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On one hand, the amount of information I can transmit is proportional to the bandwidth. The higher the frequency, the more information I can transmit. On the other hand, the number of photons is reverse proportional to the frequency. I cannot possibly transmit more information than the number of photons I send. Therefore, it appears, that at low intensity levels, a higher frequency signal may contain less information than a lower frequency signal.

For example, consider a camera sensor in a high amplification mode (known in digital photography as "high ISO"). Provided the light intensity is uniform by color, blue sensor pixels would receive fewer photons than red pixels. The photon noise in modern sensors is one of the main quality limitations. Thus, in low light conditions, blue images would be grainier than red images meaning that the amount of the transmitted information is reverse proportional to the frequency.

Considering these two competing trends, an optimum must exist. Is there a known formula or estimate for the optimal frequency to transmit the highest amount of information for a given received power? Or, stating this in reverse, is there a formula for the minimum received power required to avoid the photon quantization noise at a given frequency?

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  • $\begingroup$ "I cannot transmit more information than the number of photons" - why not? Photons have multiple attributes that could be used to encode information, polarisation (H/V/C), frequency, detection time, and so on. In other scenarios 1bit of information requires many photons. I think the answer to this question depends on how information is encoded. $\endgroup$ – JMLCarter Sep 27 '17 at 23:31
  • $\begingroup$ @JMLCarter Absolutely! However, the amount of information in such conditions would still be proportional to the number of photons, even if you count all their attributes. So I'm just interested if there indeed is an optimum or some other effects prohibit it. For example, low frequencies have many photons, but cannot transmit much information. So just the number of photons doesn't mean much alone. $\endgroup$ – safesphere Sep 27 '17 at 23:50
  • $\begingroup$ One of the key ideas here is temperature. If the temperature of the detector is such that $kT > h f$ where $f$ is the photon frequency, then the detector won't work too well because it will be thermally excited. $\endgroup$ – DanielSank Sep 28 '17 at 0:09
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    $\begingroup$ Theoretically, you could have four times as many blue photon sensors on a plate than red photon sensors, because blue photons have half the wavelength. So if you're communicating by aiming photons at sensors, rather than trying to detect images with low levels of light, blue photons have compensating advantages. $\endgroup$ – Peter Shor Sep 28 '17 at 2:58
  • $\begingroup$ @PeterShor Good point. I did think that the wavelength was a factor. For example, I can have an enormous number of photons at a low radio frequency, but detecting them would require tools of a large physical scale. $\endgroup$ – safesphere Sep 28 '17 at 3:46
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The optimal distribution of photon frequencies for sending messages, assuming no noise but quantum shot noise, is indistinguishable from thermal (blackbody) radiation at a given temperature. So find the temperature for thermal radiation corresponding to your desired power, find its entropy, convert that to bits, and you have the theoretical maximum amount of information for a given power.

Why is this true? I'll give a brief sketch of a proof. The Holevo formula for the quantum information that can be sent over a quantum channel ${\cal N}$ at a given power is

$$ \max_{\{ p_i, |\psi_i\rangle \langle \psi_i | \}} S\left({\cal N}\left({\sum_i p_i |\psi_i\rangle \langle \psi_i|}\right)\right) - \sum_i p_i S\left(\cal N( | \psi_i \rangle \langle \psi_i | )\right), $$ where the maximization is over all probability distributions of input states to the channel with the desired power constraint, and $S$ is entropy. In words, this is the entropy of the average output less the average entropy of the output.

If the channel is noiseless, then the second term on the right-hand-side is 0, and you just need to maximize the entropy of the average channel output. This maximization is the same as the maximization for determining a thermal state of the channel output given a fixed power.

Of course when you try to do this, you may run into problems, like discovering that the spatial width of your channel (which you haven't specified) makes a difference. I suspect that you would need to specify your problem quite a bit more before you could get a definite numerical answer.

If you're trying to send a signal through an environment with noise, the theorem that the optimal distribution of photon frequencies is a thermal state no longer holds, and things get much more complicated. But I think that for your question, you wanted the assumption of no noise.

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  • $\begingroup$ Interesting, bu not clear to me. Can you please elaborate? "indistinguishable from thermal radiation" - why? "find the temperature ... corresponding to ... power" - how? The second part is understood - I'm not looking for a numeric answer, just a concept. $\endgroup$ – safesphere Sep 28 '17 at 6:07
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    $\begingroup$ "Find the temperature ... corresponding to ... power". If you're sending the signal using electromagnetic radiation (say through a wave guide, which is just a long pipe), there are formulas for how much power is in a thermal state of electromagnetic radiation at a given temperature, and for how much entropy is in a thermal state at that temperature. Look these up and use them. $\endgroup$ – Peter Shor Sep 28 '17 at 10:36
  • $\begingroup$ Thanks Peter! I see now what you mean. So there indeed is an optimal frequency (or frequency distribution) for given power in a given channel, as I expected. So, if I use a narrow frequency band (like radio stations do) instead of the wide spectrum of the black body, would you agree that the optimal frequency would match (at least approximately) the peak frequency of the black body spectrum? $\endgroup$ – safesphere Sep 28 '17 at 18:43
  • $\begingroup$ If you use a narrow frequency band, then the classical information capacity is $C=B \log_2(1+S/N)$. For quantum noiseless channels at low frequencies, the quantum information capacity will be almost exactly the same, with the noise being quantum shot noise. If you reduce the frequency and keep the same power, the shot noise gets smaller, and the capacity approaches infinity as the frequency approaches $0.$ $\endgroup$ – Peter Shor Oct 1 '17 at 16:51
  • $\begingroup$ For intuition about why this happens ... the classical continuous noiseless channel has infinite capacity. (You put in a power level with an infinite number of digits, and you get exactly the same power level out.) And as you make the frequency smaller while keeping the power fixed, the quantum channel behaves more and more like the classical continuous channel. $\endgroup$ – Peter Shor Oct 1 '17 at 16:57

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