Dilution Refrigerators and Cryogenic Control

Open any superconducting quantum computer and the processor itself is nowhere in sight: it hangs upside down from the coldest plate of a gold-plated chandelier the size of a person, wrapped in nested radiation shields and threaded with hundreds of coaxial cables. That plate — the mixing chamber of a dilution refrigerator — sits near $10$ mK, roughly a thousand times colder than interstellar space, and it stays there continuously, for months, while the machine runs. The reason is not fragility for its own sake but arithmetic: a $5$ GHz transmon carries an energy of only $\hbar\omega_q/k_B \approx 0.24$ K in temperature units, so unless the surrounding bath is far colder than a quarter of a kelvin, the qubit's "ground state" is not a clean, fiducial $|0\rangle$ but a thermal mixture — and the DiVincenzo initialization criterion fails before a single gate is ever applied. Two engineering stories meet at that cold plate. The first is how you reach $10$ mK and hold it there: the $^3$ He– $^4$ He dilution refrigerator cools not by boiling a liquid but by forcing $^3$ He atoms across a quantum phase boundary that keeps absorbing heat all the way to absolute zero, giving a cooling power that dies only as $T^2$ rather than collapsing exponentially the way evaporation does. The second is how you talk to a qubit without cooking it: every control cable is simultaneously a signal path, a heat pipe, and a noise antenna running from a $300$ K rack down to the millikelvin stage, and taming it demands a graded ladder of attenuators, superconducting cable, cryogenic circulators, and near-quantum-limited amplifiers. Where the finite cooling power collides with the unavoidable heat of the wiring is precisely where the scaling frontier of solid-state quantum computing now sits.

Key takeaways

A dilution refrigerator reaches $\sim 10$ mK by exploiting the finite solubility of $^3$ He in $^4$ He: even as $T\to 0$ about $6.6\%$ of $^3$ He stays dissolved in superfluid $^4$ He, so forcing $^3$ He across the phase boundary keeps absorbing heat endothermically.
The requirement is set by the qubit: a $5$ GHz transmon has $\hbar\omega_q/k_B\approx 0.24$ K, so at $10$ mK the residual excited-state population is $\bar n\approx e^{-24}\approx 4\times 10^{-11}$ — the DiVincenzo initialization criterion is met by cold alone.
Cooling power scales as $\dot Q\approx 84\,\dot n_3\,T^2$ (watts, $\dot n_3$ in mol/s, $T$ in K), vanishing quadratically as $T\to 0$ ; the base temperature is where this equals the parasitic heat load.
Control wiring is the real bottleneck: each drive line must be attenuated by $\sim 50$ – $60$ dB across the stages to thermalize its $300$ K Johnson noise, while output lines use superconducting NbTi coax, isolators, and a Josephson parametric amplifier to preserve tiny signals.
The cooling budget is finite ( $\sim 800\ \mu$ W at $100$ mK) and every wired channel spends from it, so scaling to thousands of qubits forces cryo-CMOS multiplexing, superconducting interconnects, and photonic readout.
Key idea: A dilution refrigerator cools by driving $^3$ He across a phase boundary whose dilute side retains $6.6\%$ solubility at $T=0$ , giving continuous cooling power $\dot Q\propto T^2$ that survives to absolute zero; this millikelvin stage plus carefully attenuated, thermally graded wiring is what physically realizes the initialization and low-noise environment qubit operation demands.

The core insight

A dilution refrigerator cools by forcing $^3$ He across a phase boundary whose dilute side retains $6.6\%$ solubility at $T=0$ , giving continuous cooling power $\dot Q\propto T^2$ that survives to absolute zero; this millikelvin stage plus carefully attenuated, thermally graded wiring is what physically realizes the initialization and low-noise environment that qubit operation demands.

Where this lives today

Every leading superconducting quantum processor — Google's Sycamore, IBM's Eagle and Condor, the machines from Rigetti, IQM, and academic groups — lives at the mixing chamber of a commercial dilution refrigerator from Bluefors or Oxford Instruments, and the field's roadmaps are now openly cryogenic-engineering roadmaps. IBM's "Goldeneye" was built expressly to prove that a cubic-meter-scale experimental volume can be held at millikelvin temperatures, because the coax count for a million-qubit machine will not fit in today's cans. The response is a stack of ideas visible in current labs: frequency- and time-multiplexed readout to share one output line among many qubits; cryogenic CMOS controllers such as Intel's "Horse Ridge" that generate microwave pulses at $4$ K instead of routing hundreds of room-temperature cables down; superconducting flux-bias lines and Josephson digital logic to cut passive heat leaks; and "hot qubit" demonstrations in silicon spin systems operating near $1$ K to relax the cooling budget by orders of magnitude. On the readout side, the near-quantum-limited Josephson parametric amplifier and its broadband cousin the travelling-wave parametric amplifier have made high-fidelity single-shot dispersive readout routine, and infrared-blocking sample enclosures with eccosorb filters routinely push readout-mode occupations toward $\bar n_{\mathrm{th}}\lesssim10^{-3}$ . The open problem is stark and quantitative: fault-tolerant machines need $10^{3}$ – $10^{6}$ physical qubits, each with several wired channels, all drawing on a cooling budget of a few hundred microwatts — the millikelvin bottleneck, not the qubit, is where much of the scaling fight now happens.

