20

10

*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.

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? – agaitaarino – 2018-04-18T04:43:21.317

@agaitaarino I'm not sure. My expertise is in superconducting qubits. – DanielSank – 2018-04-18T06:03:19.070

1Nice 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? – glS – 2018-03-15T11:40:48.223

1@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. – DanielSank – 2018-03-15T15:28:11.783

1I 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. – glS – 2018-03-15T15:46:52.447

1@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 :-) – DanielSank – 2018-03-15T15:47:46.707