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Online descriptions of quantum computers often discuss how they must be kept near absolute zero $\left(0~\mathrm{K}~\text{or}~-273.15~{\left. {}^{\circ}\mathrm{C} \right.}\right)$.

Questions:

  1. Why must quantum computers operate under such extreme temperature conditions?

  2. Is the need for extremely low temperatures the same for all quantum computers, or does it vary by architecture?

  3. What happens if they overheat?


Sources : Youtube, D-Wave

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    $\begingroup$ @DIDIx13 Change it to something like: "Is it necessary that quantum computers have to be kept at around -273°C ?". Also, mention a few primary sources (textbooks/articles/lectures) from which you formed the idea that quantum computers have to be kept close to absolute 0 (-273°C). Basically mention where you've "seen" (as you have stated in the question). $\endgroup$ – Sanchayan Dutta Mar 14 '18 at 15:38
  • $\begingroup$ Alright, I've up-voted it. Hope you will get useful answers. $\endgroup$ – Sanchayan Dutta Mar 14 '18 at 15:46
  • $\begingroup$ This seems like a great question. Yes, it's intro-level, but that's perfectly fine! A lot of folks reading random news blurbs about quantum computers might see the crazy-low temperatures and be curious about it. It'd be good to have a Q&A addressing the basic topic. That said, other questions might be asked about more specific contexts. $\endgroup$ – Nat Mar 14 '18 at 15:47
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    $\begingroup$ I think "absolute zero (temperature)" is much more clear than "-273C". $\endgroup$ – dylnan Mar 14 '18 at 16:06
  • $\begingroup$ @DIDIx13 Sorry, I was responding to Blue $\endgroup$ – dylnan Mar 14 '18 at 17:34
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Well, first, not all systems must be kept near absolute zero. It depends on the realization of your quantum computer. For example, optical quantum computers do not need to be kept near absolute zero, but superconducting quantum computers do. So, that answers your second question.

To answer your first question, superconducting quantum computers (for example) must be kept at low temperatures so that the thermal environment cannot induce fluctuations in the qubits' energies. Such fluctuations would be noise/errors in the qubits.

(See Blue's question Why do optical quantum computers not have to be kept near absolute zero while superconducting quantum computers do? and Daniel Sank's answer for some follow up information.)

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  • $\begingroup$ +1. Nice short answer. But I'd request not to throw around the word "noise" like that, without defining it. Causes a lot of confusion in beginners. Also, perhaps a bit of elaboration on why superconducting quantum computers need to be kept near absolute zero, but optical quantum computers need not be, would be good. $\endgroup$ – Sanchayan Dutta Mar 14 '18 at 17:05
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    $\begingroup$ @Blue I'm still lost on how "noise" might be a confusing word, even for beginners. The above use of it looks good to me. $\endgroup$ – Nat Mar 14 '18 at 17:31
  • $\begingroup$ @DanielSank If Stack Exchange sites were explicitly meant for domain experts then ~95% of the active user-base wouldn't be present. :P It's meant for everybody who is interested in the or is studying the subject, or is an academic/expert. And that is why you had seen the options: Just Curious/Beginner/Prosumer/Avid Enthusiast/Academic/Expert while committing to the Area 51 proposal. $\endgroup$ – Sanchayan Dutta Mar 14 '18 at 17:34
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    $\begingroup$ Anyway, as someone who is unaware about the experimental aspects of quantum computing I'd like to see some elaboration on why there is no energy fluctuations in architectures like "optical quantum computers", like there is, in "super-conducting quantum computers". $\endgroup$ – Sanchayan Dutta Mar 14 '18 at 17:37
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    $\begingroup$ @Blue, sorry, I didn't see your question earlier! I added a link to your new question in my answer so readers of this can go to your q/a. $\endgroup$ – heather Mar 15 '18 at 6:06
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To understand this question (and its possible answers) properly, we need to discuss a couple of concepts related to temperature and its relation to quantum states. Since I think the question makes more sense in the solid state, this answer will assume that's what we're talking about.

For starters, I find it useful to think about Boltzmann's distribution: a probability distribution that gives the probability $p_i$ that a system will be in a certain state $i$ as a function of that state’s energy ${\varepsilon}_i$ and the temperature $T$ of the system:

$p_i={\frac{e^{- {\varepsilon}_i / k T}}{\sum_{j=1}^{M}{e^{- {\varepsilon}_j / k T}}}}$

where $k$ is Boltzmann's constant.

In a system that is in equilibrium, as defined by statistical mechanics, the population of the different quantum states is governed by this equation (the system will be in a thermal state). If we think of a single quantum system rather than a collection or "ensemble", this distribution of populations would correspond to a series of weights in a mixed state. In any case, these are not the conditions one needs for quantum computing, where at any given time we want to have a good control the wavefunction. However, note that these probabilities have an exponential dependence on ${\varepsilon}_i$. This will be important further down.

Additionally, we need to consider phonons, the collective excitations in periodic, elastic arrangements of atoms or molecules in condensed matter. These are often the carriers of energy to and from our qubits into the part of the solid where we do not have an exquisite quantum control and thich therefore is thermalized: the so-called thermal bath.

Why must quantum computers operate under such extreme temperature conditions?

We can never fully control the quantum state of a solid chunk of matter. At the same time, we do need full control over the quantum state of our quantum computer, meaning the subset of quantum states where our information resides. These will live in pure states (including quantum superpositions), surrounded by a disordered -thermalized- environment.

Think about the Boltzmann distribution described above, and about the exponential term. In practice, its equation means that we can assume $p_i=0$ when the relation between temperature and the energy of the states we're interested in (which often means the states corresponding to our qubits) is such that ${\varepsilon}_i<<kT$.

Kinetics are often hard to model, but you know that inevitably your system will tend to thermalize. So, you need to keep your quantum computer, for as long as possible, in a state such that the only excitations that occur are those corresponding to the quantum states and quantum operations that are part of your computation. If the temperature of the solid where the quantum system is residing is low, you only need to worry about your qubits uncontrollably relaxing to a lower-energy state (which is bad enough). If the temperature is high, you also need to worry about your qubits being uncontrollably excited to higher-energy states. Inevitably, this also includes states that are outside your computational basis, meaning states that, for your qubit state, are neither $|0>$ nor $|1>$, nor any complex combination thereof: harder-to-correct errors.

If you now think about the phonons, recall that they are excitations, which cost energy, and thus are more abundant at high temperature. With rising temperatures, there is a rising number of available phonons, and they will present rising energies, sometimes allowing for interaction with different kinds of excitations (accelerating the kinetics toward thermalization): eventually, those that are detrimental to our quantum computer.

Is the need for extremely low temperatures the same for all quantum computers, or does it vary by architecture?

It does vary, and dramatically so. Within the solid state, it depends on the energies of the states that constitute our qubits. Outside the solid state, as pointed out above and in a follow-up question (Why do optical quantum computers not have to be kept near absolute zero while superconducting quantum computers do?), it's a whole another story.

What happens if they overheat?

See above. In a nutshell: you lose your quantum information faster.

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