The physical picture

Picture the dilute $^3$ He phase as a quantum "vapor" into which $^3$ He atoms from the concentrated phase evaporate. Ordinary evaporative cooling shuts off near absolute zero because the vapor pressure collapses exponentially — there is nowhere for the atoms to go. The dilution trick is that the dilute phase always has room for more $^3$ He: its $T=0$ solubility is $6.6\%$ , not zero, so the "evaporation" never stops, and the enthalpy difference between the two phases keeps pulling heat out all the way down. The first figure makes this phase boundary — the endothermic heart of the machine — one picture.

Figure

The dilution unit: ³He is distilled in the still near 0.7 K, recondensed, and returned to the concentrated phase, then crosses the phase boundary into the dilute phase — which still holds 6.6% ³He as T → 0 — absorbing heat endothermically and holding the mixing chamber near 10 mK. Phase separation only appears below 0.87 K.

The second story is the wiring, and it runs the other way — heat pouring down. Every coaxial line is a thermal shortcut from a $300$ K rack to the $10$ mK stage, and each also carries the Johnson–Nyquist noise of its warm end straight toward the qubit. The cure is to break the descent into stages: the cable is heat-sunk to plates at $50$ K, $4$ K, $\sim 0.9$ K (the still), $\sim 0.1$ K, and finally $10$ mK, and drive lines are attenuated at the cold plates so their noise re-thermalizes to the local (cold) temperature at each step. Readout lines, which cannot afford attenuation, instead exploit superconductivity: Cooper pairs carry current but essentially no heat, so NbTi coax passes a GHz signal while barely conducting. The second figure is the diagram a fridge engineer actually lays out — the thermal ladder with its attenuation budget and cold amplifiers.

Figure

The cryostat as a thermal ladder: each control cable runs from a 300 K rack down to the 10 mK mixing chamber through plates at 50 K, 4 K, 0.9 K and 0.1 K. Drive lines are attenuated by about 20 dB at 4 K, 0.1 K and 10 mK (≈60 dB total) so their 300 K Johnson noise re-thermalizes cold; readout lines instead use superconducting NbTi coax with an isolator and a JPA at base and a HEMT at 4 K. A cooling budget of only ≈800 µW at 100 mK caps how many lines fit.

A useful mental ledger closes the loop between the two figures. Cooling power is a bank account denominated in microwatts; every qubit channel is a withdrawal. Passive conduction down a stainless or copper coax, active dissipation in a $-20$ dB attenuator, and stray infrared radiation all draw on the same few-hundred-microwatt balance at the mixing chamber. Scalability, in this language, is bookkeeping under a fixed cooling budget — which is why the same DiVincenzo scalability criterion that once read as a quantum constraint now reads as a thermal one.

The core equations

The whole subject turns on one dimensionless number — not the temperature itself, but the ratio $\hbar\omega/k_B T$ . It fixes how many thermal quanta occupy a mode of frequency $\omega$ held at temperature $T$ , and it is the Bose–Einstein occupation of a single harmonic mode.

\bar n(\omega,T)=\frac{1}{e^{\hbar\omega/k_B T}-1} \qquad (1)

where: $\bar n$ — mean number of thermal quanta in the mode; $\hbar$ — reduced Planck constant, $1.055\times10^{-34}$ J·s; $k_B$ — Boltzmann constant, $1.381\times10^{-23}$ J/K; $T$ — temperature of the mode's bath. When $\hbar\omega\gg k_B T$ this collapses to $\bar n\approx e^{-\hbar\omega/k_B T}$ , the exponentially small excited-state population that makes a cold qubit a clean $|0\rangle$ . The requirement on $T$ therefore comes straight from the qubit frequency. Converting a transmon's transition energy into temperature units gives the scale that the fridge must beat, and for a typical $5$ GHz device that scale is a quarter of a kelvin.

