A Why low temperature is better in many precision experiments?

[Malamala]

Hello! In many precision experiment, especially those overlapping with quantum computing techniques, such as trapping a few ions in a Paul trap, they use cryogenic systems (around 1K). I am not totally sure I fully understand the advantage of that, compared to room temperature.

For example, a blackbody spectrum at room temperature has a peak around $1000$ cm$^{-1}$. Frequencies around this value don't seem to have any effect on the transitions used in atoms. Electronic levels are usually of the order of $10000$ cm$^{-1}$, and hyperfine transitions are around $1$ cm$^{-1}$, so it doesn't seem like the blackbody radiation would produce any unwanted transitions.

Even for molecules, $1000$ cm$^{-1}$ is not particularly dangerous (at least for the lowest 2 electronic levels), as rotational spacing is of the order of $1$ cm$^{-1}$ (I assume that in that case vibrational transitions might be an issue, for some molecules?). Can the BB photons change the center of mass motion of the system (assuming we would want to work in the motional ground state)? From what I read even that frequency is far from $1000$ cm$^{-1}$. So I guess my question is, what advantage does a cryogenic Paul trap gives over a room temperature one? Thank you!

 

[anuttarasammyak]

Do you have an idea how much energy between the levels are in Paul trap? I guess it is tiny.

 

[Baluncore]

So I guess my question is, what advantage does a cryogenic Paul trap gives over a room temperature one?

Things happen slowly at low temperatures. Coherence times are longer in colder quantum computers.
If the trap was hotter, the mean-free-path-length of the contents would be bigger than the trap.

 

[Twigg]

For example, a blackbody spectrum at room temperature has a peak around 1000 cm−1. Frequencies around this value don't seem to have any effect on the transitions used in atoms.

A lot of repeated, uncorrelated interactions with far-detuned photons can still lead to decoherence. The intensity and frequency will determine the decoherence rate.

so it doesn't seem like the blackbody radiation would produce any unwanted transitions.

It's the dephasing that's a bigger problem. Your T1 could be fine but your T2 can tank from many repeated small interactions with, say, blackbody photons.

So I guess my question is, what advantage does a cryogenic Paul trap gives over a room temperature one?

The cryogenic Paul trap will have lower blackbody (BB) radiation, as discussed, and the vacuum will be dramatically better for the cryogenic system. Background gas collisions also cause decoherence, after all.

Do you have an idea how much energy between the levels are in Paul trap? I guess it is tiny.

The motional quanta (per axis of motion) are $n \hbar \omega$ where $\omega$ is the trap frequency in a given axis of motion and $n$ is the motional quantum number. A few MHz is typical for the trap frequency. That's $6.6 \times 10^{-28} \mathrm{J/MHz}$ or $48 \mathrm{\mu K/MHz}$ in temperature units. So yep, tiny

 

[Malamala]

Things happen slowly at low temperatures. Coherence times are longer in colder quantum computers.
If the trap was hotter, the mean-free-path-length of the contents would be bigger than the trap.

I guess this is what I don't understand. In some of the experiments I have in mind, you trap one (or few) ions along the axis of a Paul trap, which is kept at ultra high vacuum ($10^{-10}$ mbar or lower). So at this vacuum, even at room temperature, the trapped ions won't interact with any gas particles inside the trap, even at 300K, over the measurement time (which is of the order of 1 second). So given that I have no collisions, what exactly happens slower?

 

[Malamala]

A lot of repeated, uncorrelated interactions with far-detuned photons can still lead to decoherence. The intensity and frequency will determine the decoherence rate.It's the dephasing that's a bigger problem. Your T1 could be fine but your T2 can tank from many repeated small interactions with, say, blackbody photons.The cryogenic Paul trap will have lower blackbody (BB) radiation, as discussed, and the vacuum will be dramatically better for the cryogenic system. Background gas collisions also cause decoherence, after all. The motional quanta (per axis of motion) are $n \hbar \omega$ where $\omega$ is the trap frequency in a given axis of motion and $n$ is the motional quantum number. A few MHz is typical for the trap frequency. That's $6.6 \times 10^{-28} \mathrm{J/MHz}$ or $48 \mathrm{\mu K/MHz}$ in temperature units. So yep, tiny

Thanks a lot! I see what you mean. However I am still confused about actual transitions. For example in this paper, on the first page, right column they say: "Blackbody radiation continuously perturbs the molecule, causing rotational state jumps on a time scale of tens of milliseconds to seconds for the states that we study" (and their experiment is done at room temperature). The rotational constant of their molecule is about 150 GHz (~5 cm$^{-1}$), which is so far away from the peak of the blackbody spectrum. Is it that the tail of the distribution still has enough intensity to lead to these rotational transitions? Also, if we are to go to ~1K, the peak of the distribution would be around 10 cm$^{-1}$, and I imagine that there the transition probability would be so much higher (as the peak is almost on resonance with rotational transitions). Won't in that case the experiment be even harder to perform due to random BB transitions?

