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I have a question on an issue that already bothers me for a longer time now.

In quantum computing, one criterion that should be fulfilled is that an ensemble of qubits can be initialized in a low-entropy fiducial state. This is, amongst other things, done by cooling the system. In thermodynamics, the entropy of water is lowered when the temperature is lowered and water starts to freeze into ice. I would also argue that intuitively, it makes sense that a system's entropy is reduced if it is cooled to a lower temperature.

However, from a mathematical point of view, I could not confirm this statement for me yet. If I look at the second law of thermodynamics for a reversible process,

$dS = \frac{\delta Q}{T}$,

where $S$ is entropy, $Q$ is heat and $T$ is temperature, then it seems as if the entropy would increase if I lower the temperature of the system, in contradiction to what I wrote before. My questions therefore are now:

  • What do I not consider in my thinking that leads to this contradiction?
  • How can it be shown that a system's entropy is reduced by cooling it?

I am looking forward to interesting answers.

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You're looking at it the wrong way. If your equation was $$S=\frac{some constant}{T}$$ I would agree that it says the entropy increases when T decreases. But we definitely don't have something of that form. The stuff on top is not some fixed universal constant. And most importantly, the equation is about $dS$, not $S$.

$dS$ is the infinitesimal $\textbf{variation}$ of $S$. To know if $S$ increases, you just need to know about the sign of $dS$.

In your case, if you're cooling down by exchanging heat with your surroundings, $\delta Q$ is negative (because you are losing heat to the heat bath). So $dS≤0$, and the entropy decreases

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  • $\begingroup$ Ok thank you, in the context of variation it is now pretty clear to me. So if a temperature $T_1 < T_2$, the entropy decreases faster for $T_1$ in comparison to $T_2$, if the heat content changed the same? That makes sense. And is there a possibility to have an equation $S = S(T)$, that shows values of the entropy proportional to the temperature, rather than an equation with $dS$? $\endgroup$ Jul 1 at 11:33
  • $\begingroup$ In general no. To get an expression for S from this law, you're gonna need to integrate the above equation between your initial and final state. If you have specified your problem well enough to do that, it will give you the entropy difference between starting point and arrival. But in general, there's no equation giving us S(T). As a side note, the entropy can be impacted by a multitude of other factors than a temperature change ( a universal S(T) relation would greatly reduce the relevance of the notion of entropy itself, textbooks wouldn't bother you so much with it ;) ) $\endgroup$ Jul 1 at 12:04
  • $\begingroup$ Ok thank you. I am aware of the fundamental relation $S = S(U,V,N)$, but apparently there is no $T$ so I was curious. $\endgroup$ Jul 1 at 12:19
  • $\begingroup$ Are you referring to the fundamental relation of thermodynamics? It reads $dU=TdS-PdV$ (if you choose internal energy as your potential) and temperature does appear in it $\endgroup$ Jul 1 at 12:33

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