I 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.