Particles with small Wavelengths passing through huge slits?

[InfiniteMonkey]

I´´m confused. How can a single photon in the lightspectrum with wavelength of a few hundert nanometers go through both slits in the double slit experiment at the same time. I understand the wave- particle duality and the concepts in principle. My confusion is in the context of little wavelength spreading out on a bigger scale, because in my imagination the wavelength should determine the area in which the photon spreads.

 

[vanhees71]

You are confused, because you think of photons as if they were little (massless) localized particles, but they are far from that. If you want to think about photons in a classical way, it's much closer to the true meaning of what a photon is, to think in terms of electromagnetic waves. The wavelength has nothing to do with the spatial extension of the wave. It's just telling the spatial periodicity of the wave. The extreme case is a plane wave, which is extended over the entire space but has a sharp wavelength.

Now if you have a true single-photon state with pretty sharp momentum (i.e., pretty sharp frequency and wavelength) that means, heuristically, you have an electromagnetic wave is pretty much extended in space and thus goes through both slits. Because it's a single photon, on the other hand, it can be either registered by a detector (e.g., a photoplate or CCD cam) as a whole or it's not detected at all. The reason is, and that's what makes it in a way "particle like", the interaction of em. radiation with a frequency $\omega$ with matter is always in integer multiples of the "energy quantum" $E_{\omega}=\hbar \omega$. By definition a single-photon state is an energy eigenstate of the electromagnetic field of frequency $\omega$ with energy eigenvalue $E_{\omega}$. So if the photon interacts with matter making up the detector it can be either completely absorbed and registered at the place where the interaction took place or it just get's at most a bit scattered by the interaction but not absorbed and not registered. There's no way to detect any portion of the photon only.

That's why you need the probability interpretation of the quantum formalism: On the one hand you have the wave picture, which describes continuous phenomena on the other hand if you deal with single-photon states the photon can only be registered as a whole or not registered at all. So all you get is a single point on your detector screen (photoplate or CCD cam). So the observable feature of the quantum state is that it predicts the probability to detect a photon at the place where the detector is located.

 

[InfiniteMonkey]

Okay I understand that, but then what tells us the wavelength at all what the frequency doesn´t already tells us, and how do we know if the wavelength is a "real" thing when it is completely independend from the spreading of the wave?

 

[vanhees71]

An electromagnetic wave can be described by a superposition of plane waves, of the form $\vec{E}(t,\vec{x})=\vec{E}_0 \exp(-\mathrm{i} \omega t+\mathrm{i} \vec{k} \cdot \vec{x})$. The frequency $\omega$ tells you that the temporal period is $T=2 \pi/\omega$ and $\vec{k}$ tells you the direction the wave propagates in as well as the spatial period $\lambda=2 \pi/|\vec{k}|$. You also must fulfill the wave equation, which leads to the relation between frequency and wave number, $\omega = c |\vec{k}|=2 \pi c/\lambda$.

 

[EPR]

Google Uncertainty principle

 

[DrChinese]

Okay I understand that, but then what tells us the wavelength at all what the frequency doesn´t already tells us, and how do we know if the wavelength is a "real" thing when it is completely independend from the spreading of the wave?

Most light (the quantum particles are photons) spreads in a generally spherical fashion from a source (which need not be a point), and at the speed of light in the medium. That spread being the probability of detection/interaction. Although wavelength is a factor in the spread, it's not the main factor.

The "probability wave" may self interact (overlap) in regions such that the probability of detection/interaction is increased or decreased. That is what causes interference patterns to form, because the likelihood of detection is not uniform in all regions.

As a photon is a quantum particle, you cannot really speak of it as going through one slit or the other in the usual fashion. Better to think of it as a superposition upon detection. I.e. of having an X% chance of having gone through the left slit and Y% chance through the right slit.

 

[InfiniteMonkey]

Most light (the quantum particles are photons) spreads in a generally spherical fashion from a source (which need not be a point), and at the speed of light in the medium. That spread being the probability of detection/interaction. Although wavelength is a factor in the spread, it's not the main factor.

The "probability wave" may self interact (overlap) in regions such that the probability of detection/interaction is increased or decreased. That is what causes interference patterns to form, because the likelihood of detection is not uniform in all regions.

As a photon is a quantum particle, you cannot really speak of it as going through one slit or the other in the usual fashion. Better to think of it as a superposition upon detection. I.e. of having an X% chance of having gone through the left slit and Y% chance through the right slit.

 

[InfiniteMonkey]

Okay, as I understand so far the wavelength (in a classical sense) influences the probability distribution of the wavefunction but is not the only factor of the spread of the wavefunction. What if an electron changes energy state and emits one photon. How does this photon spreads out? Spherical or with a preferred direction?

 

[vanhees71]

It depends on the kind of transition. Usually you have dipole transitions. Then the photon probability looks like the usual pictures of electric dipole radiation in classical electrodynamics:

https://en.wikipedia.org/wiki/Dipole#Dipole_radiation

The "modulus of the Poynting vector" as tells the label in the nicely animated picture, for a single photon depicts the probability distribution to find a photon with a detector that is located at any place.