Independence of Trace-Partition function

[George444fg]

TL;DR Summary

Partition Function of a separable hamiltonian

I am trying to calculate the partition function of the system of two completely decoupled systems. Probability-wise, the decoupled nature means that the PDF is the product of the PDF of each subsystem. I just wanted to be sure that it would translate into:

$$
H = \sum_{k_i, s_i}e^{H_s(s_i)}e^{H_k(k_i)} = \sum_{k_i}(\sum_{s_i}e^{H_s(s_i)})e^{H_k(k_i)} = \sum_{s_i}e^{H_s(s_i)}\sum_{k_i}e^{H_k(k_i)} = Tr(e^{H_s(s_i)})*Tr(e^{H_k(k_i)})
$$

I know the question seems trivial, but I got a bit confused, and I would like to be 100% sure. Thank you for any help you can provide

 

[div_grad]

Your notation seems rather awkward (the "##H##" on the left-hand-side of your formula has nothing to do with the "##H##'s" in the exponents), but essentially yes - your expression is correct.

You can also see that you're right by considering, e.g., the Gibbs distribution for the subsystem ##n## (with the Hamiltonian ##H_n##) at the given temperature ##T## (##k_B## below is the Boltzmann constant):
$$ p_n(T) = \frac{e^{-H_n/k_B T}}{Z_n(T)} \rm{,}$$
where ##Z_n(T)## is the partition function for this subsystem,
$$ Z_n(T) = \text{Tr}\{e^{-H_n/k_B T}\} \rm{.}$$
Now, if you have two statistically independent subsystems ##s## and ##k##, the probability density of the total system ##s+k## is a product, as you've already noted:
$$p_{sk}(T) = p_s(T) \cdot p_k(T) \rm{.}$$
From this you immediately obtain that the partition function for the total ##s+k## system is ##Z_{sk}(T) = Z_s(T) \cdot Z_k(T)## - which is the result you got at the rightmost hand-side of your formula.