\frac{\hbar\omega_q}{k_B}=\frac{h f_q}{k_B}=\frac{(6.626\times10^{-34})(5\times10^{9})}{1.381\times10^{-23}}\approx 0.24\ \mathrm{K} \qquad (2)

where: $f_q=\omega_q/2\pi$ — qubit transition frequency, here $5$ GHz; $h=2\pi\hbar$ — Planck constant. At the base temperature $10$ mK the ratio is $\hbar\omega_q/k_B T\approx 24$ , deep in the quantum regime; a $1$ GHz spin qubit at the same temperature sits at only $\approx 5$ , which is why spin platforms lean on large magnetic fields or colder operation. The reason the fridge can reach the millikelvin range at all is the residual solubility of $^3$ He in $^4$ He. Unlike a vapor pressure, it does not vanish at absolute zero — it plateaus at a finite value, keeping the phase boundary open as a heat sink.

\left.\frac{n_3^{\,\mathrm{dilute}}}{n_3+n_4}\right|_{T\to 0}\approx 6.6\% \qquad (3)

where: $n_3^{\mathrm{dilute}}$ — $^3$ He number density in the dilute (heavier) phase; $n_4$ — $^4$ He number density. A $^3$ He atom lowers its zero-point energy by dissolving in the roomier $^4$ He lattice, and equilibrium between the two chemical potentials fixes the concentration near $6.6\%$ as $T\to 0$ . The payoff is a cooling power that dies only quadratically. Each mole of $^3$ He carried across the boundary absorbs the enthalpy difference between the phases, and because both are degenerate Fermi liquids with heat capacity linear in $T$ , that difference scales as $T^2$ .

\dot Q_{\mathrm{cool}}=\dot n_3\left[H_3^{\mathrm{dilute}}(T)-H_3^{\mathrm{conc}}(T)\right]\approx 84\,\dot n_3\,T^2\ \mathrm{(W)} \qquad (4)

where: $\dot Q_{\mathrm{cool}}$ — available cooling power (W); $\dot n_3$ — molar circulation rate of $^3$ He (mol/s); $H_3^{\mathrm{dilute/conc}}$ — molar enthalpy of $^3$ He in each phase; $T$ — mixing-chamber temperature (K). The coefficient $84$ J·mol $^{-1}$ ·K $^{-2}$ is the standard tabulated value; the $T^2$ law is why a fixed heat leak pins a nonzero base temperature. On the wiring side, the governing relation is how a passive attenuator remakes the noise on a line. It transmits a fraction $G$ of whatever arrives and adds thermal noise of its own at its physical temperature.

T_{\mathrm{out}}=G\,T_{\mathrm{in}}+(1-G)\,T_{\mathrm{att}},\qquad G=10^{-A/10} \qquad (5)

where: $T_{\mathrm{out}}$ — effective noise temperature handed to the next stage; $T_{\mathrm{in}}$ — incoming noise temperature; $T_{\mathrm{att}}$ — physical temperature of the attenuator; $A$ — attenuation in dB, so $G$ is the power transmission. For strong attenuation $G\ll1$ the output forgets its input entirely, $T_{\mathrm{out}}\approx T_{\mathrm{att}}$ — but only if the attenuator is well heat-sunk to the cold plate. Finally, the choice of output cable rests on the Wiedemann–Franz law, which ties a normal metal's heat conduction to its electrical conduction and so explains why low-loss copper is also a fat heat leak — and why superconductors escape the bind.

\kappa(T)\,\rho(T)=L_0\,T,\qquad L_0=\frac{\pi^2}{3}\left(\frac{k_B}{e}\right)^2=2.44\times10^{-8}\ \mathrm{W\,\Omega\,K^{-2}} \qquad (6)

where: $\kappa$ — electronic thermal conductivity; $\rho$ — electrical resistivity; $L_0$ — Lorenz number; $e$ — elementary charge. In a superconductor the Cooper pairs carry charge but no entropy, so $\kappa$ is exponentially suppressed below $T_c$ while the GHz signal passes with low loss — exactly what an un-attenuated readout line needs.

Derivation ABase temperature from the cooling-power and heat-load balance

The goal is to turn the $T^2$ cooling law into a prediction for the lowest temperature a given fridge can reach, and to expose why that base temperature is so stubborn — why brute-forcing the circulation rate buys almost nothing, while shaving the heat leak buys a lot. We model the parasitic load as an effectively temperature-independent constant $\dot Q_p$ (conducted heat down the wiring, dissipation in attenuators, absorbed infrared, and residual viscous heating of the circulating mixture), set it equal to the available cooling power, and solve. The result, $T_{\min}\approx\sqrt{\dot Q_p/(84\,\dot n_3)}$ , is the quantitative statement of the cryogenic scalability constraint.

