The difference between the packet length and the cohererence length

[zrek]

Please help me to understand the difference/connection between the coherence length, packet length, wavelength and their effect on the interference. I'd like to understand these terms in case of one photon.

"The degree of coherence is measured by the interference visibility..."
http://en.wikipedia.org/wiki/Coherence_(physics)

"Wave interference is strong when the paths taken by all of the interfering waves differ by less than the coherence length"
http://en.wikipedia.org/wiki/Coherence_length

I also found that the different kinds of light sources can produce different quality of coherence, even with the same frequency. Here is a strong statement about the lengths:
"Interference is only visible if the coherence length of the light is at least as long as the path-length difference that creates the interference."
http://wiki.answers.com/Q/What_are_the_coherence_lengths_of_the_different_types_of_lasers

Until this point this is very clear for me.

Now let's switch to a photon representation, first I'll try to imagine the photon as a (probability) wave packet.
http://en.wikipedia.org/wiki/Wave_packet

The wave packet is just like the wave group, the speed of it is an "envelope" speed, not completely independent of the phase speed.
http://en.wikipedia.org/wiki/Group_velocity
(The speed of a particle is the superponated phase velocity of he waves that builds up the group http://doboandor.freeweb.hu/pdfs/fazis_es_csoportsebesseg.pdf )

Now let's take a single photon, which is a wave packet. I can see no more possibility than assume that the coherence length is the same as the group length (if there is a group velocity, then we have a group length too)

Now let's imagine a "normal" light source, emitting a single photon (with a specific energy that determines the wave length "L"), that going through a double slit and acting as expected and hitting the screen on the position that (in case of lots of similar photons later on) finally will produce an interference pattern.

I assume that the group length is depending only on the energy of the photon, which depends on the wavelength (L). The interference visibility depends on the coherence length, by this there will be far places on the screen (over the point X) that will not display interference pattern.

Now let's change our light source to a "high-quality" laser and repeat the experiment with the same wavelength (L). The article above said that the coherence length can be "greater than 100 km"! The point X will be changed...

What is the difference between the photon emitted by the "normal" light source or the "high-quality" laser?

How it is possible that the coherence length is independent from the wavelength in case of single photons? Maybe the photon have a "hidden, local" quantum number that is about the coherence length?

Thank you for your help.

 

[Jolb]

Please help me to understand the difference/connection between the coherence length, packet length, wavelength and their effect on the interference. I'd like to understand these terms in case of one photon.

These are all quite different concepts, and not all of them can be understood in terms of one photon. Most of what you said is valid up to here:

Now let's take a single photon, which is a wave packet. I can see no more possibility than assume that the coherence length is the same as the group length (if there is a group velocity, then we have a group length too)

I wouldn't necessarily say "a photon is a wavepacket." Roughly that is legitimate, but typically, if we say "there is one photon present," then we mean there is a single photon with a given wavelength/frequency/energy. Quantum mechanically, a photon with a given wavelength is NOT a wavepacket; it's a plane wave. To create a localized photon requires a Fourier sum of photons with many different energies. So a wavepacket is made of a superposition of different photons with various energies. So your discussion of a "normal" light source emitting a single photon is a bit mixed up.

Now let's change our light source to a "high-quality" laser and repeat the experiment with the same wavelength (L). The article above said that the coherence length can be "greater than 100 km"! The point X will be changed...

What is the difference between the photon emitted by the "normal" light source or the "high-quality" laser?

Lasers emit photons that occur over a very small range of frequencies (normal lights have a very broad blackbody spectrum, or they at least have several emission bands which are each very wide compared to a laser's) and they are emitted with a uniform polarization and direction of propagation.

How it is possible that the coherence length is independent from the wavelength in case of single photons? Maybe the photon have a "hidden, local" quantum number that is about the coherence length?

Coherence length depends crucially on the bandwidth over which the laser emits. A good laser has a very small band and thus has a very long coherence length. The problem is that, for real light sources, you never have just a single wavelength/energy; there is always a spread in them. The smaller the spread, the better the coherence, and the better fringes you can get in an interference pattern at long distances.

As an example, imagine we had a "perfect" laser which emits at precisely one frequency. Then its coherence length would be infinite and you could always get good fringes in a michelson interference experiment. However, this is impossible for a bunch of reasons: one very fundamental theoretical reason is that, due to the time-energy uncertainty principle, if we know what time interval the photon was emitted during, then there must be uncertainty in the fequency. So if we had the perfect laser, we would have had to keep it on for ALL of history, and we would never know when the photon actually gets emitted! So a real laser, since we know when we turn it on and off, has to have a nonzero frequency bandwidth, and hence a finite coherence length. [But in a real laser, the spread in wavelength is typically much wider and is due to doppler broadening since the gas atoms in the gain medium are flying around at thermal speeds.]

 

[zrek]

Thank you for the quick response.

At least now I understand that I not really understand the way when we switch back and forth between the wave and photon representation of the light.

But now (thanks for your explanation) I feel like I'm getting closer.

One last question (group).
Here I found this:
http://en.wikipedia.org/wiki/Wave_packet
"The wave packet is thus a mathematical solution to the Schrödinger equation"
And there is an animation, showing a moving wave packet.

Is it representing a single photon? Will the solved wave function look like that (on the animation) if we count with the Schrödinger? And most important question: I see that the wave packet on the picture have a specific frequency, is this the same as the frequency of the photon itself represents? Or at least is there a connection between the photon frequency and the frequency of the calculated wave function that is enclosed inside the wave packet?

Thank you!

