# How to compute configurations (entropy) of a system?

If we have a system $X$ consisting of subsystems $X_1$ and $X_2$.

We also know that $X_1$ and $X_2$ have eigenstates $H_1 = 1 \times 10^{20}$ and $H_2 = 1 \times 10 ^{22}$.

Can we calculate the entropy given the above data? If yes, what's the basis?

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If these are perfectly independent systems, then the total number of eigenstates is a simple product of $H_1$ and $H_2$. You can imagine this as a giant square where one axis as $H_1$ positions and the other has $H_2$ positions. So the total number of positions is:

$$10^{20} \times 10^{22} = 10^{42}$$

That is under the assumption that objects can not be superposed (e.g. we are ignoring the possibility of combinations of states in $H_1$ and $H_2$)

The entropy is then just the logarithm of the number the states times a proportionality constant:

$$S=k \ln \Omega$$

If we set $k=1$ then

$$S=\ln 10^{42} = 96.7$$

We could also use $log_{10}$, in which case the entropy is just $42$.

I would add that Mark Mitchison's answer is more correct in the way that you would calculate this, but because of the simple case proposed this would work as well.

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I don't think my argument is "more correct". Actually it is the same as what you have described here, apart from the following point. Your assumption: "objects cannot be superposed (we are ignoring combinations of states in $H_1$ and $H_2$)" is surely not correct. In order to calculate the entropy, your $\Omega$ should enumerate the number of classically distinguishable states. Arbitrary combinations of states in $H_1\otimes H_2$ are not distinguishable. There are only $\mathrm{dim}(H_1\otimes H_2)$ distinguishable states, hence your answer is correct with no assumption. – Mark Mitchison Feb 17 '13 at 2:30
Agree, but I like the use of the direct product symbol, which is better 'grammatically' – user11547 Feb 17 '13 at 3:16

The thermodynamic entropy is defined as the von Neumann entropy of the quantum state that both maximises von Neumann entropy and is consistent with the experimental data. If you really know nothing about the system except the dimension of the Hilbert space, then the correct (maximum entropy) state is just the identity on $H_1 \otimes H_2$. The entropy of such a state is trivially found to be $k_B\log(\mathrm{dim}(H_1\otimes H_2)) = 42 k_B \log 10$.

Note that this situation is extremely contrived: normally you would have access to at least some macroscopic averages. For example, you would expect to be able to measure the energy of an isolated system. Restricting yourself to knowledge of such a limited (empty) set of expectation values means that your resulting theory of thermodynamics has basically no predictive power.

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