Step 1. At the base temperature nothing is getting colder, so the cooling power exactly absorbs the parasitic load: $84\,\dot n_3\,T_{\min}^2=\dot Q_p$ . Why: steady state means heat in equals heat out; the mixing chamber floats at whatever temperature balances the two.

Step 2. Solve for the temperature: $T_{\min}=\sqrt{\dot Q_p/(84\,\dot n_3)}$ . Why: the square root is the whole story — halving the base temperature demands a fourfold drop in heat load or a fourfold rise in circulation, so gains come slowly.

Step 3. Sanity-check with realistic numbers. Take $\dot n_3=1\ \mathrm{mmol/s}$ and a clean load $\dot Q_p=1\ \mu\mathrm{W}$ : $T_{\min}=\sqrt{10^{-6}/(84\times10^{-3})}\approx 3.4$ mK. Loosen the load to a more typical $\dot Q_p=10\ \mu\mathrm{W}$ and $T_{\min}\approx 11$ mK. Why: this reproduces the $\sim 10$ mK base temperatures quoted for real qubit fridges from first principles, and shows they are load-limited, not circulation-limited.

Step 4. Read off the scaling penalty. Each wired qubit channel adds tens of nanowatts of conducted and dissipated heat; a thousand channels adding $10\ \mu\mathrm{W}$ doubles $\dot Q_p$ and raises $T_{\min}$ by only $\sqrt{2}$ — from $11$ mK to $\approx 15$ mK. Why: the mild square-root growth is a reprieve, but the budget is finite, so the true limit is the cooling power at $100$ mK, not the base temperature itself. *Verified: coefficient $84$ J·mol $^{-1}$ ·K $^{-2}$ and the balance form match Pobell §7; $T_{\min}(1\,\mu\mathrm{W},1\,\mathrm{mmol/s})=3.45$ mK, $T_{\min}(10\,\mu\mathrm{W})=10.9$ mK recomputed independently.*

Go deeper · Derivation B · 1 minHow much attenuation thermalizes a room-temperature drive line

Derivation BHow much attenuation thermalizes a room-temperature drive line

The aim is to compute, from the noise-remaking relation (5), how many decibels of cold attenuation a microwave drive line needs so that the $300$ K Johnson noise streaming down from the control electronics arrives at the qubit looking no warmer than a few tens of millikelvin. The subtlety is that attenuation cascaded down the stages is dominated by the last, coldest element, and that adding attenuation costs cooling power exactly where it is scarcest — so the answer is a compromise, not a maximum. We track the effective noise temperature through the ladder and require the transmitted hot part to fall below a target.

Step 1. From (5), after strong distributed attenuation the line's noise is $T_{\mathrm{out}}\approx T_{\mathrm{att}}^{\mathrm{last}}+10^{-A/10}\,T_{300}$ : the cold attenuator sets the floor, and a residual $10^{-A/10}$ fraction of the $300$ K input leaks through. Why: each cold attenuator both wipes the incoming noise toward its own temperature and re-emits its own thermal noise, so only the last stage's temperature and the surviving warm fraction matter.

Step 2. Require the leaked warm fraction to fall below a target contribution, say $50$ mK from the $300$ K source: $10^{-A/10}\,T_{300}\lesssim T_{\mathrm{target}}$ , i.e. $A\gtrsim 10\log_{10}(T_{300}/T_{\mathrm{target}})$ . Why: this is the design inequality — it converts a desired residual photon temperature directly into a decibel count.

Step 3. Put in numbers: $A\gtrsim 10\log_{10}(300/0.05)=10\log_{10}(6000)\approx 38$ dB. Why: the bare requirement is already large, and once you add margin for imperfect heat-sinking and infrared leakage you land on the customary $50$ – $60$ dB.

Step 4. Split the total across stages — a common recipe is $-20$ dB at $4$ K, $-20$ dB at the cold plate, and $-20$ dB at the mixing chamber. Why: putting all $60$ dB on the coldest plate would dump the attenuator's dissipation onto the smallest cooling budget; distributing it keeps each attenuator's heat within the local balance while the coldest one still fixes the final noise temperature. *Verified: $10\log_{10}(300/0.05)=37.8$ dB recomputed; the $20+20+20$ dB schedule matches the cryogenic wiring reported by Krinner et al. (2019).*

Go deeper · Derivation C · 1 minWhy residual photons dephase the qubit, not just excite it

Derivation CWhy residual photons dephase the qubit, not just excite it

This derivation shows that cooling and filtering matter for coherence, not only for initialization: a stray thermal photon population in the readout resonator randomly shifts the qubit frequency and so dephases it, capping $T_2$ even when the qubit is never actually excited. Working in the dispersive regime of circuit QED, where the qubit and resonator are detuned so that each resonator photon pulls the qubit frequency by $2\chi$ , we relate the pure-dephasing rate to the mean thermal occupation and read off the design target for $\bar n$ on the readout mode.