 

[Baluncore]

So given that I have no collisions, what exactly happens slower?

Collisions.
10^-10 mbar is nowhere near zero, it is closer to a 10⁹ molecules per litre.

 

[anuttarasammyak]

The motional quanta (per axis of motion) are nℏω where ω is the trap frequency in a given axis of motion and n is the motional quantum number. A few MHz is typical for the trap frequency. That's 6.6×10−28J/MHz or 48μK/MHz in temperature units. So yep, tiny

Thank you so much @Twigg. For not ion but superconducting qubit I have read in a popular science magazine that cosmic background of 2.7 K is too hot for the qubits so cryogenic environment is cooler than the Universe. I think similar would apply to ion qubit to some extent.

 

[Twigg]

For not ion but superconducting qubit I have read in a popular science magazine that cosmic background of 2.7 K is too hot for the qubits so cryogenic environment is cooler than the Universe.

I'm no expert, but they must be using He3 or dilution fridges, because He4 boils at 4K, and would be hotter than the cosmic background. However, there's an important difference. Superconducting qubits are cooled cryogenically, while ion qubits are laser cooled. Laser cooling will get you a lot colder than cryogens. A dilution fridge will get you to 10mK or so. For the right atom, laser cooling can get you to 100's of nanokelvins, and some folks have laser cooled neutral gases straight into bose-einstein condensates (for bosons) or degenerate fermi gases (for fermions). What the OP is asking is why you need to use cryogens to cool the environment near the ion qubits, and that is because a hot environment will result in faster decoherence.

10-10 mbar is nowhere near zero, it is closer to a 109 molecules per litre.

Someone has high standards Getting a vacuum system to $10^{-10} mbar$ would cost me an inch off my hairline from stress! On a serious note, the coulomb crystals in ion quantum computers consist of 10's of ions spaced by 10's of microns. Pessimistically, say you had 100nm scattering length for the background gas <-> ions interaction. That's an interaction volume on the order of $10^{-15}$ liters. That's not to say that background pressure is irrelevant (it's very relevant). It's just to say that having $10^9$ molecules per liter isn't automatically a serious problem.

The rotational constant of their molecule is about 150 GHz (~5 cm−1), which is so far away from the peak of the blackbody spectrum. Is it that the tail of the distribution still has enough intensity to lead to these rotational transitions?

I believe that's correct. It's the only way it makes sense to me that they would see rotational excitations. If you wanted to try and estimate the rate for an arbitrary system, just multiply the transition dipole moment by the blackbody RMS electric field amplitude (derived from integrating the Planck distribution), just as you would calculate a Rabi rate. It should at least give you a ballpark for the timescale.

Also, if we are to go to ~1K, the peak of the distribution would be around 10 cm−1, and I imagine that there the transition probability would be so much higher (as the peak is almost on resonance with rotational transitions). Won't in that case the experiment be even harder to perform due to random BB transitions?

Don't forget the power (aka photon flux) drops with temperature to the fourth. Compared to room temp, that's $10^{-10}$ fewer photons to interact with.

 

[f95toli]

Another reason for cooling traps can be to reduce field noise from the surrounding materials. By cooling the whole trap you can sometimes reduce the effect of microscopic defect (impurities, lattice defects etc) in the dielectrics (and sometimes metals). In conventional electronics you would typically see these effects as 1/f noise. In ions traps this is sometimes known as "hot spots".
The noise reduction by cooling isn't huge, but in some trap designs it is worth it.

The same underlying physics is in play for superconducting qubits; defects on surfaces and at material interfaces are these days the main source of decoherence in all solid state qubits. There are two groups of defects commonly known as TLS and TLF; TLS being "quantum" and TLF "classical" with level splittings above or below KbT respectively.
The TLS (which can be highly coherent) can absorb/emit photons (and dispersively shift qubit frequencies) while the TLF cause "background noise" which in turn affects the TLS (which in turn affects the qubit).
The density of defects is extremely low (for TLS a few per um2) and energies very low (which is why TLF don't freeze out even at 10mK) meaning even figuring out what they are is extremely difficult; so far we've only identified a few sources of TLS and TLF but we know there are more.

 

[TeethWhitener]

The rotational constant of their molecule is about 150 GHz (~5 cm−1), which is so far away from the peak of the blackbody spectrum. Is it that the tail of the distribution still has enough intensity to lead to these rotational transitions?

I believe that's correct. It's the only way it makes sense to me that they would see rotational excitations.

Was going to mention Stefan-Boltzmann, but I see that @Twigg got to it in post 9. The frequency is low enough in the case of rotational transitions that you can just use the pre-Planck approximation:
https://en.m.wikipedia.org/wiki/Rayleigh–Jeans_law
Showing that the spectral radiance at a given frequency is roughly linear with temperature.