This is a follow-up question to @heather's answer to the question : Why must quantum computers be kept near absolute zero?

What I know:

  • Superconducting quantum computing: It is an implementation of a quantum computer in a superconducting electronic circuit.

  • Optical quantum computing: It uses photons as information carriers, and linear optical elements to process quantum information, and uses photon detectors and quantum memories to detect and store quantum information.

Next, this is what Wikipedia goes on to say about superconducting quantum computing:

Classical computation models rely on physical implementations consistent with the laws of classical mechanics. It is known, however, that the classical description is only accurate for specific cases, while the more general description of nature is given by the quantum mechanics. Quantum computation studies the application of quantum phenomena, that are beyond the scope of classical approximation, for information processing and communication. Various models of quantum computation exist, however the most popular models incorporate the concepts of qubits and quantum gates. A qubit is a generalization of a bit - a system with two possible states, that can be in a quantum superposition of both. A quantum gate is a generalization of a logic gate: it describes the transformation that one or more qubits will experience after the gate is applied on them, given their initial state. The physical implementation of qubits and gates is difficult, for the same reasons that quantum phenomena are hard to observe in everyday life. One approach is to implement the quantum computers in superconductors, where the quantum effects become macroscopic, though at a price of extremely low operation temperatures.

This does make some sense! However, I was looking for why optical quantum computers don't need "extremely low temperatures" unlike superconducting quantum computers. Don't they suffer from the same problem i.e. aren't the quantum phenomena in optical quantum computers difficult to observe just as for superconducting quantum computers? Are the quantum effects already macroscopic at room temperatures, in such computers? Why so?

I was going through the description of Linear optical quantum computing on Wikipedia, but found no reference to "temperature" as such.


3 Answers 3


I was looking for why optical quantum computers don't need "extremely low temperatures" unlike superconducting quantum computers.

Superconducting qubits usually work in the frequency range 4 GHz to 10 GHz. The energy associated with a transition frequency $f_{10}$ in quantum mechanics is $E_{10} = h f_{10}$ where $h$ is Planck's constant. Comparing the qubit transition energy to the thermal energy $E_\text{thermal} = k_b T$ (where $k_b$ is Boltzmann's constant), we see that the qubit energy is above the thermal energy when $$f_{10} > k_b T / h \, .$$

Looking up Boltzmann's and Planck's constants, we find $$h/k_b = 0.048 \, \text{K / GHz} \, .$$

Therefore, we can write $$f_{10} > 1 \, \text{GHz} \, \,\frac{T}{0.048 \, \text{K}}$$

So, for the highest frequency superconducting qubit at 10 GHz, we need $T < 0.48 \, \text{K}$ in order for there to be a low probability that the qubit is randomly excited or de-excited due to thermal interactions. This is why superconducting qubits are usually operated in dilution refrigerators at ~15 milliKelvin. Of course, we also need the temperature to be low enough to get the metals superconducting, but for aluminum that happens at 1 K so actually the constraint we already talked about is more important.

On the other hand, suppose the two states of the optical qubit $\left \lvert 0 \right \rangle$ and $\left \lvert 1 \right \rangle$ the presence and absence of an optical photon. An optical photon has a frequency of around $10^{14}$ Hz, which corresponds to a temperature of 14,309 Kelvin. Therefore, there's an extremely low probability of the thermal environment changing the qubit state by creating or removing a photon. This is why optical light is sort of intrinsically quantum mechanical in nature.

Don't they suffer from the same problem i.e. aren't the quantum phenomena in optical quantum computers difficult to observe just as for superconducting quantum computers?

Well, the difficulties between superconducting quantum computers and optical quantum computers are different. Optical photons essentially don't interact with each other. To get an effective interaction between two photons, you have to either put them through a nonlinear crystal, or do some kind of photodetection measurement. The challenge with nonlinear crystals is that they're very inefficient; only a very small fraction of photons that go in actually undergo the nonlinear process that causes interaction. The challenge with photodetection is that it's hard to build a photodetector that has high detection efficiency and low dark counts$^{[a]}$. In fact, the best photo-detectors actually need to be operated in cryogenic environments anyway, so some optical quantum computing architectures need cryogenic refrigeration despite the fact that the qubits themselves have very high frequency.

P.S. This answer could be expanded quite a bit. If someone has a particular aspect they'd like to know more about, please leave a comment.

$[a]$: Dark counts means the times a photodetector thinks it saw a photon even though there really wasn't one. In other words, isn't the rate that the detector counts photons when its in the dark.

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    $\begingroup$ Nice answer! With regards to your argument as to why photons are more resilient to temperature: arguably the most common way to encode q information in photons is using their internal degrees of freedom, not using a "there/not there" encoding. This is especially true as many quantum optical QC protocols work in postselection anyway. It seems to me that this line of reasoning addresses the degree of attenuation/absorption more than the degree of decoherence. Does this kind of argument work when dealing with the transition between, say, horizontal and vertical polarization states of a photon? $\endgroup$
    – glS
    Commented Mar 15, 2018 at 11:40
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    $\begingroup$ @glS whether or not photon internal degrees of freedom are more or less common, they certainly are used, so this answer should be expanded. I know your answer touches on this point, and I was thinking whether I should edit your answer to expand it, or add my own version here. $\endgroup$
    – DanielSank
    Commented Mar 15, 2018 at 15:28
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    $\begingroup$ I guess that depends on what the addition would be. If you can expand your energetic argument to the transitions between internal degrees of freedom of photons then it would probably be a better fit in your answer. $\endgroup$
    – glS
    Commented Mar 15, 2018 at 15:46
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    $\begingroup$ @glS The energetic argument doesn't really work for internet degrees of freedom. Your answer about interactions strengths is more relevant there. The only reason I didn't go into that was that there's already your answer :-) $\endgroup$
    – DanielSank
    Commented Mar 15, 2018 at 15:47
  • $\begingroup$ When you write "The challenge with nonlinear crystals is that they very inefficient; only a very small fraction of photons that go in actually undero the nonlinear process that causes interaction.", is this interaction temperature-independent? $\endgroup$ Commented Apr 18, 2018 at 4:43

Because light, at the right frequencies, interacts weakly with matter. In the quantum regime, this translates to single photons being largely free of the noise and decoherence that is the main obstacle with other QC architectures. The surrounding temperature doesn't disturb the quantum state of a photon as much as it does when the quantum information is carried by matter (atoms, ions, electrons, superconducting circuits etc.). For example, reliable transmission of photonic qubits (more precisely, a QKD protocol) between China and Austria, using a low-orbit satellite as link, was recently demonstrated (see e.g. here).

Unfortunately, light also interacts extremely weakly (as in, it basically doesn't) with other light. Different photons not interacting with each other is what makes optical quantum computation somewhat tricky. For example, basic elements like two-qubit gates, when the qubits are carried by different photons, require some form of nonlinearity, which is generally harder to implement experimentally.


DanielSank is correct, but I think the answer is actually even more subtle. If there was no loss there would also be no way the background radiation leaking into your quantum device. Even if it was initially thermally excited, one could actively reset the state of the qubits. Thus, in addition to thermal excitations of microwave qubits, the fundamental reason for them being cooled down to so low temperature is really the dielectric loss of the materials the quantum state lives in.

Air imposes almost no loss to optical photons, but the electric circuits do attenuate the microwave frequency plasmons carrying the quantum information. So far the only way to get rid of these losses is to use superconductors, and, in addition, to go to cryogenic temperatuers much lower than the critical temperature of the superconductors, but there is no fundamental reasons for not being able to use higher temperatures in future, once materials with a lower loss become available.


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