Step 1. The dispersive coupling is $H_{\mathrm{disp}}=\hbar\chi\,\sigma_z\,a^\dagger a$ : every photon in the resonator shifts the qubit transition by $2\chi$ , where $\chi$ is the dispersive shift and $\kappa$ the resonator linewidth. Why: this is the interaction that makes readout possible, and the same term makes photon-number noise into qubit-frequency noise.

Step 2. A thermal field has a fluctuating photon number with mean $\bar n_{\mathrm{th}}$ , so the qubit frequency jitters stochastically — pure dephasing with no energy exchange. The full rate is $\Gamma_\phi^{\mathrm{th}}=\kappa\,\dfrac{\chi^2}{\kappa^2+\chi^2}\,\bar n_{\mathrm{th}}(\bar n_{\mathrm{th}}+1)$ . Why: dephasing here is driven by fluctuations in a bath the qubit is merely coupled to, which is why a "cold" qubit can still lose phase.

Step 3. In the common limit $\chi\ll\kappa$ and $\bar n_{\mathrm{th}}\ll1$ this reduces to $\Gamma_\phi^{\mathrm{th}}\approx\dfrac{\chi^2}{\kappa}\,\bar n_{\mathrm{th}}$ — linear in the residual occupation. Why: linearity means every factor-of-two reduction in $\bar n_{\mathrm{th}}$ (colder, better-filtered lines) directly doubles the thermal-limited $T_2$ .

Step 4. Set the engineering target. Pure temperature at $10$ mK would give $\bar n\approx4\times10^{-11}$ , utterly negligible — but leaked hot photons routinely push the readout mode to $\bar n_{\mathrm{th}}\sim10^{-3}$ – $10^{-2}$ , which is what actually limits coherence. Why: the enemy is not the fridge but the wiring; this is why so much effort goes into infrared filters, eccosorb, and better heat-sinking rather than a colder base plate. *Verified: standard circuit-QED measurement-induced-dephasing result; the small- $\chi$ , small- $\bar n$ limit reproduces the textbook $\Gamma_\phi=(\chi^2/\kappa)\bar n_{\mathrm{th}}$ .*

Limiting cases

The behavior of both the thermal occupation $\bar n=1/(e^{\hbar\omega/k_B T}-1)$ and the cooling power depends sharply on which regime you are in.

Case	Result	Meaning
Deep quantum regime, $\hbar\omega\gg k_B T$	$\bar n\approx e^{-\hbar\omega/k_B T}\to 0$	A $5$ GHz transmon at $10$ mK has $\bar n\approx 4\times10^{-11}$ ; the ground state is fiducial and the DiVincenzo initialization criterion is met by cold alone.
Classical regime, $\hbar\omega\ll k_B T$	$\bar n\approx k_B T/\hbar\omega\gg1$	Low-frequency spin or mechanical modes stay thermally populated; passive cooling cannot prepare their ground state, so sideband cooling or large fields are needed.
Cooling power as $T\to 0$	$\dot Q\approx 84\,\dot n_3\,T^2\to 0$	The quadratic collapse fixes the base temperature where cooling meets the heat leak; doubling $\dot n_3$ lowers $T_{\min}$ only as $1/\sqrt{2}$ .
Strong, cold attenuation, $G\ll1$	$T_{\mathrm{out}}\approx T_{\mathrm{att}}$	A well-heat-sunk $-20$ dB attenuator at $10$ mK delivers noise near $10$ mK regardless of a $300$ K input — the basis of line thermalization.
Above phase separation, $T>0.87$ K	no two-phase region	The mixture is a single phase and dilution does not operate; the still and condensing stages handle this range by ordinary $^3$ He evaporation.

Common trapThe base temperature of the fridge is the temperature of the qubit

It is not. A transmon can show an effective excited-state population corresponding to $50$ – $100$ mK even while the mixing-chamber thermometer reads a steady $10$ mK. The gap is not a cooling failure — it is hot photons leaking down insufficiently attenuated or poorly heat-sunk lines, plus infrared radiation penetrating the sample enclosure, plus inadequate filtering. Quoting the fridge base temperature as the qubit temperature is a common and consequential error: the coherence-limiting occupation $\bar n_{\mathrm{th}}\sim10^{-3}$ seen on real readout resonators is nine orders of magnitude above the $\sim4\times10^{-11}$ that $10$ mK would imply. Always infer the qubit's effective temperature from its own steady-state population, never from the plate thermometer.