 

[DrDu]

Is it representing a single photon? Will the solved wave function look like that (on the animation) if we count with the Schrödinger? And most important question: I see that the wave packet on the picture have a specific frequency, is this the same as the frequency of the photon itself represents?

A photon need not have a single frequency or be an infinite extended plane wave. You are right that a photon is in praxis a wavepacket with both a spread of wavelength and frequency (although these can be made very small).
There are two kinds of coherence effects important, namely spatial coherence and temporal coherence.
When you consider an interference pattern to be built up slowly by single photons hitting the screen one after the other, you have to take into account that the photons will have a statistical variation in frequency and wavelength. This explains the restricted visibility of the interference pattern.
So the coherence length is a statistical property of an ensemble of (many) photons.

 

[zrek]

Thank you DrDu!

I like your explanaiton but now I arrived to more serious questions, please try to make these clear for me.

"A photon need not have a single frequency", ok, but now how will be there a connection between the frequency and the energy of the photon? The only thing I can imagine if I say that the low energy photon can build low frequency if hits a material (absorbs any way), only a higher energy photon can create an effect that causes higher frequency "movement" or "wave" after the elimination.
Is this correct?

In this case we can say that there is no direct connection between the wave length of the light and the wavepacket length. Also, the frequency inside the wavepacket representation of the photon have no direct connection with the light frequency.
Is this still correct?

(Maybe the packet length is the same for every kind of photons?)

But in this case I can assume that even the very long length EM wave contains low energy photons, with small wavepacket lengths.
Is this possible? To build up a very long EM wave from particle-like, short packetlength photons?
OK, I can imagine that this is possible, why not.
But in this case it would be possible to detect a very long EM wave with a photon detector (hits the detector and gives a short time effect).
Theoretically is this possible?

Thank you!

 

[DrDu]

Thank you DrDu!

I like your explanaiton but now I arrived to more serious questions, please try to make these clear for me.

"A photon need not have a single frequency", ok, but now how will be there a connection between the frequency and the energy of the photon?

Energy and frequency are synonymous for photons (well, up to h). A wavepacket is then a superposition of energy eigenstates.

 

[zrek]

Energy and frequency are synonymous for photons (well, up to h). A wavepacket is then a superposition of energy eigenstates.

OK, but if it is a superposition of frequencies and represented as "wave"packet, then is this "wave" frequency do have a connection with the frequency of the light that the photon represents?
(I feel now that the connection between the light frequency and the (average or main) frequency inside the groupwave representation of the photon is only indirect. I'd like to make sure that I understand this correctly.)

 

[Jolb]

Thank you DrDu!

I like your explanaiton but now I arrived to more serious questions, please try to make these clear for me.

"A photon need not have a single frequency", ok, but now how will be there a connection between the frequency and the energy of the photon?

Sorry we didn't make this clear before. Given either a wavelength, a fequency, or an energy, you get the others for free according to the equation E=hv=hc/λ, where E is the energy, v is the frequency, λ is the wavelength, h is Planck's constant, and c is the speed of light.

Basic stuff like this is very important for understanding your question; also Fourier transformations will probably help you to figure this out.

In this case we can say that there is no direct connection between the wave length of the light and the wavepacket length. Also, the frequency inside the wavepacket representation of the photon have no direct connection with the light frequency.
Is this still correct?

No. If a wavepacket is absorbed by some object A, and after the absorption, A's energy increases by an energy E, that means there must have been some component to the wavepacket corresponding to the energy E=hv=hc/λ. But if it were a localized wavepacket rather than a plane wave, there must be other energies corresponding to other frequencies/wavelengths that go into the wavepacket.

(Maybe the packet length is the same for every kind of photons?)

A wavepacket can be constructed as long or as narrow as you want it, regardless of the average energy/frequency/wavelength of photons in the wavepacket. A more narrow wavepacket (one that is more localized in space) means you need a wider spread of energies/frequencies/wavelengths, and a more broad wavepacket means a smaller spread of wavelengths, and if we wanted only a single wavelength/frequency/energy photon, the wavepacket would be infinitely broad--it's a plane wave. The wider the wavepacket, the shorter the coherence length.

[From here on, I'll quit saying wavelength/frequency/energy; just remember given one you get all of them.]

But in this case I can assume that even the very long length EM wave contains low energy photons, with small wavepacket lengths.

I don't exactly know what you mean by a "long length EM wave." But if you are saying that the wave is made out of photons with given energies, then each energy photon is a plane wave. A single energy component is a plane wave, and packets have to be made of multiple energies.

OK, but if it is a superposition of frequencies and represented as "wave"packet, then is this "wave" frequency do have a connection with the frequency of the light that the photon represents?

The average frequency of photons in a wavepacket will be the average frequency you measure when you measure that wave. There will be other frequencies present in a wavepacket, too, and they will be close to the average frequency--in fact they follow a gaussian distribution around the average. The wider the gaussian, the bigger the spread in the possible energies of photons in the EM wave, and the shorter the coherence length

 

[zrek]

Very good, I think now I understand it, I'll try to make a conclusion.

"A wavepacket can be constructed as long or as narrow as you want it..."
"The wider the wavepacket, the shorter the coherence length"

It is our choice that how long wavepacket we calculating with. If we try to model a high-quality long-coherence-length laser, then we have to calculate with a long wavepacket, otherwise we can use shorter ones.
The photons however have no additional (quantum) information inside about this (the coherence length), it just happens that the photons will behave like a long packet if the source produces long coherence length light, and we don't have to care about it, because it is described in the time-energy uncertainty principle.

Is this correct?