Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

This is a summary from Physics World of the paper: L J Wang et al. 2000 Nature 406 277-- "Wang and colleagues begin by using a third continuous-wave laser to confirm that there are two peaks in the gain spectrum and that the refractive index does indeed change rapidly with wavelength in between. Next they send a 3.7-microsecond long laser pulse into the caesium cell, which is 6 centimetres long, and show that, at the correct wavelength, it emerges from the cell 62 nanoseconds sooner than would be expected if it had travelled at the speed of light. 62 nanoseconds might not sound like much, but since it should only take 0.2 nanoseconds for the pulse to pass through the cell, this means that the pulse has been travelling at 310 times the speed of light. Moreover, unlike previous superluminal experiments, the input and output pulse shapes are essentially the same."

I realise that the velocity of light in a medium is comprised of the phase velocity, the group velocity, and the front velocity. While the group velocity can exceed the value of c in a vacuum, the front velocity is not supposed to. The way this sounds, the front velocity is exceeding c in a vacuum by 310 times.

Frontgroupphase.gif‎

blue dot=phase velocity, green=group, red=front (wiki)

The wavepacket seems to exit the cell well before it enters, but the negative refractive index "forward shifts" the leading edge of the pulse. It is argued that although this is superluminal, information cannot be transmitted faster than c. Gauthier and Stenner introduced a jump discontinuity into the waveform.Its max speed was c Here is a popular account from NewScientist . What if you used one photon? What if the photon itself were the information and not a carrier?

share|improve this question
2  
What did you use to create the wave animation? –  John McVirgo Feb 17 '11 at 14:30

2 Answers 2

You couldn't use only one photon, because the experiment depends on the light being in a pulse. A pulse is a sum of light waves with a range of different frequencies, which all experience different indices of refraction (i.e. phase velocities) when they travel through the medium. This is called dispersion. It is the precise combination of the frequencies in the pulse, and the dispersion and group-velocity dispersion in the caesium cell, that make this experiment possible.

In the same issue as Wang's paper, Nature printed a "News and Views" article about it. There, they say:

They use smooth, well-defined light pulses, so that the peak of the pulse at the output results from the forward rising edge of the input pulse, which occurs far earlier in time, making it consistent with causality. An abrupt feature in the light pulse would not be able to travel faster than c. This means that even if the 'effect' appears to precede the 'cause', you still can't send useful information — such as news of an impending accident — faster than c.

share|improve this answer
    
This doesn't make sense. One photon is not an "abrupt feature". It's just one quantum excitation of an oscillator mode - or quantum superposition of many oscillator modes - of the electromagnetic field. For any field configuration you can have, such as a laser pulse of arbitrary shape, you can have a single photon with that "wavefunction" too. (I only say "wavefunction" in scare quotes because there are problems with treating a photon as a classical particle with a wavefunction.) It sounds like you're thinking of a photon as a sharp spike in the field, which is just incorrect. –  Keenan Pepper Feb 17 '11 at 13:44
    
Ok, I know that a photon is not a sharp spike in the field, and I have removed the part about one photon being an "abrupt feature". However, if you have a single photon detector at the other end of the cesium cell, you certainly won't get a click before sending the photon into the cell. i.e. I still maintain that the experiment depends on the light being in a pulse. –  ptomato Feb 17 '11 at 14:07
    
I don't understand why you say that you can have a single photon with any possible features of a laser pulse of arbitrary shape. Can you explain that in more detail? Or, if I'm completely on the wrong track, then answer the question so that I can learn too? –  ptomato Feb 17 '11 at 14:10
    
Well, as I hastily explain in the last part of my answer, you actually can get a click before you might think you'd be able to, but the problem is you'll also get a lot of clicks from the unavoidable quantum noise, so statistically you don't get any information faster than $c$. –  Keenan Pepper Feb 17 '11 at 14:32
    
A photon in a definite-momentum state is a state with a single excitation of a plane-wave mode of the EM field: $a^\dagger(k)|0>$. But you can also have a superposition of these states, for example $(a^\dagger(k_1)|0> + a^\dagger(k_2)|0>)/\sqrt{2}$. This is an eigenstate of the photon number operator with eigenvalue 1, so there's definitely one photon there, but its momentum is uncertain. Similarly, you can have $\int d^3k f(\mathbf{k}) a^\dagger(\mathbf{k}) |0>$, where $f$ is some function whose square integrates to one. This is a single photon with $f$ as its "wavefunction". –  Keenan Pepper Feb 17 '11 at 14:38

The front velocity, defined as the propagation speed of the point where the field first differs (by any arbitrarily small amount) from exactly zero, is always no greater than $c$. (In fact, the front velocity is always exactly equal to $c$, no greater or less.)

The problem here is that a Gaussian pulse extends infinitely in both directions, so it simply does not have a "front" to speak of. Of course the amplitude decays super-exponentially on both sides, but that doesn't matter. There is no causality problem with the pulse emerging arbitrarily early in time, because your input pulse made the field start changing long before that.

As a thought experiment, let's imagine we have a button we can push to start the input pulse going. If any trace of the output pulse comes out before a signal at $c$ has a chance to propagate from when and where the button was pushed, that's a causality violation. But that is impossible, for the following reason:

Since a theoretically perfect Gaussian pulse has no finite start time, but has a nonzero amplitude at arbitrarily early times, it's impossible to create such a perfect Gaussian pulse by pushing a button. Of course, you can get arbitrarily close to perfection, but there will always be some distortion that gets worse and worse as you try to make the pulse fire "faster", that is, try to make a shorter separation in time between the button push and the maximum of the pulse. The theorem that the medium has a causal response to the field guarantees that the response to this distortion will always interfere with the response to the perfect Gaussian to cancel out exactly at times earlier than a signal traveling at $c$.

Of course, in a real experiment, the "button push" (actually some electronic signal to the machine that creates the pulse) happens a relatively long time before any trace of the output pulse is detected.

Here's a great book chapter I just found about this: http://books.google.com/books?id=kE8OUCvt7ecC&pg=PA26 It has lots more math than my answer (though not too high a level) and might clear up a lot of things.

You also ask about single photons, but that opens up a huge can of worms I can't really get into (not least because I don't understand it well enough myself). Let me just say that there is always a minimum amount of noise in any mode / degree of freedom of the electromagnetic field, which is equivalent to half a photon. You could make a pulse of the right shape that's so weak there's only a single photon in it (I'm sure people do things like that all the time), but the problem is if you try to "announce" too early that you've detected that photon, you'll be wrong such a large fraction of the time that you can show statistically that no information is being transmitted. It's really difficult to do quantum-limited measurement in the first place (because you have to severely limit the back-reaction of the measurement apparatus on the field you're measuring), and if you try to do it too fast it becomes literally impossible.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.