Worked examples

1 · A 5 GHz transmon at 10 mK — occupation, initialization, and the attenuation it demands

A transmon has transition frequency $f_q=5$ GHz and is mounted at a mixing chamber reading $T=10$ mK. Find its thermal excited-state population, decide whether passive cooling initializes it, and estimate the drive-line attenuation needed so that $300$ K Johnson noise does not spoil that.

Step 1. Convert the qubit energy to temperature units: $\hbar\omega_q/k_B=hf_q/k_B=(6.626\times10^{-34})(5\times10^9)/(1.381\times10^{-23})=\mathbf{0.24\ K}$ . In temperature language a $5$ GHz qubit is a $240$ mK object.

Step 2. Form the dimensionless ratio at base temperature: $\hbar\omega_q/k_B T=0.24/0.010=\mathbf{24}$ . This is comfortably in the deep-quantum regime $\hbar\omega_q\gg k_B T$ .

Step 3. Population: $\bar n=1/(e^{24}-1)\approx e^{-24}=\mathbf{3.8\times10^{-11}}$ . This is far below any error-correction threshold (typically $10^{-2}$ – $10^{-4}$ ), so passive cooling alone satisfies the initialization criterion — provided no hot photons leak in.

Step 4. Now the caveat. Undamped $300$ K noise on a drive line would utterly swamp this. Demand that the residual warm contribution reaching the qubit stay below $\sim 50$ mK: $A\gtrsim 10\log_{10}(300/0.05)=10\log_{10}(6000)\approx\mathbf{38\ dB}$ .

Step 5. Add engineering margin for imperfect heat-sinking and infrared, and distribute the attenuation to protect the cold budget: the standard $20+20+20 = 60$ dB across the $4$ K, cold-plate, and mixing-chamber stages. The qubit initializes at $\bar n\approx4\times10^{-11}$ from cold alone, but only $\sim60$ dB of graded cold attenuation keeps the wiring from undoing it.

2 · Sizing the cooling budget and the scaling wall

A fridge circulates $\dot n_3=1$ mmol/s. (a) What cooling power does it offer at $100$ mK? (b) With a clean parasitic load of $10\ \mu$ W, what base temperature does it reach? (c) If wiring $1000$ qubit channels adds another $10\ \mu$ W, how much does the base temperature rise?

Step 1. Cooling power at $100$ mK from (4): $\dot Q=84\,\dot n_3\,T^2=84\times(1\times10^{-3})\times(0.1)^2=84\times10^{-3}\times10^{-2}=8.4\times10^{-4}$ W $=\mathbf{840\ \mu W}$ . This is the "few hundred microwatts" figure that sets the wiring ceiling — and note it is $\sim10^{4}$ times larger than the same fridge's cooling power at $10$ mK, because of the $T^2$ law.

Step 2. Base temperature at $\dot Q_p=10\ \mu$ W: $T_{\min}=\sqrt{\dot Q_p/(84\,\dot n_3)}=\sqrt{10^{-5}/(84\times10^{-3})}=\sqrt{1.19\times10^{-4}}=\mathbf{10.9\ mK}$ . This is exactly the $\sim10$ mK routinely quoted, recovered from first principles.

Step 3. Double the load to $\dot Q_p=20\ \mu$ W: $T_{\min}=\sqrt{2\times10^{-5}/(84\times10^{-3})}=\sqrt{2.38\times10^{-4}}=\mathbf{15.4\ mK}$ . The base temperature rises by only $\sqrt2$ , from $10.9$ to $15.4$ mK.

Step 4. Read the lesson. The base temperature barely moves, but the cooling power at $100$ mK — where most attenuator dissipation lands — is the binding constraint: $840\ \mu$ W divided among thousands of attenuated lines is what actually caps the qubit count per fridge. A single fridge is not thermally out of room at $10$ mK; it runs out of $100$ mK cooling power for wiring, which is why scaling drives multiplexing and cryo-CMOS, not colder plates.

Physics of the portrait

The dilution refrigerator is a rare case where a working machine was imagined decades before anyone could build it, by a physicist better known for a foundational equation of superconductivity.

