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An atom was prepared in a superposition of ground state and excited states.I propose to measure the state by coupling the system to a cold enough substance.

By cold enough I mean $$kT\ll E_1,$$ where $E_1$ is the first excited energy of the atom.

Due to Boltzmann distribution, we expect the final state of the atom to be its ground state. We note the change in the temperature of the substance, which will be used to infer the state of the atom.

The change in the temperature times the specific heat will be equal to energy of the system: $$E = C_v \Delta T.$$

If the change in the temperature is zero, then we can infer that the state of the system is the ground state. If there is a change in the temperature, then that change in the temperature will tell us which excited state the atom is in. In the process however the state of the atom will be reset to its ground state.

My interest in this question comes from trying to understand the irreversible change in entropy due to measurement. It seems to me that, we need to have overwhelming odds for a forward process(formation of permanent record), over the reverse process(spontaneous erasure of record). A related question is why doesn't a photographic place spontaneously unmark itself emitting a photon in the process? I considered this method of measurement as a toy model to understand these issues.

In this case the weight associated with the odds is

$$ P = e^{-S} = e^{-E/T} $$

From these considerations it appears that the change in the temperature of the substance must be of a statistical nature. I don't have precise mathematical way stating this. But does this sound reasonable and is this system indeed capable of measurement? Any comments and clarifications will be much appreciated.

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2 Answers

It actually seems like if you put a quantum system in a "cold enough" heat bath, basically nothing will happen.

First, let's assume that the quantum system in question is in it's ground state. If the temperature $T$ of the heat bath satisfies $$kT \ll E_1$$ where $E_1$ is the first excited state of the quantum system, then by definition of the quantization of energy, the system will not transition to $E_1$, it will still be in $E_0$.

Note that the above argument is true (qualitatively) for either a single particle system or a bosonic system where all the particles are in the ground state. If it's a fermionic system, you'll have to measure the number of particles that have more energy than the Fermi energy.

Although if you're trying to get your system to transition from the ground to first excited state, you won't be measuring a significant difference in either case.

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Apologies for a very delayed response, I am not looking to excite the system into Energy state E1. I am looking to measure the state of the quantum system. Using the temprature change. The end observed effect will be either a small change in the temperature, or no change in temperature corresponding to excited state and ground state. –  Prathyush Apr 2 '13 at 21:42
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There are certain difficulties involved. To take a simple example consider the second lowest energy state of an ideal gas. We'll call that $E_1$. If you think about it, we get to this state when all the molecules except 1 are in $E_0$. This is a very very small amount of energy. Can you detect the temperature change the loss of this energy will cause in an external "cold" heat bath?

Real systems are even more complex. The difference between $E_0$ and $E_1$ for a perfect crystal is also a very small quantity.

So the problem is the attempt to detect a microscopic change in a macroscopic parameter, the temperature.

Now such changes can sometimes be measured in special systems, but in general you just can't see the change.

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I don't follow. There is a heat bath, whose energy levels are separated by small distance. And There is a system that whose properties we are interested in measuring. I am saying that KT<<E1(1st excited energy state of the Quantum System Not the heat bath). The heat bath itself can be anything thing, It must have energy levels, that are sufficiently close so that the temperature reading of the heat bath will have some validity. –  Prathyush Oct 14 '12 at 8:14
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