from.

So when the artist drew that pretty blue $E$ sine wave they were (in principle - if they had actually shown a labelled axis) restricted in the amplitude they could draw for it. That's where the particle-like nature of light appears in the picture.

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Electromagnetic waves are a fundamentally classical concept. For wave particle duality we need the domain of quantum mechanics. Consider this wikipedia definition of photon "The photon is a type of elementary particle. It is the quantum of the electromagnetic field including electromagnetic radiation such as light and radio waves..."

Modern quantum physics has not evolved very far and we still have a lot of concepts to clear but what I interpret from my study is that quantum particles do not behave like particle or wave, they behave in a radically different way which transcends our everyday behaviour of particles or waves so we try to understand them on basis of these behaviours.

Likewise Electromagnetic radiation is not wave or particle as per quantum mechanics but actually neither and you need both these concepts to understand it fully.

So how does EM wave behave like particles? It is a inherent result of its quantum nature.

What you have shown above is the wave picture of EM wave but it also has a particle picture which is equally valid.

Sometimes it behaves like particles(photoelectric effect and such) sometimes like waves(interference, diffraction ,etc) sometimes any will do(reflection,refraction) but in reality it is neither.

What it is a "quantum particle" and cannot fit perfectly in either pictures so you have to use both of them.

Wikipedia also seems to suggest "In quantum mechanics, an alternate way of viewing EMR is that it consists of photons, uncharged elementary particles with zero rest mass which are the quanta of the electromagnetic force, responsible for all electromagnetic interactions". (note: "an alternate way" not only way)

Hope this should clear this rather philosophical question!

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    $\begingroup$ An electromagnetic field is a harmonic oscillator. Quantizing it requires that it exist only at discrete, evenly-spaced energy levels. These energy levels are often thought of as different numbers of photons. A photon is "like" a particle in some ways but it is not a particle as such. $\endgroup$ – Guy Inchbald Mar 2 '20 at 17:24
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For photons in vacuum there is a simple explanation.

As an example we take Young's double slit experiment: Light is emitted at A and absorbed at B, between A and B we find interfering waves, and it seems to be not clear how particle characteristics are transmitted.

But for photons in vacuum the solution is simple to see: The spacetime interval between A and B is zero, that means that A and B are adjacent in spacetime. That means that there is no room for the travel of a particle, instead the particle characteristics are transmitted directly from a particle at A to a particle at B, for example from the emitting electron at A to the absorbing electron at B.

This explanation is working only for massless particles with lightlike propagation. For mass particles, quantum physics is required.

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  • $\begingroup$ Could you clarify what you mean by "The space time interval between A and B is zero"? $\endgroup$ – Alessio Popovic Mar 2 '20 at 12:53
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    $\begingroup$ @Alessio Popovic, The spacetime of general and special relativity is Lorentzian, that means that there are timelike, spacelike and lightlike intervals, the square of the intervals is $ds^2 = cdt^2 - dx^2 - dy^2 - dz^2$ - For lightlike intervals this equation gives zero. Photons in vacuum are lightlike. $\endgroup$ – Moonraker Mar 2 '20 at 13:12

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