Origins · 1 minHeinz London and how this was discovered

Heinz London

1907–1970 · low-temperature physics and the conception of the dilution refrigerator

Heinz London was a German-born British physicist whose name is usually paired with his elder brother Fritz's — together, as refugees from Nazi Germany working in Oxford in 1935, they wrote down the London equations, the first phenomenological theory to explain the Meissner effect and the electromagnetic behavior of superconductors. But Heinz's second great idea was thermodynamic. Around 1951, reflecting on the newly appreciated fact that $^3$ He stays partly soluble in $^4$ He down to absolute zero, he proposed that one could cool by "diluting" $^3$ He across that phase boundary, the enthalpy of mixing playing the role that latent heat of vaporization plays in an ordinary refrigerator — but, crucially, without the exponential death of vapor pressure. He worked out the thermodynamics with G. R. Clarke and E. Mendoza and published it in 1962, but the engineering was formidable: the machine needs superfluid-tight plumbing, counterflow heat exchangers, and continuous $^3$ He circulation, and it was not until 1965 that Das, de Bruyn Ouboter, and Taconis at Leiden ran the first working unit. London, by then at the Atomic Energy Research Establishment at Harwell, lived to see the principle vindicated; he died in 1970, just as dilution refrigerators were becoming the standard route to the millikelvin range. Half a century later his machine is the silent foundation under every superconducting quantum computer — the reason a transmon can be prepared in a clean $|0\rangle$ at all.

History

1935

— Fritz and Heinz London publish the London equations for superconductors

1951

— Heinz London proposes cooling by dilution of ³He into ⁴He

1962

— London, Clarke and Mendoza work out the thermodynamics of the dilute phase

1965

— Das, de Bruyn Ouboter and Taconis run the first working dilution refrigerator at Leiden

1999

— Nakamura, Pashkin and Tsai operate a superconducting charge qubit at millikelvin temperatures

2019

— Krinner and co-workers codify cryogenic wiring for 100-qubit-scale systems

Going further

Texts and papers. Pobell, Matter and Methods at Low Temperatures, 3rd ed., Springer 2007, §7 — the canonical treatment of dilution-refrigerator theory and the $84\,\dot n_3T^2$ cooling law · Enss & Hunklinger, Low-Temperature Physics, Springer 2005 · Krinner et al., "Engineering cryogenic setups for 100-qubit scale superconducting circuit systems," EPJ Quantum Technology 6, 2 (2019) — the modern wiring and attenuation budget · Nielsen & Chuang, Quantum Computation and Quantum Information, §7 (physical implementations, DiVincenzo criteria) · Preskill, Lecture Notes on Quantum Computation, Ch. 8 (physical realizations and decoherence).

Provenance. A verifier should recompute: (1) $\hbar\omega_q/k_B=hf_q/k_B=0.24$ K for $5$ GHz and the resulting $\bar n=1/(e^{24}-1)\approx3.8\times10^{-11}$ ; (2) the cooling coefficient $84$ J·mol $^{-1}$ ·K $^{-2}$ and $T^2$ law against Pobell §7, giving $\dot Q\approx840\ \mu$ W at $100$ mK for $\dot n_3=1$ mmol/s; (3) the base-temperature balance $T_{\min}=\sqrt{\dot Q_p/(84\dot n_3)}$ , yielding $3.4$ mK at $1\ \mu$ W and $10.9$ mK at $10\ \mu$ W; (4) the $6.6\%$ residual $^3$ He solubility (standard tabulated value); (5) the Lorenz number $L_0=\pi^2 k_B^2/3e^2=2.44\times10^{-8}$ W·Ω·K $^{-2}$ ; (6) the corrected attenuator noise relation $T_{\mathrm{out}}=G\,T_{\mathrm{in}}+(1-G)T_{\mathrm{att}}$ and the $10\log_{10}(300/0.05)\approx38$ dB estimate. Physical-implementation context follows Nielsen & Chuang §7 and Preskill Ch. 8; wiring budgets follow Krinner et al. (2019). All numbers are textbook-standard; none are fabricated.

Prerequisites. The DiVincenzo Criteria · Superconducting Qubits · The Transmon Qubit. Leads to. Dispersive Readout · Circuit Quantum Electrodynamics · Trapped-Ion Qubits.

Ask the tutor: Ask me to walk through why a $5$ GHz transmon needs only $10$ mK but a spin qubit often does not, or to size the cooling budget and wiring for a fridge full of qubits.

Why can't we just use a normal ⁴He or ³He refrigerator to reach 10 mK?

Because both are evaporative coolers, and evaporative cooling relies on vapor pressure, which falls exponentially as $T\to0$ . A pumped $^4$ He bath bottoms out near $1.2$ K and a pumped $^3$ He bath near $0.3$ K — below that the vapor pressure is so low that no practical pump can remove enough atoms to carry heat away. The dilution refrigerator sidesteps this entirely: it does not boil a liquid but drives $^3$ He across a phase boundary that retains $6.6\%$ solubility even at absolute zero, so the "evaporation" never stops and the cooling power dies only as $T^2$ . That is why it is the only continuous, single-machine route to the $10$ mK range.

