Is a dilution refrigerator the only way to cool superconducting qubits down to 10 millikelvin? If not, what other methods are there, and why is dilution refrigeration the primary method?


1 Answer 1


Is a dilution refrigerator the only way to cool superconducting qubits down to 10 millikelvin?

There's another type of refrigerator that can get to 10 mK: the adiabatic demagnetization refrigerator (ADR).$^{[a]}$

why is dilution refrigeration the primary method?

To understand that, let's talk about one of the main limitations of the ADR.

How an ADR works

An ADR usually gets to about 3K with a helium compressor. That compressor can run all the time, so the refrigerator can sit at 3K indefinitely. To get down to mK temperatures, the ADR works like this:

  1. Raise the magnetic field surrounding a solid with nuclear spins. This aligns the spins.
  2. Slowly turn the field off. This allows the spins to randomize their direction, which absorbs entropy from the surroundings and lowers the temperature.
  3. Once the field is back to zero, we've sucked enough heat out of the surroundings to bring them to mK temperatures.

ADR limitations

This is all great and it really works, but it's a "one-shot" process. Once the field is down to zero, you can't go any lower. Heat from the surroundings, such as the room temperature outer parts of the refrigerator, leak heat into the part you're trying to keep cold, and since we've already lowered the magnetic field to zero, we can't do anything to remove that heat. Therefore, after cooling the ADR, it starts to warm up (hopefully slowly enough to run your experiment).

It's typical for an ADR to stay below 100mK for maybe twelve hours, although that number depends a lot on how many wires you have running to the cold part of the ADR. After the temperature rises above what you want, you have to raise the magnetic field again and slowly lower it to re-cool. Raising and lowering the field takes a while and heats up the refrigerator, and that big magnetic field is often incompatible with superconducting qubit experiments, so you can't run experiments while you're in that stage of the process.

ADR vs. dilution refrigerator

The dilution refrigerator, on the other hand, runs continuously, so you have as long as you need to run your experiment. That's a pretty big reason that they're in common use. Note, however, that other refrigerators aside from the ADR are used in many superconducting qubit labs for tasks where the benefits of a dilution refrigerator aren't needed and the shorter cold time of an ADR is ok. For example, ADR's are common for experiments with superconducting resonators, which are used to test the quality of materials that may later be used for a qubit.

$[a]$: Apologies for not finding a better link. Edits on that are welcome.

  • 2
    $\begingroup$ In this image, it appears that there are very high current wires which would only make sense if electromagnets were involved; would this mean that IBM is using ADR (at least at the time of writing of the article I found the picture in)? $\endgroup$
    – auden
    Commented Mar 15, 2018 at 6:01
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    $\begingroup$ @heather That looks like a dilution refrigerator to me. Those enormous copper braids are some kind of mechanical isolators. I think they're copper to keep the chassis of the cryostat all at the same voltage and avoid ground currents. The sliver windy wire looking thingy in the center is actually a pipe full of helium-4 and helium-3 mixture. The center pipe around which it's wrapped is the cold part of a compressor that gets to ~3 Kelvin. The thinner pipe is wrapped around to pre-cool the helium mixture as it makes it's way toward the mixing chamber where it gets to 10 mK. $\endgroup$
    – DanielSank
    Commented Mar 15, 2018 at 6:08
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    $\begingroup$ @heather The cupper is not for grounding, but for heatlink between cold head of the pulsetube cooler, which cools the two outer shields down to 4K and 1K respectivly. Cupper is a very good heat conductor, flexiblity is for mechanical decoupling from vibrations the pulse-tubes have. $\endgroup$
    – Johu
    Commented Jul 17, 2018 at 18:41

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