The watcher's bill
Freezing a qubit by watching it: is that free, or quietly expensive?
Partly known
Both. The heat dumped into the qubit and the minimum cost of erasing the check records both shrink the harder you watch. The light you actually use to watch grows with the square of how long you hold. In the realistic example the light bill is about 37,000 times the physics floor.
Hold a qubit, run up the bill
The job from the entry's example: hold one qubit still for 0.5 ms against a 1 kHz nudge, allowing a 1% chance of slipping. The panel does it with sharp checks. Between checks the nudge turns the qubit; each check keeps it with odds cos²(angle/2).
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checks per hold N = (ΩT)²/(4ε), rounded down: 246 at 0.5 ms keeps the slip at 1% · probe light scales from the best plan's 341 quanta as T² (photons ≥ (ΩT)²/(4ε)) · the floor was only computed for the 0.5 ms example, so at other hold times it is left blank.
In plain words
Every check jolts the qubit a little. Add up the jolts over the whole hold and the heat comes out to the allowed slip times one quantum: 0.01 quanta here. The Landauer floor (the least energy needed to wipe a record, kT·ln2 per bit) is 0.009 quanta, because a frozen qubit's records nearly all say the same thing and carry almost no news. That floor falls like 1/N, but only for an eraser that wipes the records together. Wiping each record on its own costs about log N.
The light dwarfs both. A real check is a pulse of light bounced off the qubit, and the best plan is always many very faint pulses. The photon count obeys photons ≥ (ΩT)²/(4ε), where Ω is the nudge rate, T the hold time and ε the allowed slip. Twice the hold costs four times the light. In the example that is 341 quanta, about 37,000 times the floor. Watching is free in principle and expensive in practice.