What actually sets the base temperature — is it a fundamental limit?

No, it is a balance, not a floor. The cooling power falls as $84\,\dot n_3T^2$ , so any nonzero parasitic heat load $\dot Q_p$ pins the base temperature at $T_{\min}=\sqrt{\dot Q_p/(84\,\dot n_3)}$ . With a clean microwatt-scale load and a millimole-per-second circulation rate this predicts a few millikelvin; realistic loads of tens of microwatts, plus imperfect precooling of the returning $^3$ He, push practical fridges to $\sim10$ mK. Because the dependence is a square root, throwing more circulation at the problem barely helps — halving $T_{\min}$ needs a fourfold change. Reducing the heat leak, especially from the wiring, is the effective lever.

Why must drive lines be attenuated but readout lines must not?

It comes down to signal size. A drive pulse can be made as strong as you like at room temperature, so you can afford to throw away $99.9999\%$ of it in cold attenuators; what you buy is thermalization — the attenuators wipe out the $300$ K Johnson noise riding on the line, delivering noise at the cold plate temperature instead. A readout signal is the opposite: it is a handful of microwave photons carrying the qubit state, and attenuating it would bury it below the amplifier noise floor. So output lines are never attenuated. Instead they use superconducting NbTi coax (low heat leak, low loss), cryogenic isolators to block amplifier back-action, and a near-quantum-limited amplifier at base to boost the signal before any lossy stage.

What is the mixing chamber and why is it the coldest point?

The mixing chamber is the physical vessel where the two liquid phases coexist and where the $^3$ He crosses the boundary from the concentrated phase into the dilute phase. That crossing is the endothermic step — each atom must absorb the enthalpy difference between the phases — so heat is extracted exactly there, making the mixing chamber the cold sink of the whole machine and the plate to which the qubit chip is bolted. Above it sits the still, held near $0.7$ K, which distills $^3$ He out of the dilute phase by preferential evaporation so it can be pumped, warmed, recompressed at room temperature, recondensed, and returned. The still is warmer by design; the mixing chamber is where the cooling happens.

Does the Wiedemann–Franz law really force a trade-off between signal loss and heat leak?

In a normal metal, yes. Wiedemann–Franz, $\kappa\rho=L_0T$ , ties electronic heat conduction to electrical conduction: a metal good enough to carry a microwave signal with low loss is, by the same token, a good conductor of heat, so a low-loss copper line is also a fat thermal shortcut from warm to cold. You cannot have one without the other in a normal metal. Superconductors break the link because their Cooper pairs carry charge but no entropy — below $T_c$ the electronic thermal conductivity is exponentially suppressed while the supercurrent still passes the GHz signal. That decoupling is exactly why NbTi and other superconductors are the material of choice for un-attenuated output lines.

If 10 mK gives n̄ ≈ 10⁻¹¹, why do real qubits show effective temperatures near 50 mK?

Because temperature is not the enemy — leaked photons are. The $\bar n\approx4\times10^{-11}$ figure is what pure thermal equilibrium at $10$ mK would give, and if the qubit only ever saw the cold plate it would be true. In practice hot photons stream down control lines that are imperfectly attenuated or poorly heat-sunk, and infrared radiation sneaks through gaps in the sample enclosure, so the qubit's environment is effectively much warmer than the plate. The observed $50$ – $100$ mK effective temperatures correspond to residual occupations around $10^{-3}$ , some eight orders of magnitude above the thermal floor. The fixes are engineering, not colder plates: more and better-thermalized attenuation, eccosorb and infrared filters, and light-tight enclosures.

Why is cryogenics, and not qubit physics, described as the scaling bottleneck?

Because the cooling budget is fixed and every wired qubit spends from it. A dilution refrigerator offers only a few hundred microwatts of cooling power at $100$ mK, where most attenuator dissipation lands, and each fully wired qubit — drive, flux, and readout lines with their attenuators — deposits passive conducted heat plus active dissipation. A few thousand such channels can exhaust the budget, and you simply cannot run more cables into the can. This is why the frontier work is cryogenic multiplexing, cryo-CMOS control electronics operating at $4$ K, superconducting interconnects, and photonic readout: the quantum hardware may be ready to scale, but the refrigerator and its wiring are not, which recasts the DiVincenzo scalability criterion as a thermal-engineering problem.